Classification of polytope metrics and complete scalar-flat Kähler 4-Manifolds with two symmetries
CClassification of polytope metrics and completescalar-flat K¨ahler 4-Manifolds with two symmetries
September 2015
Abstract
We study unbounded 2-dimensional metric polytopes such as those arising as K¨ahlerquotients of complete K¨ahler 4-manifolds with two commuting symmetries and zeroscalar curvature. Under a mild closedness condition, we obtain a complete classificationof metrics on such polytopes, and as a result classify all possible metrics on on thecorresponding K¨ahler 4-manifolds. If the polytope is the plane or half-plane then onlyflat metrics are possible, and if the polytope has one corner then the 2-parameter familyof generalized Taub-NUTs (discovered by Donaldson) are indeed the only possiblemetrics. Polytopes with n ≥ n + 2)-dimensional family of possiblemetrics. We study polytope metrics of the kind arising from the reduction of complete K¨ahler 4-manifolds with a pair of commuting holomorphic Killing fields X , X and zero scalar curva-ture; we are particularly interested in the case that M is complete. In the simply-connectedcase, real-holomorphic fields are generated by potentials; here these are functions ϕ i with ω ( X i , · ) = − dϕ i . (1)This leads to the so-called moment map Φ : M (cid:55)→ R given by Φ( p ) = ( ϕ ( p ) , ϕ ( p )),whose image is a polytope called the associated moment polytope [11][28][5][21], which wecall Σ . Since the X i are Killing the map Φ : M → Σ , which is generically submersive,endows Σ with a Riemannian metric g Σ ; the result is a metric polytope (Σ , g Σ ) called theK¨ahler reduction of ( M , J, ω, X , X ). Setting V = |X | |X | − (cid:104)X , X (cid:105) = |∇ ϕ | |∇ ϕ | − (cid:10) ∇ ϕ , ∇ ϕ (cid:11) (2)and letting (cid:52) Σ be the laplacian on (Σ , g Σ ), we shall see that (cid:52) Σ √V + 12 s √V = 0 (3)1 a r X i v : . [ m a t h . DG ] S e p here s is the scalar curvature on M , expressed, with √V as a function on Σ . This is aversion of the Abreu equation from [1]; see the discussion in Section 2. When s = 0 we havea Laplace equation and √V is a natural harmonic coordinate on Σ .Our work involves only the metric polytopes themselves, whether or not they comefrom any K¨ahler reduction. For this reason we take the time to state our hypotheses ina strictly analytic framework, apart from any K¨ahler geometry it may be reduced from.Unless specifically stated otherwise, we shall always assume our polytopes (Σ , g Σ ) satisfyA) (Closedness) Σ is a closed subset of the ( ϕ , ϕ ) coordinate plane.B) (Boundary connectedness) The polytope boundary has just one component or no com-ponents.C) (Natural polytope condition) Σ is a convex polytope with finitely many edges. Themetric g Σ is C ∞ up to boundary segments and Lipschitz at corners, and each boundarysegment or ray is totally geodesic.and our functions ϕ , ϕ satisfyD) (Natural boundary conditions) The unit vector fields ∇ ϕ i |∇ ϕ i | are well-defined and covariant-constant on boundary segments.E) (Pseudo-toric condition) The functions ϕ , ϕ are C ∞ and [ ∇ ϕ , ∇ ϕ ] = 0 on (Σ , g Σ ).F) (Pseudo-ZSC condition) With V = |∇ ϕ | |∇ ϕ | − (cid:10) ∇ ϕ , ∇ ϕ (cid:11) , we have (cid:52) Σ √V = 0.Conditions (C), (D), and (E) are automatic when (Σ , g Σ ) is the K¨ahler reduction ofsome M (see § M also has zero scalar curvature.Conditions (A) and (B), however, are genuine restrictions in the sense that there existmanifolds ( M , J, ω, X , X ) with associated polytopes that violate either or both; see Ex-amples 7 and 8 below. We will not discuss such polytopes, except to say that (A) is violatedwhen some symmetry has a “zero at infinity” as in Example 7, and (B) is violated when thezero locus {X = 0 } ∪ {X = 0 } is disconnected, as in Example 8. For a discussion of whythese cases are more difficult, and remain unresolved, see the final remark in Section 5.We completely classify polytope metrics under conditions (A)-(F). Should (Σ , g Σ ) bea K¨ahler reduction, it is well-known that all data on M can be reconstructed from g Σ (see § , g Σ ) have profound implications for K¨ahlermanifolds ( M , J, ω, X , X ) with commuting symmetries. Theorem 1.1 (Compact Polytope) If (Σ , g Σ ) is compact, and even if (F) is relaxed to (cid:52) Σ √V ≥ , then g Σ is flat. (C.f. Corollary 2.6.) Corollary 1.2 (Compact M with s ≤ ) Assume ( M , J, ω, X , X ) has s ≤ . If theassociated metric polytope (Σ , g Σ ) is compact, then M is flat. (C.f. Corollary 2.7.) The simplest non-compact case is when Σ is metrically complete; this is the subject of ourfirst substantial results. Note also the different sign required of s . Theorem 1.3 (Polytope is complete)
Assume (Σ , g Σ ) is a geodesically complete met-ric polytope with (cid:52) Σ √V ≤ (this is a relaxation of (A) and (F)). Then g Σ is flat. (C.f.Theorem 3.3.) Corollary 1.4
Assume ( M , J, ω, X , X ) is simply connected and has s ≥ . If the fieldsare no-where zero and no-where equal, then M is flat. (C.f. Corollary 3.4.) Our third theorem deals with the case that Σ is a half-plane. Unlike the previous results,its proof requires the full strength of (A)-(F). Theorem 1.5 (Polytope is a half-plane) If (Σ , g Σ ) is a closed half-plane, then g Σ isflat. (C.f. Theorem 4.17.) Corollary 1.6
Assume ( M , J, ω, X , X ) is simply connected, scalar-flat, and has associ-ated polytope (Σ , g Σ ) with just one edge. Then M is flat. (C.f. Corollary 4.18.) When X , X have common zeros, the situation is more complicated. Since at least thepapers of Donaldson [13] and Abreu–Sena-Dias [3], it has been known that if M is scalar-flat and has a polytope with one or more vertices, then M admits at least a two-parameterfamily of scalar flat metrics (up to homothety). As we show, when Σ has just one vertex,these are all such metrics: one of these metrics is flat, one is the Taub-NUT metric, and therest are the achiral generalized Taub-NUT metrics of Donaldson (from section 6 of [13]). Theorem 1.7 (Polytope has one vertex)
Assume (Σ , g Σ ) has a single corner, so af-ter affine recombination of ϕ , ϕ we may assume Σ is the closed first quadrant. Up tohomothety, the metric g Σ lies within a 2-parameter family of metrics (cf. Theorem 5.2). Corollary 1.8
Under the hypotheses of Theorem 1.5, suppose also that X , X have a singlecommon zero. Up to homothety, M has precisely a 2-parameter family of metrics whichgive it a metric of zero scalar curvature (cf. Corollary 5.3). The situation of more than two edges is different still.3 heorem 1.9 (Generic case)
