GG E O M E T R Y O F 1 - C O D I M E N S I O N A LM E A S U R E S I N H E I S E N B E R G G R O U P S a n d r e a m e r l o * abstract This paper is devoted to the study of tangential properties of measures with density in the Heisen-berg groups H n . Among other results we prove that measures with ( n + ) -density have only flat tangents andconclude the classification of uniform measures in H . keywords Preiss’s rectifiability Theorem, Heisenberg groups, density problem. msc ( ) A , A , C . In the Euclidean spaces the notion of rectifiability of a measure is linked to the metric by the celebrated:
Theorem . (Preiss, [ ]) . Suppose ≤ m ≤ n are integers, φ is a Radon measure on R n and: < Θ m ( φ , x ) : = lim r → φ ( U r ( x )) r m < ∞ at φ -almost every x , ( ) where U r ( x ) is the Euclidean ball of centre x and radius r. Then φ is m-rectifiable, i.e., φ -almost all of R n can be covered bycountably many m-dimensional Lipschitz submanifolds of R n . The most difficult part of the proof of Theorem . is to show that the existence of the density, namely that ( )holds, implies that the measure φ has flat tangents, i.e.:Tan ( φ , x ) ⊆ Θ m ( φ , x ) {H m (cid:120) V : V is an m -plane } at φ -almost every point. ( )The fact that the inclusion ( ) implies Theorem . is a consequence of the Marstrand-Mattila rectifiability crite-rion, see for instance Theorem . in [ ]. The proof of such inclusion depends on the structure of the Euclideanball and it is not known whether it is possible to extend it to a general finite dimensional Banach space. The onlyprogress in this direction, to our knowledge, was done by A. Lorent, who proved that 2-locally uniform measuresin (cid:96) ∞ are rectifiable, see Theorem in [ ]. As one should expect, although the assumption of local uniformityis far stronger than the mere existence of the density, already in this strengthened hypothesis the proof is reallyintricate and depends on the shape of the ball.In this paper we investigate to what extent the local structure of 1-codimensional measures in H n is affected bythe regular behaviour of the measure of Koranyi balls. Although the Heisenberg group shares many similaritieswith the Euclidean spaces, it has Hausdorff dimension 2 n + k -purely unrectifiable metric space forany k ∈ { n +
1, . . . , 2 n + } , i.e., for any compact set K ⊆ R k and any Lipschitz function f : K → ( H n , (cid:107)·(cid:107) ) wehave: H k (cid:107)·(cid:107) ( f ( K )) = H k (cid:107)·(cid:107) is Hausdorff measure associated to the Koranyi norm, see for instance Theorem . in [ ] or Theorem . in [ ]. This degeneracy of the structure of H n poses a big obstacle to extending many Euclidean results and * Scuola Normale Superiore, Piazza dei Cavalieri, , Pisa, Italy a r X i v : . [ m a t h . M G ] N ov ntroduction 2 definitions to the context of the Heisenberg groups, or in bigger generality to Carnot groups. In particular it isnot a priori clear what the correct notion of rectifiability should be, or even if there is one. In the paper [ ] B.Franchi, R. Serapioni and F. Serra Cassano, introduced an intrinsic notion of rectifiability. A set E ⊆ H n is said tobe ( n + ) - intrinsic rectifiable if for S n + -a.e. x ∈ E there exists a ( n + ) -dimensional homogeneous subgroup V x such that: Tan n + ( S n + (cid:120) E , x ) = {S n + (cid:120) V x } ,and Θ n + ∗ ( S n + (cid:120) V x ) >
0. This definition makes the recovery of De Giorgi’s rectifiability theorem of boundariesof finite perimeter sets possible in H n .This paper addresses the question of whether or not the definition of intrinsic rectifiability given by Franchi,Serapioni and Serra Cassano can be characterised by the metric in a similar way as rectifiability is in the Euclideanspaces. In other words we are interested in determining if a result in the spirit of Theorem . is available in H n ,where rectifiability is replaced with this new intrinsic rectifiability .The main goal of this paper is to prove the analogue of the inclusion ( ) in the Heisenberg groups thereforegetting closer to a metric justification of the notion of intrinsic rectifiability: Theorem . . Suppose φ is a Radon measure in H n such that: < Θ n + ( φ , x ) : = lim r → φ ( B r ( x )) r n + < ∞ for φ -a.e. x, ( ) where B r ( x ) is the Koranyi ball. Then:Tan n + ( φ , x ) ⊆ Θ n + ( φ , x ) M ( n + ) for φ -a.e. x , where M ( n + ) is the family of the Haar measures of ( n + ) -homogeneous subgroups of H n which assign measure tothe unit ball. This is the first example beyond the Euclidean spaces, where the mere existence of the density implies theflatness of the tangents. Theorem . leaves open the very interesting problem of determining whether or not aresult in the spirit of the Marstrand-Mattila rectifiability criterion is available in the context of Heisenberg groups.An affirmative answer to such a question would imply, as in the Euclidean spaces, that φ is an intrinsic rectifiablemeasure.The study of the density problem in the Heisenberg groups was started in by V. Chousionis and J. Tysonin [ ] where they proved that if φ is a Radon measure on H n having α -density, i.e.:0 < Θ α ( φ , x ) : = lim r → φ ( B r ( x )) r α < ∞ for φ -a.e. x , ( )then Marstrand theorem holds in H n for the Koranyi norm, i.e., α is an integer in {
0, . . . , 2 n + } . This was done,very much as in the Euclidean spaces, by proving that ( ) implies that φ -almost everywhere tangent measuresto φ are α -uniform measures (see Definition . ) and that the support of such α -uniform measures are analyticmanifolds. The same strategy with minor modifications works in general Carnot groups when endowed with aleft invariant polynomial norm. The development of those ideas allowed Chousionis, Tyson and Magnani in [ ]to characterise 1 and 2 uniform measures in H and to prove that vertically ruled 3-uniform measures are flat.As a byproduct of our analysis we complete the characterisation of uniform measures in H , see Section .We present here a survey of the strategy of the proof of Theorem . , giving for each key step a brief discussionof the ideas involved. In Section , we prove that the support of a uniform measure µ is contained in a quadraticsurface. This is a result in the spirit of Theorem . of [ ] (cfr. also with Theorem . in [ ]): Theorem . . Let α ∈ {
1, . . . , 2 n + } \ { } , and suppose that µ is an α -uniform measure. Then there are b ∈ R n , T ∈ R and Q ∈
Sym ( n ) with Tr ( Q ) (cid:54) = such that: supp ( µ ) ⊆ K ( b , Q , T ) : = { ( x , t ) ∈ R n + : (cid:104) b , x (cid:105) + (cid:104) x , Q x (cid:105) + T t = } . ntroduction 3 Despite the fact that we already know (thanks to Proposition . of [ ]) that the supports of a uniform measureis an analytic variety, the algebraic simplicity of the quadrics containing the support in the 1-codimensional casewill be a fundamental simplification in our computations.The proof of Theorem . is based on an adaptation of the arguments of Section of [ ]. In particular wehave extended Preiss’s moments to this non-Euclidean context (the Heisenberg moments b µ k , s are introduced inDefinition . ) in such a way that it is possible to prove (see Proposition . ) that for any s > u is thesupport of a given uniform measure µ , we have: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∑ k = b µ k , s ( u ) − s (cid:107) u (cid:107) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ s (cid:107) u (cid:107) ( + ( s (cid:107) u (cid:107) ) ) .The left-hand side in the above expression is a polynomial of fourth degree in the coordinates of u , but with somework one can reduce (see Proposition . ) the above inequality to: |(cid:104) b ( s ) , u H (cid:105) + (cid:104)Q ( s )[ u H ] , u H (cid:105) + T ( s ) u T | ≤ s (cid:107) u (cid:107) ,where u H is the vector of the first 2 n coordinates of u , u T is the last coordinate of u and b ( s ) , Q ( s ) and T ( s ) areintroduced in Definition . . From the above expression, sending s to 0 one gets the quadric containing supp ( µ ) .The most tricky part of Theorem . is to show that Tr ( Q ) (cid:54) = . .When µ is a ( n + ) -uniform measure, one expects that the fact that the support is contained in a quadricrepresents a strong information on the structure of supp ( µ ) . This idea is exploited in Section where we prove: Theorem . . The support of a ( n + ) -uniform measure µ is the closure of a union of connected components of K ( b , Q , T ) \ Σ , where Σ is the set of those points where the tangent group to K ( b , Q , T ) does not exists. The idea behind the proof of Theorem . is the following. Suppose y ∈ K ( b , Q , T ) \ supp ( µ ) and let z bea point with minimal Euclidean distance of y from supp ( µ ) . If z (cid:54)∈ Σ , thanks to Proposition . , we knowthat Tan n + ( µ , x ) = {S n + V } where V is the tangent group to K ( b , Q , T ) at z and S n + V is its Haar measure.However, by means of careful computations (see Propositions . and . ) we show that the blowup of “the holein the support” B | y − z | ( y ) ∩ K ( b , Q , T ) is a non-empty open subset of V , which is in contradiction with the factthat the support the blowup of µ at z coincides with the whole V . This implies that the boundaries of holes ofsupp ( µ ) inside K ( b , Q , T ) must be contained in Σ and thus a standard connection argument proves Theorem . .Theorem . allows us get a better understanding of the behaviour of ( n + ) -uniform measures at infinity. Inparticular we prove that:(i) if Tan n + ( µ , ∞ ) ∩ M ( n + ) (cid:54) = ∅ then µ ∈ M ( n + ) ,(ii) the set Tan n + ( µ , ∞ ) is a singleton.In the Euclidean space these properties arise from a careful analysis of the algebraic properties of moments. In ourframework the structure of moments is much more complicated and therefore (i) and (ii) are proved by meansof an explicit construction which relies on Theorem . . Thanks to these two properties of ( n + ) -uniformmeasures, in Section , we prove the following: Theorem . . Suppose there exists a functional F : M → R , continuous in the weak- ∗ convergence of measures, and aconstant ¯ h = ¯ h ( H n ) > such that:(i) if µ ∈ M ( n + ) then F ( µ ) ≤ ¯ h /2 ,(ii) if µ is a ( n + ) -uniform cone (see Definition . ) and F ( µ ) ≤ ¯ h, then µ ∈ M ( n + ) .Then, for any φ Radon measure with ( n + ) -density and for φ -almost every x:Tan n + ( φ , x ) ⊆ Θ n + ( φ , x ) M ( n + ) . ntroduction 4 The proof of Theorem . follows closely its Euclidean counterpart, and it is a standard application of the verygeneral principle that “a tangent to a tangent is a tangent” (see Proposition . ).We are left to construct the functional F satisfying all the hypothesis of Theorem . . Suppose ϕ is a smoothfunction with support contained in B ( ) such that ϕ = B ( ) . We claim that the functional: F ( µ ) : = min m ∈ S n − ˆ ϕ ( z ) (cid:104) m , z H (cid:105) d µ ( z ) ,satisfies all the hypothesis of Theorem . and therefore Theorem . follows. The fact that F is a continuousoperator on Radon measures is easy to prove (see Proposition . ) and it is immediate to see that F is identicallynull on flat measures. The most challenging hypothesis to check, as in the Euclidean case, is the existence of ¯ h .Thanks to Theorem . there are two kinds of ( n + ) -uniform measures. The ones which are contained ina quadric for with T =
0, that in the following are called vertical , and the ones with
T (cid:54) =
0, that we will call horizontal . The first step towards the verification of hypothesis (ii) of Theorem . is the following: Theorem . . There exists a constant C ( n ) > such that for any m ∈ S n − and any horizontal ( n + ) -uniform cone µ we have: ˆ B ( ) (cid:104) m , z H (cid:105) d µ ( z ) ≥ C ( n ) .The proof of Theorem . requires the entire the entire Section , but the arguments therein contained all relyon Theorem B. , which is the main result of Appendix B. Since Theorem . requires so much work, we wish todiscuss its proof more carefully, in order to help the reader keep in mind what the final goal of the Section andAppendix B is.If µ is a horizontal ( n + ) -uniform cone, we can find D ∈
Sym ( n ) \ { } such that supp ( µ ) ⊆ K ( D , − ) . InTheorem B. , we prove that such D must satisfy the algebraic constraint ( ) which implies that the operatorialnorm |||D||| of D is bounded from above and below by universal positive constants C ( n ) and C ( n ) , respectively(see Propositions . and . ) and thus Theorem . follows. We refer to the proof of Theorem . for furtherdetails.While the bound from below easily follows from Theorem B. , obtaining the bound from above is quitecomplicated. Suppose { µ i } is a sequence of ( n + ) -uniform measures invariant under dilations and assumethat supp ( µ i ) ⊆ K ( D i , − ) . If the sequence |||D i ||| diverges, then the limit points of the sequence { µ i } can onlybe vertical ( n + ) -uniform cones. Defined Q to be one of the limit points of the sequence D i / |||D i ||| , one canshow that the algebraic constraints given by Theorem B. on D i imply that for any h (cid:54)∈ Ker ( Q ) we have:2 ( Tr ( Q ) − (cid:104) n , Q n (cid:105) + (cid:104) n , Q n (cid:105) ) − ( Tr ( Q ) − (cid:104) n , Q n (cid:105) ) = n : = Q h / |Q h | . We refer to Proposition . for further details. By this key observation, via Proposition . we prove that the sequence { µ i } can only have a flat measure as limit points. The fact that the limit mustbe flat together with the fact that all the eigenvalues of the D i except one (see Proposition . , which is again aconsequence of Theorem B. ) must be bounded, implies that the assumption that such a sequence { µ i } existswas absurd. Indeed the boundedness of all eigenvalues except one would prevent the limit of the µ i ’s from beingflat. See the proof of Proposition . for further details.The above argument shows that the functional F disconnects horizontal ( n + ) -uniform cones and flatmeasures. The last piece of information we need to apply Theorem . is that F disconnects vertical non-flat ( n + ) -uniform cones from flat measures: Theorem . . There exists a constant C ( n ) > such that if µ is a vertical ( n + ) -uniform cone for which: min m ∈ S n − ˆ B ( ) (cid:104) m , z H (cid:105) d µ ( z ) ≤ C ( n ) , then µ is flat. The proof of the above theorem relies on Theorems . , . and the representation formulas of Appendix Ato get a very explicit and simple expression for the quadric containing supp ( µ ) (see Proposition . ). Thanksto the structural similarities of these quadrics to their Euclidean counterparts We were able to rearrange Preiss’soriginal disconnection argument to conclude the proof of Theorem . (see Theorem . and cfr. with the proofof Theorem . in [ ]). ntroduction 5 notation We add below a list of frequently used notations, together with the page of their first appearance: |·|
Euclidean norm, (cid:107)·(cid:107) Koranyi norm, |||·||| operatorial norm of matrices, (cid:104)· , ·(cid:105) scalar product in R n , V ( · , · ) polarisation function of the Koranyi norm, π H ( · ) projection of R n + onto the first 2 n coordinates, π T ( · ) projection of R n + onto the last coordinate, x H shorthand for π H ( x ) , x T shorthand for π T ( x ) , τ x left translation by x , D λ anisotropic dilations, V centre of H n , U r ( x ) open Euclidean ball of radius r > x , B r ( x ) open Koranyi ball of radius r > Θ α ( φ , x ) α -dimensional density of the Radon measure φ at the point x , φ x , r dilated of a factor r > φ at the point x ∈ H n , Tan α ( φ , x ) set of α -dimensional tangent measures to the measure φ at x , U H n ( α ) set of α -uniform measures, M set of Radon measures in H n , supp ( µ ) support of the measure µ , (cid:42) weak convergence of measures, V ( n ) the vertical hyperplane orthogonal to n ∈ R n , K ( b , Q , T ) the quadric (cid:104) b + Q x , x (cid:105) + T t = ( x , t ) ∈ R n + , Σ ( f ) characteristic set of the function f : R n → R Σ ( F ) set where the horizontal gradient of the function F : H n → R is null, J standard symplectic matrix, Sym ( n ) set of symmetric matrices on R n , S ( n ) subset of orthogonal matrices on R n inducing a linear isometry on H n , C c set of continuous functions with compact support, S α E α -dimensional spherical Hausdorff measure centred on the Borel set E , H keu Euclidean k -dimensional Hausdorff measure, Γ ( · ) Gamma function, b µ k , s k -th moment of the measure µ , The symbol φ will always denote a measure with density and the symbols µ , ν uniform measures. reliminaries 6 In this preliminary section we recall many well known facts and introduce some notations. In case the proof ofa Proposition is not present in literature, but the Euclidean argument applies verbatim, we will reduce ourselvesto cite a reference where the Euclidean proof can be found. . The Heisenberg group H n In this subsection we briefly recall some notations and very well known facts on the Heisenberg groups H n .Let π H : R n + → R n be the projection onto the first 2 n coordinates and π T : R n + → R be the projection onto thelast one. The Lie groups H n are the manifolds R n + endowed with the product: x ∗ y : = ( x H + y H , x T + y T + (cid:104) x H , Jy H (cid:105) ) ,where x H and x T are shorthands for π H ( x ) and π T ( x ) while J is the standard sympletic matrix on R n : J : = (cid:18) − id 0 (cid:19) .We metrize the group ( H n , ∗ ) with the Koranyi distance d ( · , · ) : H n × H n → H n defined as: d ( x , y ) : = (cid:16) | y H − x H | + | y T − x T − (cid:104) x H , Jy H (cid:105)| (cid:17) .Moreover we let (cid:107) x (cid:107) : = d ( x , 0 ) be the so called Koranyi norm and B r ( x ) : = { z ∈ H n : d ( z , x ) ≤ r } the Koranyiball. The geometry of H n is quite rich and it is well known that d ( · , · ) is left invariant, i.e., for any z ∈ H n onehas: d ( z ∗ x , z ∗ y ) = d ( x , y ) .As a consequence, left translations τ x ( y ) : = x ∗ y are isometries and we have that d ( x , y ) = (cid:107) x − ∗ y (cid:107) = (cid:107) y − ∗ x (cid:107) . Moreover, defined the anisotropic dilations D λ : H n → H n as D λ ( x ) : = ( λ x H , λ x T ) , we also have that d ( · , · ) is homogeneous with respect to D λ , i.e.: d ( D λ ( x ) , D λ ( y )) = λ d ( x , y ) .Besides the left translations, we have some other isometries of ( H n , d ) . Define:S ( n ) : = { U ∈ O ( n ) : U T JU = J } ∪ { U ∈ O ( n ) : U T JU = − J } , ( )and let s be the function s : S ( n ) → {−
1, 1 } which satisfies U JU = s ( U ) J . It is easy to check that S ( n ) is agroup under multiplication and that s ( · ) is a homomorphism between ( S ( n ) , · ) and ( {−
1, 1 } , · ) . Furthermorewe have: Proposition . . Let U ∈ S ( n ) . The map Ξ U : R n × R → R n × R defined as: Ξ U : ( x , t ) (cid:55)→ ( Ux , s ( U ) t ) , ( ) is an isometry of H n . For further references and a much more comprehensive account on the Heisenberg groups we refer to themonographs [ ] and [ ]. . Measures with density and their blowups
We recall in this subsection some very well known facts about measures with density and their blowups. reliminaries 7
Definition . . A Radon measure φ on H n is said to have α -density, for some α >
0, if the limit: Θ α ( φ , x ) : = lim r → φ ( B r ( x )) r α ,exists finite and non-zero, for φ -almost every x ∈ H n .Assume { µ k } is a sequence of measures in M . We say that { µ k } converges to µ and we write µ k (cid:42) µ , if:lim k → ∞ ˆ f ( x ) d µ k ( x ) = ˆ f ( x ) d µ ( x ) for any f ∈ C c ( R n ) .Since the paper is concerned with the study of the tangents to measures with α -density, we need a meaningfulconcept of tangent for a measure. Let φ be a Radon measure on H n with α -density and denote by φ x , r the measurethat satisfies: φ x , r ( A ) = φ ( x ∗ D r ( A )) , ( )for any Borel set A ⊆ H n . The set of tangent measures to φ at x is denoted by Tan α ( φ , x ) and it consists of theRadon measures µ for which there exists a sequence r i → φ x , r i r α i (cid:42) µ .The set Tan α ( φ , x ) is non-empty for φ -almost every x ∈ H n . Indeed, fix a point x ∈ H n for which Θ α ( φ , x ) < ∞ .Then for every ρ > r − α φ x , r ( B ρ ( )) = r − α φ ( B ρ r ( x )) ≤ Θ α ( φ , x ) ρ α ,for sufficiently small r . Therefore the family of measures r − α φ x , r is uniformly bounded on compact sets andcompactness of measures yields the existence of a limit for a suitable subsequence. Definition . . We say that a Radon measure µ is an α -uniform measure if:(i) 0 ∈ supp ( µ ) ,(ii) µ ( B r ( x )) = r α for any r > x ∈ supp ( µ ) .We will denote the set of α -uniform measures with the symbol U H n ( α ) .The following two propositions are of capital importance as Proposition . insures that the tangent measuresto a measure φ with α -density are φ -almost everywhere α -uniform and Proposition . that for φ -almost every x ∈ H n , the tangent measures to any element of Tan ( φ , x ) are still in Tan ( φ , x ) . The latter stability property isusually summarized in the effective but imprecise expression tangent to tangents are tangents . Proposition . . Assume φ is a measure with α -density on H n . Then for φ almost every x ∈ H n we have:Tan α ( φ , x ) ⊆ Θ α ( φ , x ) U H n ( α ) . Proof.
The proof of this proposition follows almost without modifications the one given in the Euclidean case inProposition . of [ ]. Proposition . . Let φ be a Borel measures having α -density in H n . Then for φ -a.e. x if µ ∈ Tan α ( φ , x ) we have:r − α µ y , r ∈ Tan α ( φ , x ) , for every y ∈ supp ( µ ) and r > Proof.
For a proof in generic metric groups see for instance Proposition . in [ ]. Proposition . . Let φ be a Radon measure and µ ∈ Tan α ( φ , x ) be such that r − α i φ x , r i (cid:42) µ for some r i → . Ify ∈ supp ( µ ) , there exists a sequence { z i } i ∈ N ⊆ supp ( φ ) such that D r i ( x − z i ) → y.Proof. A simple argument by contradiction yields the claim, the proof follows verbatim its Euclidean analogue,Proposition . in [ ]. reliminaries 8 If µ is an α -uniform measure, we can also define its blowups at infinity, or blowdowns. Such tangents at infinityare Radon measures ν for which there exists a sequence { R i } → ∞ such that: R − α i φ R i (cid:42) ν .We will denote with Tan α ( µ , ∞ ) , the set of tangent measures at infinity of µ . The following proposition is astrengthened version of Proposition . for uniform measures. Proposition . . Assume µ is a α -uniform measure. Then for any z ∈ supp ( µ ) ∪ { ∞ } we have: ∅ (cid:54) = Tan α ( µ , z ) ⊆ U H n ( α ) . Proof.
A straightforward adaptation of the proof of Lemma . in [ ] yields the desired conclusion.The following is a compactness result for uniform measure and for their supports. Lemma . . If { µ } i ∈ N is a sequence of α -uniform measures converging in the weak topology to some ν then:(i) ν is an α -uniform measure,(ii) if y ∈ supp ( ν ) then there exists a sequence { y i } ⊆ H n such that y i ∈ supp ( µ i ) and y i → y,(iii) if there exists a sequence { y i } ⊆ H n such that y i ∈ supp ( µ i ) and y i → y, then y ∈ supp ( ν ) .Proof. The proof of this lemma is an almost immediate adaptation of the proofs of Proposition . and Proposition . .The following Theorem was proved by V. Chousionis, J.Tyson in [ ]. They proved that if H n is endowed withthe Koranyi metric, the density problem reduces to integer exponents only, as in the Euclidean case: Theorem . . The set U H n ( α ) is non-empty if and only if α ∈ {
0, 1, . . . , 2 n + } . In particular if φ is a measure with α -density in H n , then: α ∈ {
0, . . . , 2 n + } . Remark . . Note that U H n ( ) = { δ } . Moreover arguing as in Proposition . of [ ], we can also deduce that U H n ( n + ) = {L n + } . From now on we will always assume α ∈ {
1, . . . , 2 n + } . Remark . . The stratification of zeros of holomorphic functions is the tool used by B. Kirchheim and D. Preiss in[ ] and later by V. Chousionis and J.Tyson in [ ] to prove Marstrand’s theorem in the Euclidean spaces and inthe Heisenberg groups, respectively. It is easy to see that the same argument yields Marstrand’s theorem for anyCarnot group endowed with a polynomial norm. . Basic properties of uniform measures in H n Proposition . . Let Σ : ( H n , (cid:107)·(cid:107) ) → ( H n , (cid:107)·(cid:107) ) be a surjective isometry of ( H n , (cid:107)·(cid:107) ) into itself. If µ ∈ U H n ( α ) andthere exist u ∈ supp ( µ ) such that Σ ( u ) = , then Σ ( φ ) ∈ U H n ( α ) and: supp ( Σ ( φ )) = Σ ( supp ( φ )) . Proof.
Since Σ − ( B r ( g )) = B r ( Σ − ( g )) , for any g ∈ H n and any r > Σ φ ( B r ( g )) = φ ( Σ − ( B r ( g ))) = φ ( B r ( Σ − ( g ))) .If g ∈ Σ ( supp ( φ )) , then there exists h ∈ supp ( φ ) such that g = Σ ( h ) and then: Σ φ ( B r ( g )) = φ ( B r ( h )) = r α .In particular Σ ( supp ( φ )) ⊆ supp ( Σ ( φ )) . The other inclusion can be obtained similarly. reliminaries 9 Definition . . Let E ⊆ H n be a Borel set. For any 0 ≤ α ≤ n + δ >
0, define: S αδ , E ( A ) : = inf (cid:40) ∞ ∑ j = r sj : A ∩ E ⊆ ∞ (cid:91) j = cl ( B r j ( x j )) , r j ≤ δ and x j ∈ E (cid:41) ,whenever ∅ (cid:54) = A ⊆ H n is Borel and S αδ , E ( ∅ ) =
0. We define the E -centred spherical Hausdorff measure as: S α E ( A ) : = sup B ⊆ A sup δ > S αδ , E ( B ) .In the case E : = H n , our definition reduces to the standard spherical Hausdorff measure (see Definition . . ( )in [ ]). In such a case we let S α : = S α H n . Remark . . Let E ⊆ A be Borel subset of H n . It is easy to see that S α A ( S ) ≤ S α E ( S ) ≤ α S α A ( S ) for any S ⊆ E Borel.In particular the measures S α A (cid:120) E and S α E are equivalent.The following characterization of uniform measures easily follows from Federer’s spherical differentiationtheorems: Proposition . . If µ is a α -uniform measure on H n , then µ = S α supp ( µ ) .Proof. First of all note that since µ is uniform, µ ( ∂ B r ( x )) = r ≥ x ∈ supp ( µ ) . If we let F : = { cl ( B r ( y )) : y ∈ E and r > } , since µ is uniform, we also deduce that:lim r → sup (cid:8) r − α µ ( cl ( B r ( y ))) : x ∈ B r ( y ) ∈ F (cid:9) =
1. ( )The equality ( ) together with Theorem . of [ ], imply that whenever V ⊆ H n is an open set, we have: µ ( V ) = S α supp ( µ ) ( V ) .In particular thanks to Lemma . . of [ ] the above identity implies that µ ( E ) = S α supp ( µ ) ( E ) , for any Borel set E . Remark . . The above proposition implicitly says that α -uniform measures are uniquely determined by theirsupport. Remark . . The above proposition could also be proved using the fact that we know thanks to [ ] that supportsof uniform measures are analytic varieties together with the area formulas proved in [ ]. Definition . (Radially symmetric functions) . We say that a function ϕ : H n → R is radially symmetric if thereexists a profile function g : [ ∞ ) → R such that ϕ ( z ) = g ( (cid:107) z (cid:107) ) .Integrals of radially symmetric functions with respect to uniform measures are easy to compute and we havethe following change of variable formula, which will be extensively used throughout the paper: Proposition . . Let µ ∈ U H ( α ) and suppose ϕ : H n → R is a radially symmetric non-negative function. Then for anyu ∈ supp ( µ ) : ˆ H n ϕ ( u − ∗ z ) d µ ( z ) = α ˆ ∞ r α − g ( r ) dr , where g is the profile function associated to ϕ .Proof. First one proves the formula for simple functions of the form: ϕ ( z ) : = k ∑ i = a i χ B ri ( ) ,where 0 ≤ a i , r i for any i =
1, . . . , k . The result for a general ϕ follows by Beppo Levi convergence theorem. niform measures have support contained in quadrics 10 Proposition . . Let p > and µ ∈ U H ( α ) . Then: ˆ H n (cid:107) z (cid:107) p e − s (cid:107) z (cid:107) d µ ( z ) = α s α + p Γ (cid:18) α + p (cid:19) . Proof.