Assume Σ has n ≥ many edges. Under conditions (A)-(F), Σ admits precisely an ( n + 2) -parameter family of metrics. (C.f. Theorem 5.4.) Corollary 1.10
Suppose (Σ , g Σ ) is the K¨ahler reduction of a manifold ( M , J, ω, X , X ) ,and Σ has n ≥ edges, has no edges at infinity, and has connected boundary. If M is scalar flat, its metric is one in a ( n + 2) -parameter family of possible metrics.F (C.f.Corollary 5.5.) We give a detailed construction of these polytope metrics in the proof of Corollary5.5, and a detailed recipe for constructing metrics on M from the metric on Σ in § M whosepolytope has two corners; these include the metrics on T C P such as the Eguchi-Hansonmetric, and the metrics of LeBrun [29]. Of course obtaining the M metric from its polytopeis nothing new, see [20] [1], but the method we find most helpful is a variation on what usuallyappears in the literature. In particular we avoid the “K¨ahler potential” techniques of [20]and [1], which lead to a 4th order scalar PDE, and instead use essentially equivalent systemof second order PDEs.The proofs of Theorems 1.1 and 1.3 are reasonably self-contained and easy. Theorem1.5 requires the preparatory work of § § §
4, whichclassifies non-negative solutions of x ( ϕ xx + ϕ yy ) − ϕ x = 0 on the half-plane H = { x > } with zero boundary values. The geometric classifications in Theorems 1.7 and 1.10 areessentially corollaries of the analytic classification. The equation x ( ϕ xx + ϕ yy ) − ϕ x = 0 isin fact a geometric PDE that appeared first in a paper of Donaldson’s [12]; see Proposition2.5 and the discussion at the beginning of Section 4.The main result of Section § x ( f xx + f yy ) + 3 f x = 0 by combining the techniques of blow-ups, barriers, and Fourieranalysis. A certain duality relation exists between x ( f xx + f yy ) + 3 f x = 0 with arbitraryboundary conditions on the half-plane H and the PDE x ( ϕ xx + ϕ yy ) − ϕ x = 0 with zeroboundary conditions on H , so the Liouville-type theorem for the first gives the desiredclassification of all non-negative solutions for the second: these all have the form Cx for C > x ( g xx + g yy ) + νg x = 0 and the corresponding parabolic equation g t = x ( g xx + g yy ) + νg x have been studied in a variety of contexts, almost exclusively for ν ≥ ν > ν <
1: if g solves theequation for ν then ˜ g = x ν − g solves it for ˜ ν = 2 − ν ).The operator (cid:52) µ = x (cid:16) ∂ ∂x + ∂ ∂y (cid:17) can be considered a Laplace operator for the righthalf-plane with a hypoerbolic metric. Thus from a naive point of view, g t = (cid:52) µ g + νg x canbe considered a heat flow through an isotropic but inhomogeneous medium of hyperbolicspecific heat density, with a directional transfer bias indicated by ν . Less naively, thisequation (or similar equations such as the Heston equation, equation (104)) has been studied4n connection with mathematical finance, stochastic PDE and Feynman-Kac theory [22],mathematical biology [14], and the porous medium equation [10]. The change of variables s = x , t = √ y turns x ( g xx + g yy ) − g x = 0 into sf ss + f tt = 0, which has been studiedin connection with population dynamics [14].However, much of this study has been related to existence/uniqueness of flows or bound-ary value problems, H¨older estimates, admissibility of boundary values, and so on. Muchof this work has been local or confined to pre-compact domains, although there are globalresults such as global H¨older estimates. To this author’s knowledge, this paper contains thefirst Liouville-type result for these equations. Remark.
Our 4-manifolds need not be “toric” exactly, but only have commutingholomorphic Killing fields. Our work takes place mostly on the moment polytope where thevertex angles are irrelevant. There is no particular need that the momentum constructionrebuild any actual 4-manifold.
Remark . Our conclusions will hold directly on certain orbifolds with commutingholomorphic Killing fields. However we do not explore the significance of our results to thesituation of “marked polytopes” [30] and the like.
Remark.
Notice that Theorem 1.1 requires s ≤ s ≥ s = 0 is necessary for the other three theorems. We conjecture that Theorem 1.5 (thatthe half-plane is flat) continues to hold if s ≥
0. Further, we conjecture that if s ≥ § s ≡
0, and a more sophisticated analysisis required. See the discussion at the beginning of § Remark.
It is important to mention two issues not addressed in this paper. The firstis the question of how many of the metrics in our ( n + 2)-dimensional family are repeats.Conceivably even the dimension of this family could be reduced by the action of somediffeomorphism group, although we do not believe this to be the case.The second is that, although the polytopes Σ with n ≥ n + 2)-dimensional family of metrics, we do not address the issue of which of these may actuallycome from the reduction of some M . This is a another way the family of scalar-flat metricson M may possibly be smaller than ( n + 2)-dimensional. Acknowledgements . The author would like to thank Xiuxiong Chen, Ryan Hynd,Philip Gressman, and Camelia Pop for a number of helpful conversations that providedsome valuable insights. 5
K¨ahler Reduction
Here we discuss the detailed relationship between the K¨ahler reduction (Σ , g Σ ) and ( M , J, ω, X , X ).The philosophy is “Guillemin’s principle,” to wit, sympelctic coordinates are useful in K¨ahlergeometry [20], [2]. This principle has been excellently developed by a number of authors[20] [1] [12] [3] [6], but begging the knowledgeable reader’s forbearance, we redevelop someaspects as suits the present interest. First we indicate this section’s primary milestones.In § ϕ , ϕ , θ , θ ) to the complex-analytic coordinates ( z , z ). The resulting holomorphic volume form is a quarter of theparallelochoron volume, which is V (cid:44) |X | |X | − (cid:104)X , X (cid:105) = |∇ ϕ | |∇ ϕ | − (cid:10) ∇ ϕ , ∇ ϕ (cid:11) , (4)which implies that the scalar curvature of ( M , J, ω ) is s = −(cid:52) log V . Our notational set-upthis well-known construction will be useful later.Even with this elliptic relation, and even with a sign on s , the behavior of log V on M may be hard to understand, and so in § , g Σ ) itself, which is naturally a Riemannian manifold that may or may nothave boundary. In the inherited metric we show (cid:52) Σ V + s V = 0, which is a version ofthe Abreu equation: see (10) of [1]. Of course when M is scalar-flat then √V is harmonicon (Σ , g Σ ), and we have