The profile function associated to (cid:107) z (cid:107) p e − s (cid:107) z (cid:107) is r p e − sr , thus by Proposition . we have that: ˆ H n (cid:107) z (cid:107) p e − s (cid:107) z (cid:107) d µ ( z ) = α ˆ ∞ r α − r p e − sr dr = α s α + p ˆ ∞ t α + p − e − t dt = α s α + p ˆ ∞ x α + p − e − x dx = α s α + p Γ (cid:18) α + p (cid:19) . First of all we introduce some notation:
Definition . . Let b ∈ R n , Q ∈
Sym ( n ) and T ∈ R . We define K ( b , Q , T ) to be the set of ( x , t ) ∈ R n × R forwhich: (cid:104) b , x (cid:105) + (cid:104) x , Q x (cid:105) + T t = Theorem . . Let m ∈ {
1, . . . , 2 n + } \ { } . For any µ ∈ U H n ( m ) there exist b ∈ R n , T ∈ R and Q ∈
Sym ( n ) withTr ( Q ) (cid:54) = such that: supp ( µ ) ⊆ K ( b , Q , T ) .The proof of the above theorem is divided into 2 main steps. First we construct b ∈ R n , Q ∈
Sym ( n ) , T ∈ R for which: (cid:104) b , u H (cid:105) + (cid:104) u H , Q u H (cid:105) + T u T = u ∈ supp ( µ ) . Secondly we prove that Tr ( Q ) (cid:54) = in [ ] (for a more detailed explanation see thebeginning of Subsection . below). The basic idea behind all these computations is that the identity µ ( B r ( u )) = µ ( B r ( )) for any u ∈ supp ( µ ) and any r > (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) B r ( u ) ( r − (cid:107) u − ∗ z (cid:107) ) d µ ( z ) − B r ( ) ( r − (cid:107) z (cid:107) ) d µ ( z ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C (cid:107) u (cid:107) / r ,for any u ∈ H n and any r >
0, from which it is not hard to build a quadric containing supp ( µ ) (for details seeSubsection . ).The second part (contained in Subsection . ) is devoted to prove that the quadric is non-degenerate. In theEuclidean case this is almost free, however in the sub-Riemannian context it requires some effort. In particularwe are able to prove that if the support of µ is far away from the vertical axis V : = { x H = } , then the quadric isnon-degenerate. The reason for which we have to avoid the case m = . is that S V is a 2-uniformmeasure and indeed in that case the matrix Q of our construction is 0. . Moments in the Heisenberg group and their algebraic structure
One of the fundamental tools introduced by Preiss in [ ] are moments of uniform measures. If µ is an m -uniform meausure in R n , for any k ∈ N and s > k -th moment of µ : b µ k , s ( u , . . . , u k ) : = ( s ) k + m I ( m ) k ! ˆ R n k ∏ i = (cid:104) z , u i (cid:105) e − s | z | d µ ( z ) , niform measures have support contained in quadrics 11 where I ( m ) : = ´ R n e −| z | d µ ( z ) . Using these functions Preiss is able to prove the following expansion formula: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q ∑ k = b k , s ( u ) − q ∑ k = s k (cid:107) u (cid:107) k k ! (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ n + ( s (cid:107) u (cid:107) ) , ( )for any u ∈ supp ( µ ) , which allows him to find algebraic equations for points in supp ( µ ) .The problem we tackle in this subsection is to prove an analogue of the inequality ( ) in a context where thereis no scalar product inducing the metric. The strategy of our choice is to use a suitable polarization V ( · , · ) ofthe Koranyi norm, which is a 4-th degree polynomial (see Proposition . ) for which a weak form of the Cauchy-Schwarz inequality holds (see Proposition . ). This will allow us to prove in the next subsection an inequalityof the type ( ) with our modified moments (see Proposition . ).It is possible to prove an inequality of the type ( ) even in Banach spaces with the suitable polarisation of thenorm and with the proper definition of moments. The problem is that such an expansion would not yield manyinformation on the structure of the support of measures as one really needs an explicit algebraic expression forthe substitute of the scalar product in order to be able to push further the argument. In Carnot groups howeverone always have a smooth polynomial norm which can be used as in Heisenberg case, computations would bejust much more complicated. Definition . (Substitute for the scalar product) . Let V : H n × H n → R be the polarisation of the Koranyi norm,i.e.: V ( u , z ) : = (cid:107) u (cid:107) + (cid:107) z (cid:107) − (cid:107) u − ∗ z (cid:107) u , z ∈ H n . Proposition . . The function V ( z , u ) can be decomposed as: V ( u , z ) = L ( u , z ) + Q ( u , z ) + T ( u , z ) , where:(i) L ( u , z ) : = (cid:104) u H , 4 | z H | z H + z T Jz H (cid:105) ,(ii) Q ( u , z ) : = − (cid:104) z H , u H (cid:105) − | z H | | u H | − (cid:104) Jz H , u H (cid:105) + z T u T ,(iii) T ( u , z ) : = (cid:104) z H , 4 | u H | u H + u T Ju H (cid:105) .Remark . . Note that L ( u , z ) , Q ( u , z ) , T ( u , z ) are 1, 2, 3-intrinsic homogeneous in u , respectively, and moreoverwe have L ( z , u ) = T ( u , z ) . Proof.
Thanks to the definition of V and of (cid:107)·(cid:107) , we have:2 V ( u , z ) = (cid:107) u (cid:107) + (cid:107) z (cid:107) − (cid:107) u − ∗ z (cid:107) = | u H | + | u T | + | z H | + | z T | − | z H − u H | − | z T − u T − (cid:104) u H , Jz H (cid:105)| = − (cid:104) u H , z H (cid:105) − | u H | | z H | + | u H | (cid:104) u H , z H (cid:105) + | z H | (cid:104) u H , z H (cid:105)− (cid:104) u H , Jz H (cid:105) + z T (cid:104) u H , Jz H (cid:105) + u T z T − u T (cid:104) u H , Jz H (cid:105) .Recognising L ( u , z ) , Q ( u , z ) , T ( u , z ) in the computation above proves the claim. Proposition . . For any z , u ∈ H n the following estimates hold:(i) | L ( u , z ) | ≤ (cid:107) u (cid:107)(cid:107) z (cid:107) ,(ii) | Q ( u , z ) | ≤ (cid:107) z (cid:107) (cid:107) u (cid:107) ,(iii) | T ( u , z ) | ≤ (cid:107) z (cid:107)(cid:107) u (cid:107) . niform measures have support contained in quadrics 12 Proof.
We start with proving the estimate for L ( u , z ) : | L ( u , z ) | = |(cid:104) u H , 4 | z H | z H + z T Jz H (cid:105)| ≤ (cid:107) u (cid:107) (cid:12)(cid:12)(cid:12) | z H | z H + z T Jz H (cid:12)(cid:12)(cid:12) .Since z and Jz are orthogonal: || z H | z H + z T Jz H | = | z H | + z T | z H | = | z H | (cid:107) z (cid:107) .From which we get: | L ( u , z ) | ≤ (cid:107) u (cid:107)(cid:107) z (cid:107) | z H | .By Remark . , we have that L ( z , u ) = T ( u , z ) and hence point (iii) follows. At last we prove the bound for Q ( u , z ) : | Q ( u , z ) | ≤ (cid:104) z H , u H (cid:105) + | z H | | u H | + (cid:104) Jz H , u H (cid:105) + | z T || u T | ≤ | z H | | u H | + | z T || u T | ≤ (cid:107) z (cid:107) (cid:107) u (cid:107) . Proposition . (Cauchy-Schwarz inequality for V ( · , · ) ) . For any u , z ∈ H n the following holds: | V ( u , z ) | ≤ (cid:107) u (cid:107)(cid:107) z (cid:107) ( (cid:107) u (cid:107) + (cid:107) z (cid:107) ) . Proof.
By the triangle inequality we have that (cid:107) u − ∗ z (cid:107) ≥ |(cid:107) u (cid:107) − (cid:107) z (cid:107)| . Therefore: (cid:107) u − ∗ z (cid:107) ≥|(cid:107) z (cid:107) − (cid:107) u (cid:107)| = (cid:107) u (cid:107) − (cid:107) u (cid:107) (cid:107) z (cid:107) + (cid:107) u (cid:107) (cid:107) z (cid:107) − (cid:107) u (cid:107)(cid:107) z (cid:107) + (cid:107) z (cid:107) .By the definition of V we conclude that:4 (cid:107) u (cid:107) (cid:107) z (cid:107) − (cid:107) u (cid:107) (cid:107) z (cid:107) + (cid:107) u (cid:107)(cid:107) z (cid:107) ≥(cid:107) u (cid:107) + (cid:107) z (cid:107) − (cid:107) u − ∗ z (cid:107) = V ( u , z ) .Collecting terms, we have: 2 (cid:107) u (cid:107)(cid:107) z (cid:107) ( (cid:107) u (cid:107) + (cid:107) z (cid:107) ) ≥ V ( u , z ) .The bound from below for V ( u , z ) is obtained similarly.The following definition extends from the Euclidean spaces to the Heisenberg group the notion of moment ofa uniform measure given by Preiss in [ ]: Definition . (Preiss’ moments) . For any k ∈ N , s > u , . . . , u k ∈ H n , we define: b µ k , s ( u , . . . , u k ) : = s k + m k ! C ( m ) ˆ H n k ∏ i = V ( u i , z ) e − s (cid:107) z (cid:107) d µ ( z ) ,where C ( m ) : = Γ (cid:0) m + (cid:1) . Moreover, if u = . . . = u k , we let: b µ k , s ( u ) : = b µ k , s ( u , . . . , u ) . Proposition . . For any u ∈ H n the following estimate holds: | b µ k , s ( u ) | ≤ k ( (cid:107) u (cid:107) s ) k k ! Γ ( m + k ) Γ (cid:0) m (cid:1) (( (cid:107) u (cid:107) s ) k + ) . niform measures have support contained in quadrics 13 Proof.
Thanks to Proposition . , we have the following preliminary estimate: | b µ k , s ( u ) | ≤ s m ( s ) k k ! C ( m ) ˆ H n | V ( u , z ) | k e − s (cid:107) z (cid:107) d µ ( z ) ≤ s m ( s ) k k ! C ( m ) ˆ H n k (cid:107) u (cid:107) k (cid:107) z (cid:107) k ( (cid:107) u (cid:107) + (cid:107) z (cid:107) ) k e − s (cid:107) z (cid:107) d µ ( z ) .Moreover, Jensen inequality (used in the first line) and Proposition . (used in the third line to explicitlycompute the integrals) imply that: | b µ k , s ( u ) | ≤ k s m ( s ) k k ! C ( m ) ˆ H n (cid:107) u (cid:107) k (cid:107) z (cid:107) k ( (cid:107) u (cid:107) k + (cid:107) z (cid:107) k ) e − s (cid:107) z (cid:107) d µ ( z ) ≤ k s m ( s ) k k ! C ( m ) (cid:18) ˆ H n (cid:107) u (cid:107) k (cid:107) z (cid:107) k e − s (cid:107) z (cid:107) d µ ( z ) + ˆ H n (cid:107) u (cid:107) k (cid:107) z (cid:107) k e − s (cid:107) z (cid:107) d µ ( z ) (cid:19) = k m (cid:107) u (cid:107) k s k k ! C ( m ) (cid:18) (cid:107) u (cid:107) k s k Γ (cid:18) m + k (cid:19) + Γ (cid:18) m + k (cid:19)(cid:19) ≤ k ( (cid:107) u (cid:107) s ) k k ! Γ (cid:16) m + k (cid:17) Γ (cid:0) m (cid:1) (( (cid:107) u (cid:107) s ) k + ) . Definition . . Let α ∈ N \ { (
0, 0, 0 ) } , s > u ∈ H n . We define the functions c α , s : (cid:78) | α | i = H n → R as: c α , s ( u ) : = α ! α ! α ! s | α | + m C ( m ) ˆ H n L ( u , z ) α Q ( u , z ) α T ( u , z ) α e − s (cid:107) z (cid:107) d µ ( z ) ,where | α | : = α + α + α . Moreover, for any l ∈ N , we let: A ( l ) : = { α ∈ N \ { (
0, 0, 0 ) } : α + α + α ≤ l } .The moments b µ k , s can be expressed by means of the functions c α , s defined above: b µ k , s ( u ) = s k + m k ! C ( m ) ˆ H n ( V ( u , z )) k e − s (cid:107) z (cid:107) d µ ( z ) = s k + m k ! C ( m ) ˆ H n ( L ( u , z ) + Q ( u , z ) + T ( u , z )) k e − s (cid:107) z (cid:107) d µ ( z )= s k + m k ! C ( m ) ˆ H n ∑ | α | = k k ! α ! α ! α ! L ( u , z ) α Q ( u , z ) α T ( u , z ) α e − s (cid:107) z (cid:107) d µ ( z )= ∑ | α | = k α ! α ! α ! s k + m C ( m ) ˆ H n L ( u , z ) α Q ( u , z ) α T ( u , z ) α e − s (cid:107) z (cid:107) d µ ( z ) = ∑ | α | = k c α , s ( u ) . ( ) Proposition . . Let α ∈ N \ { (
0, 0, 0 ) } , s > and u ∈ H n . Then: | c α , s ( u ) | ≤ D ( α )( s (cid:107) u (cid:107) ) α + α + α , for some constant < D ( α ) .Proof. Proposition . allows us to estimate the integrand in the definition of c α , s in the following way (as it givesbounds on | L ( u , z ) | , | Q ( u , z ) | and | T ( u , z ) | ): | c α , s ( u ) | ≤ α ! α ! α ! s | α | + m C ( m ) ˆ H n | L ( u , z ) | α | Q ( u , z ) | α | T ( u , z ) | α e − s (cid:107) z (cid:107) d µ ( z ) ≤ α + α α α ! α ! α ! s | α | + m C ( m ) (cid:107) u (cid:107) α + α + α ˆ H n (cid:107) z (cid:107) α + α + α e − s (cid:107) z (cid:107) d µ ( z ) .Moreover Proposition . yields an explicit value for the integrals in the last line of the above computations,therefore we get: | c α , s ( u ) | ≤ α + α α α ! α ! α ! s | α | + m C ( m ) (cid:107) u (cid:107) α + α + α ˆ H n (cid:107) z (cid:107) α + α + α e − s (cid:107) z (cid:107) d µ ( z )= α + α α α ! α ! α ! Γ (cid:16) m + α + α + α (cid:17) Γ (cid:0) m (cid:1) ( (cid:107) u (cid:107) s ) α + α + α . niform measures have support contained in quadrics 14 With the choice: D ( α ) : = α + α α α ! α ! α ! Γ (cid:16) m + α + α + α (cid:17) Γ ( m ) , we get the desired conclusion. Proposition . . Assume µ is also invariant under dilations, i.e., for any λ > we have µ λ λ m = µ , where µ λ was definedin ( ) . Then: c α , s ( u ) = s α + α + α c α ,1 ( u ) , for any s > , α ∈ N \ { (
0, 0, 0 ) } and any u ∈ H n .Proof. For any 0 < λ , we have that: L ( u , D λ ( z )) α Q ( u , D λ ( z )) α T ( u , D λ ( z )) α = λ α + α + α L ( u , z ) α Q ( u , z ) α T ( u , z ) α .Therefore, defining λ : = s , using the fact that µ λ / λ m = µ , we conclude that: c α , s ( u ) = α ! α ! α ! s | α | + m C ( m ) ˆ H n L ( u , z ) α Q ( u , z ) α T ( u , z ) α e − s (cid:107) z (cid:107) d µ ( z )= α ! α ! α ! s α + α + α C ( m ) ˆ H n L ( u , z ) α Q ( u , z ) α T ( u , z ) α e −(cid:107) z (cid:107) d µ λ ( z ) λ m . . Expansion formulas for moments
This subsection is devoted to the proof of the expasion formula for the moments of uniform measures ( ).Moreover in Proposition . , we start to flesh out the complex algebra of the inequality ( ), in order to buildthe desired quadric containing supp ( µ ) . We start with a technical lemma which will be required in the proof ofProposition . : Lemma . . For any m , k ∈ N we have the following estimate: Γ (cid:18) k + m (cid:19) ≤ m (cid:18) k (cid:19) k e − k Γ (cid:16) m (cid:17) . Proof.
By definition of the Γ function we have: Γ (cid:18) k + m (cid:19) = ˆ ∞ t k + m − e − t dt ≤ (cid:107) g (cid:107) ∞ ˆ ∞ t m − e − t /8 dt = m (cid:107) g (cid:107) ∞ Γ (cid:16) m (cid:17) ,where g ( t ) : = t k e − t /8 . The function g attains its maximum at t ∗ : = k and thus: (cid:107) g (cid:107) ∞ ≤ (cid:18) k (cid:19) k e − k .This yields Γ (cid:16) k + m (cid:17) ≤ m (cid:16) k (cid:17) k e − k Γ (cid:0) m (cid:1) .The following proposition is the technical core of this section. As we already remarked, ( ) will allow us toconstruct the algebraic surfaces containing supp ( µ ) . The proof follows closely its Euclidean analogue which canbe found in Section . of [ ] or in Lemma . of [ ]. Proposition . (Expansion formula) . There exists a constant < G ( m ) such that for any s > , q ∈ N and u ∈ supp ( µ ) we have: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q ∑ k = b µ k , s ( u ) − q ∑ k = s k (cid:107) u (cid:107) k k ! (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ G ( m )( s (cid:107) u (cid:107) ) q + ( + ( s (cid:107) u (cid:107) ) q ) . ( ) niform measures have support contained in quadrics 15 Proof.
First consider the case s (cid:107) u (cid:107) ≥
1: triangle inequality and Proposition . (used to get the bound in thesecond line) imply that: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q ∑ k = b µ k , s ( u ) − q ∑ k = s k (cid:107) u (cid:107) k k ! (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q ∑ k = b µ k , s ( u ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q ∑ k = s k (cid:107) u (cid:107) k k ! (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ q ∑ k = k ( (cid:107) u (cid:107) s ) k k ! Γ (cid:16) m + k (cid:17) Γ (cid:0) m (cid:1) (( (cid:107) u (cid:107) s ) k + ) + q ∑ k = s k (cid:107) u (cid:107) k k ! ≤ ( (cid:107) u (cid:107) s ) q + (cid:16)(cid:0) ( (cid:107) u (cid:107) s ) q + (cid:1) E ( m ) + q ∑ k = k ! (cid:17) ,where E ( m ) : = ∑ qk = k k ! Γ ( m + k ) Γ ( m ) . In order to prove the proposition in this case we are left to prove that E ( m ) isfinite. To do this, we use Lemma . to get an upper bound on E ( m ) : E ( m ) = ∞ ∑ k = k k ! Γ (cid:16) m + k (cid:17) Γ (cid:0) m (cid:1) ≤ m ∞ ∑ k = k e − k k ! (cid:18) k (cid:19) k .The series on the right-hand side in the above inequality converges by the ratio testand thus by comparison E ( m ) is also finite. Defined: G ( m ) : = max { E ( m ) , e } ,we have that the proposition is proved in the case s (cid:107) u (cid:107) ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q ∑ k = b µ k , s ( u ) − q ∑ k = s k (cid:107) u (cid:107) k k ! (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ G ( m )( (cid:107) u (cid:107) s ) q + (cid:16) ( (cid:107) u (cid:107) s ) q + (cid:17) .We have to prove the thesis in the case that s (cid:107) u (cid:107) <
1. The well known identity ∑ ∞ k = s k (cid:107) u (cid:107) k k ! = e s (cid:107) u (cid:107) implies that: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q ∑ k = s k (cid:107) u (cid:107) k k ! − e s (cid:107) u (cid:107) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ ∑ k = q + s k (cid:107) u (cid:107) k k ! (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ( s (cid:107) u (cid:107) ) q + ∞ ∑ k = q + k ! ≤ e ( s (cid:107) u (cid:107) ) q + .For any fixed s >
0, we prove that for any u ∈ supp ( µ ) such that s (cid:107) u (cid:107) <
1, the series ∑ ∞ k = b k , s ( u ) convergesabsolutely: ∞ ∑ k = | b k , s ( u ) | ≤ ∞ ∑ k = k ( (cid:107) u (cid:107) s ) k k ! Γ (cid:16) m + k (cid:17) Γ (cid:0) m (cid:1) (( (cid:107) u (cid:107) s ) k + ) ≤ ∞ ∑ k = k k ! Γ (cid:16) m + k (cid:17) Γ (cid:0) m (cid:1) = E ( m ) .We can therefore estimate its tail: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ ∑ k = b k , s ( u ) − q ∑ k = b k , s ( u ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ ∑ k = q + b k , s ( u ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ∞ ∑ k = q + k ( (cid:107) u (cid:107) s ) k k ! Γ (cid:16) m + k (cid:17) Γ (cid:0) m (cid:1) (cid:0) ( (cid:107) u (cid:107) s ) k + (cid:1) ≤ ( (cid:107) u (cid:107) s ) q + (( (cid:107) u (cid:107) s ) q + + ) ∞ ∑ k = q + k k ! Γ (cid:16) m + k (cid:17) Γ (cid:0) m (cid:1) ≤ ( (cid:107) u (cid:107) s ) q + (( (cid:107) u (cid:107) s ) q + + ) E ( m ) . ( )The next step in the proof is to prove the following equality: ∞ ∑ k = b k , s ( u ) = e s (cid:107) u (cid:107) ( ) niform measures have support contained in quadrics 16 for every s > u ∈ supp ( µ ) such that s (cid:107) u (cid:107) <
1. Note that by definition: ∞ ∑ k = b k , s ( u ) = lim q → ∞ q ∑ k = s m k ! C ( m ) ˆ H n ( sV ( u , z )) k e − s (cid:107) z (cid:107) d µ ( z ) .We would like to exchange integral and the limit above using dominated convergence. To do so we first have tofind a dominating function: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q ∑ k = ( sV ( u , z )) k k ! e − s (cid:107) z (cid:107) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ e − s (cid:107) z (cid:107) q ∑ k = ( s (cid:107) u (cid:107)(cid:107) z (cid:107) ( (cid:107) u (cid:107) + (cid:107) z (cid:107) ) ) k k ! = e − s (cid:107) z (cid:107) + s (cid:107) u (cid:107)(cid:107) z (cid:107) ( (cid:107) u (cid:107) + (cid:107) z (cid:107) ) ,where in the first in the first inequality we applied Proposition . . The function f ( · ) : = e − s (cid:107)·(cid:107) + s (cid:107) u (cid:107)(cid:107)·(cid:107) ( (cid:107) u (cid:107) + (cid:107)·(cid:107) ) isin L ( µ ) thanks to Proposition . . Thus applying the dominated convergence theorem (pointwise convergenceis obvious), we get: ∞ ∑ k = b k , s ( u ) = s m C ( m ) ˆ H n (cid:32) ∞ ∑ k = ( sV ( u , z )) k k ! (cid:33) e − s (cid:107) z (cid:107) d µ ( z ) = s m C ( m ) ˆ H n e sV ( u , z ) e − s (cid:107) z (cid:107) d µ ( z )= s m C ( m ) e s (cid:107) u (cid:107) ˆ H n e − s (cid:107) z (cid:107) + sV ( u , z ) − s (cid:107) z (cid:107) d µ ( z ) .Thus by definition of V ( u , z ) , using Proposition . and Proposition . we get the desired equality ( ).: ∞ ∑ k = b k , s ( u ) = s m C ( m ) e s (cid:107) u (cid:107) ˆ H n e − s (cid:107) u − ∗ z (cid:107) d µ ( z ) = e s (cid:107) u (cid:107) .Triangle inequality and the bound on the tail of the series ∑ ∞ k = b k , s ( u ) (see equation ( )) conclude the proof: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q ∑ k = b k , s ( u ) − q ∑ k = s k (cid:107) u (cid:107) k ! (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q ∑ k = b k , s ( u ) − e s (cid:107) u (cid:107) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) e s (cid:107) u (cid:107) − q ∑ k = s k (cid:107) u (cid:107) k ! (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ( (cid:107) u (cid:107) s ) q + (( (cid:107) u (cid:107) s ) q + + ) E ( m ) + e ( s (cid:107) u (cid:107) ) q + ≤ G ( m )( (cid:107) u (cid:107) s ) q + (( (cid:107) u (cid:107) s ) q + ) .Recall that in equation ( ), we showed how b k , s are sum of functions c α , s : b k , s ( u ) = ∑ | α | = k c α , s ( u ) .Proposition . implies that already for q =
1, in the left-hand side of inequality ( ): (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∑ k = ∑ | α | = k c α , s ( u ) − s (cid:107) u (cid:107) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ G ( m )( s (cid:107) u (cid:107) ) ( + ( s (cid:107) u (cid:107) ) ) ,there are a lot of terms bounded by ( s (cid:107) u (cid:107) ) . In the next proposition we get rid of those terms pushing them tothe right-hand side. Proposition . . For any s > and any u ∈ supp ( µ ) we have: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∑ α ∈ A ( ) c α , s ( u ) − s (cid:107) u (cid:107) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ( s (cid:107) u (cid:107) ) B ( s (cid:107) u (cid:107) ) , ( ) where B ( · ) is a suitable polynomial whereas c α , s ( · ) and A ( ) where defined in Definition . . niform measures have support contained in quadrics 17 Proof.
With the choice q =
1, the formula ( ) turns into: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∑ k = b k , s ( u ) − s (cid:107) u (cid:107) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ G ( m )( s (cid:107) u (cid:107) ) ( + ( s (cid:107) u (cid:107) ) ) .By the triangle inequality, we have: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∑ α ∈ A ( ) c α , s ( u ) − s (cid:107) u (cid:107) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∑ k = ∑ | α | = k c α , s ( u ) − s (cid:107) u (cid:107) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∑ α (cid:54)∈ A ( ) | α |≤ c α , s ( u ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) .By equation ( ) we deduce that the first term in the right-hand side of the above inequality coincides with (cid:12)(cid:12)(cid:12) ∑ k = b µ k , s ( u ) − s (cid:107) u (cid:107) (cid:12)(cid:12)(cid:12) and thus we are just left to estimate the second one. Proposition . implies that: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∑ α (cid:54)∈ A ( ) | α |≤ c α , s ( u | α | ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∑ α (cid:54)∈ A ( ) | α |≤ D ( α )( s (cid:107) u (cid:107) ) α + α + α (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) .Therefore the claim holds true with the choice: B ( t ) : = G ( m ) t + ∑ α (cid:54)∈ A ( ) | α |≤ D ( α ) t α + α + α − . . Construction of the candidate quadric containing the support
Before describing the content of this subsection we give the following:
Definition . . For any s ∈ ( ∞ ) we let:(i) the horizontal barycentre of the measure µ at time s to be the vector in R n : b ( s ) : = s + m C ( m ) ˆ ( | z H | z H + z T Jz H ) e − s (cid:107) z (cid:107) d µ ( z ) ,(ii) the symmetric matrix Q ( s ) associated to the measure µ at time s to be the element of Sym ( n ) : Q ( s ) : = − s + m C ( m ) ˆ H n | z H | e − s (cid:107) z (cid:107) d µ ( z ) id n − s + m C ( m ) ˆ ( z H ⊗ z H + Jz H ⊗ Jz H ) e − s (cid:107) z (cid:107) d µ ( z )+ s + m C ( m ) ˆ ( | z H | z H ⊗ z H + z T Jz H ⊗ Jz H ) e − s (cid:107) z (cid:107) d µ ( z )+ s + m C ( m ) ˆ H n | z H | z T ( Jz H ⊗ z H + z H ⊗ Jz H ) e − s (cid:107) z (cid:107) d µ ( z ) ,(iii) the vertical barycentre of the measure µ at time s to be the real number: T ( s ) : = s + m C ( m ) ˆ z T e − s (cid:107) z (cid:107) d µ ( z ) .The first half of this Subsection is devoted to the proof of Proposition . , where from Proposition . we areable to further simplify the algebra of inequality ( ) proving the existence of constant 0 < C such that: |(cid:104) b ( s ) , u H (cid:105) + (cid:104)Q ( s ) u H , u H (cid:105) + T ( s ) u T | ≤ Cs (cid:107) u (cid:107) . niform measures have support contained in quadrics 18 In the second half of this Subsection we prove that b ( · ) , Q ( · ) and T ( · ) are bounded curves as s goes to 0 andtherefore by compactness we can find b , Q and T for which for any u ∈ supp ( µ ) we have: (cid:104) b , u H (cid:105) + (cid:104) u H , Q u H (cid:105) + T u T = . is that as s →
0, we can find a limit Q for which Tr ( Q ) (cid:54) = Proposition . . For any s > and any u ∈ supp ( µ ) we have that the following inequality holds: |(cid:104) b ( s ) , u H (cid:105) + (cid:104)Q ( s )[ u H ] , u H (cid:105) + T ( s ) u T | ≤ s (cid:107) u (cid:107) B (cid:48) ( s (cid:107) u (cid:107) ) , where B (cid:48) ( · ) is a suitable polynomial and b ( · ) , Q ( · ) and T ( · ) where introduced in Definition . .Proof. First of all note that if α (cid:54)∈ A ( ) (see Definition . ) then α + α + α ≥
3. Therefore Proposition . implies that: ∑ | α |≤ α (cid:54)∈ A ( ) | c α , s ( u ) | ≤ ( s (cid:107) u (cid:107) ) ∑ | α |≤ α (cid:54)∈ A ( ) D ( α )( s (cid:107) u (cid:107) ) α + α + α − = ( s (cid:107) u (cid:107) ) B (cid:48)(cid:48) ( s (cid:107) u (cid:107) ) .where B (cid:48)(cid:48) ( t ) : = ∑ | α |≤ α (cid:54)∈ A ( ) D ( α ) t α + α + α − . Hence Proposition . yields: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∑ α ∈ A ( ) c α , s ( u ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∑ α ∈ A ( ) c α , s ( u ) − s (cid:107) u (cid:107) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + s (cid:107) u (cid:107) + ∑ | α |≤ α (cid:54)∈ A ( ) | c α , s ( u ) |≤ ( s (cid:107) u (cid:107) ) B ( s (cid:107) u (cid:107) ) + s (cid:107) u (cid:107) + ( s (cid:107) u (cid:107) ) B (cid:48)(cid:48) ( s (cid:107) u (cid:107) ) ,where B (cid:48) ( t ) : = t B ( t ) + t + B (cid:48)(cid:48) ( t ) . Since A ( ) = { (
1, 0, 0 ) , (
2, 0, 0 ) , (
0, 1, 0 ) } we just have to prove that: c ( ) , s ( u ) + c ( ) , s ( u ) + c ( ) , s ( u ) = (cid:104) b ( s ) , u H (cid:105) + (cid:104) u H , Q ( s ) u H (cid:105) + T ( s ) u T .The expansion of c ( ) , s ( u ) yields: c ( ) , s ( u ) = s + m C ( m ) ˆ H n L ( u , z ) e − s (cid:107) z (cid:107) d µ ( z ) = s (cid:42) u H , s + m C ( m ) ˆ H n | z H | z H + z T Jz H e − s (cid:107) z (cid:107) d µ ( z ) (cid:43) = s (cid:104) u H , b ( s ) (cid:105) ,Expanding c ( ) , s ( u ) we get the first part of the quadric Q ( s ) : c ( ) , s ( u ) = s + m C ( m ) ˆ H n L ( u , z ) e − s (cid:107) z (cid:107) d µ ( z )= s + m C ( m ) ˆ H n (cid:104) u H , | z H | z H (cid:105) + (cid:104) u H , z T Jz H (cid:105) e − s (cid:107) z (cid:107) d µ ( z ) + s + m C ( m ) ˆ H n (cid:104) u H , | z H | z H (cid:105)(cid:104) u H , z T Jz H (cid:105) e − s (cid:107) z (cid:107) d µ ( z )= s (cid:104) u H , Q ( s )[ u H ] (cid:105) + s (cid:104) u H , Q ( s )[ u H ] (cid:105) ,where: Q ( s ) : = s + m C ( m ) ˆ H n | z H | z H ⊗ z H + z T Jz H ⊗ Jz H e − s (cid:107) z (cid:107) d µ ( z ) , Q ( s ) : = s + m C ( m ) ˆ H n | z H | z T ( z H ⊗ Jz H + Jz H ⊗ z H ) e − s (cid:107) z (cid:107) d µ ( z ) . ( )At last c ( ) , s ( u ) contains the vertical barycentre and the second half of Q ( s ) : c ( ) , s ( u ) = s + m C ( m ) ˆ H n Q ( u , z ) e − s (cid:107) z (cid:107) d µ ( z )= − s + m C ( m ) ˆ H n ( (cid:104) z H , u H (cid:105) + (cid:104) Jz H , u H (cid:105) ) e − s (cid:107) z (cid:107) d µ ( z ) + s + m C ( m ) ˆ H n ( − | z H | | u H | + z T u T ) e − s (cid:107) z (cid:107) d µ ( z ) . niform measures have support contained in quadrics 19 From which we deduce that: c ( ) , s ( u ) = − s (cid:104)Q ( s )[ u H ] , u H (cid:105) − s (cid:104)Q ( s )[ u H ] , u H (cid:105) + s T ( s ) u T ,where: Q ( s ) : = s + m C ( m ) ˆ H n ( z H ⊗ z H + Jz H ⊗ Jz H ) e − s (cid:107) z (cid:107) d µ ( z ) , Q ( s ) : = s m + C ( m ) ˆ H n | z H | e − s (cid:107) z (cid:107) d µ ( z ) id n . ( )Noticing that Q ( s ) = Q ( s ) + Q ( s ) − Q ( s ) − Q ( s ) id, the claim is proven. Remark . . Define V to be the span of the horizontal projection of supp ( µ ) , i.e.: V : = span { u H : u ∈ supp ( µ ) } .Then with a small abuse of notation, we will always make the identification: b ( s ) = B ( s ) : = s + m C ( m ) ˆ ( | z H | z H + z T π V ( Jz H )) e − s (cid:107) z (cid:107) d µ ( z ) ,where the function π V : R n → R n is the orthogonal projection onto the subspace V . The reason for which wemake this identification is that: (cid:104) b ( s ) , u H (cid:105) = (cid:104) B ( s ) , u H (cid:105) ,for any u ∈ supp ( µ ) which explains why we take this freedom. Proposition . . Both Q ( s ) and T ( s ) are bounded functions on ( ∞ ) . To be precise:(i) Endowed Sym ( n ) with a norm |·| there exists a constant < C , such that sup s ∈ ( ∞ ) | Q ( s ) | ≤ C .(ii) There exists a constant < C , such that sup s ∈ ( ∞ ) |T ( s ) | ≤ C .Remark . . In particular the function s (cid:55)→ Tr ( Q ( s )) is bounded. Proof.
Proposition . implies that there exists a constant 0 < ˜ G for which:˜ Gs (cid:107) u (cid:107) ≥| c ( ) , s ( u ) + c ( ) , s ( u ) | = s |(cid:104) u H , Q ( s ) u H (cid:105) + T ( s ) u T | ≥ s ||(cid:104) u H , Q ( s ) u H (cid:105)| − |T ( s ) || u T || .Thus, it suffices to give a bound for T ( s ) and the other will follow: |(cid:104) u H , Q ( s )[ u H ] (cid:105)| ≤ (cid:32) ˜ G + sup s ∈ [ ∞ ) |T ( s ) | (cid:33) (cid:107) u (cid:107) .The estimate on the supremum norm of T ( s ) follows by its definition and by Proposition . : |T ( s ) | ≤ s + m C ( m ) ˆ (cid:107) z T (cid:107) e − s (cid:107) z (cid:107) d µ ( z ) ≤ s + m C ( m ) ˆ (cid:107) z (cid:107) e − s (cid:107) z (cid:107) d µ ( z ) = Γ (cid:0) m + (cid:1) Γ (cid:0) m (cid:1) .From the above proposition we deduce that for any sequence { s j } j ∈ N such that s j tends to zero, by compactnesswe can extract a subsequence { s j k } k ∈ N , such that Q ( s j k ) and T ( s j k ) are converging to some ˜ Q , ˜ T . Therefore byProposition . :0 ≤ lim k → ∞ (cid:12)(cid:12) (cid:104) b ( s j k ) , u H (cid:105) + (cid:104)Q ( s j k ) u H , u H (cid:105) + T ( s j k ) u T (cid:12)(cid:12) ≤ lim k → ∞ s j k (cid:107) u (cid:107) B (cid:48) ( s j k (cid:107) u (cid:107) ) = u ∈ supp ( µ ) :lim k → ∞ (cid:104) b ( s j k ) , u H (cid:105) = −(cid:104) ˜ Q u H , u H (cid:105) − ˜ T u T . ( ) niform measures have support contained in quadrics 20 Proposition . . There exists a B ∈ V such that lim k → ∞ b ( s j k ) = B.Proof.
First of all we note that (cid:104) b ( s ) , v (cid:105) = v ∈ V ⊥ , since b ( s j k ) ∈ V for any k ∈ N (see Remark . ).Choose { u , . . . , u l } ⊆ supp ( µ ) , such that ( u ) H , . . . , ( u l ) H is a basis for V . Thus by equation ( ), we have that: B = − l ∑ i = (cid:0) (cid:104) ˜ Q ( u i ) H , ( u i ) H (cid:105) + ˜ T ( u i ) T (cid:1) ( u i ) H .If the measure µ is invariant under dilations, finding a candidate (non-degenerate) quadric containing supp ( µ ) is quite easy: Proposition . . If µ λ = µ for any < λ , then b ( s ) = for any s > and: (cid:104) u H , Q ( ) u H (cid:105) + T ( ) u T = for any u ∈ supp ( µ ) .Proof. In the proof of Proposition . we defined: s (cid:104) u H , b ( s ) (cid:105) : = c ( ) , s ( u ) ,and: s ( (cid:104) u H , Q ( s ) u H (cid:105) + T ( s ) u T ) : = c ( ) , s ( u ) + c ( ) , s ( u ) .Therefore Proposition . implies that: s (cid:104) u H , b ( s ) (cid:105) = s (cid:104) b ( ) , u H (cid:105) .and: s ( (cid:104) u H , Q ( s ) u H (cid:105) + T ( s ) u T ) = s ( (cid:104) u H , Q ( ) u H (cid:105) + T ( ) u T ) ,Therefore by Proposition . we have that: (cid:12)(cid:12)(cid:12) s − (cid:104) b ( ) , u H (cid:105) + (cid:104) u H , Q ( ) u H (cid:105) + T ( ) u T (cid:105) (cid:12)(cid:12)(cid:12) ≤ s (cid:107) u (cid:107) B (cid:48) ( s (cid:107) u (cid:107) ) ,for any u ∈ supp ( µ ) , any s (cid:107) u (cid:107) ≤
1. Sending s to 0 we deduce that b ( ) = (cid:104) u H , Q ( ) u H (cid:105) + T ( ) u T = Remark . . Note that as we already mentioned, if µ = S V where V is the vertical axis { z ∈ R n + : z H = } , wehave that µ is a 2-uniform measure but Q ( ) =
0. Therefore in this case our constructed quadric becomes trivialand thus not meaningful. . Non-degeneracy of the candidate quadric
The main result of this subsection can be stated as follows. Assume µ is a m -uniform measure in H n for which:lim s → Tr ( Q ( s )) =
0, ( )where Q ( s ) is the curve of symmetric matrices built in Proposition . . Then:Tan m ( µ , ∞ ) = {S V } .It is not hard to show that the condition ( ) is actually equivalent to: b ( s ) → Q ( s ) → T ( s ) → s → of [ ]). niform measures have support contained in quadrics 21 Proposition . . Let Q ( s ) be the matrix defined in Proposition . . For any s > the following equality holds:Tr ( Q ( s )) = s m + C ( m ) ˆ H n | z H | ( s (cid:107) z (cid:107) − ( + n )) e − s (cid:107) z (cid:107) d µ ( z ) . Proof.
Let { e , . . . , e n } be an orthonormal basis of R n . Then:Tr ( Q ( s )) = n ∑ i = (cid:104) e i , Q ( s ) e i (cid:105) + (cid:104) e i + n , Q ( s ) e i + n (cid:105) .Using the explicit expression for Q ( s ) we can compute both (cid:104) e i , Q ( s ) e i (cid:105) for any i =
1, . . . , 2 n . If 1 ≤ i ≤ n wehave: (cid:104) e i , Q ( s ) e i (cid:105) = − s + m C ( m ) ˆ ( z i + | z H | + z i + n ) e − s (cid:107) z (cid:107) d µ ( z ) + s + m C ( m ) ˆ ( | z H | z i + z T z i + n ) e − s (cid:107) z (cid:107) d µ ( z )+ s + m C ( m ) ˆ H n | z H | z T ( z i + n z i ) e − s (cid:107) z (cid:107) d µ ( z ) .On the other hand if n + ≤ i ≤ n another similar expression holds: (cid:104) e i + n , Q ( s )[ e i + n ] (cid:105) = − s + m C ( m ) ˆ ( z i + n + | z H | + z i ) e − s (cid:107) z (cid:107) d µ ( z ) + s + m C ( m ) ˆ ( | z H | z i + n + z T z i ) e − s (cid:107) z (cid:107) d µ ( z ) − s + m C ( m ) ˆ H n | z H | z T ( z i + n z i ) e − s (cid:107) z (cid:107) d µ ( z ) .Putting together the above computations with the definition of Tr ( Q ( s )) we have:Tr ( Q ( s )) = n ∑ i = − s + m C ( m ) ˆ ( z i + n + | z H | + z i ) e − s (cid:107) z (cid:107) d µ ( z ) = s m + C ( m ) ˆ H n | z H | ( s (cid:107) z (cid:107) − ( + n )) e − s (cid:107) z (cid:107) d µ ( z ) . Proposition . . Let f : ( ∞ ) → [ ∞ ) be the function:f ( s ) : = ˆ H n | z H | e − s (cid:107) z (cid:107) d µ ( z ) . ( ) (i) If supp ( µ ) (cid:54)⊆ V one has < f ( s ) for any s > ,(ii) f is a smooth function and its derivatives are:f ( i ) ( s ) = ( − ) i ˆ H n (cid:107) z (cid:107) i | z H | e − s (cid:107) z (cid:107) d µ ( z ) . ( ) (iii) s m + + i f ( i ) ( · ) is a bounded function on ( ∞ ) for any i ∈ N .Proof. Assume there exists w ∈ supp ( µ ) such that w H (cid:54) =
0. Then:0 < | w H | e − s (cid:16) (cid:107) w (cid:107) (cid:17) µ ( B | wH | /2 ( w )) ≤ ˆ H n | z H | e − s (cid:107) z (cid:107) d µ ( z ) = f ( s ) , niform measures have support contained in quadrics 22 for any s >
0. The fact that f is smooth is proven showing that ( ) holds and this is a standard application ofthe dominated convergence theorem. The last point is a direct consequence of Proposition . and the formulafor f ( i ) : | s m + + i f ( i ) ( s ) | ≤ s m + + i ˆ H n (cid:107) z (cid:107) i + e − s (cid:107) z (cid:107) d µ ( z ) ≤ m Γ (cid:18) m + + i (cid:19) . Remark . . Proposition . and Proposition . imply that:(i) If supp ( µ ) (cid:54)⊆ V , we have ( − ) i f ( i ) ( s ) > i ∈ N .(ii) The expression of the trace can be rewritten as follows:Tr ( Q ( s )) = s m + C ( m ) ˆ H n | z H | ( s (cid:107) z (cid:107) − ( + n )) e − s (cid:107) z (cid:107) d µ ( z ) = − s m + C ( m ) f (cid:48) ( s ) − ( + n ) s m + C ( m ) f ( s ) . ( )In particular this implies by Proposition . that Tr ( Q ( s )) is a smooth, bounded function on ( ∞ ) . Proposition . . The function f defined in ( ) has the following representation by means of Tr ( Q ( · )) :f ( s ) = − C ( m ) s n + ˆ s λ n − − m Tr ( Q ( λ )) d λ , for any s > .Proof. Since f is smooth on ( ∞ ) by Proposition . , the following equality holds true: dds (cid:16) s + m f ( s ) (cid:17) = (cid:18) + m (cid:19) s + m f ( s ) + s + m f (cid:48) ( s ) ,The expression for Tr ( Q ( s )) in terms of f and f (cid:48) given in Remark . (ii) togheter with the above identity imply: C ( m ) Tr ( Q ( s )) = − s m + f (cid:48) ( s ) − ( + n ) s m + f ( s ) = − dds (cid:16) s + m f ( s ) (cid:17) + ( + m − n ) s + m f ( s ) . ( )Define now g ( s ) : = s + m f ( s ) , and note that equation ( ) becomes: C ( m ) Tr ( Q ( s )) = − g (cid:48) ( s ) + ( + m − n ) g ( s ) s .For any δ > g ( s ) solves the following Cauchy problem: (cid:40) g (cid:48) ( s ) + ( n − m − ) s g ( s ) = − C ( m ) Tr ( Q ( s )) , g ( δ ) = δ + m f ( δ ) .Such Cauchy Problem has an explicit unique solution on ( δ , ∞ ) (as coeffcients are smooth, Lipschitz and thevector field is sublinear in g ), which is: h δ ( s ) = − C ( m ) s n − m − (cid:20) ˆ s δ λ n − m − Tr ( Q ( λ )) d λ + δ n + f ( δ ) (cid:21) ,and coincides by uniqueness with g on ( δ , ∞ ) . Point (iii) of Proposition . implies that: (cid:12)(cid:12)(cid:12) δ n + f ( δ ) (cid:12)(cid:12)(cid:12) ≤ C δ n + − m ,for some 0 < C . Moreover, since m ≤ n + δ n + f ( δ ) → δ →
0. This implies that for any fixed s > g ( s ) = lim δ → h δ ( s ) = − C ( m ) s n − m − lim δ → (cid:20) ˆ s δ λ n − m − Tr ( Q ( λ )) d λ + δ n + f ( δ ) (cid:21) = − C ( m ) s n − m − ˆ s λ n − m − Tr ( Q ( λ )) d λ ,where the last equality comes from the fact that |·| n − m − Tr ( Q ( · )) ∈ L ([
0, 1 ]) as m ≤ n + niform measures have support contained in quadrics 23 Remark . . Since Tr ( Q ( s )) is bounded (see Remark . (ii)), if:lim s → Tr ( Q ( s )) ,does not exists or exists non-zero, there is a sequence { s j } j ∈ N such that the trace of the matrices Q ( s ) convergesto a non-zero value. Up to passing to a subsequence (for which b ( · ) , Q ( · ) and T ( · ) converge) we can build B , ˜ Q and ˜ T as in Proposition . for which Tr ( Q ) (cid:54) =
0. This would imply that the quadric K ( b , ˜ Q , ˜ T ) would containsupp ( µ ) and would be non-degenerate. Therefore without loss of generality, in what follows we should alwaysassume lim s → Tr ( Q ( s )) = Proposition . . Suppose that lim s → Tr ( Q ( s )) = . Then lim s → s m + f ( s ) = .Proof. For any (cid:101) there exists a δ > s ∈ ( δ ) we have | Tr ( Q ( s )) | ≤ (cid:101) . In particular Tr ( Q ( s )) > − (cid:101) , and Proposition . implies that for s ∈ ( δ ) : f ( s ) = − C ( m ) s n + ˆ s λ n − − m Tr ( Q ( λ )) d λ < (cid:101) C ( m ) s n + ˆ s λ n − − m d λ < (cid:101) C ( m ) s n + s n + − m n + − m = (cid:101) C ( m ) ( n + − m ) s m + .Summing up, we proved that for any s ∈ ( δ ) we have 0 < s m + f ( s ) < (cid:101) C ( m ) ( n + − m ) . Proposition . . The following are equivalent:(i) lim s → s m + f ( s ) = ,(ii) for any α > there exists an R ( α ) > such that if R > R ( α ) , then supp ( µ ) \ B R ( ) ⊆ { z : | z H | ≤ α (cid:107) z (cid:107)} .Proof. Suppose (ii) fails. Then there exists an α (cid:48) > j ∈ N there exists y j ∈ supp ( µ ) \ B j ( ) forwhich | ( y j ) H | ≥ α (cid:48) (cid:107) y j (cid:107) . We prove that along the sequence s j : = | ( y j ) H | − , the function s m + f ( s ) is bounded awayfrom 0, which contradicts (i): s m + j ˆ H n | z H | e − s j (cid:107) z (cid:107) d µ ( z ) ≥ s m + j | ( y j ) H | e − s j ( α (cid:48) | ( y j ) H | /2 ) µ (cid:16) B | ( y j ) H | /2 ( y j ) (cid:17) = s m + j m + e − s j ( α (cid:48) | ( y j ) H | /2 ) | ( y j ) H | m + ≥ e − ( α (cid:48) ) m + ,where we used the fact that for any z ∈ B | ( y j ) H | /2 ( y j ) one has that | ( y j ) H | /2 ≤ | z H | and (cid:107) z (cid:107) ≤ α (cid:48) | ( y j ) H | /2.Viceversa suppose (ii) holds. This implies that for any α > s m + ˆ H n | z H | e − s (cid:107) z (cid:107) d µ ( z ) ≤ s m + ˆ B R ( α ) ( ) (cid:107) z (cid:107) e − s (cid:107) z (cid:107) d µ ( z ) + s m + α ˆ B cR ( α ) ( ) (cid:107) z (cid:107) e − s (cid:107) z (cid:107) d µ ( z ) ,The above computation, Proposition . and Proposition . imply that: s m + ˆ H n | z H | e − s (cid:107) z (cid:107) d µ ( z ) ≤ m ˆ s R ( α ) t m + e − t dr + α m Γ (cid:18) m + (cid:19) Therefore: 0 ≤ lim sup s → s m + f ( s ) ≤ lim sup s → m ˆ s R ( α ) t m + e − t dr + α m Γ (cid:18) m + (cid:19) ≤ α m Γ (cid:18) m + (cid:19) .The arbitrariness of α concludes the proof. Proposition . . If supp ( µ ) ⊆ V , then µ = S V . n + ) - uniform measures have no holes 24 Proof.
Since supp ( µ ) ⊆ V , then µ ( B r ( z )) = µ ( B r ( z ) ∩ V ) = r m , for any z ∈ supp ( µ ) and any r >
0. Note that: B r ( z ) ∩ V = { ( s ) ∈ R n + : | s − z T | < r } = U r ( x ) ∩ V ,where as usual U r ( z ) denotes the Euclidean ball of radius r and centre x . This implies that µ ( U r ( z )) = r m and hence µ is a m /2-uniform measure with respect to Euclidean balls and which support is contained in theline V . Mastrand theorem implies that m /2 must be an integer and since V is 1-dimensional, we deduce bydifferentiation that m /2 is either 0 or 1. As we escluded by hypoothesis m =
0, we deduce by the classification of1-uniform measures in R n proved in [ ] that µ = H eu (cid:120) V . Since m = ( µ ) = V , the result follows byProposition . . Proposition . . Suppose that for any α > there exists an R ( α ) > such that if R > R ( α ) , then: supp ( µ ) \ B R ( ) ⊆ { z : | z H | ≤ α (cid:107) z (cid:107)} . Then m = and Tan ( µ , ∞ ) = {S V } .Proof. Let η ∈ C ∞ c ( R n + ) such that 0 ≤ η ≤ η ( B ( )) = η ( B ( ) c ) =
0. For r > η R ( z ) : = η ( D R ( z )) .Let ν ∈ Tan ( µ , ∞ ) and { λ l } l ∈ N be such that λ l → ∞ and µ λ l λ m (cid:42) ν . Then we have: ˆ H n η R ( z ) | z H | e −(cid:107) z (cid:107) d ν ( z ) = lim l → ∞ ˆ H n η R ( z ) | z H | e −(cid:107) z (cid:107) d µ λ l ( z ) λ ml ≤ lim l → ∞ ˆ H n (cid:12)(cid:12) ( D λ l ( z )) H (cid:12)(cid:12) e − (cid:13)(cid:13)(cid:13) D λ l ( z ) (cid:13)(cid:13)(cid:13) d µ ( z ) λ ml = lim l → ∞ λ m + l ˆ H n | z H | e − (cid:107) z (cid:107) λ l d µ ( z ) = . and our hypothesis on µ . Thearbitrariness of r > ˆ H n | z H | e −(cid:107) z (cid:107) d ν ( z ) = . implies that supp ( ν ) ⊆ V and Proposition . implies that ν = S V . Therefore by Proposition . µ is a 2-uniform measure.As an immediate concequence we get: Corollary . . Let µ ∈ U H n ( m ) . If lim s → Tr ( Q ( s )) = , then m = and:Tan ( µ , ∞ ) = {S V } . In particular for any m ∈ {
1, . . . , 2 n + } \ { } , there exist b ∈ R n , T ∈ R and Q ∈
Sym ( n ) with Tr ( Q ) (cid:54) = such that: supp ( µ ) ⊆ K ( b , Q , T ) . ( n + ) - uniform measures have no holes Let µ be a ( n + ) -uniform measure (which should be considered fixed throughout the section) and K ( b , Q , T ) be the non-degenerate quadric in which supp ( µ ) is contained. The existence of such a quadric has been shownin Section . What is left to understand is whether supp ( µ ) is some kind of very irregular set inside K ( b , Q , T ) orif it has a better structure. To state our main result we need some notation. Let F : R n + → R be the quadraticpolynomial: F ( z ) : = (cid:104) b , z H (cid:105) + (cid:104) z H , Q z H (cid:105) + T z T , ( ) n + ) - uniform measures have no holes 25 whose zero-set is the quadric K ( b , Q , T ) . We define the set of singular points of K ( b , Q , T ) as: Σ ( F ) : = { x ∈ K ( b , Q , T ) : b + ( Q − T J ) x H = } . ( )Usually Σ ( F ) is called characteristic set if T (cid:54) = singular set if T =
0. We should not bother ourselves withsuch distinctions, and regar Σ ( F ) as the set of points where K ( b , Q , T ) behaves like a cone tip. The followingtheorem is the main result of this section and it should be regarded as an analogue of Proposition . in [ ]. Weshow not only that supp ( µ ) is not a fractal inside K ( b , Q , T ) but also that it can be viewed as a quadratic surfacewith no holes: Theorem . . The support of µ is (the closure of) a union of connected components of K ( b , Q , T ) \ Σ ( F ) . We give now a short account of the content for each subsection. Subsection . is split in two parts. Themain result of the first part is Proposition . , where we show that everywhere outside Σ ( F ) , ( n + ) -uniformmeasures have flat blowups. Therefore, even though in principle µ may have holes, they are not inherited bytangents and therefore locally supp ( µ ) behaves exactly as the whole surface K ( b , Q , T ) . The second part ofSubsection . is devoted to the proof of Proposition . , where we show that T is an invariant for µ : if T = K ( b (cid:48) , Q (cid:48) , T (cid:48) ) containing supp ( µ ) we have T (cid:48) =
0. This implies that there are twotypes of ( n + ) -uniform measures which are qualitatively different. If T =
0, Theorem . implies that supp ( µ ) is vertically ruled, and hence invariant by translations with elements of the centre, while if T (cid:54) = ( µ ) coincides with K ( b , Q , T ) , which is a t -graph, for a precise statement see Proposition . .Subsection . and Subsection . are completely devoted to the proof of Theorem . in the cases T (cid:54) = T =
0, respectively. The idea behind the proof is the same in both situations: pick a point x ∈ supp ( µ ) \ Σ ( F ) forwhich for any r > B r ( x ) ∩ supp ( µ ) c ∩ K ( b , Q , T ) (cid:54) = ∅ ,and show that the holes in the support pass to the blowup, contradicting the mentioned fact that at points ofsupp ( µ ) outside Σ ( F ) the tangent measure are planes. . Regularity of ( n + ) -uniform measures The first part of this subsection is devoted to the study of the local properties of µ . First of all we introduce thedefinition of vertical hyperplane and flat measure: Definition . . For any n ∈ R n we define: V ( n ) : = { z ∈ H n : (cid:104) n , z H (cid:105) = } ,and we say that V ( n ) is the vertical hyperplane orthogonal to n . A ( n + ) -uniform measure ν is said to be a flatmeasure , or simply flat , if: ν = S n + V ( n ) ,for some n ∈ R n . Remark . . The measure S n + V ( n ) is invariant under translation by the elements of V ( n ) . Let E be a Borel subset of V ( n ) and assume that the balls { B i } i ∈ N , centred at points of V ( n ) , are a countable cover of E . For any v ∈ V ( n ) the balls { v ∗ B i } i ∈ N are still centred at points of V ( n ) , are a countable cover of v ∗ E and: ∑ i ∈ N diam ( v ∗ B i ) = ∑ i ∈ N diam ( B i ) ,since translations are isometries for the Koranyi metric. This in particular implies that S n + V ( n ) ( E ) = S n + V ( n ) ( v ∗ E ) .The measure S n + V ( n ) is also ( n + ) -uniform, indeed Proposition A. together with Corollary . in [ ] implythat for any r > S n + V ( n ) ( B r ( )) = H n − eu ( B r ( x )) ω n = r n + , ( ) n + ) - uniform measures have no holes 26 where ω n is the volume of the unitary 2 n -dimensional Euclidean ball. In ( ) the constant in the last equalitydoes not coincide with the one of Corollary . in [ ]. This is due to the fact that we are not using the samemetric.The following proposition shows that ( n + ) -uniform measure with support contained in a vertical hyper-plane are flat. It is an adaptation of Remark . in [ ]. Proposition . . Suppose that µ is a ( n + ) -uniform measure for which there exists an n ∈ R n for which supp ( µ ) ⊆ V ( n ) . Then: µ = S n + V ( n ) . Proof.