natural isothermal coordinates ( x, y ) on Σ where x = √V and y is defined by J Σ dy = dx . We compute the Gauss curvature K Σ of Σ in these coordinates.In § g Σ , g simply by knowing ϕ , ϕ asfunctions of ( x, y ). We also compute the intrinsic Gaussian curvature K Σ of (Σ , g Σ ) ina second, independent way, and show that although K Σ is not necessarily signed, after acanonical conformal change it is indeed signed.In § x, y ) actually has no critical points,and that the map X = ( x, y ) T sends Σ onto the right half-plane { x ≥ } in a one-to-onefashion, and the isothermal coordinates are therefore global. This fundamentally relies oncondition (B), connectedness of the polytope boundary.Finally in § ϕ , ϕ , x , and y near the polytopeedges. This provides us with boundary conditions that are necessary to our analysis of § We have a simply connected K¨ahler 4-manifold with commuting Killing fields ( M , J, ω, X , X ).The momentum construction consists of finding potentials ϕ , ϕ satisfying ω ( X i , · ) = − dϕ i ,traditionally called momentum variables or action coordinates . Two additional coordinates6 , θ , called cyclic variables or angle coordinates , are defined by taking a transversal to the {X , X } distribution and then pushing the natural R variables forward along the actionof {X , X } . Obviously the values of θ , θ are not canonical, but the fields ∂∂θ , ∂∂θ arecanonical. This construction gives a full coordinate system { ϕ , ϕ , θ , θ } with fields ∂∂ϕ = |X | V ∇ ϕ − (cid:104)X , X (cid:105)V ∇ ϕ , ∂∂θ = X , ∂∂ϕ = − (cid:104)X , X (cid:105)V ∇ ϕ + |X | V ∇ ϕ , ∂∂θ = X . (5)Letting G − be the matrix G − = (cid:18) |X | (cid:104)X , X (cid:105)(cid:104)X , X (cid:105) |X | (cid:19) , (6)then the ordered frame ∂∂ϕ , ∂∂ϕ , ∂∂θ , ∂∂θ produces the metric, complex structure, and sym-plectic form g = (cid:18) G G − (cid:19) , J = (cid:18) − G − G (cid:19) , ω = (cid:18) − IdId (cid:19) . (7) Remark . Compare (7) to compare (4.8) of [20] or (2.2), (2.3) of [2]. The more usualconstruction sets G = ∇ u for a function u called the K¨ahler potential. In the present workwe do not find this formulation useful. In our formulation, of course it may be objected thatexpressing the metric in terms of inner products is redundant. Still, this formulation lendsitself to the study of second order instead of fourth order PDEs, and is useful in relatingthe polytope metric g Σ with the original metric g . Lemma 2.1 (Symplectic and holomorphic coordinates on M ) The complex-valuedfunctions z = f ( ϕ , ϕ ) + √− θ z = f ( ϕ , ϕ ) + √− θ (8) form a holomorphic coordinate chart provided f , f satisfy df = Jdθ = |X | V dϕ − (cid:104)X , X (cid:105)V dϕ df = Jdθ = − (cid:104)X , X (cid:105)V dϕ + |X | V dϕ . (9) Further, dJdθ i = 0 , so indeed such functions f , f can be found locally, and thereforeholomorphic coordinates of the form (8) exist on ( M , J, ω ) .Proof . Using that ¯ ∂ = (cid:0) d + √− Jd (cid:1) on functions gives that ¯ ∂z i = 0 if and only if (9)holds. That dJdθ i = 0 follows from the preservation of J under the action of X , X (alengthy though elementary calculation can give dJdθ i = 0 directly). (cid:3) ∂∂z = 12 (cid:18) |X | ∂∂ϕ + (cid:104)X , X (cid:105) ∂∂ϕ − √− ∂∂θ (cid:19) = 12 (cid:0) ∇ ϕ − √− X (cid:1) ∂∂z = 12 (cid:18) (cid:104)X , X (cid:105) ∂∂ϕ + |X | ∂∂ϕ − √− ∂∂θ (cid:19) = 12 (cid:0) ∇ ϕ − √− X (cid:1) dz = |X | V dx − (cid:104)X , X (cid:105)V dx + √− dθ = Jdθ + √− dθ dz = − (cid:104)X , X (cid:105)V dx + |X | V dx + √− dθ = Jdθ + √− dθ (10)so in the holomorphic frame the Hermitian metric and associated volume element are simply h i ¯ = 12 (cid:18) |X | (cid:104)X , X (cid:105)(cid:104)X , X (cid:105) |X | (cid:19) = 12 G − , det h i ¯ = 14 V . (11) Proposition 2.2
The Ricci form and scalar curvature of ( M , J, ω ) are ρ = −√− ∂ ¯ ∂ log V s = −(cid:52) log V . (12) Proof . Textbook formulas. (cid:3)
Remark . We have given, in Lemma 2.1 and (10), a very prosaic version of what isusually a expressed as a Legendre transform between the so-called K¨ahler and symplecticpotentials, as developed in [20] (see also [1]). As we have chosen not to work with K¨ahlerpotentials, these Legendre transform methods won’t apply. In any case the workaday for-mulation above is probably more useful in the present context. M to a Metric Polytope Let (Σ , g Σ ) be the Riemannian quotient (leaf space) of M by the Killing actions of X , X . The commutativity of X , X imply the functions ϕ , ϕ pass to the quotient and thereconstitute a coordinate system. As is well-known, the differentiable map Φ : Σ → R ,Φ( p ) = ( ϕ ( p ) , ϕ ( p )) T is 1-1 with differentiable inverse, so Σ may be identified with itsimage in the ( ϕ , ϕ )-coordinate plane. This constitutes the moment polytope of M .For clarity, objects on Σ will be indicated with a subscript, so for instance s Σ and s indicate the scalar curvatures on (Σ , g Σ ) and ( M , J, ω ), respectively. The inherited metric8nd natural complex structure are g Σ = V − (cid:18) |X | − (cid:104)X , X (cid:105)− (cid:104)X , X (cid:105) |X | (cid:19) = G, (13) J Σ = V − (cid:18) (cid:104)X , X (cid:105) −|X | |X | − (cid:104)X , X (cid:105) (cid:19) . (14)Of course J Σ is not inherited from J , but is the Hodge star of (Σ , g Σ ). Proposition 2.3 ( Σ and M Laplacian relationship) If f : M → C is any functionon M that is X - and X -invariant, then (cid:52) f is X , X invariant, so f , (cid:52) f are naturallyfunctions on Σ . As functions on Σ , the two Laplacians are related by (cid:52) f = (cid:52) Σ f + (cid:68) ∇ Σ log V , ∇ Σ f (cid:69) Σ (15) Proof . A function f : M → C is invariant under X , X if and only if it is a function of ϕ , ϕ only. Noting that det( g ) = 1 and det( g Σ ) = V − we have (cid:52) f = ∂∂ϕ i (cid:18) g ij ∂f∂ϕ j (cid:19) = ∂∂ϕ i (cid:18) g ij Σ ∂f∂ϕ j (cid:19) = V ∂∂ϕ i (cid:18) g ij Σ V − ∂f∂ϕ j (cid:19) + g ij Σ ∂ log V ∂ϕ i ∂f∂ϕ j = (cid:52) Σ f + 12 (cid:104)∇ Σ log V , ∇ Σ f (cid:105) Σ . (16) (cid:3) Corollary 2.4 (Second order Abreu equation)
The scalar curvature s on M passesto a function on Σ , and (cid:52) Σ V + 12 s V = 0 . (17) Proof . This follows from s = −(cid:52) log V and Proposition 2.3. (cid:3) Proposition 2.5 (The ϕ i elliptic equations) On Σ we have d (cid:16) V − J Σ dϕ i (cid:17) = 0 .Proof . If Φ : M → Σ is the moment map projection onto the polytope, then from (7) and(14) we obtain df i = Φ ∗ ( V − J Σ dϕ i ), where the f i are from Lemma 2.1. Finally d Φ ∗ = Φ ∗ d and the fact that Φ is a submersion gives d ( V − J Σ dϕ i ) = 0 (cid:3) Remark.