By Proposition . for any x ∈ supp ( µ ) and any r > S n + ( µ ) ( B r ( x )) = µ ( B r ( x )) = r n + .Therefore, thanks to ( ), for any r > S n + ( µ ) ( B r ( )) = S n + V ( n ) ( B r ( )) .Since supp ( µ ) is closed in V ( n ) , we deduce that supp ( µ ) = V ( n ) : if a ball was missing somewhere the aboveequality would not be possible.The following proposition shows that outside Σ ( F ) tangents to µ are flat measures. Proposition . . For any x ∈ supp ( µ ) \ Σ ( F ) we have:Tan n + ( µ , x ) = {S n + V ( n ( x )) } , where n ( x ) : = b + ( Q − T J ) x H .Proof. By Proposition . for any x ∈ supp ( µ ) the set Tan n + ( µ , x ) is non-empty and it is contained in U H n ( n + ) . Pick any ν ∈ Tan n + ( µ , x ) and recall that by definition of tangent, there exist r i → µ x , r i / r n + i (cid:42) ν .Therefore Proposition . implies that for any y ∈ supp ( ν ) there exists a sequence { x i } ⊆ supp ( µ ) , for which D r i ( x − x i ) → y . Defined y i : = D r i ( x − x i ) , we have that x i = xD r i ( y i ) and thus for any i ∈ N :0 = (cid:104) b , ( x i ) H (cid:105) + (cid:104) ( x i ) H , Q ( x i ) H (cid:105) + T ( x i ) T = r i (cid:104) b , ( y i ) H (cid:105) + r i (cid:104) ( Q − T J ) x H , ( y i ) H (cid:105) + r i (cid:104) ( y i ) H , Q ( y i ) H (cid:105) .Since y i → y , dividing by r i and taking the limit as i → ∞ we deduce:0 = (cid:104) b + ( Q − T J ) x H , y H (cid:105) .This implies that for any ν ∈ Tan n + ( µ , x ) , we have supp ( ν ) ⊆ V ( n ( x )) . The claim follows by Proposition . .In the upcoming proposition we show that U H n ( n + ) is split in two families that are characterized by thecoefficient T . Proposition . . Suppose that there are b ∈ R n , Q ∈
Sym ( n ) \ { } and T ∈ R such that: supp ( µ ) ⊆ K ( b , Q , T ) . Then T = if and only if T = .Proof. Assume by contradiction that
T (cid:54) = T =
0. Since:supp ( µ ) ⊆ S : = K ( b , Q , T ) ∩ K ( b , Q , 0 ) ,by Proposition . and Remark . , we have that: µ ( B r ( )) = S n + ( µ ) ( B r ( )) ≤ n + S n + K ( b , Q , T ) ( S ∩ B r ( )) .Note that the projection π H ( S ) has L n -measure 0 in R n . Therefore proposition A. implies that S n + K ( b , Q , T ) ( S ∩ B r ( )) =
0, which contradicts the fact that 0 ∈ supp ( µ ) . n + ) - uniform measures have no holes 27 Definition . . If there exist b ∈ R n and Q ∈
Sym ( n ) \ { } such that:supp ( µ ) ⊆ K ( b , Q , 0 ) ,then µ is said to be a vertical uniform measure . If such b and Q do not exist µ is said to be a horizontal uniformmeasure . . Structure of the support of horizontal uniform measures
In this subsection we prove Theorem . in case µ is a horizontal uniform measure and therefore throughoutthis subsection we assume T (cid:54) =
0. Let f : R n → R be the smooth function: f ( h ) : = − (cid:104) h , Q h (cid:105) + (cid:104) b , h (cid:105)T . ( )Since supp ( µ ) ⊆ gr ( f ) , we deduce that:supp ( µ ) ∩ Σ ( F ) = { ( h , f ( h )) ∈ R n + : h ∈ Σ ( f ) } ,where Σ ( f ) is the set of characteristic points of f (see Subsection A in the Appendix for a more extensiveexplanation): Σ ( f ) : = { x ∈ R n : b + ( Q − T J ) h = } . ( )Thanks to Proposition A. , if n > Σ ( f ) cannot disconnect R n , as Σ ( f ) is an affine space of dimensionless then n . However if n = R \ Σ ( f ) is split in two connected components,which we will denote in the following by C i , with i =
1, 2. The following proposition is Theorem . in case µ isa horizontal measure. The idea behind the proof is to tuck cylinders inside holes of supp ( µ ) in such a way thatthey are also tangent to supp ( µ ) is some point outside Σ ( F ) and to show with some careful computations thatthe tangents to µ at the point of tangency cannot be flat. Proposition . . If n > then supp ( µ ) = K ( b , Q , T ) . If n = we have two cases:(i) if dim ( Σ ( f )) = , then supp ( µ ) = K ( b , Q , T ) ,(ii) if dim ( Σ ( f )) = , then either supp ( µ ) = K ( b , Q , T ) or it coincides with the closure of the graph of f | C or f | C .Proof. Since supp ( µ ) is a closed set and it is contained in K ( b , Q , T ) , then S : = { p ∈ R n : ( p , f ( p )) ∈ supp ( µ ) } is closed in R n . For any y ∈ K ( b , Q , T ) , there exists z ( y ) ∈ S (possibly coinciding with y H ) such that: | z ( y ) − y H | = dist eu ( y H , S ) .As a consequence the (possibly empty) cylinder: c | z ( y ) − y H | ( y ) : = { ( x , t ) ∈ R n × R : | x − y H | < | z ( y ) − y H |} ,does not intersect supp ( mu ) .We claim that for any y ∈ K ( b , Q , T ) \ ( Σ ( F ) ∪ supp ( µ )) the tangency point z ( y ) is contained in Σ ( f ) .Let us prove the proposition assuming that the above claim holds. If Σ ( f ) = ∅ , then supp ( µ ) = K ( b , Q , T ) (as y cannot exist). Hence, without loss of generality we can assume Σ ( f ) (cid:54) = ∅ .If n > Ω : = R n \ Σ ( f ) is a connected open set and S is relatively closed inside it. Suppose that thereexists a p ∈ S ∩ Ω such that for any r > q r ∈ S c ∩ U r ( p ) . This would imply (for a sufficiently small r > eu ( S , q r ) < r < dist eu ( Σ ( f ) , q r ) ,contradicting the fact that if z ∈ S satisfies | z − p | = dist eu ( S , q r ) then z ∈ Σ ( f ) . Note that the same argumentworks in the remaining cases in which n =
1. If dim ( Σ ( f )) =
1, where we apply the reasoning above to theconnected components C and C . n + ) - uniform measures have no holes 28 Let us now prove the claim. In order to ease notation in the following we define ζ : = ( z , f ( z )) ∈ supp ( µ ) . Ifby contradiction z (cid:54)∈ Σ ( f ) , then ζ (cid:54)∈ Σ ( F ) and thus Proposition . implies that:Tan n + ( µ , ζ ) = {S n + (cid:120) V ( n ( ζ )) } ,where n ( ζ ) = − ( b + ( Q − T J ) z ) / T . Moreover Proposition . implies that if w ∈ V ( n ( ζ )) and r i →
0, thereexists a sequence { v i } i ∈ N ⊆ supp ( µ ) such that: w i : = D r i ( ζ − v i ) → w . ( )Since v i (cid:54)∈ c | z − y H | ( y ) by construction, we deduce that: | z − y H | ≤ | ( ζ D r i ( w i )) H − y H | ,which reduces to: 0 ≤ (cid:104) z − y H , ( w i ) H (cid:105) + r i | ( w i ) H | .Taking the limit as i → ∞ we deduce that:supp ( ν ) ⊆ V ( n ( ζ )) ∩ { w ∈ R n + : (cid:104) z − y H , w H (cid:105) ≥ } .If z − y H is not parallel to n ( ζ ) this would contradict Proposition . as supp ( ν ) would be contained in a propersubset of V ( n ( ζ )) . This implies that there exists λ (cid:54) = z − y H = λ n ( ζ ) . From ( ) , we deduce that:( α ) ( v i ) H = z + r i w H + R i , with | R i | = o ( r i ) ,( β ) r i w T + o ( r i ) = ( v i ) T − ζ T − (cid:104) z , J ( v i ) H (cid:105) .Hence putting together ( α ) and ( β ), we get: ( v i ) T = ζ T + r i (cid:104) z , Jw H (cid:105) + (cid:104) z , JR i (cid:105) + r i w T + o ( r i ) .Since supp ( µ ) ⊆ gr ( f ) , using the definition of f and ( α ), we deduce that: T ( v i ) T = T ζ T − r i (cid:104) b + Q z , w H (cid:105) − (cid:104) b + Q z , R i (cid:105) − r i (cid:104) w H , Qw H (cid:105) + o ( r i ) .The above computations, ( β ) and the fact that w ∈ V ( n ( ζ )) imply: r i w T + o ( r i ) = r i (cid:104) n ( ζ ) , w H (cid:105) + (cid:104) n ( ζ ) , R i (cid:105) − r i (cid:104) w H , Q w H (cid:105)T + o ( r i )= (cid:104) n ( ζ ) , R i (cid:105) − r i (cid:104) w H , Q w H (cid:105)T + o ( r i ) .Therefore recollecting terms the above equality to: (cid:104) n ( ζ ) , R i (cid:105) = r i (cid:18) (cid:104) w H , Q w H (cid:105)T + w T (cid:19) + o ( r i ) . ( )On the other hand, ( α ) and the definition of w i imply that ( w i ) H = w i + R i r i . Since w i (cid:54)∈ c | z − y H | ( y ) , we have:0 ≤ (cid:104) z − y H , ( w i ) H (cid:105) + r i | ( w i ) H | = λ (cid:28) n ( ζ ) , R i r i (cid:29) + r i (cid:12)(cid:12)(cid:12)(cid:12) w H + R i r i (cid:12)(cid:12)(cid:12)(cid:12) ,where the last line comes from the fact that w ∈ V ( n ( ζ )) . Using ( ) and dividing by r i , we deduce:0 ≤ λ (cid:18) (cid:104) w H , Q w H (cid:105)T + w T (cid:19) + (cid:12)(cid:12)(cid:12)(cid:12) w H + R i r i (cid:12)(cid:12)(cid:12)(cid:12) + o ( r i ) r i = λ (cid:18) (cid:104) w H , Q w H (cid:105)T + w T (cid:19) + | w H | + o ( r i ) r i .Sending i → ∞ , we get: − λ w T ≤ λ (cid:104) w H , Q w H (cid:105) T + | w H | ,which constitutes a non-trivial bound on w T and this contradicts Proposition . . This implies that z ∈ Σ ( f ) . n + ) - uniform measures have no holes 29 . Structure of the support of vertical uniform measures
In this subsection we prove Theorem . in case µ is a vertical uniform measure. First of all we need to establishthe relation between the centre of an Euclidean ball tangent to supp ( µ ) and the point of tangency. Proposition . . Let y ∈ K ( b , Q , 0 ) \ supp ( µ ) and ζ ∈ supp ( µ ) \ Σ ( F ) be such that: | ζ − y | = dist eu ( y , supp ( µ )) . Then:(i) ζ T = y T ,(ii) there exists λ (cid:54) = such that λ ( ζ H − y H ) = Q ζ H + b.Proof. Since ζ (cid:54)∈ Σ ( F ) , Proposition . implies that:Tan n + ( µ , ζ ) = {S n + V ( Q ζ H + b ) } .By Proposition . , for any w ∈ V ( Q ζ H + b ) and any r i →
0, there exists a sequence { v i } i ∈ N ⊆ supp ( µ ) suchthat w i : = D r i ( ζ − v i ) → w . Therefore writing this convergence componentwise, we have:( α ) ( v i ) H = ζ H + r i w H + R i , with | R i | = o ( r i ) ,( β ) r i w T + o ( r i ) = ( v i ) T − ζ T − (cid:104) ζ H , J ( v i ) H (cid:105) .Putting together conditions ( α ) and ( β ) we deduce that: ( v i ) T = ζ T + r i (cid:104) ζ H , Jw H (cid:105) + (cid:104) ζ H , JR i (cid:105) + r i w T + o ( r i ) .The fact that v i (cid:54)∈ U | ζ − y | ( y ) and the above expression for ( v i ) T imply that:0 ≤ r i (cid:104) ζ H − y H − ( ζ T − y T ) J ζ H , w H (cid:105) + (cid:104) ζ H − y H , R i (cid:105) + | r i w H + R i | + ( ζ T − y T )( (cid:104) ζ H , JR i (cid:105) + r i w T + o ( r i ))+( r i (cid:104) ζ H , Jw H (cid:105) + (cid:104) ζ H , JR i (cid:105) + r i w T + o ( r i )) , ( )for any i ∈ N . Define N : = ζ H − y H − ( ζ T − y T ) J ζ H . If N =
0, dividing the above inequality by r i and sending i → ∞ , we get: 0 ≤| w H | + ( ζ T − y T ) w T + (cid:104) ζ H , Jw i (cid:105) . ( )If N = ζ T (cid:54) = y T , otherwise we would have that ζ = y and this is not possible by the choice of ζ and y . Therefore if N =
0, inequality ( ) constitutes a non-trivial bound on w T which is in contradiction withProposition . .On the other hand, if N (cid:54) =
0, dividing by r i inequality ( ) and sending i → ∞ , we deduce that 0 ≤ (cid:104) N , w H (cid:105) .This implies that: V ( Q ζ H + b ) ⊆ { ( x , t ) : (cid:104) N , x (cid:105) ≥ } ,and therefore there exists λ (cid:54) = λ N = Q ζ H + b . Therefore ( ):0 ≤ (cid:104) N , R i (cid:105) + | r i w H + R i | + ( ζ T − y T )( r i w T + o ( r i ))+( r i (cid:104) ζ H , Jw H (cid:105) + (cid:104) ζ H , JR i (cid:105) + r i w T + o ( r i )) . ( )The condition v i ∈ K ( b , Q , 0 ) implies:0 = (cid:104) Q ζ H + b , R i (cid:105) + r i (cid:104) w H , Q w H (cid:105) + r i (cid:104) w H , QR i (cid:105) + (cid:104) R i , Q R i (cid:105) , n + ) - uniform measures have no holes 30 which, together with the fact that λ N = Q ζ H + b , yields: − λ (cid:104) N , R i (cid:105) = r i (cid:104) w H , Q w H (cid:105) + r i (cid:104) w H , QR i (cid:105) + (cid:104) R i , Q R i (cid:105) .Using the above information, ( ) becomes:0 ≤ − λ ( r i (cid:104) w H , Q w H (cid:105) + r i (cid:104) w H , QR i (cid:105) + (cid:104) R i , Q R i (cid:105) ) + | r i w H + R i | + ( ζ T − y T )( r i w T + o ( r i ))+( r i (cid:104) ζ H , Jw H (cid:105) + (cid:104) ζ H , JR i (cid:105) + r i w T + o ( r i )) .Dividing the above inequality by r i and sending i → ∞ , we deduce that:0 ≤ − λ (cid:104) w H , Q w H (cid:105) + | w H | + ( ζ T − y T ) w T + (cid:104) ζ H , Jw H (cid:105) ,which if ζ T (cid:54) = y T isa non-trivial bound on w T . This contradicts Proposition . . Therefore ζ T = y T and thus: λ ( ζ H − y H ) = λ N = Q ζ H + b .In the following proposition we prove Theorem . in case µ is a vertical measure. The idea behind the proof isthe following. Let ζ and y be as in the statement of Proposition . . If | ζ H − y H | is small then the vector ζ H − y H roughly lies in the tangent space of K ( b , Q , 0 ) at ζ . Therefore (ii) of Proposition . implies that such a vectorin the tangent space to K ( b , Q , 0 ) at ζ should be parallel to the normal to K ( b , Q , 0 ) at ζ , which is clearly notpossible. Proposition . . Assume C is a connected component of K ( b , Q , 0 ) \ Σ ( F ) . Then:(i) either supp ( µ ) ∩ C = ∅ ,(ii) or C ⊆ supp ( µ ) .Proof. If µ is flat there is nothing to prove. Therefore thanks to Proposition A. we can assume without loss ofgenerality that: S n + ( Σ ( F ) ∩ K ( b , Q , 0 )) = C ∩ supp ( µ ) is relatively closed in C , thus if it is also relatively open in C , by connectedness either C ∩ supp ( µ ) = ∅ or C ∩ supp ( µ ) = C . By contradiction suppose that this is not the case, and thus there exist x ∈ supp ( µ ) ∩ C and a r > α ) for any 0 < r < r there exists y r ∈ supp ( µ ) c ∩ C ,( β ) cl ( U r ( x )) ∩ K ( b , Q , 0 ) = cl ( U r ( x )) ∩ C .Thus, for any 0 < r < r there exists ζ r ∈ supp ( µ ) ∈ B r ( x ) such that: | ζ r − y r | = dist eu ( y r , supp ( µ )) .By Proposition . , we deduce that for any 0 < r < r :(i) ( ζ r ) T = ( y r ) T ,(ii) there exists λ r (cid:54) = λ r (( ζ r ) H − ( y r ) H ) = Q ( ζ r ) H + b . ( )As ζ r , y r ∈ K ( b , Q , 0 ) we have that:0 = (cid:104) ( y r ) H − ( ζ r ) H , Q [( y r ) H − ( ζ r ) H ] (cid:105) + (cid:104) Q [( ζ r ) H ] + b , π H y r − ( ζ r ) H (cid:105) .Therefore for a sufficiently small r >
0, equation ( ) implies:1 = (cid:12)(cid:12)(cid:12)(cid:12)(cid:28) ( ζ r ) H − ( y r ) H | ( ζ r ) H − ( y r ) H | , 2 Q ( ζ r ) H + b | Q ( ζ r ) H + b | (cid:29)(cid:12)(cid:12)(cid:12)(cid:12) = |(cid:104) ( y r ) H − ( ζ r ) H , Q [( y r ) H − ( ζ r ) H ] (cid:105)|| ( y r ) H − ( ζ r ) H || Q ( ζ r ) H + b | ≤ (cid:107)Q(cid:107)| ( y r ) H − ( ζ r ) H || Q ( ζ r ) H + b | .However, since | y r − ( ζ r ) H | converges to 0 and 2 Q ( ζ r ) H + b converges to 2 Q x H + b (cid:54) = r tends to zero, wehave a contradiction. isconnectedness of ( n + ) - uniform cones implies rigidity of tangents 31 ( n + ) - uniform cones implies rigidity of tangents In this section we reduce the problem of establishing the flatness of blowups of measures with ( n + ) -densityto the study of some properties of ( n + ) -uniform cones, that we introduce in the following: Definition . . An m -uniform measure µ on H n is said to be an m -uniform cone if µ λ = µ , for any λ >
0. Wedenote by C H n ( m ) the set of m -uniform cones.Such reduction consists in constructing a continuous functional on Radon measures which ”disconnects" ( n + ) -flat measures to the non-flat ( n + ) -uniform cones in the following way: Theorem . . Suppose that there exists a functional F : M → R , continuous in the weak- ∗ convergence of measures, anda constant ¯ h = ¯ h ( H n ) > such that:(i) if µ ∈ M ( n + ) then F ( µ ) ≤ ¯ h /2 ,(ii) if µ ∈ C H n ( n + ) and F ( µ ) ≤ ¯ h, then µ ∈ M ( n + ) .Then, for any φ Radon measure with ( n + ) -density and for φ -almost every x:Tan n + ( φ , x ) ⊆ Θ n + ( φ , x ) M ( n + ) .The proof of Theorem . relies on the following two properties of ( n + ) -uniform measures: P. if Tan n + ( µ , ∞ ) ∩ M ( n + ) (cid:54) = ∅ then µ ∈ M ( n + ) , P. the set Tan n + ( µ , ∞ ) is a singleton.In the Euclidean case, these properties are algebraic consequences of the expansion of moments. For instancethe proof of P. in R n is quite immediate (see Theorem . ( ) of [ ]), but it really relies on the fact that momentsare symmetric multilinear functions. In H n the structure of moments is much more complicated because they arenot multilinear. This is the reason why we could prove these properties only in the codimension 1 case, wherefairly strong structure results for supp ( µ ) are available (see Section ).In Subsections . and . we will establish properties P. and P. , respectively, while in Subsection . we willprove Theorem . . . Flatness at infinity implies flatness
In this section we prove P. . As a first step, we show that if µ is a uniform measure whose support is containedin K ( b , Q , T ) , then any ν ∈ Tan n + ( µ , ∞ ) has support contained in K ( Q , T ) . This implies that if µ has a flattangent at infinity then K ( Q , T ) must contain a hyperplane. This is only possible when rk ( Q ) =
1, 2 (see theproof of Theorem . ) and T =
0. In Proposition . we prove that if rk ( Q ) =
1, then µ must be flat, while inProposition . , we show that if rk ( Q ) =
2, then either µ is flat or it has a unique non-flat tangent at infitity. Proposition . . Let µ be a ( n + ) -uniform measure for which supp ( µ ) ⊆ K ( b , Q , T ) . Then for any ν ∈ Tan n + ( µ , ∞ ) we have supp ( ν ) ⊆ K ( Q , T ) .Proof. Proposition . implies for any w ∈ supp ( ν ) there is are sequences { v i } i ∈ N ⊆ supp ( µ ) and R i → ∞ forwhich w i : = D R i ( v i ) → w . The condition v i = D R i ( w i ) ∈ K ( b , Q , T ) reads: R i (cid:104) ( w i ) H , Q [( w i ) H ] (cid:105) + R i (cid:104) b , ( w i ) H (cid:105) + R i T ( w i ) T = R i and sending i to infinity, we get that: (cid:104) w H , Q [ w H ] (cid:105) + T w T = ( ν ) ⊆ K ( Q , T ) . isconnectedness of ( n + ) - uniform cones implies rigidity of tangents 32 Proposition . . Let E , F be closed sets in H n and suppose that S n + E is a ( n + ) -uniform measure and S n + ( E ∩ F ) = . The measure S n + E ∪ F is ( n + ) -uniform measure if and only if S n + ( F ) = .Proof. Let x ∈ E be such that ρ : = dist ( x , F ) > r >
0. For any δ > S n + E ∪ F (cid:120) E ( B r ( x )) = S n + E ∪ F ( B r ( x ) ∩ B ( F , δ ) c ∩ E ) + S n + E ∪ F ( B r ( x ) ∩ B ( F , δ ) ∩ E )= S n + E ( B r ( x ) ∩ B ( F , δ ) c ) + S n + E ∪ F ( B r ( x ) ∩ B ( F , δ ) ∩ E ) ,where the last equality comes from the following observation. If { B i } i ∈ N is a covering of B r ( x ) ∩ B ( F , δ ) c ∩ E with balls of radii smaller than δ /2 and centred at E ∪ F , then the centres must be contained in E . Sending δ to 0,since F is closed, we deduce that: S n + E ∪ F (cid:120) E ( B r ( x )) = S n + E ( B r ( x ) ∩ F c ) + S n + E ∪ F ( F ∩ E ) = S n + E ( B r ( x )) ,where the last equality comes from the fact that S n + (cid:120) E and S n + E are mutually absolutely continuous and thatby hypothesis S n + ( E ∩ F ) = r < ρ /2, we have S n + E ∪ F ( B r ( x )) = S n + E ( B r ( x )) = r n + , and on the other hand if S n + ( F ) > r > ρ such that S n + E ∪ F ( B r ( x )) > r n + . Corollary . . Suppose the quadric K ( b , Q , T ) is connected and that it supports a ( n + ) -uniform measure µ . If Σ ( F ) ∩ K ( b , Q , T ) = ∅ , then µ = S n + K ( b , Q , T ) .Proof. The corollary is an immediate consequence of Propositions . (i) (for the case T (cid:54) =
0) and . (for the case T = ( Q ) =
1, i.e., there exists a non-zero vector n such that Q = n ⊗ n . Proposition . . Let µ be a ( n + ) -uniform measure supported on K ( b , n ⊗ n , 0 ) . Then µ is flat.Proof. By scaling we can assume that n is unitary. If b = . directlyimplies that µ ∈ M ( n + ) . Therefore we can assume without loss of generality that b (cid:54) =
0. A consequence ofProposition . and the discussion of the case in which b =
0, is that Tan ( µ , ∞ ) = (cid:8) S n + V ( n ) (cid:9) .There are two possibilities for b : either it is parallel to n or it is not. We begin with the simpler case in which b is parallel to n . In such a case, there exists λ ∈ R \ { } for which b = λ n and: K ( λ n , n ⊗ n , 0 ) = V ( n ) ∪ (cid:18) − λ n | n | + V ( n ) (cid:19) .The points w of the singular set Σ ( F ) (see ( )) must satisfy the equation: ( λ + (cid:104) w H , n (cid:105) ) n = Σ ( F ) = ∅ . Since S n + V ( n ) is ( n + ) -uniform, Proposition . together with Proposition . ,imply that µ = S n + V ( n ) .We are left to discuss the case in which b is not parallel to n . Since Tan n + ( µ , ∞ ) = {S n + V ( n ) } , Proposition . implies that for any w ∈ V ( n ) there exists a sequence { v i } i ∈ N ⊆ supp ( µ ) such that D i ( v i ) → w . Let u ∈ R n be a unitary vector, orthogonal to n and such that (cid:104) b , u (cid:105) >
0. Moreover, let W be the the orthogonal in R n ofthe span of the vectors u and n and denote by P W the orthogonal projection on W . Recall that for every i the v i ’smust satisfy the equation: (cid:104) b , ( v i ) H (cid:105) + (cid:104) n , ( v i ) H (cid:105) = v i along u , n and W , becomes:0 = (cid:104) b , n (cid:105)(cid:104) ( v i ) H , n (cid:105) + (cid:104) b , u (cid:105)(cid:104) ( v i ) H , u (cid:105) + (cid:104) b , P W [( v i ) H ] (cid:105) + (cid:104) n , ( v i ) H (cid:105) ≥(cid:104) b , n (cid:105)(cid:104) ( v i ) H , n (cid:105) + (cid:104) b , u (cid:105)(cid:104) ( v i ) H , u (cid:105) + (cid:104) b , P W [( v i ) H ] (cid:105) , isconnectedness of ( n + ) - uniform cones implies rigidity of tangents 33 for any i ∈ N . If we divide by i the above inequality and let i → ∞ , we get: (cid:104) b , u (cid:105)(cid:104) w H , u (cid:105) ≤ −(cid:104) b , P W [ w H ] (cid:105) , ( )since w H is orthogonal to n . By the arbitrariness of w ∈ V ( n ) , inequality ( ) must be satisfied for any w H orthogonal to n . Therefore, since ( ) holds for both w H and − w H , then: (cid:104) b , u (cid:105)(cid:104) w H , u (cid:105) = −(cid:104) b , P W [ w H ] (cid:105) .However, the above identity cannot be satisfied for any w H orthogonal to n , proving that K ( b , Q , 0 ) in this casecannot support a uniform measure. Proposition . . Let µ be a ( n + ) -uniform measure supported on K ( b , Q , 0 ) . If Q is semidefinite, then rk ( Q ) = .Proof. For any ν ∈ Tan n + ( µ , ∞ ) , Proposition . implies that supp ( ν ) ⊆ K ( Q , 0 ) . Suppose by contradictionthat rk ( Q ) ≥
2, then: supp ( ν ) ⊆ rk ( Q ) (cid:92) i = V ( n i ) .where n i are the eigenvectors relative to non-zero eigenvalues of Q . This would imply by Proposition A. that S n + ( supp ( ν )) =
0, which is a contradiction.The following proposition will be useful in the rest of the section as it provides an efficient way to describe thestructure of the support of tangent measures at infinity to those ( n + ) -uniform measures which are supportedon graphs. Proposition . . Let n ∈ S n − and suppose that g : R n − → R is a continuous function such that: lim λ → ∞ g ( λ h ) λ = g ∞ (cid:18) h | h | (cid:19) | h | , ( ) for any z ∈ R n − where g ∞ : S n − → R is a continuous function. Moreover we define: Γ : = { z + g ( z ) n : z ∈ n ⊥ } and Γ ∞ : = { z + g ∞ ( z / | z | ) | z | n : z ∈ n ⊥ } . Let µ be a ( n + ) -uniform measure for which Γ × R ⊆ supp ( µ ) . Then for any ν ∈ Tan n + ( µ , ∞ ) we have: Γ ∞ × R ⊆ supp ( ν ) . Proof.
Let ( w , τ ) ∈ Γ ∞ × R and define the curve γ : [ ∞ ) → Γ × R as: γ ( t ) : = ( tP n ( w ) + g ( tP n ( w )) n , t τ ) ,where P n : R n → n ⊥ is the orthogonal projection on n ⊥ . The curve γ is contained in supp ( µ ) andlim t → ∞ D t ( γ ( t )) = (cid:18) P n ( w ) + g ∞ (cid:18) P n ( w ) | P n ( w ) | (cid:19) | P n ( w ) | n , τ (cid:19) = ( w , τ ) ,where the last equality comes from the fact that w ∈ Γ ∞ . Let ν ∈ Tan n + ( µ , ∞ ) and R i → ∞ be the sequence forwhich R − ( n + ) i µ R i (cid:42) ν . Then we have that: D R i ( γ ( R i )) ∈ supp ( R − ( n + ) i µ R i ) .Therefore ( ) implies by Proposition . that ( w , τ ) ∈ supp ( ν ) . By the arbitrariness of ( w , τ ) and of ν , we havethat Γ ∞ × R ⊆ supp ( ν ) for any ν ∈ Tan n + ( µ , ∞ ) .The following proposition establishes both properties P. and P. in the case the quadric K ( b , Q , 0 ) withrk ( Q ) = isconnectedness of ( n + ) - uniform cones implies rigidity of tangents 34 Proposition . . Suppose µ is a ( n + ) -uniform measure supported on K ( b , Q , 0 ) . If rk ( Q ) = , then one of thefollowing two mutually exclusive conditions holds:(i) µ ∈ M ( n + ) ,(ii) Tan n + ( µ , ∞ ) = { ν } and ν is not flat.Proof. Since Q is symmetric, has rank 2 and has no sign (by Proposition . ), there are e , e ∈ R n orthonormalvectors and λ , λ > Q = − λ e ⊗ e + λ e ⊗ e . We define n : = − λ e + λ e and m : = λ e + λ e .If b = K ( Q , 0 ) = V ( n ) ∪ V ( m ) and that the singular set Σ ( F ) coincides with V ( n ) ∩ V ( m ) . In particular K ( Q , 0 ) is disconnected by Σ ( F ) in four half planes which we denote by C i , with i =
1, . . . , 4.We claim that supp ( µ ) can coincide with the closure of the union of just two of the C i ’s. First of all, Proposition . implies that supp ( µ ) must coincide with the closure of the union of some of these half-planes and on the otherhand Proposition . implies that supp ( µ ) cannot coincide with the closure of just one half-plane. MoreoverProposition . shows that there cannot be more than 3 half-planes contained in the support of µ , and thus theonly remaining possibility is that there are only two half-planes contained in supp ( µ ) . If these half-planes arecontained in the same plane, then µ is flat, while if they are one contained in V ( n ) and one in V ( m ) , then µ is notflat and its tangent at infinity is unique and coincide with µ itself (as µ is invariant under dilations).If b (cid:54) = b is contained in the image of Q or it is not. First we discuss the simplercase in which b (cid:54)∈ span ( e , e ) (note that this implies that n > K ( b , Q , 0 ) is a graph of a quadraticpolynomial. Indeed, complete e , e to an orthonormal basis { e , . . . , e n } of R n and assume without loss ofgenerality that (cid:104) b , e (cid:105) (cid:54) =
0. Then K ( b , Q , 0 ) is an e -graph, indeed if h ∈ R n satisfies (cid:104) h , b (cid:105) + (cid:104) h , Q h (cid:105) =
0, then: (cid:104) h , e (cid:105) = − − λ (cid:104) h , e (cid:105) + λ (cid:104) h , e (cid:105) + ∑ i (cid:54) = (cid:104) b , e i (cid:105)(cid:104) h , e i (cid:105)(cid:104) b , e (cid:105) .This implies that the quadric K ( b , Q , 0 ) is an e -graph and thus it is a connected set. Moreover, since the equation b + Q h = h ∈ R n , the singular set Σ ( F ) is empty. Therefore Proposition . impliesthat supp ( µ ) = K ( b , Q , 0 ) . This by Proposition . implies that supp ( ν ) = K ( Q , 0 ) but this is not possible bythe study of the case b =
0, and thus µ cannot be uniform.Thus, we are left to study the case where b (cid:54) = b = b e + b e for some b , b ∈ R . For any x ∈ K ( b , Q , 0 ) ,once completed the squares we have that:0 = − (cid:18) λ (cid:104) x , e (cid:105) − b λ (cid:19) + (cid:18) λ (cid:104) x , e (cid:105) + b λ (cid:19) + b λ − b λ , ( )For any x ∈ Σ ( F ) (see ( )), we have: − λ (cid:104) x H , e (cid:105) e + λ (cid:104) x H , e (cid:105) e + b e + b e = b = λ (cid:104) x H , e (cid:105) and b = − λ (cid:104) x H , e (cid:105) . This in particular implies by ( ) that x cannot becontained in K ( b , Q , 0 ) if b /4 λ − b /4 λ (cid:54) = b /4 λ − b /4 λ > Σ ( F ) = ∅ . Thanks to the identity ( ), thequadric K ( b , Q , 0 ) is easily seen to be the disjoint union of two e -graphs Γ (which we assume contains 0) and Γ . The functions g , g : e ⊥ → e which define Γ and Γ respectively, satisfy the hypothesis of Proposition . .Indeed: lim t → ∞ g ( th ) t = ± λ |(cid:104) h , e (cid:105)| λ e = g ∞ (cid:18) h | h | (cid:19) | h | e .Proposition . implies that Γ must be contained in supp ( µ ) and therefore the graph of g ∞ is contained in thesupport of any tangent measure at infinity to µ by Proposition . . Suppose now by contradiction that supp ( µ ) contains also Γ . Again by Proposition . we would have that the graph of g ∞ is contained in the support of anytangent measure at infinity to µ . However, since the union the graphs of g ∞ and g ∞ coincides with K ( Q , 0 ) , bythe discussion of the case b =
0, this is not possible. Therefore the support of any tangent measure ν at infinity to µ coincides with the graph of g ∞ . Therefore by Proposition . Tan n + ( µ , ∞ ) is a singleton and its only element isconnectedness of ( n + ) - uniform cones implies rigidity of tangents 35 cannot be flat (as the graph of g ∞ is not a hyperplane). The case b /4 λ − b /4 λ < e and e reversed.If b /4 λ − b /4 λ =
0, the quadric K ( b , Q , 0 ) coincides with the solutions of the equation: (cid:18) λ x − b λ (cid:19) = (cid:18) λ x + b λ (cid:19) .Let τ : = ( b /2 λ ) e + ( b /2 λ ) e and note that K ( b , Q , 0 ) coincides with τ ∗ ( V ( n ) ∪ V ( m )) . Since left translationsare isometries of H n , the discussion of this case reduces to the one in which b =
0, concluding the proof of theproposition.Eventually, the following theorem concludes the proof of P. . Theorem . . If µ is a ( n + ) -uniform measure for which there exists n ∈ S n − such that S n + V ( n ) ∈ Tan n + ( µ , ∞ ) .Then µ = S n + V ( n ) .Proof. If supp ( µ ) ⊆ K ( b , Q , T ) , then for any ν ∈ Tan n + ( µ , ∞ ) we have supp ( ν ) ⊆ K ( Q , T ) by Proposition . .In particular V ( n ) ⊆ K ( Q , T ) and this implies that T =
0. Complete n to an orthonormal basis { n , e , . . . , e n } of R n and note that, since V ( n ) ⊆ K ( Q , 0 ) , we have that: n ∑ i = (cid:104) e i , Q e i (cid:105)(cid:104) w H , e i (cid:105) + ∑ ≤ i < j ≤ n (cid:104) e i , Q e j (cid:105)(cid:104) w H , e i (cid:105)(cid:104) w H , e j (cid:105) = w ∈ V ( n ) and thus (cid:104) e i , Q e j (cid:105) = ≤ i , j ≤ n . This implies that for any x ∈ H n we have: (cid:104) x H , Q x H (cid:105) = (cid:104) x H , n (cid:105) (cid:16) (cid:104) n , Q n (cid:105)(cid:104) x H , n (cid:105) + n ∑ i = (cid:104) n , Q e i (cid:105)(cid:104) x H , e i (cid:105) (cid:17) = (cid:104) x , n (cid:105)(cid:104) m , x (cid:105) ,where m : = (cid:104) n , Q n (cid:105) n + ∑ ni = (cid:104) n , Q e i (cid:105) e i . In particular rk ( Q ) ≤ m is parallel to n , Proposition . impliesthat µ is flat. On the other hand if m is not parallel to n , since rk ( Q ) = µ has a flat tangent at infinity,Proposition . implies that µ is flat. . Uniqueness of the tangent at infinity
This subsection is devoted to prove P. . The uniqueness of the tangents at infinity also implies that they are ( n + ) -uniform cones. Indeed, if for the sequence R i → ∞ we have R − ( n + ) i µ R i (cid:42) ν , for any λ > ( λ R i ) − ( n + ) µ λ R i (cid:42) λ − ( n + ) ν λ .Therefore the uniqueness of the tangent implies that λ − ( n + ) ν λ = ν for any λ >
0. Thus ν is a ( n + ) -uniformcone by Definition . .The idea behind the proof of the uniqueness of the tangent at infinity is the following. Let ν ∈ Tan n + ( µ , ∞ ) and fix a point w ∈ supp ( ν ) . If we can find a continuous curve γ : [ ∞ ) → H n contained in supp ( µ ) for which:lim t → ∞ D t ( γ ( t )) = w ,then w ∈ ξ for any ξ ∈ Tan n + ( µ , ∞ ) by Proposition . (iii), since D t ( γ ( t )) ∈ t − ( n + ) µ t .In the various cases, the curve γ will always be constructed inside the quadric K ( b , Q , T ) supporting µ andits initial point γ ( ) will always be a point of supp ( µ ) . In order to make sure that the whole γ is containedin supp ( µ ) , we force γ to avoid the singular set Σ ( F ) , so that continuity implies that γ contained in just oneconnected component of K ( b , Q , 0 ) \ Σ ( F ) . Since the staring point was contained in supp ( µ ) by hypothesis, thisimplies that γ must be contained in supp ( µ ) by Proposition . . Theorem . . Suppose µ is a ( n + ) -uniform measure. Then Tan n + ( µ , ∞ ) is a singleton. isconnectedness of ( n + ) - uniform cones implies rigidity of tangents 36 Proof.
Assume that supp ( µ ) ⊆ K ( b , Q , T ) for some non-zero Q ∈
Sym ( n ) . We can assume without loss ofgenerality that b (cid:54) =
0. Indeed if b =
0, the singular set Σ ( F ) and the quadric K ( Q , T ) are dilation invariant.Therefore the measure µ by Proposition . is dilation invariant too. This implies that the tangent at infinity to µ coincides with µ itself and in particular Tan n + ( µ , ∞ ) is a singleton. In the following the measure ν will be alwaysconsidered an arbitrary element of Tan n + ( µ , ∞ ) . We also let { R i } be the sequence for which R − n − i µ R i (cid:42) ν with R i → ∞ .First of all we study the simpler case in which T (cid:54) =
0. For the reader’s convenience we recall that f and Σ ( f ) were introduced in the Section in ( ) and ( ), respectively.If T (cid:54) = ( µ ) = K ( b , Q , T ) , for any w ∈ K ( Q , T ) , the curve: t (cid:55)→ ( tw H , −(cid:104) tb + t Q w H , w H (cid:105) / T ) = : γ w ( t ) ,is contained in supp ( µ ) and we have lim t → ∞ D t ( γ w ( t )) = w . This by the arbitrariness of w and Proposition . implies that K ( Q , T ) ⊆ supp ( ν ) . On the other hand Proposition . implies the previous inclusion holds as anequality and thus Proposition . implies that Tan n + ( µ , ∞ ) is a singleton.If T (cid:54) = ( µ ) (cid:32) K ( b , Q , T ) , Proposition . implies that n = ( µ ) is contained in theimage under f (see ( )) of one of the two connected components in which R is splitted by Σ ( f ) (see ( )). Let p , v ∈ R be such that p + span ( v ) = Σ ( f ) and µ = S f ( C ) where C : = p + H + ( v ) and H + ( v ) : = { z ∈ R n : (cid:104) v , z (cid:105) ≥ } . For any w ∈ H + ( v ) the curve: t (cid:55)→ ( p + tw H , −(cid:104) p + tw H , Q ( p + tw H ) + b (cid:105) / T ) = : γ w ( t ) ,is contained in supp ( µ ) and lim t → ∞ D t ( γ w ( t )) = w . This implies by Proposition . that { ( z , −(cid:104) z , Q z (cid:105) / T ) : z ∈ H + ( v ) } = supp ( ν ) .We are left to prove the thesis in the case in which T =
0. In this case it is harder to build the curves wementioned above because of the presence of the singular set Σ ( F ) . We will need to distinguish two cases in orderto rule out this problem.If T = n ∈ R n for which ν ( V ( n )) >
0, then V ( n ) ⊆ K ( Q , 0 ) . Arguing as in the proof ofTheorem . , we deduce that either rk ( Q ) = ( Q ) =
2. Proposition . and Proposition . prove the thesisof the proposition in these cases. We can therefore assume without loss of generality that ν ( V ( n )) = n ∈ R n . This in particular implies that rk ( Q ) ≥ . we deduce that Q is not semidefinite.For the remainder of the proof we should consider w ∈ supp ( ν ) fixed. By Proposition . we can also find asequence { v i } ⊆ supp ( µ ) for which D R i ( v i ) → w . We can also assume that these { v i } are contained in thesame connected component of K ( b , Q , 0 ) \ Σ ( F ) .At first assume that Σ ( F ) (cid:54) = ∅ . Since ν ( V ( b )) =
0, we can also assume without loss of generality that (cid:104) b , w H (cid:105) (cid:54) =
0. Let ˜ x ∈ Σ ( F ) and define: γ ( t ) : = (cid:0) ( v i ) H + tw H + t θ ( t )( ˜ x ) H , ( v i ) T + t w T (cid:1) where θ ( t ) : = (cid:104) Q [( v i ) H ] + b , w H (cid:105) / (cid:104) b , ( v i ) H + tw H (cid:105) . If i is big enough then (cid:104) b , ( v i ) H + tw H (cid:105) (cid:54) = D R i ( v i ) → w ) and thus θ ( t ) is well defined for t ≥
0. First we check that lim t → ∞ D t ( γ ( t )) = w . Indeed:lim t → ∞ D t ( γ ( t )) = lim t → ∞ (cid:18) ( v i ) H + tw H + t θ ( t ) ˜ xt , ( v i ) T + t w T t (cid:19) = (cid:16) w H + lim t → ∞ θ ( t )( ˜ x ) H , w T (cid:17) = w ,since lim t → ∞ θ ( t ) =
0. Secondly, we check that γ ( t ) ∈ K ( b , Q , 0 ) for any t >
0. Since ˜ x ∈ Σ ( F ) we have:2 Q ( ˜ x ) H + b = (cid:104) ( ˜ x ) H , Q ( ˜ x ) H + b (cid:105) =
0. ( )The identities in ( ) imply:0 = (cid:104) ( ˜ x ) H , 2 Q ( ˜ x ) H + b (cid:105) − (cid:104) ( ˜ x ) H , Q ( ˜ x ) H + b (cid:105) = (cid:104) ( ˜ x ) H , Q ( ˜ x ) H (cid:105) = −(cid:104) b , ( ˜ x ) H (cid:105) , ( )Thus ( ) together with fact that w ∈ K ( Q , 0 ) imply: (cid:104) ( γ ( t )) H , b + Q ( γ ( t )) H (cid:105) = t (cid:104) w H , b (cid:105) + t (cid:104) ( v i ) H , 2 Q w H (cid:105) + t θ ( t ) (cid:104) ( v i ) H , 2 Q ˜ x (cid:105) + t θ ( t ) (cid:104) w H , 2 Q ˜ x (cid:105) = t (cid:104) w H , b + Q [( v i ) H ] (cid:105) − t θ ( t ) (cid:104) ( v i ) H + tw H , b (cid:105) = isconnectedness of ( n + ) - uniform cones implies rigidity of tangents 37 where the last equality comes from the definition of θ . Thanks to ( ) and ( ) we can also prove that γ does notintersect Σ ( F ) , indeed: (cid:104) Q γ ( t ) + b , ˜ x (cid:105) = −(cid:104) γ ( t ) , b (cid:105) = −(cid:104) ( v i ) H + tw H , b (cid:105) (cid:54) = t >
0. This implies that for any t ≥
0, the curve γ is contained in the same connected component of K ( Q , 0 ) \ Σ ( F ) and since the initial point of γ is contained in supp ( µ ) , the whole curve γ by continuity iscontained in the support of µ . and lim t → ∞ D t ( γ ( t )) = w . Since D t ( γ ( t )) is contained in the support of t − ( n + ) µ t and: lim t → ∞ D t ( γ ( t )) = w ,we deduce that w ∈ supp ( ξ ) for any ξ ∈ Tan n + ( µ , ∞ ) by Proposition . . Thanks to the arbitrariness of w and ξ we deduce that by Proposition . that the tangent at infinity is unique.Finally suppose that Σ ( F ) = ∅ . For any x ∈ K ( b , Q , 0 ) we have that:0 = λ (cid:104) x H , e (cid:105) + (cid:104) b , e (cid:105)(cid:104) x H , e (cid:105) + (cid:104)Q [ P ( x H )] + b , P ( x H ) (cid:105) , ( )where e is a unitary eigenvector of Q relative to a positive eigenvalue λ and P is the orthogonal projection on e ⊥ . Since ν ( V ( e )) =
0, we can assume without loss of generality that (cid:104) w H , e (cid:105) >
0, and since w ∈ K ( Q , 0 ) wehave that: (cid:104)Q [ P ( w H )] , P ( w H ) (cid:105) = − λ (cid:104) w H , e (cid:105) <
0. ( )Since D R i ( v i ) → w , defined s ( t ) : = P [( v i ) H + tw H ] and provided i is sufficiently big, we have: (cid:104)Q s ( t ) + b , s ( t ) (cid:105) < t ≥ γ w ( t ) : = (cid:18) (cid:112) (cid:104) b , e (cid:105) − λ (cid:104)Q s ( t ) + b , s ( t ) (cid:105) − (cid:104) b , e (cid:105) λ e + s ( t ) , ( v i ) T + t w T (cid:19) ,is well defined for any t ≥
0. The component of γ w along e is by construction a solution to the equation: λ ζ + (cid:104) b , e (cid:105) ζ + (cid:104)Q s ( t ) + b , s ( t ) (cid:105) = ) implies that γ w is contained in K ( b , Q , 0 ) . Since γ w is continuous, γ w ( ) = v i ∈ supp ( µ ) and Σ ( F ) = ∅ , by Proposition . we deduce that γ w ( t ) ∈ supp ( µ ) for any t ≥
0. We are left to com-pute the limit lim t → ∞ D t ( γ ( t )) . In the case of the vertical component the computation is immediate, indeedlim t → ∞ ( γ w ( t )) T / t = w T . The limit for the horizontal components is:lim t → ∞ ( γ w ( t )) H t = lim t → ∞ (cid:112) (cid:104) b , e (cid:105) − λ (cid:104)Q s ( t ) + b , s ( t ) (cid:105) − (cid:104) b , e (cid:105) λ t e + P [ w H ]= (cid:112) − λ (cid:104)Q w H + b , w H (cid:105) λ e + P [ w H ] = w H ,where the last equality comes from the identity in ( ) and the fact that (cid:104) w H , e (cid:105) >
0. Thus lim t → ∞ D t ( γ ( t )) = w ,which implies that w ∈ supp ( ξ ) for any ξ ∈ Tan n + ( µ , ∞ ) . The same argument we used in the previous caseimplies that the tangent at infinity is unique. . Proof of Theorem . The following lemma insures that if φ is a measure with ( n + ) -density, at φ -almost every x there is a flatmeasure contained in Tan n + ( µ , x ) . Lemma . . If φ is a Radon measure with ( n + ) -density, then for φ almost every x there exists a w ∈ R n for which: S n + V ( w ) ∈ Tan n + ( φ , x ) . isconnectedness of ( n + ) - uniform cones implies rigidity of tangents 38 Proof.
Proposition . and Proposition . directly imply the claim.We are ready to finish the proof of Theorem . . The argument we will use follows closely the proof ofProposition . of [ ]. By contradiction, suppose there exists a point x such that:(i) Tan n + ( φ , x ) ⊆ Θ ( φ , x ) U H n ( n + ) ,(ii) there are ζ , ν ∈ Tan n + ( φ , x ) such that ν is flat and ζ is not flat,(iii) Proposition . holds at x .We can also assume without loss of generality that Θ ( φ , x ) =
1. Since ζ is not flat, its tangent at infinity χ cannotbe flat otherwise Theorem . would imply that ζ is flat. In particular by the assumption on the functional F wehave F ( χ ) > ¯ h . Fix r k → s k → φ x , r k r n + k (cid:42) ν and φ x , s k s n + k (cid:42) χ .We can further suppose that s k < r k . Define for any r > f ( r ) : = F ( r − ( n + ) φ x , r ) , and note thatsince F is continuous with respect to the weak- ∗ convergence of measures, f is continuous in r . Since ν is flat,then: lim r k → f ( r k ) = F ( ν ) ≤ ¯ h /2,Thus for sufficiently small r k we have f ( r k ) < ¯ h . On the other hand, since:lim s k → f ( s k ) = F ( χ ) > ¯ h ,for suffciently small s k we have that f ( s k ) > ¯ h . Fix σ k ∈ [ s k , r k ] such that f ( σ k ) = ¯ h and f ( r ) ≤ ¯ h for r ∈ [ σ k , r k ] .By compactness there exists a subsequence of { σ k } k ∈ N , not relabeled, such that σ − ( n + ) k φ x , σ k converges weakly- ∗ to a measure ξ ∈ U H n ( n + ) . Clearly by continuity: F ( ξ ) = lim σ k → f ( σ k ) = ¯ h .Note that r k / σ k → ∞ , otherwise if for some subsequence not relabeled, we had that r k / σ k converged to a constant C (larger than 1) we would conclude that ξ C C n + = ν since: ν = lim k → ∞ φ x , r k r n + k = lim k → ∞ (cid:18) σ k r k (cid:19) n + (cid:32) φ x , σ k σ n + k (cid:33) r k .In particular ξ would be flat, which is not possible as F ( ξ ) = ¯ h . Note that for any given R > ( R σ k ) − ( n + ) φ x , R σ k (cid:42) R − ( n + ) ξ R .This by continuity of F implies that: F ( R − ( n + ) ξ R ) = lim k → ∞ f ( R σ k ) .Moreover, since r k / σ k → ∞ we conclude that for any R > R σ k ∈ [ σ k , r k ] whenever k is largeenough. This, by our choice of σ k and r k , implies that: F ( R − ( n + ) ξ R ) = lim k → ∞ f ( R σ k ) ≤ ¯ h , ( )for every R ≥
1. Let ψ be the tangent measure at infinity to ξ , which by Theorem . is unique and it is a cone.Therefore, thanks to ( ) we have that: F ( ψ ) = lim R → ∞ F ( R − ( n + ) ξ R ) ≤ ¯ h ,and in particular thanks to the assumptions on the functional F , we have that ψ ∈ M ( n + ) . This is incontradiction with Theorem . , which would imply that ξ is flat. imits of sequences of horizontal ( n + ) - uniform cones 39 ( n + ) - uniform cones The main result of Section , Theorem . , implies that if we can prove that flat measures are quantitativelydisconnected from the other ( n + ) -uniform cones, measures with ( n + ) -density have only flat tangents. Afirst step towards this direction is to show that horizontal ( n + ) -uniform cones are disconnected from ( n + ) -vertical cones.Let { µ i } i ∈ N be a sequence of horizontal ( n + ) -uniform cones supported on the quadrics K ( D i , − ) . Sup-pose these measures are weakly converging to a measure ν which by Proposition . is a ( n + ) -uniform cone.We define also the following sequence of matrices: Q i : = − D i |||D i ||| ,where with the symbol ||| A ||| we denote the operatorial norm of the matrix A . Furthermore we can assume, upto non-relabeled subsequences, that −D i / |||D i ||| converges to some matrix Q ∈
Sym ( n ) with |||Q||| = ν is vertical if and only if the sequenceof the norms |||D i ||| diverges. Secondly, we prove that the only possible vertical limits of sequences of horizontal ( n + ) -uniform cones are flat measures. This second step of the section is where the results of Appendix Bcome into play. Indeed Theorem B. applies to µ i for any i ∈ N and it forces D i to satisfy the following identityfor any z ∈ R n for which ( D i + J ) z (cid:54) = ( D i ) − (cid:104) n i , D i n i (cid:105) + (cid:104) n i , D i n i (cid:105) ( n − ) + n − n − − + (cid:104)D i J n i , n i (cid:105) n − − ( Tr ( D i ) − (cid:104) n i , D i n i (cid:105) ) ( n − ) =
0. ( )where n i ( z ) : = ( D i + J ) z | ( D i + J ) z | is the so called horizontal normal to the graph of f ( x ) = (cid:104) x , D i x (cid:105) at the point ( x , f ( x )) .Since identity ( ) holds for any i ∈ N , we can deduce some constraints (see Proposition . ) for the limit matrix Q . Using a similar argument we are able to prove in Proposition . that the sequence of the second biggest (inmodulus) eigenvalue of D i is bounded. Putting together these information, we are able to prove in Proposition . that there must exists a constant depending only on n which bounds the biggest eigenvalue of the matrices D i associated quadrics K ( D i , − ) support a horizontal ( n + ) -uniform measure.The following proposition shows the equivalence between the geometric condition for ν to be vertical ( n + ) -uniform cone and the algebraic condition on the divergence of the sequence (cid:8) |||D i ||| (cid:9) i ∈ N . This will be very usefulin the forthcoming computations. Proposition . . Let { µ i } i ∈ N and ν as above. The following are equivalent:(i) ν is supported on K ( Q , 0 ) where Q is the limit of D i / |||D i ||| as above,(ii) lim i → ∞ |||D i ||| = ∞ .Proof. Proposition . implies that for any y ∈ supp ( ν ) there exists a sequence { y i } i ∈ N such that y i ∈ supp ( µ i ) and y i → y . Assume at first the sequence |||D i ||| is bounded. This implies that we can find a subsequence D i (notrelabeled) such that the matrices D i converge to some D ∈
Sym ( n ) . Thanks to our assumption on y i we knowthat ( y i ) T = (cid:104) ( y i ) H , D i [( y i ) H ] (cid:105) . Thus taking the limit as i to infinity, we get: y T = (cid:104) y H , D y H (cid:105) .Therefore if |||D i ||| is bounded, we have that ν is supported on both the quadrics K ( Q , 0 ) and K ( D , − ) ,however this is not possible thanks to Proposition . which implies that either T = T (cid:54) = ( µ ) . Viceversa, if we suppose that |||D i ||| diverges, with some iterations of the triangle inequality,we deduce the follwoing bound: |(cid:104) y H , Q y H (cid:105)| ≤(cid:107)Q − Q i (cid:107)| y H | + | y H − ( y i ) H | ( | y H | + | ( y i ) H | ) + | ( y i ) T ||||D i ||| . ( )The right-hand side of the above inequality goes to 0 as i → ∞ since we can assume without loss of generalitythat (cid:107) y i (cid:107) ≤ (cid:107) y (cid:107) . Therefore for any y ∈ supp ( ν ) we have that (cid:104) y H , Q y H (cid:105) = ( ν ) ⊆ K ( Q , 0 ) . imits of sequences of horizontal ( n + ) - uniform cones 40 Remark . . For later convenience we remark that up to considering an isometric copy of the sequence { µ i } i ∈ N ,we can assume that the biggest eigenvalue of Q is 1. Without loss of generality we can assume that up to anon-relabeled subsequence, the biggest eigevalues have the same sign. Let U ∈ S ( n ) and recall that the map Ξ U introduced in Proposition . is a surjective isometry. In particular Propositions . and . imply that ( Ξ U ) µ isa ( n + ) -uniform measure. Moreover a routine computation shows that:supp (( Ξ U ) µ ) ⊆ K ( Ub , U Q U T , s ( U ) T ) . ( )Suppose now that the biggest eigenvalue in modulus of Q is − U ∈ S ( n ) for which s ( U ) = − s was defined after ( )). Since µ i (cid:42) ν and supp ( µ i ) ⊆ K ( D i , − ) , we have that: ( Ξ U ) µ i (cid:42) ( Ξ U ) ν and supp (( Ξ U ) µ i ) ⊆ K ( U D i U T , 1 ) = K ( − U D i U T , − ) by ( )In particular, if |||D i ||| → ∞ , Proposition . implies that supp (( Ξ U ) ν ) ⊆ K ( − U Q U T , 0 ) . Let v be an eigenvec-tor of Q relative to the eigenvalue −
1, and note that: − U Q U T ( Uv ) = − U Q v = Uv .This argument also shows that if µ is a horizontal ( n + ) -uniform cone, we can always assume without lossof generality that supp ( µ ) ⊆ K ( D , − ) where the biggest eigenvalue of D is positive.From now on we shall always assume that |||D i ||| → ∞ , or in other words that the limit measure ν is a vertical ( n + ) -uniform cone, and that 1 is an eigenvalue of Q . In the following proposition we show how the quadric K ( Q , 0 ) “remembers” that the measure ν is the limit of a sequence of horizontal ( n + ) -uniform cones. Proposition . . Let Q be the limit of the matrices D i / |||D i ||| as above. Then, for any h (cid:54)∈ Ker ( Q ) we have: ( Tr ( Q ) − (cid:104) n , Q n (cid:105) + (cid:104) n , Q n (cid:105) ) − ( Tr ( Q ) − (cid:104) n , Q n (cid:105) ) =
0, ( ) where n : = Q h |Q h | .Proof. Since h (cid:54)∈ ker ( Q ) , there exists N ∈ N for which for any i ≥ N we have: ( D i + J ) h (cid:54) =
0. ( )Indeed, if there was a (non-relabeled) subsequence of indeces for which ( D i + J ) h =
0, dividing the above equalityby |||D i ||| and sending i to infinity, we would deduce that Q h =
0. Note that defined n i ( h ) : = ( D i + J ) h / | ( D i + J ) h | , we have that: lim i → ∞ n i ( h ) = Q h |Q h | = n .Therefore Theorem B. implies that for any i for which ( ) holds, we have:Tr ( D i ) − (cid:104) n i , D i n i (cid:105) + (cid:104) n i , D i n i (cid:105) ( n − ) + n − n − − + (cid:104)D i J n i , n i (cid:105) n − − ( Tr ( D i ) − (cid:104) n i , D i n i (cid:105) ) ( n − ) =
0. ( )As already remarked, equation ( ) holds definitely, therefore dividing ( ) by |||D i ||| and taking the limit as i goes to infinity, we obtain ( ).The following corollary is an easy application of Proposition . in the case h is a non-zero eigenvector of Q . Corollary . . Let Q be the limit of the matrices D i / |||D i ||| as above. Then, every non-zero eigenvalue λ of Q satisfies theequation: − λ + Tr ( Q ) λ + ( Tr ( Q ) − Tr ( Q ) ) =
0. ( ) In particular Q has at most two distinct non-zero eigenvalues. imits of sequences of horizontal ( n + ) - uniform cones 41 Proof.