When s = 0, the equations V d (cid:16) V − Jdϕ i (cid:17) = 0 are just the equations ϕ ixx + ϕ iyy − x − ϕ ix = 0 on the right half-plane. See Section 4.9he following simple theorem illustrates a contrast between the compact case, where s ≤ s ≤ Corollary 2.6 If (Σ , g Σ ) is topologically compact, it is impossible that (cid:52) Σ √V ≥ .Proof . The maximum principle gives V = const . But V = 0 on ∂ Σ , so V ≡ (cid:3) Corollary 2.7 If ( M , J, ω, X , X ) is topologically compact, it is impossible that s ≤ .Proof . If s ≤ (cid:52) Σ √ V ≤
0, so the previous corollary applies. But
V ≡ X and X are co-linear throughout, an impossibility. (cid:3) M provided s = 0 We show how to reconstruct the metrics on ( M , J, ω ) and (Σ , g Σ ) from knowledge of thefunctions ϕ , ϕ , V on Σ . Proposition 2.5 tells us the coordinates ϕ i are not harmonic on(Σ , g Σ ); indeed if ϕ , ϕ are harmonic then g Σ and g are both flat metrics; see Lemma3.1. Instead, when s = 0, the equation (cid:52)V = 0 shows V and its dual harmonic functionconstitute a natural harmonic coordinate system. We denote these by ( x , x ), where x and x are defined as the solutions to x = √V , dx = J Σ dx . (18)We shall use the notations ( x, y ) and ( x , x ) interchangeably. Regarding ϕ , ϕ as functionsof ( x, y ), the formula for J Σ in (14) gives1 x (cid:10) ∇ ϕ i , ∇ ϕ j (cid:11) dϕ ∧ dϕ = − dϕ i ∧ J Σ dϕ j . (19)Changing variables to express both sides in terms of ( x , x ) we have dϕ ∧ dϕ = dϕ dx i dϕ dx j dx i ∧ dx j = (cid:18) dϕ dx dϕ dx − dϕ dx dϕ dx (cid:19) dx ∧ dx , − dϕ i ∧ J Σ dϕ j = (cid:18) dϕ i dx dϕ j dx + dϕ i dx dϕ j dx (cid:19) dx ∧ dx , (20)and so g ij Σ = (cid:10) ∇ ϕ i , ∇ ϕ j (cid:11) = x · dϕ i dx dϕ j dx + dϕ i dy dϕ j dydϕ dx dϕ dy − dϕ dx dϕ dy . (21)10e can succinctly express this by letting A = (cid:16) ∂ϕ i ∂x j (cid:17) and B = A − = (cid:16) ∂x i ∂ϕ j (cid:17) be thecoordinate transition matrices, and writing (cid:16) g ij Σ (cid:17) = x det( A ) A A T , ( g Σ ij ) = 1 x det( B ) B T B. (22)Explicitly, g Σ = det( A ) x δ kl ∂x k ∂ϕ i ∂x l ∂ϕ j dϕ i ⊗ dϕ j = det( A ) x δ ij dx i ⊗ dx j . (23)So the coordinate transition matrices fully determine g and g Σ . A simple expression for theGaussian curvature of (Σ , g Σ ) is K Σ = − x det( A ) (cid:32)(cid:18) ∂∂x (cid:19) + (cid:18) ∂∂y (cid:19) (cid:33) log det( A ) x . (24)Compare (29). Clearly K Σ is invariant under any SL (2 , R ) recombination of ϕ , ϕ . Buttransforming ϕ , ϕ by a GL (2 , R ) combination will usually involve a rescaling of the metric,given by the inverse of the determinant of the transformation. This will be an importantpoint in § Remark . It is important to point out the scaling properties of (21). The right-handside is scale-invariant in the { ϕ i } functions, whereas g ij Σ clearly scales linearly in each; itmight be objected that this appears nonsensical. However the rules we have set up, in (18)and (4), show that if the metric remains unchanged, the x , y variables must actually scalequadratically with the { ϕ i } , so indeed scale-invariance is retained. That aside, certainly( ϕ , ϕ ) can be considered functions of ( x, y ), without reference to any already-existingmetric, and from this point of view ( ϕ , ϕ ) can obviously be scaled without scaling ( x, y ).Via the metric-reconstruction procedure above, the effect is homothetic rescaling of themetric. Remark . For this section, particularly (23), see also [12], or § u from our functions x , y , ϕ , ϕ by using the recipe of Theorem 1.1of [12]. Here we make a second computation of the curvature of (Σ , g Σ ), and show the effects of acertain conformal change of the metric. In this section only, for convenience, we use both g and g Σ to indicate the metric on Σ. But the “ s ” in (27), (30) and (31) still refers to the M scalar curvature.The metric on (Σ , g ) has the “pseudo-K¨ahler” property ∂g ij ∂ϕ k = ∂g ik ∂ϕ j . (25)11his was noted in [3], and is equivalent to both [ ∇ ϕ i , ∇ ϕ j ] = 0 and d ( V − J Σ dϕ i ) = 0. TheChristoffel symbols areΓ kij = 12 ∂g ij ∂ϕ s g sk , Γ k (cid:44) g ij Γ kij = − g ks ∂∂ϕ s log V . (26)The divergence of Γ k has a nice interpretation: using (cid:112) det( g ) = V − we get ∂ Γ k ∂ϕ k = −V − (cid:112) det( g ) ∂∂ϕ k (cid:18)(cid:112) det( g ) g sk ∂∂ϕ s V (cid:19) = −V − (cid:52) Σ V = 12 s. (27)The usual formula for scalar curvature in terms of Christoffel symbols is s Σ = g ij ∂ Γ sij ∂ϕ s − g ij ∂ Γ ssj ∂ϕ i + g st Γ s Γ t − g is g jt g kl Γ kij Γ lst , (28)and using ∂∂ϕ s (cid:0) g ij Γ sij (cid:1) = ∂∂ϕ i (cid:0) g ij Γ ssj (cid:1) this transforms to s Σ = g is g jt g kl Γ kij Γ lst − g st Γ s Γ t = (cid:12)(cid:12) Γ kij (cid:12)(cid:12) − (cid:12)(cid:12) Γ k (cid:12)(cid:12) = (cid:12)(cid:12) Γ kij (cid:12)(cid:12) − (cid:12)(cid:12)(cid:12) ∇ log V (cid:12)(cid:12)(cid:12) . (29)Now modify the metric by (cid:101) g Σ = V g Σ . The usual conformal-change formula gives (cid:101) s Σ = V − (cid:16) s Σ − (cid:52) Σ log V (cid:17) = V − (cid:18)(cid:12)(cid:12) Γ kij (cid:12)(cid:12) − (cid:12)(cid:12)(cid:12) ∇ log V (cid:12)(cid:12)(cid:12) − (cid:52) Σ log V (cid:19) . (30)But (cid:52) Σ log V + (cid:12)(cid:12)(cid:12) ∇ log V (cid:12)(cid:12)(cid:12) = V − (cid:52) Σ V so the conformally related scalar is simply (cid:101) s Σ = V − (cid:16)(cid:12)(cid:12) Γ kij (cid:12)(cid:12) − V − (cid:52) Σ V (cid:17) = V − (cid:16)(cid:12)(cid:12) Γ kij (cid:12)(cid:12) + s (cid:17) . (31)Therefore if s ≥
0, then (cid:101) s Σ ≥
0. This will be used decisively in the proof Theorem 1.3.
Fundamental to our paper is the use of the harmonic coordinates ( x, y ) as defined in (18).But we need some a priori facts about this system, stemming from either the geometricalsituation or from properties (A)-(E), in order to utilize our analytic theorems of §
4. Specif-ically, we need to know that ( x, y ) is actually a global coordinate system, and doesn’t haveany branch points.
Proposition 2.8
Assume the polytope (Σ , g Σ ) is closed and the boundary is connected.Then map X = ( x, y ) T is -to- and onto the right half-plane. roof . The function z = x + √− y is a holomorphic function on Σ , and maps the right-halfplane { ϕ ≥ } into the right half-plane { x ≥ } . From ( ix ) below, we have y = ϕ + O ( r ),so that the line { x = 0 } is indeed in the image of z . Definition of pseudoboundary point . We define the notion of a pseudo-boundary point:a point X = ( x , y ) is a pseudoboundary point if the following holds: given any neighbor-hood U ⊂ R of X , there is some component U of X − ( U ) ⊆ Σ such that X ( U ) (cid:54) = U (this essentially means U is not evenly covered, excluding the phenomenon of ramification).Letting U ∈ R be any neighborhood of a point (0 , y ) of the actual boundary { x = 0 } , wesee the boundary locus { x = 0 } is contained in the pseudoboundary locus. All pseudoboundary points lie on { x = 0 } . Let X be a pseudoboundary point, and let U be a neighborhood that is not evenly covered, and let U ⊆ X − ( U ) be a componentthat does not completely cover U . We can assume that U itself is precompact, and that X : U → U is 1-1 but not onto. We can assume this because the pre-image of X in U cannot be a branch point, or else U → U would certainly be onto. By shrinking U ifnecessary, we can assume U it contains no branch points at all.The pre-image U ⊂ Σ is either pre-compact or not. If it is precompact, then itmust intersect the boundary of Σ . This is because U has no critical points, so either U intersects the boundary of Σ or else X is a diffeomorphism in a neighborhood of U ,meaning U must evenly cover U , an impossibility. Thus X is actually in { x = 0 } , and sois actually part of the boundary.Now assume the closure U is not compact. All points on ∂ ( X ( U )) \ ∂U are in factpseudoboundary points. Thus X is not an isolated pseudoboundary point.Because U is not compact, it extends to infinity in ( ϕ , ϕ ) coordinates. Then considerthe pushforwards of ϕ i (cid:12)(cid:12) U to U . We see that ϕ is infinite at all of the pseudoboundarypoints on ∂U . But the ϕ i satisfy the equation ϕ ixx − x − ϕ ix + ϕ iyy = 0, which is uniformlyelliptic away from { x = 0 } , and therefore each has isolated singularities away from { x = 0 } .Thus, again, we see that the pseudoboundary points ∂ ( X ( U )) \ ∂U must lie on the { x = 0 } locus. Pre-images of domains containing points of { x = 0 } are pre-compact. Let X ∈ { x = 0 } , and let U be a sufficiently small neighborhood around (0 , ϕ A ( x, y ) = A (cid:0) − y + x log( x ) − x (cid:1) , where A > . (32)This solves the equation x ( x − ϕ x ) x + ϕ yy = 0. Let Ω be the component of { ϕ > }∩{ x ≥ } that contains the point (0 , ϕ back along X : Σ → R , we obtain a function on Σ , still denoted ϕ . On Σ this function still satisfies d (cid:16) V − J Σ dϕ (cid:17) = 0, which is an elliptic equation (see Proposition2.5 above and Equation (54) from Section 4).13onsider the domains Ω (cid:15) = Ω ∩ { x ≥ (cid:15) } , and the images X − (Ω (cid:15) ) of these domainson Σ . We have that ϕ = 0 on X − (Ω (cid:15) \ { x = (cid:15) } ), so therefore ϕ A < ϕ i on this boundarycomponent. As (cid:15) →
0, we have that ϕ i → ∞ , because the { x = 0 } locus occurs at coordinateinfinity. Thus, given any A , there is some (cid:15) so that (cid:15) < (cid:15) implies ϕ A < ϕ i on Ω (cid:15) . Thus ϕ A < ϕ i on all of X − (Ω). Sending A → ∞ implies that ϕ i = ∞ on all of X − (Ω), animpossibility. The pre-image of { x = 0 } has a single component. The pre-image of { x = 0 } is theboundary of the polytope: this follows trivially from the fact that x (cid:54) = 0 on the interior ofΣ , and from the fact, proved in the paragraph above, that points at infinite do not map to { x = 0 } . Conversely, because it is assumed that the polytope contains all of its edges, theboundary of the polytope is precisely the locus on Σ on which { x = 0 } . By Proposition2.11 below (or by (D)), the function X : Σ → R is Lipschitz, therefore continuous. Thus,if X is k -to-1 on { x = 0 } , the pre-image of { x = 0 } must have k components. The map ( x, y ) : Σ → R is one-to-one. If X = ( x, y ) T is not one-to-one, then theremust be pseudoboundary points, or else x is not one-to-one as a map onto { x = 0 } . Bothof these possibilities were disproven above. (cid:3) Remark . For an example of how Proposition 2.8 fails if the polytope’s boundary isdisconnected, see Example 8.