For any h ∈ R n unitary eigenvector relative to the eigenvalue λ (cid:54) =
0, we have n ( h ) = Q h |Q h | = sgn ( λ ) h .Thanks to Proposition . we have that:2 ( Tr ( Q ) − λ + λ ) − ( Tr ( Q ) − λ ) = λ , proves the corollary.The following proposition shows that the non-zero eigenvalue of Q different from 1 does not exists. Thisimplies that Q is semidefinite and ν is flat. Proposition . . Suppose lim i → ∞ |||D i ||| = ∞ and lim i → ∞ D i / |||D i ||| = Q . Then rk ( Q ) = and in particular the limit ν of the sequence { µ i } is flat.Proof. Since 1 is an eigenvalue of Q , Corollary . implies that 1 solves the equation ( ). Therefore we have: − + ( Q ) + ( ( Q ) − Tr ( Q ) ) =
0. ( )Using the above equality we deduce that ( ) becomes: − λ + ( Q ) λ + ( − ( Q )) = λ = λ = ( Q ) −
1. ( )If λ ≥
0, the matrix Q would be semi-definite. Since K ( Q , 0 ) supports a ( n + ) -uniform measure, Proposition . implies that rk ( Q ) = . implies that µ must be flat. Therefore we can assume withoutloss of generality that λ <
0. Let now k be the dimension of the eigenspace relative to the eigenvalue 1 and k be the dimension of the eigenspace relative to the eigenvalue λ . By assumption k ≥ k ≥
1, otherwise Q would be semidefinite and Proposition . and Proposition . imply that µ is flat. This implies, by ( ) that:Tr ( Q ) = k + λ k = k + ( Q ) k − k .The above equation together with ( ), allows us to express Tr ( Q ) , Tr ( Q ) as functions of k , k :Tr ( Q ) = ( k − k ) − k and Tr ( Q ) = k + (cid:18) k − − k (cid:19) k . ( )Substituting in ( ) the above identities, we deduce that: − + ( k − k ) − k + k + (cid:18) k − − k (cid:19) k − ( k − k ) ( − k ) = = k ( k − ) + k ( k − k + ) + ( − k + k − ) ,which allow us to express k in terms of k . Indeed the solutions of the above quadratic equation are: k = k − k − k = − k .The only couple of natural numbers for which k = k − k − holds is ( k , k ) = (
1, 0 ) , which has been already takenin consideration. On the other hand, there are four couples of natural numbers for which k = − k , whichare (
3, 0 ) ; (
2, 1 ) ; (
2, 1 ) ; (
0, 3 ) . However the only couples of natural numbers for which Q is not semi-defined are (
1, 2 ) and (
2, 1 ) . In both these cases ( ) implies that Tr ( Q ) = ) and therefore λ = ), which is incontradiction with the assumption that λ < imits of sequences of horizontal ( n + ) - uniform cones 42 We introduce here further notation. For any i ∈ N we let λ i ( ) , . . . , λ i ( n ) to be the eigenvalues of D i . Sucheigenvalues are ordered in the following way: |||D i ||| = λ i ( ) ≥ | λ i ( ) | ≥ . . . ≥ | λ i ( n ) | ,where we can assume without loss of generality that the biggest eigenvalue in modulus is positive thanks toRemark . . We further let { e i ( ) , . . . , e i ( n ) } be an orthonormal basis of R n for which e i ( j ) is an eigenvectorrelative to the eigenvalue λ i ( j ) .The following proposition will allow us to show in Proposition . that ν must be a horizontal ( n + ) -uniformcone. If this is not the case, we know by Proposition . that supp ( ν ) is a vertical hyperplane V . On the otherhand, Proposition . implies that while λ ( j ) is diverging, the other eigenvalues remain bounded. Therefore thelimit of the quadrics K ( D i , − ) must coincide with supp ( µ ) by Proposition . but on the other hand it mustbe a proper subset of V :Figure : The image shows why a sequence of quadrics having only one eigenvalue diverging cannot convergeto a full plane. Proposition . . Let the sequence of matrices {D i } be as above, i.e.:(i) K ( D i , − ) for any i ∈ N supports a ( n + ) -uniform cone µ i which converges to some ν ,(ii) lim i → ∞ |||D i ||| = lim i → ∞ λ i ( ) = ∞ . Then the sequence { λ i ( ) } i ∈ N is bounded.Proof. By contradiction assume that | λ i ( ) | → ∞ and let e i : = e i ( ) and λ i : = λ i ( ) . We want to apply TheoremB. to the points h = e i . In order to do so we need to give an explicit expression to the quantities involved in( ) of Theorem B. . First we compute the horizontal normal of K ( D i , − ) at the point ( e i , (cid:104) e i D i e i (cid:105) ) : n i : = n ( e i ) : = D i e i + Je i |D i e i + Je i | = λ i e i + Je i | λ i e i + Je i | = λ i e i + Je i ( + λ i ) Secondly we compute the quantities (cid:104) n i , D i n i (cid:105) , (cid:104) n i , D i n i (cid:105) and (cid:104) n i , D i J n i (cid:105) : (cid:104) n i , D ki n i (cid:105) = (cid:42) λ i e i + Je i ( + λ i ) , D ki (cid:32) λ i e i + Je i ( + λ i ) (cid:33)(cid:43) = λ ki − λ ki + (cid:104) e i , J D ki Je i (cid:105) + λ i , (cid:104) n i , D i J n i (cid:105) = (cid:42) λ i e i + Je i ( + λ i ) , D i J (cid:32) λ i e i + Je i ( + λ i ) (cid:33)(cid:43) = λ i − λ i (cid:104) e i , J D i Je i (cid:105) + λ i . ( ) imits of sequences of horizontal ( n + ) - uniform cones 43 In order to further simplify the notation, we define: ( I ) i : = Tr ( D i ) − (cid:104) n i , D i n i (cid:105) + (cid:104) n i , D i n i (cid:105) , ( I I ) i : = Tr ( D i ) − (cid:104) n i , D i n i (cid:105) .With these notations, Theorem B. applied to h = e i turns into: ( I ) i ( n − ) + n − n − − + λ i − λ i (cid:104) e i , J D i Je i (cid:105) + λ i − ( I I ) i ( n − ) =
0. ( )We give now a more explicit description of both ( I ) i and ( I I ) i using the expressions for (cid:104) n i , D i n i (cid:105) , (cid:104) n i , D i n i (cid:105) and (cid:104) n i , D i J n i (cid:105) we found in ( ): ( I ) i = Tr ( D i ) − (cid:104) n i , D i n i (cid:105) + (cid:104) n i , D i n i (cid:105) = n ∑ j = λ i ( j ) − λ i − λ i + (cid:104) e i , J D i Je i (cid:105) + λ i + (cid:32) λ i − λ i + (cid:104) e i , J D i Je i (cid:105) + λ i (cid:33) = ∑ j (cid:54) = λ i ( j ) − λ i + (cid:104) e i , J D i Je i (cid:105) + λ i − λ i λ i + (cid:104) e i , J D i Je i (cid:105) + λ i + ( λ i + (cid:104) e i , J D i Je i (cid:105) ) ( + λ i ) , ( I I ) i = Tr ( D i ) − (cid:104) n i , D i n i (cid:105) = ∑ j (cid:54) = λ i ( j ) − λ i + (cid:104) e i , J D i Je i (cid:105) + λ i . ( )The absurd assumption that λ i → ∞ has the following consequences. First of all:lim i → ∞ (cid:104) e i , J ( D i / |||D i ||| ) k Je i (cid:105) ( + λ i ) =
0, ( )for any k ∈ N . Secondly, thanks to ( ) and ( ) we deduce that:lim i → ∞ ( I ) i |||D i ||| = i → ∞ ( I I ) i |||D i ||| =
1, ( )where we used the fact that by definition λ i ( ) = |||D i ||| and that lim i → ∞ λ i ( j ) / |||D i ||| = j ∈ {
2, . . . , 2 n } thanks to Proposition . . Dividing by |||D i ||| the left-hand side of the identity ( ), and sending i → ∞ yields,thanks to ( ):lim i → ∞ ( I ) i / |||D i ||| ( n − ) − (( I I ) i / |||D i ||| ) ( n − ) + (cid:16) n − n − − (cid:17) |||D i ||| + ( λ i / (cid:107)D i (cid:107) ) − ( λ i / |||D i ||| ) (cid:104) e i , J ( D i / |||D i ||| ) Je i (cid:105) + λ i = ( n − ) ,which shows that the assumption λ i → ∞ is absurd thanks to fact that identity ( ) holds for every i ∈ N . Proposition . . There exists a constant C ( n ) > such that if K ( D , − ) supports a ( n + ) -uniform measure, then |||D||| ≤ C ( n ) .Proof. By contradiction assume that there exists a sequence { µ i } i ∈ N of ( n + ) -uniform measures such that K ( D i , − ) supports µ i and |||D i ||| → ∞ . By the compactness of measures we can extract a converging subse-quence and by Propositions . and . we deduce that the limit ν is flat. On the other hand Proposition . implies that with the exception of λ ( i ) , the eigenvalues are bounded in modulus by some constant C > { µ i } ). For any y ∈ supp ( ν ) , Proposition . implies that we can find a sequence { y i } ⊆ H n such that y i ∈ supp ( µ i ) and y i → y . For such a sequence { y i } , we have: ( y i ) T = (cid:104) ( y i ) H , D i ( y i ) H (cid:105) = n ∑ j = λ i ( j ) (cid:104) e i ( j ) , ( y i ) H (cid:105) ≥ n ∑ j = λ i ( j ) (cid:104) e i ( j ) , ( y i ) H (cid:105) ≥ − ( n − ) C | ( y i ) H | .Sending i to infinity we deduce that y T ≥ − ( n − ) C | y H | , which constitutes a non-trivial bound on y T . Thiscomes in contradiction with the fact that supp ( ν ) must contain a vertical hyperplane as ν is flat. imits of sequences of horizontal ( n + ) - uniform cones 44 Proposition . . There exists a constant C ( n ) > such that if K ( D , − ) supports a ( n + ) -uniform measure, then |||D||| ≥ C ( n ) .Proof. Repeatedly applying the triangle inequality, and using the following trivial bounds:Tr ( D ) ≤ n |||D||| , | Tr ( D ) | ≤ n |||D||| , |(cid:104) v , D k v (cid:105)| ≤ |||D||| k | v | , |(cid:104) v , DJv (cid:105)| ≤ |||D|||| v | ,for any v ∈ S n − we have: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Tr ( D ) − (cid:104) v , D v (cid:105) + (cid:104) v , D v (cid:105) ( n − ) + (cid:104)D Jv , v (cid:105) n − − ( Tr ( D ) − (cid:104) v , D v (cid:105) ) ( n − ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ( n + n + ) |||D||| + |||D||| ( n − ) .Theorem B. implies that if K ( D , − ) supports a ( n + ) -uniform measure, identity ( ) must be satisfiedfor any e ∈ S n − for which ( D + J ) e (cid:54) =
0. In particular we have that: (cid:12)(cid:12)(cid:12)(cid:12) n − n − − (cid:12)(cid:12)(cid:12)(cid:12) ≤ ( n + n + ) |||D||| + |||D||| ( n − ) . ( )With some algebraic computations, which we omit, we see that the positive solutions in ||| D ||| to ( ) are containedin [ √ n + n + ∞ ) . Theorem . . There exists a constant C ( n ) > such that for any horizontal m ∈ S n − and any ( n + ) -uniform cone µ we have: ˆ B ( ) (cid:104) m , z H (cid:105) d µ ( z ) ≥ C ( n ) . Proof.
We can suppose without loss of generality that supp ( µ ) ⊆ K ( D , − ) . Therefore Proposition . impliesthat µ = S n + ( µ ) and by Proposition A. for any positive Borel function h : H n → R we have: ˆ h ( z ) µ ( z ) = ˆ h ( z ) d S n + ( µ ) ( z ) = c n ˆ π H ( supp ( µ )) h ( z ) | ( D + J ) y | dy .If n >
1, Proposition . implies that π H ( supp ( µ )) = R n , and thus: c n ˆ B ( ) (cid:104) m , z H (cid:105) d µ ( z ) = ˆ | y | + (cid:104) y , D y (cid:105) ≤ (cid:104) m , y (cid:105) | ( D + J ) y | dy = ˆ S n − (cid:104) m , v (cid:105) | ( D + J ) v | ˆ ( + (cid:104) v , D v (cid:105) ) − r n + drd σ ( v )= ( n + ) ˆ S n − (cid:104) m , v (cid:105) | ( D + J ) v | ( + (cid:104) v , D v (cid:105) ) n + d σ ( v ) , ( )where σ : = H n − eu (cid:120) S n − and in the second equality we performed the change of variables y = rv in R n . Since K ( D , 1 ) supports a uniform measure, Proposition . implies that |||D||| ≤ C ( n ) , and thus thanks to ( ), wehave: ˆ B ( ) (cid:104) m , z H (cid:105) d µ ( z ) ≥ ´ S n − (cid:104) m , v (cid:105) | ( D + J ) v | d σ ( v )( n + ) c n ( + C ( n ) ) n + . ( )Suppose e ∈ S n − is the vector at which D attains the operatorial norm, i.e., |D e | = |||D||| . Then e is aneigenvector of D and thus | ( D + J ) e | = ||| D ||| +
1. Therefore for any u ∈ S n − ∩ B ( e ) , we have: | ( D + J ) u | ≥| ( D + J ) e | + (cid:104) ( D + J ) e , ( D + J )( u − e ) (cid:105) ≥ ( |||D||| + ) − |||D + J ||| | u − e |≥ ( |||D||| + )( − | u − e | ) ≥ ( C ( n ) + ) /2 = : C ( n ) , ( )where in the second last inequality we used the Jensen inequality |||D + J ||| ≤ ( |||D||| + ) and in the last onethe bound on |||D||| yielded by Proposition . . Putting together ( ) and ( ) we deduce that: ˆ B ( ) (cid:104) m , z H (cid:105) d µ ( z ) ≥ C ( n ) ´ S n − ∩ U ( e ) (cid:104) m , v (cid:105) d σ ( v )( n + ) c n ( + C ( n ) ) n + ≥ C ( n ) C ( n )( n + ) c n ( + C ( n ) ) n + = : C ( n ) , ( ) isconnection of vertical non - flat cones and flat measures 45 where C ( n ) : = min m ∈ S n − ´ S n − ∩ U ( e ) (cid:104) m , v (cid:105) d σ ( v ) and as usual U ( e ) denotes the Euclidean ball of radius1/8 and centre e in R n .The case in which n = π H ( supp ( µ )) is a half plane, the argument is the same. The only difference is thatwe have be careful to choose the vector e at which D attains the operatorial norm in π H ( supp ( µ )) . This choice of e together with the computations in ( ) imply that U ( e ) ⊆ π H ( supp ( µ )) and therefore ( ) holds. - flat cones and flat measures This section is devoted to prove that non-flat vertical ( n + ) -uniform cones are quantitatively disconnectedfrom flat measures. To be precise, we prove that there is a universal constant C ( n ) > µ is avertical ( n + ) -uniform cone and: min m ∈ S n − ˆ B ( ) (cid:104) m , z H (cid:105) d µ ( z ) ≤ C ( n ) , ( )then µ is flat. The first step towards the proof is to use the study of the support of vertical ( n + ) -uniform conescarried on in Subsection . and the representation formulas of the perimeter given in Appendix A to obtain thefollowing more explicit expression for the quadric containing µ (see Proposition . ). For any w ∈ supp ( µ ) , wehave that: | w H | − ( n − ) (cid:104) w H , u (cid:105) d ω µ ( u ) =
0, ( )where ω µ : = H n − eu (cid:120) π H ( supp ( µ )) ∩ S n − . The existence of C ( n ) , as showed in Proposition . , is a directconsequence of ( ) by means of few algebraic manipulations.The following technical lemma is a consequence of the coarea formula and the representation formulas for theintrinsic perimeter we proved in Appendix A. . Lemma . . Let f : R → R and g : R n → R be positive Borel functions and µ a vertical ( n + ) -uniform cone. Defined ω µ : = H n − eu (cid:120) S n − ∩ π H ( supp ( µ )) , we have that:(i) ˆ B ( ) g ( z H ) d µ ( z ) = c n ˆ r n − (cid:112) − r ˆ g ( rv ) d ω µ ( v ) dr , (ii) ˆ f ( z T ) g ( z H ) d µ ( z ) = c n ˆ f ( t ) dt · ˆ ∞ r n − ˆ g ( rv ) d ω µ ( v ) dr . Proof.
Suppose supp ( µ ) ⊆ K ( D , 0 ) . As a first step we prove that for any positive Borel function h : R n → R ,defined S : = π H ( supp ( µ )) , we have that: ˆ h ( y ) d H n − eu (cid:120) S = ˆ ∞ r n − ˆ h ( rv ) d ω µ ( v ) dr , ( )where ω µ is as above. To do so, for any δ > U ( ker ( D ) , δ ) be the open Euclidean neighbourhood of radius δ of the set ker ( D ) and define K δ : = π H ( K ( D , 0 )) \ U ( ker ( D ) , δ ) . The set K δ is a smooth submanifold of R n and since the function u ( x ) : = | x | is smooth on R n \ U ( ker ( D ) , δ ) , we can apply the coarea formula in Remark . of Chapter of [ ] with the choice g = χ A ∩ S , where A is any Borel set of H n . We therefore obtain: ˆ A ∩ S ∩ K δ J ∗ K δ u ( x ) d H n − eu = ˆ R H n − eu ( A ∩ S ∩ K δ ∩ r S n − ) dr ,where J ∗ K δ u = |∇ T u | is the Jacobian of the tangetial gradient of u along K δ . We claim that J ∗ K δ u = K δ . Toprove this, we note that the vector x / | x | is contained in the tangent to K δ for any x ∈ K δ . This is due to thefact that the curve γ ( s ) : = x + sx | x | is contained in K δ provided s ≥
0. Complete x / | x | to an othonormal basis of isconnection of vertical non - flat cones and flat measures 46 Tan ( K δ , x ) with vectors v , . . . , v n − and note that ∇ T u [ v i ] = (cid:104)∇ u , v i (cid:105) = i =
2, . . . , 2 n −
1. This impliesthat |∇ T u | = |(cid:104)∇ u ( x ) , x / | x |(cid:105)| = H n − eu ( A ∩ S ∩ K δ ) = ˆ R H n − eu ( A ∩ S ∩ K δ ∩ r S n − ) dr ,Therefore sending δ to 0, by Beppo-Levi convergence theorem we get: H n − eu (cid:120) S ( A ) = ˆ R H n − eu ( A ∩ S ∩ r S n − ) dr = ˆ R r n − H n − eu (cid:18) Ar ∩ S n − ∩ S (cid:19) dr = ˆ R r n − ˆ χ A ( rv ) d H n − eu (cid:120) S n − ∩ S ( v ) dr .A standard approximation procedure with simple function proves the claim thanks to Beppo-Levi convergencetheorem for a general positive measurable function.We are ready to prove the identities (i) and (ii) of the statement. Thanks to Lemma A. we have that for anypositive Borel function G : H n → R we have: ˆ G ( z ) d µ ( z ) = c n ˆ G ( z H , z T ) d H n − eu (cid:120) S ( z H ) ⊗ dz T . ( )Therefore, in order to prove identity (i), we choose G ( z ) : = χ B ( ) ( z ) g ( z H ) and note that combining ( ) and ( ),we get: ˆ B ( ) G ( z ) d µ ( z ) = c n ˆ | y | + t ≤ g ( y ) d H n − eu (cid:120) S ( y ) ⊗ L ( t ) = c n ˆ | y |≤ g ( y ) (cid:113) − | y | d H n − eu (cid:120) S ( y )= c n ˆ r n − (cid:112) − r ˆ g ( rv ) d ω µ ( v ) dr .Moreover, (ii) follows immediately with the choice G ( z ) : = f ( z T ) g ( z H ) : ˆ B ( ) G ( z ) d µ ( z ) = c n ˆ f ( t ) g ( y ) d H n − eu (cid:120) S ( y ) ⊗ dt = c n ˆ f ( t ) dt · ˆ g ( y ) d H n − eu (cid:120) S ( y )= c n ˆ f ( t ) dt · ˆ ∞ r n − ˆ g ( rv ) d ω µ ( v ) dr ,where in the first identity we applied ( ), in the second Tonelli’s theorem and in the last one ( ).The following proposition is a refinement of Lemma . in case µ is a vertical ( n + ) -uniform cone. In sucha case T ( ) = Q ( ) ( T and Q were introduced in Definition . ) can be explicitlycomputed thanks to Lemma . . Proposition . . Suppose µ is a vertical ( n + ) -uniform cone. For any w ∈ supp ( µ ) we have: | w H | = ( n − ) (cid:104) w H , u (cid:105) d ω µ ( u ) , ( ) where ω µ = H n − eu (cid:120) S n − ∩ π H ( supp ( µ )) .Proof. Since µ is a cone, Proposition . implies that for any w ∈ supp ( µ ) we have: (cid:104) w H , Q ( ) w H (cid:105) + T ( ) w T =
0. ( )Moreover, since by assumption µ is a vertical ( n + ) -uniform cone, thanks to Proposition . we have T ( ) = Q ( ) . isconnection of vertical non - flat cones and flat measures 47 Let ϕ : R n → R be a k -homogeneous function and i ∈ N . Lemma . (ii) and the fact that z iT ϕ ( z ) e −(cid:107) z (cid:107) ∈ L ( µ ) imply that: ˆ z iT ϕ ( z H ) e −(cid:107) z (cid:107) d µ ( z ) = c n ˆ t i e − t dt · ˆ ∞ r n − ˆ ϕ ( rv ) e − r d ω µ ( v ) dr = c n ˆ t i e − t dt · ˆ ∞ r n + k − e − r dr · ˆ ϕ ( v ) d ω µ ( v ) Thanks to the above identity, we deduce that if i is odd ´ z iT ϕ ( z H ) e −(cid:107) z (cid:107) d µ ( z ) =
0, while if i is even: ˆ z iT ϕ ( z H ) e −(cid:107) z (cid:107) d µ ( z ) = Γ (cid:16) n − + k (cid:17) Γ (cid:16) i + (cid:17) c n ˆ ϕ ( v ) d ω µ ( v ) . ( )In order to compute the matrix Q ( ) , it will be convenient to make use of the notation introduced in the proofof Proposition . , where we defined the matrices Q , Q and Q , Q in ( ) and ( ) respectively. We define C ( n ) : = Γ (cid:16) n + (cid:17) Γ (cid:16) (cid:17) / c n C ( n + ) , where the constant C ( n + ) was introduced in Definition . . Thanksto ( ) and few algebraic calculations, we deduce that: Q ( ) C ( n ) = C ( n + ) ˆ | z H | z H ⊗ z H + z T Jz H ⊗ Jz H e −(cid:107) z (cid:107) d µ ( z ) = ( n + ) ˆ v ⊗ vd ω µ ( v ) + ˆ Jv ⊗ Jvd ω µ ( v ) , Q ( ) C ( n ) = C ( n + ) ˆ | z H | z T ( z H ⊗ Jz H + Jz H ⊗ z H ) e −(cid:107) z (cid:107) d µ ( z ) = Q ( ) C ( n ) = C ( n + ) ˆ | z H | e −(cid:107) z (cid:107) d µ ( z ) id = ω µ ( S n − ) id, Q ( ) C ( n ) = C ( n + ) ˆ ( z H ⊗ z H + Jz H ⊗ Jz H ) e −(cid:107) z (cid:107) d µ ( z ) = ˆ ( v ⊗ v + Jv ⊗ Jv ) d ω µ ( v ) .Summing up, since Q , . . . , Q were constructed in such a way that Q ( ) = Q ( ) + Q ( ) − Q ( ) − Q ( ) (seethe proof of Proposition . ), we deduce that: Q ( ) C ( n ) ω µ ( S n − ) = ( n + ) v ⊗ vd ω µ ( v ) + Jv ⊗ Jvd ω µ ( v ) − − ( v ⊗ v + Jv ⊗ Jv ) d ω µ ( v )= ( n − ) u ⊗ ud ω µ ( u ) − ), we deduce that for any w ∈ supp ( µ ) , we have:0 = (cid:104) w H , Q ( ) w H (cid:105) = C ( n ) ω µ ( S n − ) (cid:16) ( n − ) (cid:104) v , w H (cid:105) d ω µ ( v ) − | w H | (cid:17) . Definition . . For any vertical ( n + ) -uniform cone µ , define: M : = u ⊗ ud ω µ ( u ) ,where ω µ : = H n − eu (cid:120) π H ( supp ( µ )) ∩ S n − . Moreover, let α ≥ . . . ≥ α n ≥ M and e , . . . , e n their relative eigenvectors. Remark . . The trace of the matrix M can be explicitly computed:Tr ( M ) = n ∑ i = α i = n ∑ i = (cid:104) e i , Me i (cid:105) = n ∑ i = (cid:104) u , e i (cid:105) d ω µ ( u ) = isconnection of vertical non - flat cones and flat measures 48 The following proposition links the matrix M defined above to the functional min m ∈ S n − ´ B ( ) (cid:104) m , z H (cid:105) d µ ( z ) ,which will play a fundamental role in the proof of our main result. Proposition . . There exists a constant C ( n ) > for which for any vertical ( n + ) -uniform cone µ and any m ∈ S n − we have: ˆ B ( ) (cid:104) m , z H (cid:105) d µ ( z ) = C ( n ) (cid:104) m , M m (cid:105) . Proof.
Thanks to Lemma . (i) we have that: ˆ B ( ) (cid:104) m , z H (cid:105) d µ ( z ) = c n ˆ r n (cid:112) − r dr ˆ S n − (cid:104) m , v (cid:105) d ω µ ( v ) = C ( n ) c n ˆ S n − (cid:104) m , v (cid:105) d ω µ ( v ) ,where C ( n ) : = ´ r n √ − r dr and ω µ = H n − eu (cid:120) S n − ∩ π H ( supp ( µ )) . In order to prove the proposition, weare left to prove that ω µ ( S n − ) is a constant depending only on n . Thanks to Lemma . implies that:1 = µ ( B ( )) = ω µ ( S n − ) c n ˆ r n − (cid:112) − r dr .Therefore, defined C : = ´ r n − √ − r dr we have that ω µ ( S n − ) = c n /2 C ( n ) and in particular: ˆ B ( ) (cid:104) m , z H (cid:105) d µ ( z ) = C ( n ) C ( n ) (cid:104) m , v (cid:105) d ω µ ( v ) = C ( n ) C ( n ) (cid:104) m , M m (cid:105) .An immediate consequence of Proposition . is that:min m ∈ S n − ˆ B ( ) (cid:104) m , z H (cid:105) d µ ( z ) = C ( n ) α n = ˆ B ( ) (cid:104) e n , z H (cid:105) d µ ( z ) . ( )In particular we can link the value of ´ B ( ) (cid:104) e n , z H (cid:105) d µ ( z ) to the geometric structure of the measure µ thanks tothe following: Proposition . . Suppose µ is a vertical ( n + ) -uniform cone. For any δ > there exists an (cid:101) ( δ , n ) > such that, if: ˆ B ( ) (cid:104) e n , z H (cid:105) d µ ( z ) ≤ (cid:101) ( δ , n ) , then for any x ∈ cl ( U ( )) ∩ e ⊥ n there is z ∈ supp ( µ ) for which | z H − e | ≤ δ .Proof. Assume by contradiction that there exists a δ >
0, an infinitesimal sequence { (cid:101) i } i ∈ N and a sequence of ( n + ) -uniform cones { µ i } i ∈ N for which:( α ) defined e n ( i ) be the minimum eigenvalue of the matrix ´ z H ⊗ z H e −(cid:107) z (cid:107) d µ i ( z ) , we have: ˆ B ( ) (cid:104) z H , e n ( i ) (cid:105) d µ i ( z ) ≤ (cid:101) i ,( β ) there exists x i ∈ cl ( U ( )) ∩ e ⊥ n ( i ) for which | z H − x i | ≥ δ for any z ∈ supp ( µ ) .By compactness, up to non-relabeled subsequences, we can assume that µ i (cid:42) ν , e n ( i ) → e and x i → x . Since (cid:104)· , e n ( i ) (cid:105) χ B ( ) ( · ) is uniformly converging to (cid:104)· , e (cid:105) χ B ( ) ( · ) , we have: ˆ B ( ) (cid:104) z H , e (cid:105) d ν ( z ) = . that ν is a ( n + ) -measure which support contained in V ( e ) . Therefore applyingProposition . we deduce that ν is flat.On the other hand, since supp ( µ i ) ∩ cl ( U δ /2 ( x i )) × R e n + = ∅ for i sufficiently big, Proposition . implesthat supp ( ν ) ∩ cl ( U δ /2 ( x )) × R = ∅ . This and the fact that x ∈ e ⊥ (see the assumptions on x i in ( β )) comes incontradiction with the fact that supp ( ν ) = V ( e ) . isconnection of vertical non - flat cones and flat measures 49 The following proposition shows that non-flat vertical ( n + ) -uniform cones are quantitatively disconnectedfrom flat measures. The proof of this Therorem follows closely its Euclidean counterpart (see for instance Propo-sition . in [ ]). The reason for this is the following.Let µ be an m -uniform cone in R n . Then for any w ∈ supp ( µ ) we have:2 π − m /2 ˆ (cid:104) w , z (cid:105) e −| z | d µ ( z ) = | w | ,see for instance identity ( . ) of Lemma . in [ ]. The structure of the Euclidean quadric containing supp ( µ ) hasthe same structure of the quadric in ( ) which contains the support of vertical ( n + ) -uniform cones, and thusthe same kind of algebraic computations work. Theorem . . There exists a constant C ( n ) > such that if µ is a vertical ( n + ) -uniform cone for which: min m ∈ S n − ˆ B ( ) (cid:104) m , z H (cid:105) d µ ( z ) ≤ C ( n ) , then µ is flat.Proof. Fix some δ > m ∈ S n − ´ B ( ) (cid:104) m , z H (cid:105) d µ ( z ) ≤ (cid:101) ( δ , n ) / C ( n ) . Identity ( ) impliesthat ´ B ( ) (cid:104) e n , z H (cid:105) d µ ( z ) ≤ (cid:101) ( δ , n ) and thus, thanks to Proposition . , there exists a z ∈ supp ( µ ) such that | z H − e n − | ≤ δ . Thanks to the order imposed on the α i ’s, we have that α i + ( n − ) α n − ≤ Tr ( M ) = i ≤ n −
2. This in particular implies that: α i − n − ≤ ( n − ) (cid:18) n − − α n − (cid:19) , for every i ≤ n −
2. ( )Since z ∈ supp ( µ ) , Proposition . implies (once ( ) has been written in the notation of Definition . ) that:0 = n ∑ i = (cid:18) α i − n − (cid:19) (cid:104) e i , z H (cid:105) . ( )We observe that since M is positive semidefinite and Tr ( M ) =
1, we have that α n ≤ n . Therefore, puttingtogether ( ), ( ) and the fact that α n ≤ n , we deduce that:0 = n ∑ i = (cid:18) α i − n − (cid:19) (cid:104) e i , z H (cid:105) ≤ n − ∑ i = (cid:18) α i − n − (cid:19) (cid:104) e i , z H (cid:105) + (cid:18) α n − − n − (cid:19) (cid:104) e n − , z H (cid:105) . ( )Summing up, inequalities ( ) and ( ) together with some algebraic manipulations imply:0 ≤ ( n − ) (cid:16) n − − α n − (cid:17) n − ∑ i = (cid:104) e i , z H (cid:105) − (cid:16) n − − α n − (cid:17) (cid:104) e n − , z H (cid:105) =( n − ) (cid:16) n − − α n − (cid:17) n − ∑ i = (cid:104) e i , z H (cid:105) − (cid:16) n − − α n − (cid:17) ( + (cid:104) e n − , z H − e n − (cid:105) ) ≤ (cid:16) n − − α n − (cid:17)(cid:16) ( n − ) n − ∑ i = | z H − e n − | − ( − | z H − e n − | ) (cid:17) ≤ (cid:16) n − − α n − (cid:17) (( n − ) δ − ( − δ ) ) , ( )where the last inequality comes from the choice of z . Therefore if δ is small enough, thanks to the inequality ( ),we have that α n − ≥ ( n − ) . In this case, since M is positive semidefinite and Tr ( M ) =
1, we have that: α = . . . = α n − = n − α n = ) implies that supp ( µ ) ⊆ V ( e n ) and thus by Proposition . , µ must be flat. onclusions 50 In this section we complete the proof of Theorem . . In order to conclude the proof we need to construct thecontinuous functional F on Radon measures which fulfills the hypothesis of Theorem . . Proposition . . Let η be a non-negative smooth function such that η = on B ( ) and η = on B c ( ) . Then thefunctional F : M → R defined by: F ( µ ) : = min m ∈ S n − ˆ η ( z ) (cid:104) z H , m (cid:105) d µ ( z ) , is continuous with respect to the weak convergence of measures.Proof. Suppose that µ i (cid:42) ν and m i ∈ S n − are such that: ˆ η ( z ) (cid:104) z H , m i (cid:105) d µ i ( z ) = min m ∈S n − ˆ η ( z ) (cid:104) z H , m (cid:105) d µ i ( z ) .Up to passing to a (non-relabeled) subsequence we can also suppose that m i converges to some m ∈ S n − . Thusthe function η ( · ) (cid:104) π H ( · ) , m i (cid:105) is uniformly converging to η ( · ) (cid:104) π H ( · ) , m (cid:105) . Therefore we deduce that:lim i → ∞ ˆ η ( z ) (cid:104) z H , m i (cid:105) d µ i ( z ) = ˆ η ( z ) (cid:104) z H , m (cid:105) d µ ( z ) ,from which we infer that lim inf i → ∞ F ( µ i ) ≥ F ( µ ) . On the other hand, let F ( ν ) = ´ η ( z ) (cid:104) z H , m (cid:105) d µ ( z ) for some m ∈ S n − . Since µ i (cid:42) ν , we deduce that:lim i → ∞ ˆ η ( z ) (cid:104) z H , m (cid:105) d µ i ( z ) = ˆ η ( z ) (cid:104) z H , m (cid:105) d µ ( z ) .This implies that lim sup i → ∞ F ( µ i ) ≤ F ( µ ) and this concludes the proof.The following proposition shows that F satisfies the hypothesis (ii) of Theorem . . Proposition . . There exists a constant ¯ h ( n ) > such that if µ is a ( n + ) -uniform cone and F ( µ ) ≤ ¯ h ( n ) , then µ isflat.Proof. Thanks to the definition of F , we have that min m ∈ S n − ´ (cid:104) z H , m (cid:105) d µ ( z ) ≤ F ( µ ) . Therefore thanks toTheorems . and . , if F ( µ ) ≤ min { C ( n ) , C ( n ) } /2 = : ¯ h ( n ) the measure µ is flat.Eventually we are ready to prove our main result Theorem . . Theorem . . If φ is a measure with ( n + ) -density, then Tan n + ( φ , x ) ⊆ M ( n + ) for φ -almost all x ∈ H n .Proof. Since F ( µ ) =
0, whenever µ is flat, Propositions . and . imply that we are in the hypothesis of Theorem . . which proves the claim.A byproduct of our analysis is the conclusion of the classification of uniform measures in H , which wassystematically carried out in [ ]. Our contribution is to prove that 3-uniform measures in H are flat. The finalresult reads: Theorem . . In H we have the following complete classification of uniform measures:(i) If µ ∈ U H ( ) , then µ = S (cid:120) L, where L is a horizontal line.(ii) If µ ∈ U H ( ) then µ = S (cid:120) V , where V is the vertical axis.(iii) If µ ∈ U H ( ) then µ = S (cid:120) W, where W is a -dimensional vertical plane.Proof. Points (i) and (ii) are Theorems . and . of [ ]. Corollary . implies that if µ is supported on a quadric.Proposition . in [ ] implies that such quadric cannot be a t -graph and therefore by Theorem . of the samepaper we conclude the proof. epresentations of sub - riemannian spherical hausdorff measure 51 a representations of sub - riemannian spherical hausdorff measure In the following we will adopt the notations introduced in Section and as usual we assume b ∈ R n , Q ∈
Sym ( n ) and T ∈ R . The main goal of this section is to find a representation of the spherical Hausdorff measures S n + A , where A is a Borel subset of the quadric K ( b , Q , T ) , in terms of the Euclidean Hausdorff measure H neu .To do so, we will study separately the case where T (cid:54) = T = H n . For any x ∈ H n and any i ∈ {
1, . . . , n } we define the vector fields: X i ( x ) : = e i − x i + n e n + , Y i ( x ) : = e i + n + x i e n + ,where { e , . . . , e n + } is the standard othonormal basis of R n + . We introduce here the horizontal perimeter of aset E ⊆ H n : Definition A. . The horizontal perimeter in an open set Ω ⊆ H n of a Lebesgue measurable set E ⊆ H n is: P ( E , Ω ) : = sup (cid:26) ˆ E div H ϕ dx : ϕ ∈ C c ( Ω , R n ) , (cid:107) ϕ (cid:107) ∞ ≤ (cid:27) ,where div H ϕ = ∑ ni = ( X i ϕ i + Y i ϕ n + i ) and (cid:107)·(cid:107) ∞ is the usual supremum norm. If P ( E , Ω ) < ∞ we say that E hasfinite horizontal perimeter in Ω .Thanks to Riesz’ representation theorem, since ´ E div H ϕ ≤ P ( E , Ω ) (cid:107) ϕ (cid:107) ∞ , there is a positive Radon measure | ∂ E | Ω and a | ∂ E | Ω -measurable function n H : Ω → R n such that: ˆ E div H ϕ dx = ˆ Ω (cid:104) ϕ , n H (cid:105) d | ∂ E | Ω ,for any ϕ ∈ C c ( Ω , R n ) . Denoted with n the Euclidean unit normal to the quadric K ( b , Q , T ) at x , we let n H ( x ) be the vector: n H ( x ) : = ( (cid:104) X ( x ) , n ( x ) (cid:105) , . . . , (cid:104) X n ( x ) , n ( x ) (cid:105) , (cid:104) Y ( x ) , n ( x ) (cid:105) , . . . , (cid:104) Y n ( x ) , n ( x ) (cid:105) ) ,which is commonly known as the horizontal normal to K ( b , Q , T ) at x . The vector n H ( x ) allows us to get thefollowing useful: Proposition A. (Monti, [ ]) . Let E ⊆ H n be a set with Euclidean Lipschitz boundary and Ω a fixed open set in H n .Then: | ∂ E | Ω ( A ) = ˆ χ A | n H | d H n (cid:120) ∂ E ∩ Ω , for any open set A ⊆ H n . The above proposition together with Proposition . . of [ ] implies that the measures | ∂ E | Ω and | n H |H n (cid:120) ∂ E ∩ Ω coincide on Borel sets. Remark A. . Note that if T =
0, then n ( x ) is parallel to the vector N ( x ) : = ( Q x + b , 0 ) . Moreover, since (cid:104) N ( x ) , e n + (cid:105) =
0, the projection of N ( x ) on the plane spanned by X , . . . , Y n coincides with N ( x ) itself. There-fore in the case T = n H = n . a . Area on t-quadratic graphs
In this subsection we deal with the representation of spherical Hausdorff measure concentrated on subsetsof horizontal quadrics. Therefore we can suppose without loss of generality that T = − f ( x ) : = (cid:104) b , x (cid:105) + (cid:104) x , Q x (cid:105) be the function defined in ( ) and Σ ( f ) : = { x ∈ R n : 2 ( Q + J ) x + b = } be the characteristicset of f , which was introduced in ( ). Thanks to the simple algebraic context in which we are, the followingcharacterisation of Σ ( f ) is available: Proposition A. . The set Σ ( f ) is an affine plane in R n of dimension at most n. epresentations of sub - riemannian spherical hausdorff measure 52 Proof.