Remark.
The preceding theorem is far from a general statement on non-negativeharmonic functions on manifolds. Of course functions on virtually all manifolds must havecritical points. Absolutely critical to Proposition 2.8 here is the interplay between theharmonic functions and the moment functions.
Of course all functions ϕ , ϕ , x = √V , and y are themselves generically C ∞ , but at { x = 0 } it is possible that ϕ , ϕ are not C ∞ with respect to the ( x, y ) system.In this subsection, we prove the essential facts that near polytope edges, the functions ϕ , ϕ are of class at least C with respect to the variables ( x, y ), and that near vertices,these functions are Lipschitz. Of course the two coordinate systems ( ϕ , ϕ ) and ( x, y ) are C ∞ , but a priori their inverses might not be, so ϕ i might not even be C as measured in( x, y ) coordinates.Via the momentum reduction, we have the metric polytope (Σ , g Σ ). For the purposesof this section, define r to be the g Σ -distance from the { x = 0 } locus. This lifts to M ,where r is still a distance function. Note that ∇ r is well-defined on { x = 0 } on Σ , but notat { x = 0 } on M .We focus our attention on a segment of the polytope, and explore the behavior of x , y , and ϕ , and ϕ near that segment. In what follows, we may either assume the14olytope satisfies the conditions (A)-(E), or else that the polytope metric reduction of some( M , J, ω, X , X ). Lemma 2.9
Assume (Σ , g Σ ) comes from a K¨ahler reduction of some ( M , J, ω, X , X ) .After picking any boundary segment and possibly re-combining the ϕ , ϕ , assume ∇ ϕ isparallel to that segment. Then ∇ ϕ |∇ ϕ | is convariant-constant.Proof . We are trying to show ∇ ∇ ϕ |∇ ϕ | = 0. First note that (cid:10) ∇ r, ∇ ϕ (cid:11) = 0 on the boundarysegment, and that (cid:10) ∇ ∇ ϕ ∇ ϕ , ∇ r (cid:11) = (cid:10) ∇ ϕ , ∇ ∇ ϕ ∇ r (cid:11) = 0 (33)by total geodesy. This easily implies ∇ ∇ ϕ ∇ ϕ |∇ ϕ | = 0. To evaluate ∇ ∇ r ∇ ϕ |∇ ϕ | , we lift thesituation to M , and compute (cid:10) ∇ ∇ r ∇ ϕ , ∇ r (cid:11) = (cid:104)∇ ∇ r X , J ∇ r (cid:105) = (cid:104)∇ X ∇ r, J ∇ r (cid:105) = 0 (34)again by total geodesy. Once again, this easily implies ∇ ∇ ϕ ∇ ϕ |∇ ϕ | = 0. (cid:3) Lemma 2.10 (Pseudo-Jacobi identity for the ∇ ϕ i ) If γ is a geodesic on (Σ , g Σ ) then ∇ ϕ i obeys a second order equation along γ given by (cid:10) ∇ ˙ γ ∇ ˙ γ ∇ ϕ i , ∇ ϕ j (cid:11) = ˙ γ ˙ γ (cid:10) ∇ ϕ i , ∇ ϕ j (cid:11) − (cid:10) ∇ ˙ γ ϕ i , ∇ ˙ γ ϕ j (cid:11) . In the case that (Σ , γ Σ ) is a K¨ahler reduction of some ( M , J, ω, X , X ) then ∇ ˙ γ ∇ ˙ γ ∇ ϕ i + Φ ∗ J (cid:0) Rm( J ∇ ϕ i , Φ − ∗ ˙ γ )Φ − ∗ ˙ γ (cid:1) = 0 (35) where Φ : M → Σ is the reduction map, and Φ − ∗ indicates a choice of horizontal lift.Proof . The first claim is obvious. The second claim follows from noting that, while [ ∇ ϕ i , ˙ γ ]might not equal zero, certainly [ X i , ˙ γ ] = 0, so the derivation of the Jacobi equation canproceed as normal after multiplication of ∇ ˙ γ ∇ ˙ γ ϕ i by J . (cid:3) Our condition (D) assumes also ∇ ∇ ϕ |∇ ϕ | = 0. This is actually a consequence of ( vii )below, so we do not prove it separately.We prove the following: i ) The { x = 0 } locus is totally geodesic on both Σ and M (this is essentially obvious). ii ) We may make an SL (2 , R )-recombination of ϕ , ϕ , followed by possible addition ofa constant, so that ϕ = 0 on the segment under consideration, while leaving x = √V unchanged. iii ) [ ∇ r, ∇ ϕ ] = 0 at { x = 0 } on Σ and (cid:10) ∇ r, ∇ ϕ (cid:11) = O ( r ).15 v ) In the ( r, ϕ )-coordinate system on Σ , we have Taylor series ϕ = C ( ϕ ) r + O ( r ). v ) On each segment of the polytope, we have C ( ϕ ) = const ; we take C = . vi ) We have (cid:10) ∇ ϕ , ∇ ϕ (cid:11) = O ( r ). vii ) |∇ ϕ | = r + O ( r ) and |∇ ϕ | = C ( ϕ )(1 + 2 Kr ) + O ( r ), where K = K ( ϕ ) isthe g Σ -curvature of the polytope at the point ( r, ϕ ) = (0 , ϕ ). Here, C ( ϕ ) is simply |∇ ϕ | evaluated at ( r, ϕ ) = (0 , ϕ ). viii ) x = √V = √ C r (cid:0) Kr (cid:1) + O ( r ) ix ) y = ϕ + O ( r ) x ) ϕ , ϕ are at least C on segments and are Lipschitz everywhere in the ( x, y )-coordinateplane Proof of (i) : x = (cid:113) |∇ ϕ | |∇ ϕ | − (cid:104)∇ ϕ , ∇ ϕ (cid:105) , so { x = 0 } precisely when {∇ ϕ , ∇ ϕ } is a dependent set, so there are constants with c X + c X = 0 on M . But a combinationof Killing fields is Killing, and has totally geodesic zero locus. Thus { x = 0 } is totallygeodesic. Proof of (ii) : An SL (2 , R )-recombination of { ϕ , ϕ } gives rise to an SL (2 , R )-recombinationof ∇ ϕ , ∇ ϕ , which obviously leaves x = √V unchanged. If ∇ ϕ , ∇ ϕ are co-linear at apoint, we can certainly make such a transformation to obtain ∇ ϕ = 0 at that point. Butthe total geodesy of the segment and the fact that the ∇ ϕ i are Jacobi fields, we have that ∇ ϕ = 0 on the entire segment. Addition a constant, we can certainly make ϕ = 0 on thatsegment as well. Proof of (iii) : On Σ , the gradient ∇ ϕ is parallel to the polytope boundary segment.By total geodesy (vanishing of the second fundamental form), we have ∇ ∇ ϕ ∇ r = 0 so[ ∇ r, ∇ ϕ ] = −∇ ∇ r ∇ ϕ . In case we already know that condition (D) is satisfied, this isalready zero. In case (Σ , g Σ ) is the reduction of some M , so we perhaps do know (D)apriori. But we can lift back to M (choosing a horizontal lift, of course), where we cansee (using J -invariance and