The proof is a simple argument by contradiction which yields the following stronger result. If A is asymmetric matrix and B is an invertible antisymmetric matrix then dim ( ker ( A − B )) ≤ n .It is worth noting that Proposition A. is a very simple case of the results obtained by Z. Balogh in [ ]. Proposition A. . Let Ω be an open set in R n and define Ω (cid:48) : = Ω × R . For any positive Borel function h : H n → R wehave: ˆ h ( z ) d S n + Ω (cid:48) ∩ K ( b , Q , T ) ( z ) = c n ˆ Ω h ( x , f ( x )) | b + ( D + J ) x | dx , where c n = H neu ( B ( ) ∩ V ( e )) for any e ∈ R n .Proof. Let E : = { z ∈ R n + : z T − f ( z H ) ≥ } and note that by Proposition . of [ ] we have that E is a set ofintrinsic finite perimeter and: | ∂ E | Ω (cid:48) ( A ) = ˆ Ω ∩ π H ( A ) |∇ f ( x ) + Jx | dx = ˆ Ω ∩ π H ( A ) | b + ( Q + J ) x | dx ,for any Borel set A ⊆ H n . It is immediate to see that the measure theoretic boundary of E (see for instanceDefinition . in [ ]) coincides with K ( b , Q , − ) and thus, thanks to Corollary . of [ ], we deduce that: | ∂ E | Ω (cid:48) = c n S n + Ω (cid:48) ∩ K ( b , Q , − ) . ( )So far we proved that: S n + Ω (cid:48) ∩ K ( b , Q , − ) ( A ) = c n ˆ Ω χ A ( x , f ( x )) | b + ( Q + J ) x | dx .The standard approximation procedure of positive measurable functions together with Beppo-Levi convergencetheorem concludes the proof. Remark A. . The constant c n differs from the one of Corollary . of [ ] since we defined the spherical Hausdorffin such a way that S n + ( B ( ) ∩ V ( e )) =
1. Furthermore we are using the Koranyi norm, while in [ ] anequivalent, but different metric is used. We refer to the aformentioned article for further details. a . Area on vertical quadric
In this subsection we deal with the representation of spherical Hausdorff measure concentrated on subsets ofvertical quadrics. Therefore we let F ( x , t ) : = (cid:104) b , x (cid:105) + (cid:104) x , Q x (cid:105) be the function defined in ( ) which zeroes coincidewith K ( b , Q , 0 ) , and Σ ( F ) : { z ∈ K ( b , Q , 0 ) : 2 Q x + b = } be the set of singular points of K ( b , Q , 0 ) , which wasintroduced in ( ). Proposition A. . Let a , b be two non-parallel vectors in R n . Then S n + ( V ( a ) ∩ V ( b )) = , where V ( · ) were introducedin Definition . .Proof. The set V ( a ) can be seen as the bounduary of the Lipschitz domain { x ∈ H n : (cid:104) a , x H (cid:105) ≥ } . ThereforeProposition A. and the equivalence between the measures S n + A and S n + (cid:120) A , implies that: S n + ( V ( a ) ∩ V ( b )) = H neu ( V ( a ) ∩ V ( b )) / c n = Proposition A. . One of the following two holds:(i) b = and Q = a ⊗ a for some a ∈ R n \ { } ,(ii) S n + ( Σ ( F )) = . epresentations of sub - riemannian spherical hausdorff measure 53 Proof.
If dim ( ker ( Q )) ≤ n −
2, then Σ ( F ) is contained in an affine subspace of dimension at most 2 n −
2. ThusProposition A. implies that S n + ( Σ ( F )) =
0. On the other hand, if dim ( ker ( Q )) = n −
1, we have that Q = a ⊗ a for some a ∈ R n and the expression for F boils down to F ( x , t ) = (cid:104) a , x (cid:105) + (cid:104) b , x (cid:105) . If b (cid:54) = a , the equation 2 (cid:104) a , z H (cid:105) a + b = Σ ( F ) = ∅ . At last, if a = λ b , then K ( b , Q , 0 ) ⊆ V ( a ) ∪ λ a + V ( a ) , while Σ ( F ) ⊆ λ a /2 + V ( a ) , which implies Σ ( F ) = ∅ . The only left out case iswhen b = Σ ( F ) = V ( a ) , which proves the claim.Finally we can prove the representation formula for the spherical Hausdorff measure concentrated on verticalquadrics: Proposition A. . Let Ω be an open set in H n . For any positive Borel function h : H n → R we have: ˆ h ( x ) d S n + K ( b , Q ,0 ) ∩ Ω ( x ) = c n ˆ h ( x ) d H neu (cid:120) ( K ( b , Q , 0 ) ∩ Ω )( x ) , where c n is the constant introduced in Proposition A. .Proof. Let E be the open set E : = { z ∈ R n + : F ( z ) < } and note that ∂ E = { F = } = K ( b , Q , 0 ) . Thanks toProposition A. , since E has a Lipschitz boundary, we have that E is a set of intrinsic finite perimeter and: | ∂ E | Ω ( A ) = ˆ ∂ E ∩ Ω χ A | n H | d H neu = H neu ( K ( b , Q , 0 ) ∩ Ω ∩ A ) ,for any Borel set A of H n , since | n H | =
1. It is easy to see that the measure theoretic boundary of E , see forinstance Definition . of [ ], coincides with K ( b , Q , 0 ) . Therefore, thanks to Corollary . of [ ] we have that: | ∂ E | Ω = c n S n + K ( b , Q ,0 ) ∩ Ω ,Summing up, we have that for any Borel set A in H n : | ∂ E | Ω ( A ) = H neu ( K ( b , Q , 0 ) ∩ Ω ∩ A ) / c n .The usual approximation of positive measurable functions together with Beppo-Levi convergence theorem con-cludes the proof.Since the set supp ( µ ) \ Σ ( F ) is relatively open set in K ( b , Q , 0 ) thanks to Proposition . , we can find an openset Ω in H n such that K ( b , Q , 0 ) ∩ Ω = supp ( µ ) \ Σ ( F ) . Therefore Proposition A. implies that: S n + ( µ ) \ Σ ( F ) = H neu (cid:120) supp ( µ ) \ Σ ( F ) c n .Note the above identity in the case in which b = Q = a ⊗ a , can be rewritten as S n + ( µ ) = H neu (cid:120) supp ( µ ) / c n since the Euclidean normal is well defined on the whole V ( a ) despite the fact that Σ ( F ) = V ( a ) . Thanks toProposition A. we can assume that S n + ( Σ ( F )) = S n + ( µ ) = H neu (cid:120) supp ( µ ) c n . ( ) Lemma A. . If µ is a vertical ( n + ) -uniform cone then: µ = c n H n − eu (cid:120) π H ( supp ( µ )) ⊗ H eu (cid:120) R e n + . Proof.
First of all, thanks to the identity ( ) used in the first and second equality respectively, we deduce that: µ = S n + ( µ ) = c n H neu (cid:120) supp ( µ ) .Since by Proposition . we have that supp ( µ ) = π H ( supp ( µ )) × R e n + , Proposition . . of [ ] implies: H neu (cid:120) supp ( µ ) = H n − eu (cid:120) π H ( supp ( µ )) ⊗ H eu (cid:120) R e n + . aylor expansion of area on quadratic t - cones 54 b taylor expansion of area on quadratic t - cones Before giving a short account on the content of this appendix, we introduce some notation. Let
D ∈
Sym ( n ) \{ } and define the function f : R n → R as: f ( h ) : = (cid:104) h , D h (cid:105) .Furthermore, we let | ∂ K | be the intrinsic perimeter measure of the epigraph of f in H n , which is well definedsince f is a smooth function, see Proposition A. . Throughout this entire section, we fix a point x (cid:54)∈ Σ ( f ) (recallthat Σ ( f ) was introduced in ( )), and we let X : = ( x , f ( x )) .The main goal of this section is to determine an asymptotic expansion of | ∂ K | ( B r ( X )) for r small. Moreprecisely, written: | ∂ K | ( B r ( X )) = c ( X ) r n + + d ( X ) r n + + e ( X ) r n + + O ( r n + ) ,we want to find an expression for the coefficients c , d , e in terms of x , D and n . The coefficient c will be quite easyto study and we will show that it is a constant depending only on n . On the other hand the coefficients d and e will need much more work and they will play a fundamental role in the study of the geometric properties of1-codimensional uniform measures carried on in Section . Definition B. . Let D , X and f be as above. We denote as: n : = ( D + J ) x | ( D + J ) x | ,the horizontal normal at X to gr ( f ) and moreover we let c : = | ( D + J ) x | .The following proposition gives a first characterisation of the shape of the intersection between B r ( X ) withgr ( f ) . In particular we construct a function G at the point x whose sublevel sets are the horizontal projection of B r ( X ) ∩ gr ( f ) . Proposition B. . Following the notations introduced above, we have: π H ( B r ( X ) ∩ gr ( f )) = x + (cid:110) w ∈ R n : G ( w ) ≤ r (cid:111) , where G ( w ) : = | w | + | c (cid:104) n , w (cid:105) + (cid:104) w , D w (cid:105)| .Proof. By definition of f and of the Koranyi norm, we have: B r ( X ) : = { z ∈ R n + : | z H − x | + | z T − f ( x ) − (cid:104) x , Jz H (cid:105)| ≤ r } .Therefore, the intersection of B r ( X ) with gr ( f ) is: B r ( X ) ∩ gr ( f ) = { ( y , f ( y )) ∈ R n + : | x − y | + | f ( x ) − f ( y ) + (cid:104) x , Jy (cid:105)| ≤ r } = x + { ( w , f ( w )) ∈ R n + : | w | + | f ( x ) − f ( x + w ) + (cid:104) x , Jw (cid:105)| ≤ r } ,where in the last line we have performed the change of variable y = x + w . By definition of f , we have: f ( x ) − f ( x + w ) = (cid:104) x , D x (cid:105) − (cid:104) x + w , D ( x + w ) (cid:105) = − (cid:104) x , D w (cid:105) − (cid:104) w , D w (cid:105) .In particular, this implies that: π H ( B r ( X ) ∩ gr ( f )) = { w ∈ R n : | w | + |− (cid:104) x , D w (cid:105) − (cid:104) w , D w (cid:105) + (cid:104) x , Jw (cid:105)| ≤ r } = { w ∈ R n : | w | + | c (cid:104) n , w (cid:105) + (cid:104) w , D w (cid:105)| ≤ r } . ( )Identity ( ) and the definition of G conclude the proof. aylor expansion of area on quadratic t - cones 55 The following proposition introduces a special set of polar coordinates, which are going to be very useful inthe study of the intersection B r ( X ) ∩ gr ( f ) when r is small. Proposition B. . For any w ∈ R n \ x + span ( n ) there exists a unique triple ( ϑ , ρ , v ) ∈ C : = [ − π , π ) × ( ∞ ) × S n − ∩ n ⊥ such that: w = x + sin ϑ c ρ n + cos ϑρ v = : x + P ( ϑ , ρ , v ) . ( ) Proof.
Any w ∈ R n \ x + span ( n ) can be uniquely written as w = x + λ n + u for some λ ∈ R and u ∈ n ⊥ (cid:54) = ρ : = (cid:113)(cid:0) (cid:107) u (cid:107) + (cid:112) (cid:107) u (cid:107) + λ c (cid:1) /2, we have that ρ (cid:54) = (cid:18) c λρ (cid:19) + (cid:18) (cid:107) u (cid:107) ρ (cid:19) =
1, ( )since ρ solves the equation ζ − (cid:107) u (cid:107) ζ − c λ =
0. By ( ), there is a unique ϑ ∈ [ − π /2, π /2 ) for whichsin ϑ = c λ / ρ and cos ϑ = (cid:107) u (cid:107) / ρ . Eventually, if we let v : = u / (cid:107) u (cid:107) , thanks to the definition of θ , we have: w = x + λ n + (cid:107) u (cid:107) v = x + sin ϑ c ρ n + cos ϑρ v .In order to simplify the notations in the forthcoming propositions, we define: α n : = (cid:104) n , D n (cid:105) , β n ( v ) : = (cid:104) v , D n (cid:105) , γ ( v ) : = (cid:104) v , D v (cid:105) for v ∈ R n . ( )In the following proposition we give an explicit expression of G in the new polar coordinates P ( ϑ , ρ , v ) introducedin Proposition B. . Proposition B. . Let us define the function H : C → R as:H ( ϑ , ρ , v ) : = G ( P ( ϑ , ρ , v )) , where G was introduced in Proposition B. and both C and P ( ϑ , ρ , v ) were defined in Proposition B. . Then H has thefollowing explicit expression:H ( ϑ , ρ , v ) : = A ( ϑ , v ) ρ + B ( ϑ , v ) c ρ + C ( ϑ , v ) c ρ + D ( ϑ , v ) c ρ + E ( ϑ , ρ ) c ρ , where:(i) A ( ϑ , v ) : = ( cos ϑ + ( cos ϑγ ( v ) + sin ϑ ) ) ,(ii) B ( ϑ , v ) : = cB ( ϑ , v ) : = ϑ cos ϑβ n ( v )( cos ϑγ ( v ) + sin ϑ ) ,(iii) C ( ϑ , v ) : = c C ( ϑ , v ) : = sin ϑ ( cos ϑ ( + β n ( v ) + γ ( v ) α n ) + ϑα n ) ,(iv) D ( ϑ , v ) : = c D ( ϑ , v ) : = α n β n ( v ) sin ϑ cos ϑ ,(v) E ( ϑ ) : = c E ( ϑ ) : = ( + α n ) sin ϑ ,and ( ϑ , v ) varies in [ − π /2, π /2 ) × S n − ∩ n ⊥ .Proof. For any x + w ∈ R n \ x + span ( n ) , by Proposition B. we can find a unique ( ϑ , ρ , v ) ∈ C such that w = P ( ϑ , ρ , v ) . Note that the following identities hold: | w | = |P ( ϑ , ρ , v | = sin ϑ c ρ + ϑ cos ϑ c ρ + cos ϑρ , c (cid:104) n , w (cid:105) = c (cid:104) n , P ( ϑ , ρ , v ) (cid:105) = (cid:28) c n , sin ϑ c ρ n + cos ϑρ v (cid:29) = sin ϑρ , (cid:104) w , D w (cid:105) = (cid:104)P ( ϑ , ρ , v ) , D [ P ( ϑ , ρ , v )] (cid:105) = sin ϑ c ρ α n + ϑ cos ϑ c ρ β n ( v ) + cos ϑρ γ ( v ) . ( ) aylor expansion of area on quadratic t - cones 56 Therefore, thanks to the definition of G and H and the three identities in ( ), we have: H ( ρ , ϑ , v ) = G ( P ( ϑ , ρ , v )) = |P ( ϑ , ρ , v ) | + | c (cid:104) n , P ( ϑ , ρ , v ) (cid:105) + (cid:104)P ( ϑ , ρ , v ) , D [ P ( ϑ , ρ , v )] (cid:105)| = sin ϑ c ρ + ϑ c cos ϑρ + cos ϑρ + ( cos ϑγ ( v ) + sin ϑ ) ρ + ϑ cos ϑ c β n ( v ) ρ + sin ϑ c α n ρ + ϑ cos ϑ c ( cos ϑγ ( v ) + sin ϑ ) β n ( v ) ρ + ( cos ϑγ ( v ) + sin ϑ ) sin ϑ c α n ρ + ϑ cos ϑ c α n β n ( v ) ρ .The claim follows recollecting the various powers of ρ .We summarize in the following lemma some algebraic properties of the functions A , . . . , E introduced in Propo-sition B. , as they will be very useful in the forthcoming computations. Lemma B. . Consider ( ϑ , v ) ∈ [ − π /2, π /2 ) × S n ∩ n ⊥ fixed. Then:(i) A ( ϑ , v ) = A ( ϑ , − v ) and C ( ϑ , v ) = C ( ϑ , − v ) ,(ii) B ( ϑ , v ) = − B ( ϑ , − v ) and D ( ϑ , v ) = − D ( ϑ , − v ) ,(iii) B ( ϑ , v ) = − B ( − ϑ , v ) and D ( ϑ , v ) = − D ( − ϑ , v ) ,(iv) E does not depend on v,(v) A is bounded away from , i.e.: ω : = min [ − π /2, π /2 ) × S n ∩ n ⊥ A ( ϑ , v ) > Proof.
The first four points are direct consequence of the definition of A , B , C , D , E , and of α n , β n ( v ) , γ ( v ) (see ( )). We are left to prove the last point. Since A ( π /2, v ) = A ( − π /2, v ) , the function A has minimumin ( − π /2, π /2 ) × S n ∩ n ⊥ . Suppose such minimum is 0 and it is attained at ( ϑ , v ) . This would imply that0 = cos ϑ + ( cos ϑγ ( v ) + sin ϑ ) , but this is not possible as it would force sin ϑ = cos ϑ = π H ( B r ( X ) ∩ gr ( f )) when r issmall: Proposition B. . There exists an < r ( X ) = r such that for any < r < r , if ρ ( r ) is a solution to the equation:H ( ρ ( r ) , ϑ , v ) = r , ( ) then: ρ ( r ) = P ϑ , v ( r ) + O ( r ) : = rA − Br A + (cid:18) B A − C A (cid:19) r + O ( r ) , ( ) and the remainder O ( r ) is independent on v and ϑ .Proof. By Proposition B. , equation ( ) turns into: A ρ ( r ) + B ρ ( r ) + C ρ ( r ) + D ρ ( r ) + E ρ ( r ) = r , ( )where we dropped the dependence on v and ϑ from A , B , C , D and E to simplify the notation.We claim that there are r > c >
0, independent on ϑ and v , such that ρ ( r ) ≤ c r for any 0 < r < r .Suppose by contradiction there exists a sequence ( r i , ϑ i , v i ) such that G ( ρ ( r i ) , ϑ i , v i ) = r i for any i ∈ N , { r i } i ∈ N isinfinitesimal and ρ ( r i ) > r i / ω , where ω was introduced in Proposition B. (v). By ( ), we have that for any i ∈ N : 1 = A (cid:18) ρ ( r i ) r i (cid:19) + B (cid:18) ρ ( r i ) r i (cid:19) r i + C (cid:18) ρ ( r i ) r i (cid:19) r i + D (cid:18) ρ ( r i ) r i (cid:19) r i + E (cid:18) ρ ( r i ) r i (cid:19) r i > A (cid:18) ω (cid:19) + B (cid:18) ω (cid:19) r i + C (cid:18) ω (cid:19) r i + D (cid:18) ω (cid:19) r i + E (cid:18) ω (cid:19) r i . ( ) aylor expansion of area on quadratic t - cones 57 Define the constant: M : = (cid:18) + ω (cid:19) max [ − π /2, π /2 ] × S n ∩ n ⊥ | B | + | C | + | D | + | E | ,and note that provided r i < min (
1, 1/ M ) , the inequality ( ) implies:1 > A (cid:18) ω (cid:19) − Mr i > A (cid:18) ω (cid:19) − A / ω ≥ ), together with theuniform bound on ρ proved above, we deduce that: ρ ( r ) = rA + R ( r ) , ( )where | R ( r ) | ≤ c r for any 0 < r < r and for some constant c > ϑ and v . Plugging theabove expression for ρ ( r ) inside ( ), we can find a more explicit expression for R . Indeed, with some algebraiccomputations, which we omit, we get: r = r + A r R ( r ) + Br A + R ( r ) ,where | R | ≤ c r for any 0 < r < r and for some constant c independent on ϑ and v , from which we deduce: R ( r ) = − Br A − R ( r ) A r .Substituting the above identity in ( ), we have: ρ ( r ) = rA − Br A + R ( r ) , ( )where R ( r ) : = R ( r ) /4 A r . In particular | R ( r ) | ≤ c /4 ω r for any 0 < r < r . Once again, substituting thenewfound expression for ρ ( r ) given by ( ) in ( ), we get the following expression for R : R ( r ) : = (cid:18) B A − C A (cid:19) r + R ( r ) ,where | R ( r ) | ≤ c r for any 0 < r < r and for some constant c > ϑ and v .Proposition B. has the following almost immeiate consequence: Corollary B. . For any r > and any δ ∈ R sufficiently small, define: B r , δ : = P (cid:16)(cid:110) ( ρ , ϑ , v ) ∈ C : ρ ≤ P ϑ , v ( r ) + δ r (cid:111)(cid:17) . There exists an (cid:101) ( X ) = (cid:101) > such that for any < (cid:101) < (cid:101) , there exists < r ( (cid:101) ) = r such that for any < r < r ,we have: B r , − (cid:101) ⊆ π H ( B r ( X ) ∩ gr ( f )) ⊆ B r , (cid:101) . Proof.