taking derivatives) that (cid:10) ∇ r, ∇ ϕ (cid:11) = 0 and ∇ r (cid:10) ∇ r, ∇ ϕ (cid:11) = (cid:10) ∇ r, ∇ ∇ r ∇ ϕ (cid:11) = 0 as we limit to { x = 0 } . Proof of (iv) : On M we have that X is a Jacobi field along trajectories of ∇ r withan initial condition of X = 0. Then ∇ ϕ is also Jacobi, with initial conditions ∇ ϕ = 0and ∇ ∇ r ∇ ϕ ∝ ∇ r . With ϕ = 0 on r = 0, this is sufficient to establish the claim. Proof of (v) : Using ϕ = C ( ϕ ) r + O ( r ) we have0 = [ ∇ ϕ , ∇ ϕ ]= [ C (cid:48) r ∇ ϕ + 2 Cr ∇ r + O ( r ) ∇ r + O ( r ) ∇ ϕ , ∇ ϕ ]= C (cid:48)(cid:48) r ∇ ϕ + 2 C (cid:48) r (cid:10) ∇ ϕ , ∇ r (cid:11) ∇ ϕ + 2 C (cid:48) r ∇ r + 2 C (cid:10) ∇ ϕ , ∇ r (cid:11) ∇ r + 3 Cr [ ∇ r, ∇ ϕ ] + O ( r ) (36)16sing that [ ∇ r, ∇ ϕ ] = O ( r ) and (cid:10) ∇ ϕ , ∇ r (cid:11) = O ( r ) we have the Taylor series0 = 2 C (cid:48) ( ϕ ) r ∇ r + O ( r ) (37)and so therefore C (cid:48) = 0. Proof of (vi) : Of course (cid:10) ∇ ϕ , ∇ ϕ (cid:11) = 0 on the boundary so the inner product is atleast O ( r ). We take one, two, and three derivatives along ∇ r , and show that each is at least O ( r ). Taking one derivative ∇ r (cid:10) ∇ ϕ , ∇ ϕ (cid:11) = (cid:10) ∇ ∇ ϕ ∇ ϕ , ∇ r (cid:11) (38)which is O ( r ) because ∇ ϕ = Cr ∇ r + O ( r ), and using ( iii ) above. Taking a secondderivative,12 ∇ r ∇ r (cid:10) ∇ ϕ , ∇ ϕ (cid:11) = (cid:10) ∇ ∇ r ∇ ∇ ϕ ∇ ϕ , ∇ r (cid:11) = (cid:10) ∇ ∇ ϕ ∇ ∇ r ∇ ϕ , ∇ r (cid:11) + (cid:10) ∇ [ ∇ r, ∇ ϕ ] ∇ ϕ , ∇ r (cid:11) + (cid:10) Rm( ∇ r, ∇ ϕ ) ∇ ϕ , ∇ r (cid:11) (39)is zero because ∇ ϕ = O ( r ) and [ ∇ r, ∇ ϕ ] = C ∇ r + O ( r ) and (cid:10) ∇ ∇ r ∇ ϕ , ∇ r (cid:11) = O ( r ).Finally we take a third derivative to obtain12 ∇ r ∇ r ∇ r (cid:10) ∇ ϕ , ∇ ϕ (cid:11) = ∇ r (cid:10) ∇ ∇ ϕ ∇ ∇ r ∇ ϕ , ∇ r (cid:11) + ∇ r (cid:10) ∇ [ ∇ r, ∇ ϕ ] ∇ ϕ , ∇ r (cid:11) + ∇ r (cid:10) Rm( ∇ r, ∇ ϕ ) ∇ ϕ , ∇ r (cid:11) (40)The first term can be re-written ∇ r (cid:10) ∇ ∇ ϕ ∇ ∇ r ∇ ϕ , ∇ r (cid:11) = ∇ r ∇ ϕ (cid:10) ∇ ∇ r ∇ ϕ , ∇ r (cid:11) − ∇ r (cid:10) ∇ ∇ r ∇ ϕ , ∇ ∇ ϕ ∇ r (cid:11) = ∇ ϕ ∇ r (cid:10) ∇ ∇ r ∇ ϕ , ∇ r (cid:11) + [ ∇ r, ∇ ϕ ] (cid:10) ∇ ∇ r ∇ ϕ , ∇ r (cid:11) − ∇ r (cid:10) ∇ ∇ r ∇ ϕ , ∇ ∇ ϕ ∇ r (cid:11) (41)Of these three terms, the first is zero because ∇ ϕ = O ( r ). The second is zero be-cause [ ∇ r, ∇ ϕ ] = C ∇ r + O ( r ) and ∇ ϕ satisfies the Jacobi equation ∇ ∇ r ∇ ∇ r ∇ ϕ +Rm( ∇ ϕ , ∇ r ) ∇ r = 0. The third term is zero for this same reason, along with using aJacobi equation for ∇ ϕ .The second term is zero because [ ∇ r, ∇ ϕ ] = C ∇ r + O ( r ), and because [ ∇ r, [ ∇ r, ∇ ϕ ]]is proportional to ∇ ϕ on { x = 0 } , and because ∇ ϕ is totally geodesic on { x = 0 } andsatisfies a Jacobi equation.The third term is zero simply by noting ∇ ∇ r ∇ r ≡
0, by ∇ ϕ = Cr ∇ r + O ( r ), andusing the anti-symmetry of the Riemann tensor in the first two positions. Proof of (vii) : With C = we have ∇ ϕ = r ∇ r + O ( r ) so |∇ ϕ | = r + O ( r ). To seethat O ( r ) can be improved to O ( r ), take two more derivatives, and use the Jacobi equationfor ∇ ϕ . Then we look at the Taylor series for |∇ ϕ | . The r -term is, of course, given by17ome C ( ϕ ). To see there is no r -term, note that ∇ r |∇ ϕ | = (cid:10) ∇ ∇ r ∇ ϕ , ∇ ϕ (cid:11) = 0 bythe proof of ( iii ). Taking two derivatives, we see that12 ∇ r ∇ r |∇ ϕ | = (cid:10) ∇ ∇ r ∇ ∇ r ∇ ϕ , ∇ ϕ (cid:11) + |∇ ∇ r ∇ ϕ | (42)We have that ∇ ∇ r ∇ ϕ = 0 on { x = 0 } and that (cid:10) ∇ ∇ r ∇ ∇ r ∇ ϕ , ∇ ϕ (cid:11) = ˜ K |∇ ϕ | , where (cid:101) K is a bisectional curvature on M (by Lemma 2.10), or else comes from second order dataon the metric of (Σ , g Σ ) if we don’t know Σ is a K¨ahler reduction. We therefore obtainthat the second-order term indeed has coefficient 2 C ( ϕ ) (cid:101) K . Proof of (viii) : Using our Taylor expansions for |∇ ϕ | , |∇ ϕ | and using that (cid:10) ∇ ϕ , ∇ ϕ (cid:11) = O ( r ), it is very simple to compute x = √V = (cid:112) C · r · (cid:16) (cid:101) Kr (cid:17) + O ( r ) . (43) Proof of (ix) : Using that ∇ x = √ C ∇ r + O ( r ), we have that ∇ y = J ∇ x = √ C ∇ r + O ( r ). Of course (cid:112) C ( ϕ ) = |∇ ϕ | (cid:12)(cid:12) ( r,ϕ )=(0 ,ϕ ) , so |∇ y | = |∇ ϕ | along { x = 0 } . Since ∇ y is parallel to ∇ ϕ , we can take y = ϕ along { x = 0 } . This justifies y = ϕ + O ( r ).We re-express assertion ( x ) as a proposition. Proposition 2.11
Assume (Σ , g Σ ) is a closed polytope that either obeys (A)-(F) or elseis the reduction of ( M , J, ω, X , X ) . As measured in the ( x, y ) -coordinate system, thefunctions ϕ , ϕ are at least C along any polytope edge, and are Lipschitz everywhere.Proof . The series expansions of x and y from the proofs of ( viii ) and ( ix ) easily imply that ∂∂x = ∂∂r + O ( r ) ∂∂y = ∂∂ϕ + O ( r ) ∂∂r + O ( r ) (44)so that ∂ ∂x = ∂ ∂r + O ( r ) ∂ ∂y = (cid:18) ∂∂ϕ (cid:19) + O (1) ∂∂r + O ( r ) (45)This implies ϕ , ϕ are at least C on edges. (Note that we cannot immediately assert C due to the fact that the “ O (1)” term might conceivably not have a well-defined derivativein the ϕ direction.)To consider the case on corners, note that the function r is Lipschitz, so x = √ C r + O ( r ) is also Lipschitz, and so is y . Thus the map X : M → R given by X ( p ) =18 x ( p ) , y ( p )) T is a Lipschitz map. Therefore pushing forward the invariant C ∞ functions ϕ , ϕ along X does indeed give Lipschitz functions on the ( x, y )-plane. (cid:3) Remark . A key part of the above theorem is that the locus { x = 0 } actually exists on M or on Σ and so no edges that are “infintiely far away.” Proposition 2.11, which assertsLipschitzness of ϕ , ϕ on the { x = 0 } locus, will not apply to polytope edges that areinfinitely far away; see Example 7. The prototype is the standard matric on M = S × H where H is the pseudosphere with constant curvature −