Proposition B. and the definition of P (recall that the function P was defined in ) imply: π H ( B r ( X ) ∩ gr ( f )) = x + P (cid:16)(cid:110) ( ρ , ϑ , v ) ∈ C : H ( ρ , ϑ , v ) ≤ r (cid:111)(cid:17) . ( )The function ρ (cid:55)→ H ( ρ , ϑ , v ) is a polynomial of 8th degree in ρ , and thus the equation: H ( ρ , ϑ , v ) = r , ( ) aylor expansion of area on quadratic t - cones 58 has at most 8 solutions in ρ . Assume ρ < . . . < ρ k are the positive distinct solutions of the above equation,where k ∈ {
1, . . . , 8 } . If ρ > ρ k then H ( ρ , ϑ , v ) > r and on the other hand, since H ( ϑ , v ) =
0, if 0 ≤ ρ < ρ then G ( ρ , ϑ , v ) < r . This implies that: P ( { ( ρ , ϑ , v ) ∈ C : ρ ≤ ρ } ) ⊆ π H ( B r ( X ) ∩ gr ( f )) ⊆ P ( { ( ρ , ϑ , v ) ∈ C : ρ ≤ ρ k } ) .Proposition B. concludes the proof since ρ and ρ k coincide up to an error of order r .The following technical lemma will be needed in the computations of Proposition B. , and it is a Taylorexpansion formula for the sub-Riemmanian area element at a non-characteristic point of a horizontal quadric. Lemma B. . For any ( ρ , ϑ , v ) ∈ C we have: | ( D + J )[ x + P ( ρ , ϑ , v )] | = c + A ρ + B ρ + R ( ρ ) , where:(i) A : = A ( ϑ , v ) : = ϑ ( β n ( v ) + (cid:104) Jv , n (cid:105) ) ,(ii) B = c B : = c B ( ϑ , v ) : = (cid:0) α n sin ϑ + | P n [( D + J ) v ] | cos ϑ (cid:1) , and P n is the orthogonal projection on n ⊥ .(iii) | R ( ρ ) | ≤ c ρ for any < ρ < r and for some constant c > , independent on ϑ and v.Proof. In order to simplify the notation we let M : = ( D + J ) . First of all we want to find an explicit expressionfor the vector M [ x + P ( ρ , ϑ , v )] . M [ x + P ( ρ , ϑ , v )] = M (cid:20) x + sin ϑ c ρ n + cos ϑρ v (cid:21) = c n + ϑ c ρ M n + ϑρ Mv Secondly, we compute the squared norm of the vector M [ x + P ( ρ , ϑ , v )] : | M [ x + P ( ρ , ϑ , v )] | = c + c cos ϑρ (cid:104) Mv , n (cid:105) + ϑρ (cid:104) M n , n (cid:105) + θρ | Mv | + ϑ cos θ c ρ (cid:104) M n , Mv (cid:105) + ϑ c ρ | M n | .Note that by definition of α n and β n ( v ) , we have that (cid:104) Mv , n (cid:105) = β n ( v ) + (cid:104) Jv , n (cid:105) and thus 4 c cos ϑ (cid:104) Mv , n (cid:105) = c A .Moreover the fact that (cid:104) M n , n (cid:105) = α n and that | Mv | − (cid:104) Mv , n (cid:105) = | P n ( Mv ) | imply, by means of few algebraicomputation: 4 sin ϑ (cid:104) M n , n (cid:105) + ϑ | Mv | = c B + A .Therefore, with some algebraic computation that we omit, we can show that: | M [ x + P ( ρ , ϑ , v )] | − c − A ρ − B ρ = | M [ x + P ( ρ , ϑ , v )] | − ( c + A ρ + B ρ ) | M [ x + P ( ρ , ϑ , v )] | + c + A ρ + B ρ = (cid:16) ϑ cos θ c (cid:104) M n , Mv (cid:105) − AB (cid:17) ρ + (cid:16) ϑ c | M n | − B (cid:17) ρ | M [ x + P ( ρ , ϑ , v )] | + c + A ρ + B ρ .Therefore there are a sufficiently small r > c that can be chosen independent on ϑ and v , forwhich: (cid:12)(cid:12) | M [ x + P ( ρ , ϑ , v )] | − c − A ρ − B ρ (cid:12)(cid:12) ≤ c ρ , for any 0 < ρ < r Remark B. . For any ( ϑ , v ) ∈ [ − π /2, π /2 ] × S n − ∩ n ⊥ , we have that: aylor expansion of area on quadratic t - cones 59 (i) A ( ϑ , v ) = −A ( ϑ , − v ) ,(ii) B ( ϑ , v ) = B ( ϑ , − v ) . Proposition B. . The functions A ( · , · ) and B ( · , · ) defined in the statement of Proposition B. have the following symme-tries. For any < r < dist ( x , Σ ( f )) , we have that: | ∂ K | ( B r ( X )) = ˆ S n − ∩ n ⊥ ˆ π − π ˆ { H ( ρ , ϑ , v ) ≤ r } Ξ ( ρ , ϑ , v ) d ρ d ϑ d σ ( v ) , where:(i) Ξ ( ρ , ϑ , v ) : = ρ n c cos n − ϑ ( + sin ϑ ) (cid:12)(cid:12) ( D + J )[ x + P ( ρ , ϑ , v )] (cid:12)(cid:12) ,(ii) σ : = H n − eu (cid:120) S n − ∩ n ⊥ .Proof. Proposition . in [ ] implies that | ∂ K | ( B r ( X )) = ˆ U r |∇ f ( w ) + Jw | dw = ˆ U r | ( D + J ) w | dw . ( )We want to perform in the right-hand side of the above equation, the following change of variables: w = x + sin ϑρ c n + cos ϑρ v = x + P ( ρ , ϑ , v ) , where ( ρ , ϑ , v ) ∈ C . ( )If n >
1, we can parametrize the sphere S n − ∩ n ⊥ with the usual polar coordinates and therefore we let v = v ( ψ ) where ψ ∈ Ψ : = [ − π , π ) × [ π /2, π , 2 ] n − . The Jacobian determinant (obtained using the Laplace formula, weomit the computations) of the change of variables ( ) is: (cid:12)(cid:12)(cid:12)(cid:12) det ∂ w ( ρ , ϑ , v ) ∂ρ∂ϑ∂ψ (cid:12)(cid:12)(cid:12)(cid:12) = ρ n c ( + sin ϑ ) cos n − ϑ (cid:12)(cid:12)(cid:12)(cid:12)(cid:18) v ( ψ ) , ∂ v ( ψ ) ∂ψ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) .Therefore ( ) becomes: | ∂ K | ( B r ( X )) = ˆ Ψ ˆ π − π ˆ χ U r ( x + P ( ρ , ϑ , ψ )) ρ n c ( + sin ϑ ) cos n − ϑ (cid:12)(cid:12)(cid:12)(cid:12)(cid:18) v ( ψ ) , ∂ v ( ψ ) ∂ψ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) d ρ d ϑ d ψ = ˆ Ψ (cid:12)(cid:12)(cid:12)(cid:12)(cid:18) v ( ψ ) , ∂ v ( ψ ) ∂ψ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ˆ π − π ˆ χ U r ( x + P ( ρ , ϑ , ψ )) Ξ ( ρ , ϑ , v ( ψ )) d ρ d ϑ d ψ = ˆ S n − ∩ n ⊥ ˆ π − π ˆ χ U r ( x + P ( ρ , ϑ , ψ )) Ξ ( ρ , ϑ , v ) d ρ d ϑ d σ ( v )= ˆ S n − ∩ n ⊥ ˆ π − π ˆ { H ( ρ , ϑ , v ) ≤ r } Ξ ( ρ , ϑ , v ) d ρ d ϑ d σ ( v ) ,where the last line comes from the identity ( ). If n =
1, the computation is simpler since S n − ∩ n ⊥ = {± J n } .The following two lemmas will allow us to compute some integrals in Propositions B. and B. . Lemma B. . For any k ∈ N , any α > k + we have: ˆ ∞ − ∞ x k ( + x ) α dx = if k is odd, Γ (cid:0) k + (cid:1) Γ (cid:0) α − k + (cid:1) Γ ( α ) if k is even. aylor expansion of area on quadratic t - cones 60 Proof. If k is odd, then ´ x k / ( + x ) α =
0. If on the other hand k is even, the change of variable t = ( + x ) α implies: ˆ ∞ − ∞ x k ( + x ) α dx = ˆ ( − t ) k + − t ( α − k + ) − dt = β (cid:18) k +
12 , α − k + (cid:19) = Γ (cid:0) k + (cid:1) Γ (cid:0) α − k + (cid:1) Γ ( α ) ,where β ( · , · ) is the Euler’s beta function. The last equality follows from a well known property of β , see forinstance Theorem . in [ ]. Lemma B. . Suppose f : R → R is a measurable function such that f ( x ) / ( + x ) α ∈ L ( R ) and let d ( θ ) : = cos n − ϑ ( cos ϑ + ϑ ) . Then the following identity holds: ˆ π − π d ( ϑ ) cos α − n − ϑ f (cid:16) sin ϑ cos ϑ (cid:17) A α d ϑ = ˆ ∞ − ∞ f ( x ) (cid:0) + (cid:0) x + γ ( v ) (cid:1) (cid:1) α dx , where γ ( v ) was defined in ( ) .Proof. Thanks to the definition of A , we have that: ˆ π − π d ( ϑ ) cos α − n − ( ϑ ) f (cid:16) sin ϑ cos ϑ (cid:17) A α d ϑ = ˆ π − π d ( ϑ ) cos α − n − ϑ f (cid:16) sin ϑ cos ϑ (cid:17)(cid:0) cos ϑ + ( cos ϑ + γ ( v ) sin ϑ ) (cid:1) α d ϑ = ˆ π − π cos ϑ + ϑ cos ϑ f (cid:16) sin ϑ cos ϑ (cid:17)(cid:0) + ( sin ϑ cos ϑ + γ ( v )) (cid:1) α d ϑ = ˆ ∞ − ∞ f ( x ) (cid:0) + (cid:0) x + γ ( v ) (cid:1) (cid:1) α dx ,where the last equality is obtained with the change of variable x = sin ϑ / cos ϑ .Proposition B. is the technical core of this appendix. It gives a first description of the structure of thecoefficients c , d and e . Proposition B. . For any (cid:101) > there exists r = r ( (cid:101) ) > such that for any < r < r : | ∂ K | ( B r ( X )) = c n r n + + e ( X ) r n + + (cid:101) R ( r ) , ( ) where defined S ( n ) : = S n − ∩ n ⊥ , we have:(i) c ( X ) = c n : = √ π Γ ( n − ) σ ( S ( n ))( n + ) Γ ( n + ) ,(ii) e ( X ) = ˆ S ( n ) ˆ π − π d ( ϑ ) (cid:18)(cid:0) + n (cid:1) B A − C A − A B A + B ( n + ) (cid:19) c A n + d ϑ d σ . (iii) | R ( r ) | ≤ C ( n ) r n + for any < r < r and for some constant C ( n ) depending only on X .Proof. Thanks to Lemma B. and Proposition B. , we deduce that: | ∂ K | ( B r ( X )) = ˆ χ U r ( x + P ( ρ , ϑ , ψ )) Ξ ( ρ , ϑ , v ) d ρ d ϑ d σ ( v )= ˆ S ( n ) ˆ π − π ˆ { H ( ρ , ϑ , v ) ≤ r } d ( ϑ ) (cid:18) ρ n + A c ρ n + + B c ρ n + + R ( ρ ) ρ n c (cid:19) d ρ d ϑ d σ ( v ) .We now proceed giving estimates of each term in the last line of the above identity. In the proof of PropositionB. we showed that { H ( ρ , ϑ , v ) ≤ r } ⊆ { ρ ≤ c r } whenever 0 < r < r and c was a constant independent on ϑ and v . Therefore if r ≤ r : ˆ { H ( ρ , ϑ , v ) ≤ r } R ( ρ ) ρ n d ρ ≤ ˆ c r R ( ρ ) ρ n d ρ ≤ c ˆ c r ρ n + d ρ ≤ c c n + r n + ( n + ) , ( ) aylor expansion of area on quadratic t - cones 61 where the second inequality comes from Lemma B. provided c r ≤ r . Moreover, by Proposition B. for any (cid:101) > r > < r < r we have: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ˆ { H ( ρ , ϑ , v ) ≤ r } ρ j d ρ − ˆ P ϑ , v ( r ) ρ j d ρ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ˆ P ϑ , v ( r )+ (cid:101) r ρ j d ρ − ˆ P ϑ , v ( r ) − (cid:101) r ρ j d ρ = ( P ϑ , v ( r ) + (cid:101) r ) j + − ( P ϑ , v ( r ) − (cid:101) r ) j + j + ≤ j + (cid:101) P ϑ , v ( r ) j r j + )Thanks to the bounds ( ) and ( ), for any 0 < r < min { r , r , r } we have: | R ϑ , v ( r ) | : = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ˆ { H ( ρ , ϑ , v ) ≤ r } (cid:18) ρ n + A c ρ n + + B c ρ n + + R ( ρ ) ρ n c (cid:19) d ρ − ˆ P ϑ , v ( r ) (cid:18) ρ n + A c ρ n + + B c ρ n + (cid:19) d ρ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:101) n + P ϑ , v ( r ) n r n + + (cid:101) n + (cid:107)A(cid:107) ∞ P ϑ , v ( r ) n + r ( n + ) c + (cid:101) n + (cid:107)B(cid:107) ∞ P ϑ , v ( r ) n + r ( n + ) c + c c n + r n + ( n + ) , ( )where (cid:107)A(cid:107) ∞ and (cid:107)B(cid:107) ∞ are the maximum of the functions A and B on [ − π /2, π /2 ] × S ( n ) . Thanks to thedefinition of P ϑ , v , which was given in ( ), we can find r > (cid:101) and X , such that for any0 < r < r we have | R ϑ , v | ≤ (cid:101) C ( n ) r n + , where the constant C ( n ) depends only on X . Finally, there exists an r > C ( n ) > X such that: | ∆ ϑ , v ( r ) | : = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) P ϑ , v ( r ) n + − r n + A n + + ( n + ) Br n + A n + − n + A n + (cid:18) ( n + ) B A − C A (cid:19) r n + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C ( n ) r n + , | ∆ ϑ , v ( r ) | : = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) P ϑ , v ( r ) n + − r n + A n + + ( n + ) Br n + A n + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C ( n ) r n + , | ∆ ϑ , v ( r ) | : = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) P ϑ , v ( r ) n + − r n + A n + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C ( n ) r n + . ( )We omit the computations proving the three bounds above since they can easily obtained by explicitly computing P ϑ , v ( r ) j and truncating the expression to the desired order of r . Therefore, ( ) and ( ) imply that for any0 < r < min ( r , r ) we have: ˆ { H ( ρ , ϑ , v ) ≤ r } (cid:18) ρ n + A c ρ n + + B c ρ n + + R ( ρ ) ρ n c (cid:19) d ρ = ˆ P ϑ , v ( r ) (cid:18) ρ n + A c ρ n + + B c ρ n + (cid:19) d ρ + R ϑ , v ( r )= P ϑ , v ( r ) n + n + + A P ϑ , v ( r ) n + ( n + ) c + B P ϑ , v ( r ) n + ( n + ) c + R ϑ , v ( r )= r n + ( n + ) A n + + (cid:18) A ( n + ) c − B A (cid:19) r n + A n + + (cid:18) ( n + ) B A − C A − A B A + B ( n + ) (cid:19) r n + A n + + R ϑ , v ( r ) ,where R ϑ , v ( r ) : = R ϑ , v ( r ) + ∆ ϑ , v ( r ) + ∆ ϑ , v ( r ) + ∆ ϑ , v ( r ) and the functions B , . . . , E and A , B where introduced inthe statement of Proposition B. and in Lemma B. respectively. Thanks to the definitions of R ϑ , v and ∆ ϑ , vi , forany (cid:101) > r : = r ( (cid:101) ) > < r < r , we have | R ϑ , v ( r ) | ≤ (cid:101) C ( n ) r n + , where C ( n ) is a constant which depends only on (cid:101) and X . Therefore, using what we have deduced so far we get: | ∂ K | ( B r ( x )) = r n + ( n + ) ˆ S ( n ) ˆ π − π d ( ϑ ) A n + d ϑ d σ + r n + ˆ S ( n ) ˆ π − π d ( ϑ ) A n + (cid:18) A ( n + ) c − B A (cid:19) d ϑ d σ + r n + ˆ S ( n ) ˆ π − π d ( ϑ ) A n + (cid:32) ( n + ) B A − C A − A B A + B ( n + ) (cid:33) d ϑ d σ + R ( r ) , aylor expansion of area on quadratic t - cones 62 where R ( r ) : = ´ S ( n ) ´ π − π R ϑ , v ( r ) d ϑ d σ and for any 0 < r < r we have | R ( r ) | ≤ (cid:101) c r n + for some constant c depending only on X . Moreover, since B and A are odd functions on S ( n ) (see Proposition B. and Remark B. )we deduce that: ˆ S ( n ) A A n + d σ = ˆ S ( n ) B A n + d σ = | ∂ K | ( B r ( x )) = r n + ( n + ) ˆ S ( n ) ˆ π − π d ( ϑ ) A n + d ϑ d σ + r n + ˆ S ( n ) ˆ π − π d ( ϑ ) A n + (cid:32) ( n + ) B A − C A − A B A + B ( n + ) (cid:33) d ϑ d σ + R ( r ) ,We are left to prove that ´ S ( n ) ´ π − π d ( ϑ ) A n + d ϑ d σ is a constant depending only on n . This is done by explicit compu-tation: ˆ π − π d ( ϑ ) A n + d ϑ = ˆ ∞ − ∞ dx ( + ( x + γ ( v )) ) n + = √ π Γ (cid:16) n − (cid:17) Γ (cid:16) n + (cid:17) ,where we used Lemma B. in the first equality and Lemma B. in the second . Remark B. . The constant c n is the same constant appearing in Propositions A. and A. . Although we coulddeduce from Proposition A. that the leading term in the expansion ( ) is constant and its value, we neverthelessdecided to carry out the explicit computation (the last line in the proof of Proposition B. ) of the coefficient c ( X ) .In the previous propositions we gave a first characterisation of the coefficient of the Taylor expansion of theperimeter of quadratic surfaces. The coefficient relative r n + has been proved to be a constant depending onlyon n and the one of r n + has been proved null. In the following proposition we investigate more carefully thestructure of the coefficient relative to r n + : Proposition B. . In the notation of the previous propositions we have: e ( X ) = Tr ( D ) − (cid:104) n , D n (cid:105) + (cid:104) n , D n (cid:105) n − + n − n − − + (cid:104)D J n , n (cid:105) n − − ( Tr ( D ) − (cid:104) n , D n (cid:105) ) n − Proof.
Thanks to Proposition B. , we deduce that: e ( X ) = ˆ S ( n ) (cid:32) n + ˆ π − π d ( ϑ ) B A n + d ϑ (cid:124) (cid:123)(cid:122) (cid:125) (I) − ˆ π − π d ( ϑ ) C A n + d ϑ (cid:124) (cid:123)(cid:122) (cid:125) (II) − ˆ π − π d ( ϑ ) A B A n + d ϑ (cid:124) (cid:123)(cid:122) (cid:125) (III) + ˆ π − π d ( ϑ ) B ( n + ) A n + d ϑ (cid:124) (cid:123)(cid:122) (cid:125) (IV) (cid:33) d σ . ( )Now we study each term separately. Let us start studying (I). Since B = cos ϑ (cid:16) β n ( v ) ( sin ϑ cos ϑ ) ( sin ϑ cos ϑ + γ ( v )) (cid:17) ,Lemmas B. and B. imply:2 n + ˆ π − π d ( ϑ ) B A n + d ϑ = n + ˆ ∞ − ∞ β n ( v ) x ( γ ( v ) + x ) ( + ( γ ( v ) + x ) ) n + d ϑ = ( n + ) β n ( v ) ˆ ∞ − ∞ x ( x − γ ( v )) ( + x ) n + dx = ( n + ) β n ( v ) (cid:32) ˆ ∞ − ∞ x ( + x ) n + dx + γ ( v ) ˆ ∞ − ∞ x ( + x ) n + dx (cid:33) = C n β n ( v ) (cid:18) + ( n + ) γ ( v ) (cid:19) ,( )where C n : = √ π Γ ( n + ) n + Γ ( n + ) . aylor expansion of area on quadratic t - cones 63 We turn now our attention to (II). Since C = cos ϑ ( sin ϑ cos ϑ ) (cid:16) ( + β n ( v ) + γ ( v ) α n ) + α n sin ϑ cos ϑ (cid:17) , Lemmas B. and B. imply: ˆ π − π d ( ϑ ) C A n + d ϑ = ˆ ∞ − ∞ x (( + β n ( v ) + γ ( v ) α n ) + α n x )( + ( x + γ ( v )) ) n + dx = ˆ ∞ − ∞ ( x − γ ( v )) (( + β n ( v ) ) + α n x )( + x ) n + dx = ( + β n ( v ) − α n γ ( v )) ˆ ∞ − ∞ x ( + x ) n + dx + ( + β n ( v ) ) γ ( v ) ˆ ∞ − ∞ dx ( + x ) n + = C n ( n + )( + β n ( v )) γ ( v ) + + β n ( v ) − γ ( v ) α n )Since A B = ϑ ( β n ( v ) + (cid:104) Jv , n (cid:105) ) β n ( v ) sin ϑ cos ϑ (cid:16) γ ( v ) + sin ϑ cos ϑ (cid:17) , Lemmas B. and B. imply: ˆ π − π d ( ϑ ) A B A n + d ϑ = ( β n ( v ) + (cid:104) Jv , n (cid:105) ) β n ( v ) ˆ ∞ − ∞ x ( γ ( v ) + x )( + ( γ ( v ) + x ) ) n + d ϑ = ( β n ( v ) + (cid:104) Jv , n (cid:105) ) β n ( v ) ˆ ∞ − ∞ x ( x − γ ( v ))( + x ) n + dx = C n ( β n ( v ) + (cid:104) Jv , n (cid:105) ) β n ( v ) , ( )which concludes the discussion of the integral (III). Finally, we are left with the discussion of (IV). Thanks to thefact that B = ϑ (cid:16) α n sin ϑ cos ϑ + | P n ( D + J ) v | (cid:17) Lemmas B. and B. , imply that: ˆ π − π d ( ϑ ) A n + B ( n + ) d ϑ = n + ˆ ∞ − ∞ α n x + | P n ( D + J ) v | ( + ( x + γ ( v ))) n + dx = C n − α n γ ( v ) + | P n ( D + J ) v | )Plugging the identities ( ), ( ), ( ), ( ) into ( ), we get: e ( X ) C n σ ( S ( n )) = S ( n ) (cid:20) β n ( v ) (cid:18) + n + γ ( v ) (cid:19) − ( n + )( + β n ( v )) γ ( v ) + + β n ( v ) − γ ( v ) α n (cid:21) d σ ( v )+ S ( n ) (cid:20) − ( β n ( v ) + (cid:104) Jv , n (cid:105) ) β n ( v ) + − α n γ ( v ) + | P n ( D + J ) v | (cid:21) d σ ( v )= − n + S ( n ) γ ( v ) d σ (cid:124) (cid:123)(cid:122) (cid:125) (V) − − S ( n ) β n ( v ) (cid:104) Jv , n (cid:105) d σ (cid:124) (cid:123)(cid:122) (cid:125) (VI) + S ( n ) | P n ( D + J ) v | d σ (cid:124) (cid:123)(cid:122) (cid:125) (VII) .Eventually, in order to make the expression for e ( X ) explicit, we need to compute the integrals (V), (VI) and(VII). To do so, we let E : = { m , . . . , m n } be an orthonormal basis of R n such that m = n and: Jm i = (cid:40) m n + i if i ∈ {
1, . . . , n } , − m i − n if i ∈ { n +
1, . . . , 2 n } . ( )With respect to the basis E , the points v ∈ S ( n ) are written as v = ∑ ni = v i m i where v i : = (cid:104) v , m i (cid:105) . This is due tothe fact that v ∈ n ⊥ by definition of S ( n ) . With these notations, the integral (V) becomes: S ( n ) γ ( v ) d σ ( v ) = S ( n ) (cid:32) n ∑ i , j = (cid:104) m i , D m j (cid:105) v i v j (cid:33) d σ ( v ) = n ∑ i , j , k , l = (cid:104) m i , D m j (cid:105)(cid:104) m k , D m l (cid:105) S ( n ) v i v j v k v l d σ ( v )= n ∑ i = (cid:104) m i , D m i (cid:105) S ( n ) v i d σ ( v ) + ∑ ≤ i , j ≤ ni (cid:54) = j (cid:104) m i , D m i (cid:105)(cid:104) m j , D m j (cid:105) S ( n ) v i v j d σ ( v ) + ∑ ≤ i , j ≤ ni (cid:54) = j (cid:104) m i , D m j (cid:105) S ( n ) v i v j d σ ( v ) , aylor expansion of area on quadratic t - cones 64 where the last equality comes from the fact that integrals of odd functions on S ( n ) are null. By direct computationor using formulas stated at the beginning of section 2 c in [ ], we have that: S ( n ) γ ( v ) d σ ( v ) = n − n ∑ i = (cid:104) m i , D m i (cid:105) + n − n ∑ i (cid:54) = k = (cid:104) m i , D m i (cid:105)(cid:104) m k , D m k (cid:105) + n − n ∑ i (cid:54) = j = (cid:104) m i , D m j (cid:105) = n − n ∑ i , j = (cid:104) m i , D m j (cid:105) + n − (cid:32) n ∑ i = (cid:104) m i , D m i (cid:105) (cid:33) .Since the matrix D is symmetric, we have that Tr ( D ) = ∑ ni , j = (cid:104) m i , D m j (cid:105) . Thanks to this identity we deduce that: S ( n ) γ ( v ) d σ ( v ) = ( D ) − ∑ ni = (cid:104) m i , D n (cid:105) − (cid:104) n , D n (cid:105) + ( Tr ( D ) − (cid:104) n , D n (cid:105) ) n − = ( D ) − (cid:104) n , D n (cid:105) + (cid:104) n , D n (cid:105) + ( Tr ( D ) − (cid:104) n , D n (cid:105) ) ( n − )( n + ) ,where the last identity comes from the fact that |D n | = (cid:104) n , D n (cid:105) and some algebraic manipulations. Thecomputation of the integral (VI) is much easier. Indeed: S ( n ) (cid:104) Jv , n (cid:105) β n ( v ) d σ ( v ) = n − ∑ i , j = (cid:104) Jm i , n (cid:105)(cid:104)D m j , n (cid:105) S ( n ) v i v j d σ ( v ) = − (cid:104)D J n , n (cid:105) n − (cid:104) Jm j , n (cid:105) (cid:54) = j = n +
1, by the choice of the basis E ,and that: S ( n ) v i v j d σ ( v ) = (cid:40) i (cid:54) = j , n − if i = j . ( )We are left to study the integral (VII). Since P n is the orthogonal projection on n ⊥ , we have that: | P n ( Mv ) | = n ∑ i = (cid:104) m i , Mv (cid:105) = n ∑ i , j , k = v j v k (cid:104) m i , Mm j (cid:105)(cid:104) m i , Mm k (cid:105) .Furthermore, thanks to ( ), we deduce that: S ( n ) | P n (( D + J ) v ) | d σ ( v ) = n ∑ i , j , k = (cid:104) m i , ( D + J ) m j (cid:105)(cid:104) m i , ( D + J ) m k (cid:105) S ( n ) v j v k d σ ( v )= n ∑ i , j = (cid:104) m i , ( D + J ) m j (cid:105) S ( n ) v j d σ ( v ) = n − n ∑ i , j = (cid:104) m i , ( D + J ) m j (cid:105) .We wish now to make ∑ ni , j = (cid:104) m i , ( D + J ) m j (cid:105) more explicit. To do so note that by definition of E , we have: (cid:104) m i , Jm j (cid:105) = i ∈ {
1, . . . , n } , − i ∈ { n +
1, . . . , 2 n } ,0 otherwise. ( )The identities in ( ) imply that ∑ ni , j = (cid:104) m i , Jm j (cid:105) = n − D is symmetric, we also have that: n ∑ i , j = (cid:104) m i , D m j (cid:105)(cid:104) m i , Jm j (cid:105) = eferences Summing up what we have proved up to this point, we get: S ( n ) | P n (( D + J ) v ) | d σ ( v ) = n − n ∑ i , j = (cid:104) m i , D m j (cid:105) + (cid:104) m i , D m j (cid:105)(cid:104) m i , Jm j (cid:105) + (cid:104) m i , Jm j (cid:105) = ∑ ni , j = (cid:104) m i , D m j (cid:105) + n − n − = Tr ( D ) − (cid:104) n , D n (cid:105) + (cid:104) n , D n (cid:105) n − + n − n − ( D ) = ∑ ni , j = (cid:104) m i , D m j (cid:105) and a few algebraic manipulations. Finally putting togheter the expressions of the integrals (V), (VI) and (VII),we get: c e ( X ) C n σ ( S ( n )) = − n + (cid:18) ( D ) − (cid:104) n , D n (cid:105) + (cid:104) n , D n (cid:105) + ( Tr ( D ) − (cid:104) n , D [ n ] (cid:105) ) ( n − )( n + ) (cid:19) − + (cid:104)D J n , n (cid:105) n − + (cid:18) Tr ( D ) − (cid:104) n , D n (cid:105) + (cid:104) n , D n (cid:105) n − + n − n − (cid:19) =
14 Tr ( D ) − (cid:104) n , D n (cid:105) + (cid:104) n , D n (cid:105) n − + n − n − − + (cid:104)D J n , n (cid:105) n − − ( Tr ( D ) − (cid:104) n , D n (cid:105) ) ( n − ) ,where the last identity is obtained from the previous ones with few algebraic computations. Theorem B. . Assume µ is a ( n + ) -uniform measure supported on K ( D , − ) . For any h ∈ R n \ Σ ( f ) (see ( ) )we have: = Tr ( D ) − (cid:104) n ( h ) , D n ( h ) (cid:105) + (cid:104) n ( h ) , D n ( h ) (cid:105) ( n − ) + n − n − − + (cid:104)D J n ( h ) , n ( h ) (cid:105) n − − ( Tr ( D ) − (cid:104) n ( h ) , D n ( h ) (cid:105) ) ( n − ) ,( ) where n ( h ) : = ( D + J ) h |D + J ) h | .Proof. Suppose that X : = ( h , f ( h )) ∈ supp ( µ ) . Then Proposition . implies that there exists r > B r ( X ) ∩ K ( D , 1 ) = B r ( X ) ∩ supp ( µ ) ,for any 0 < r < r . Therefore Propositions . , A. and B. imply that: r n + = µ ( B r ( X )) = S n + ( µ ) ( B r ( X )) = | ∂ K | ( B r ( X )) c n = r n + + e ( Z ) c n r n + + R ( r ) c n ,whenever 0 < r < min ( r , r ) . In particular we deduce that e ( X ) =
0, since | R ( r ) | ≤ C ( n ) r n + for any0 < r < r and the constant C ( n ) depends only on X .If n > ( Σ ( f )) =
0, Proposition . and Proposition B. prove the claim since π H ( supp ( µ )) = R n . Onthe other hand if n = ( Σ ( f )) =
1, Proposition . implies that one of the two connected components of R \ Σ ( f ) (which are halfspaces which bonduaty pass through 0) is contained in π H ( supp ( µ )) and thus equation( ) holds for any z contained in such connected component. However, the structure of the coefficient e ( X ) andthe fact that n ( − h ) = − n ( h ) imply that equation ( ) holds on R \ Σ ( f ) . references [ ] Luigi Ambrosio and Bernd Kirchheim. “Rectifiable sets in metric and Banach spaces”. In: Math. Ann. . ( ), pp. – . issn : - .[ ] Zoltán M. Balogh. “Size of characteristic sets and functions with prescribed gradient”. In: J. Reine Angew.Math. ( ), pp. – . issn : - . eferences [ ] V. I. Bogachev. Measure theory. Vol. I, II . Springer-Verlag, Berlin, , Vol. I: xviii+ pp., Vol. II: xiv+ . isbn : - - - - .[ ] Luca Capogna, Donatella Danielli, Scott D. Pauls, and Jeremy T. Tyson. An introduction to the Heisenberggroup and the sub-Riemannian isoperimetric problem . Vol. . Progress in Mathematics. Birkhäuser Verlag,Basel, , pp. xvi+ . isbn : - - - - .[ ] Vasilis Chousionis, Valentino Magnani, and Jeremy T. Tyson. “On uniform measures in the Heisenberggroup”. In: arXiv e-prints , arXiv: . (Aug. ).[ ] Vasilis Chousionis and Jeremy T. Tyson. “Marstrand’s density theorem in the Heisenberg group”. In: Bull.Lond. Math. Soc. . ( ), pp. – . issn : - .[ ] Camillo De Lellis. Rectifiable sets, densities and tangent measures . Zurich Lectures in Advanced Mathematics.European Mathematical Society (EMS), Zürich, , pp. vi+ . isbn : - - - - .[ ] Herbert Federer. Geometric measure theory . Die Grundlehren der mathematischen Wissenschaften, Band .Springer-Verlag New York Inc., New York, , pp. xiv+ .[ ] Bruno Franchi, Raul Paolo Serapioni, and Francesco Serra Cassano. “Area formula for centered Hausdorffmeasures in metric spaces”. In: Nonlinear Anal. ( ), pp. – . issn : - X.[ ] Bruno Franchi, Raul Serapioni, and Francesco Serra Cassano. “Rectifiability and perimeter in the Heisen-berg group”. In: Math. Ann. . ( ), pp. – . issn : - .[ ] Bernd Kirchheim and David Preiss. Uniformly distributed measures in Euclidean spaces . .[ ] Oldˇrich Kowalski and David Preiss. “Besicovitch-type properties of measures and submanifolds”. In: J.Reine Angew. Math. ( ), pp. – . issn : - .[ ] Andrew Lorent. “Rectifiability of measures with locally uniform cube density”. In: Proc. London Math. Soc.( ) . ( ), pp. – . issn : - .[ ] Valentino Magnani. “A new differentiation, shape of the unit ball, and perimeter measure”. In: Indiana Univ.Math. J. . ( ), pp. – . issn : - .[ ] Valentino Magnani. “Unrectifiability and rigidity in stratified groups”. In: Arch. Math. (Basel) . ( ),pp. – . issn : - X.[ ] Pertti Mattila. Geometry of sets and measures in Euclidean spaces . Vol. . Cambridge Studies in AdvancedMathematics. Fractals and rectifiability. Cambridge University Press, Cambridge, , pp. xii+ . isbn : - - - .[ ] Pertti Mattila. “Measures with unique tangent measures in metric groups”. In: Math. Scand. . ( ),pp. – . issn : - .[ ] Roberto Monti. “Isoperimetric problem and minimal surfaces in the Heisenberg group”. In: Geometric mea-sure theory and real analysis . Vol. . CRM Series. Ed. Norm., Pisa, , pp. – .[ ] David Preiss. “Geometry of measures in R n : distribution, rectifiability, and densities”. In: Ann. of Math. ( ) . ( ), pp. – . issn : - X.[ ] Leon Simon. Lectures on geometric measure theory . Vol. . Proceedings of the Centre for Mathematical Anal-ysis, Australian National University. Australian National University, Centre for Mathematical Analysis,Canberra, , pp. vii+ . isbn : - - - .[ ] E. T. Whittaker and G. N. Watson. A course of modern analysis . Cambridge Mathematical Library. An in-troduction to the general theory of infinite processes and of analytic functions; with an account of theprincipal transcendental functions, Reprint of the fourth ( ) edition. Cambridge University Press, Cam-bridge, , pp. vi+ . isbn : - - -3