1. This has the polytope given bya half-strip {− ≤ ϕ ≤ } ∩ { ϕ > } in the ( ϕ , ϕ )-plane, whose vertical edge rays are part of the polytope, but whose vertical edge segment is not part of the polytope. R (cid:52) Σ √V ≤ is metricallycomplete (not necessarily coordinate-complete), then it is flat. The outline of the proof isas follows. First we prove that if V is a constant function, then (cid:101) g Σ is flat. Second, we showthat if (Σ , g Σ ) is complete, the “canonical” conformal change from § (cid:101) g Σ that in fact remains complete. Using this along with (cid:52) Σ √V ≤ V is indeed constant, completing the proof. Lemma 3.1
Assume (Σ , g Σ ) is metrically metric. If V is constant, or else if (cid:52) Σ √V ≤ and (Σ , J Σ ) is biholomorphic to C , then (Σ , g Σ ) is flat; indeed g Σ has constant coefficientswhen expressed in ϕ , ϕ coordinates.Proof . First assume V is constant. Then the expression d ( V − J Σ dϕ k ) = 0 from Propo-sition 2.5 reduces to dJ Σ dϕ k = 0. Thus ϕ and ϕ are harmonic, so each determines acomplex coordinate: either z = ϕ + √− η or w = ϕ + √− η (where dη k = − J Σ dϕ k ).Then ddz = 12 |∇ ϕ | (cid:0) ∇ ϕ + √− J Σ ∇ ϕ (cid:1) ddw = 12 |∇ ϕ | (cid:0) ∇ ϕ + √− J Σ ∇ ϕ (cid:1) (46)and we easily compute the transition function: dwdz = g (cid:18) ddz , ∇ w (cid:19) = (cid:10) ∇ ϕ , ∇ ϕ (cid:11) |∇ ϕ | − √− √V|∇ ϕ | . (47)But dwdz is holomorphic, so its real and imaginary parts are harmonic. Therefore |∇ ϕ | − = −V − Im ( dwdz ) is harmonic. In the z -coordinate the Hermitian metric is h Σ = (cid:12)(cid:12) ddz (cid:12)(cid:12) = |∇ ϕ | − , and therefore the curvature K Σ = − h − (cid:52) Σ log h Σ = 8 |∇|∇ ϕ || (48)19s non-negative, forcing the complete manifold (Σ , g Σ ) to be parabolic (this is due to theCheng-Yau condition for parabolicity; see [9] or the remark below). Equivalently the com-plete manifold (Σ , J Σ ) is biholomorphic to C . But then the Liouville theorem implies |∇ ϕ | − is actually constant, as it is an harmonic function on C bounded from below. Sim-ilarly |∇ ϕ | − and (cid:10) ∇ ϕ , ∇ ϕ (cid:11) are constant. Expression (13) implies g Σ are constant in( ϕ , ϕ ) coordinates, and therefore have zero curvature.For the second assertion, assume (Σ , J Σ ) is biholomorphic to C and (cid:52) Σ √V ≤
0. Sincethe function V is superharmonic and bounded from below, it is constant, and we may usethe argument from above. (cid:3) Lemma 3.2
Assume (Σ , g Σ ) is complete and (cid:52) Σ √V ≤ . Setting (cid:101) g Σ = V g Σ , then (Σ , (cid:101) g Σ ) is also complete.Proof . See § (cid:101) g Σ = V g Σ . Assume γ :[0 , R ] → Σ is a geodeisc in the (cid:101) g Σ metric that gives a shortest distance to ∂ Σ . For acontradiction, we show that if γ has finite length in (cid:101) g Σ , it also has finite length in g Σ .Set p = γ (0) and r = dist( p, · ). We are free to re-choose p along γ to make R smaller ifconvenient.As a first step we verify that, for R sufficiently small, we have V is strictly mono-tone along γ ; in particular (cid:12)(cid:12)(cid:12) ∂∂r (cid:16) T ◦ γ (cid:17)(cid:12)(cid:12)(cid:12) ≥ δ for some δ >
0. The standard method (eg.Lemma 3.4 of [18]) is the “Hopf Lemma” applied to the equation (cid:101) (cid:52) Σ V ≤
0, but in oursituation there is an issue concerning how the elliptic operator itself might degenerate nearthe boundary.To emulate the standard Hopf lemma proof we must first use Laplacian comparison[36]. On the annulus A = B p ( R ) \ B p ( R/
2) we have V > ∂B p ( R ), sowe have the existence of some (cid:15) > (cid:15) (cid:0) r − − R − (cid:1) ≤ V on ∂ A . On the other hand,since ˜ s Σ > (cid:101) s Σ gives (cid:101) (cid:52) Σ r ≤ r − so (cid:101) (cid:52) Σ r − ≥ r − . Thus V − (cid:15) (cid:0) r − − R − (cid:1) ≥ ∂ A (cid:101) (cid:52) Σ (cid:16) V − (cid:15) (cid:0) r − − R − (cid:1)(cid:17) ≤ − (cid:15)r − < . (49)We have super-harmonicity within A and non-negativity on ∂ A , so V − (cid:15) (cid:0) r − − R − (cid:1) > A . The using the fact that lim t → R V ◦ γ ( t ) = 0 and lim r → R (cid:0) r − − R − (cid:1) ◦ γ ( t ) = 0, wemust have lim t → R ∂∂t (cid:104)(cid:16) V − (cid:15) (cid:0) r − − R − (cid:1)(cid:17) ◦ γ ( t ) (cid:105) ≤ , solim t → R ∂∂t (cid:16) V ◦ γ (cid:17) ( t ) ≤ (cid:15) lim t → R ∂∂t (cid:0)(cid:0) r − − R − (cid:1) ◦ γ (cid:1) ( t ) = − (cid:15)R − . (50)20herefore, for some small δ , we must have (cid:12)(cid:12)(cid:12) ∂∂t V ◦ γ (cid:12)(cid:12)(cid:12) > δ for t ∈ ( R − δ, R ). Thus also (cid:16) V ◦ γ (cid:17) ( t ) ≥ δ ( R − t ). Then to estimate the g Σ -length of γ as t goes to its limiting value R , we use | ˙ γ | g Σ = (cid:16) V − ◦ γ (cid:17) · | ˙ γ | ˜ g Σ ≤ δ − ( R − t ) − (51)for t ∈ ( R − δ, R ) to get Length g Σ ( γ ) = (cid:90) RR − δ | ˙ γ | g Σ dt ≤ (cid:90) RR − δ δ − · ( R − t ) − dt = 2 < ∞ . (52)This is impossible, and completes the proof. (cid:3) Theorem 3.3 (cf. Theorem 1.3)
Assume (Σ , g Σ ) is metrically complete and (cid:52) Σ √V ≤ . Then (Σ , (cid:101) g Σ ) and (Σ , g Σ ) are flat Riemannian manifolds; indeed g Σ has constant coef-ficients when expressed in ϕ , ϕ coordinates.Proof . By Lemma 3.2, (Σ , (cid:101) g Σ ) is also complete, and by (31) also ˜ s Σ ≥
0. Thus (Σ , (cid:101) g Σ ) isparabolic, therefore biholomorphically equivalent to C . Then V is a positive superharmonicfunction on C , so it is constant. Lemma 3.1 gives the conclusion. (cid:3) Corollary 3.4 (cf. Corollary 1.4)
Assume ( M , J, ω, X , X ) has s ≥ , and assume X , X have no zeros and are nowhere parallel (in other words V is nowhere zero). Then M isa flat Riemannian manifold.Proof . The corresponding metric polytope (Σ , g Σ ) is complete, and since V = 0 if and onlyif {X , X } is a linearly dependent set, the polytope contains no edges. By equation (31) wehave (cid:52) Σ √V ≤
0. The theorem now implies g Σ is flat; in fact g Σ is a constant matrix. Inthe context of the equations (7), we have that G is a constant matrix, so g , J are constant.Thus M is flat. (cid:3) Remark.
In higher dimensions, the methods here fail at almost every stage.
Remark.
A crucial part of the proofs of Lemma 3.1 and Theorem 3.3 is the fact that acomplete Σ with K Σ ≥ (cid:82) ∞ tV ol B t dt = ∞ . The assertion thatparabolicity of a simply connected Riemann surface implies the surface is actually C is asimple consequence of uniformization, or even just the Riemann mapping theorem. Thesubject of parabolicity has received a great deal of attention: see for example [32] [33] [23][19] [37] and references therein. 21 Global analysis of the coordinate functions
The purpose of this section is proving that any ϕ ≥ x ( ϕ xx + ϕ yy ) − ϕ x = 0 on { x > } with zero boundary conditions, then ϕ ( x, y ) = Ax for some constant A ≥ Equation (2.5) is dJ Σ dϕ + V d V − ∧ J Σ dϕ = 0, which, with (cid:52) Σ V = − s V in (17), isthe elliptic system (cid:52) Σ ϕ − (cid:68) ∇ Σ log V , ∇ Σ ϕ (cid:69) Σ = 0 , (cid:52) Σ V + 12 s V = 0 . (53)If s = 0 then x = V is an harmonic coordinate. Therefore with H = { x > } being theopen half-plane in ( x, y ) coordinates, (53) reduces to ϕ xx + ϕ yy − x − ϕ x = 0 , or x ( x − ϕ x ) x + ϕ yy = 0 (54)on H (see equation (3) of [12] or equation (2) of [3]). By Proposition 2.8, the map X =( x, y ) T is a bijection from the polytope Σ to the half-plane { x ≥ } , so therefore thisequation is valid not just locally but globally on H (for an example of what happens if X is not one-to-one, see Example 1).We are considering the situation where one moment function, say ϕ , is zero at x = 0;geometrically this is the situation that the Σ = { ϕ ≥ } is a half-plane. For conveniencewe will simply denote this variable by ϕ . The fundamental result is that ϕ ( x, y ) = Ax forsome constant A . Now the function x is unbounded, so ϕ is likely difficult to work with.Setting f = ϕ · x − and noting that f solves f xx + f yy + 3 x − f x = 0 , or x − ( x f x ) x + f yy = 0 , (55)we expect f to be bounded, and so, presumably, easier to work with. Using the coordinatetransformation s = x − , t = √ y , equation (55) is s f ss + f tt = 0 . (56)We shall use both ( x, y ) and ( s, t ) coordinates in our proofs below. Note that the { x = ∞} locus becomes the { s = 0 } locus, and vice-versa, so we begin with no information whateveron the behavior of f at s = 0. We prove, perhaps unexpectedly, that f is bounded at s = 0,that it must actually be constant there, and that this constant is the global minimum of f ;see Corollary 4.8. This is a crucial step toward our ultimate conclusion that f is constant;see Theorem 4.15. 22 emma 4.1 ( f > even on the boundary) Assume ϕ ≥ , ϕ ∈ C ( H ) ∩ C ∞ ( H ) ,and ϕ (0 , y ) = 0 . Then f is C ( H ) ∩ C ∞ ( H ) , and f > on H .Proof . We can express ϕ as a partial Taylor series with remainder: ϕ ( x, y ) = C ( y ) + C ( y ) x + C ( y ) x + R where R = o ( x ). In other words, R ∈ C ( H ) ∩ C ∞ ( H ) and x − R ∈ C ( H ) ∩ C ∞ ( H ).The boundary conditions force C ( y ) = 0. Plugging into the PDE gives, near x = 0,lim x → (cid:0) − x − C ( y ) + R xx − x − R x (cid:1) = 0. Because lim x → x − R x = lim x → R xx = 0, thennecessarily C ( y ) = 0. Thus f = x − · ϕ = C ( y ) + x − R ∈ C ( H ) ∩ C ∞ ( H ).For the assertion that f > H , note the maximum principle implies f > H , so we have to prove that f is non-zero on { x = 0 } = ∂H . We do this using alower barrier function. The barrier will be the function ψ ( x, y ) = (cid:15) ( x − y − ¯ y ) )+ (cid:15) (cid:48) − δx − solves x − ( x ψ x ) x + ψ yy = 0. On the strip { < x < } , the region on which ψ is positiveis pre-compact. Thus to show that indeed ψ < f on { < x < } we must only show ψ < f on { x = 1 } . But since f > H we may choose (cid:15) > (cid:15) (cid:48) > (cid:15) (1 − y − ¯ y ) ) + (cid:15) (cid:48) < f along { x = 1 } regardless of δ >
0. By sending δ → f (0 , ¯ y ) > (cid:15) (cid:48) . Therefore f is not zero anywhere on the boundary { x = 0 } . (cid:3) Remark . The barrier used in this lemma will be used again in Corollary 4.6 below.See Figure (2b) for its depiction.
Remark . The condition that ϕ = 0 on the boundary is actually sufficient, and we donot need the additional assumption that ϕ ∈ C ( H ). However we shall not explore this atpresent, as our geometric situation implies actually ϕ ∈ C ∞ ( H ). In Proposition 2.11, weproved simply ϕ ∈ C ( H ). See Example 4 for a solution that is only C ,α ( H ) ∩ C ∞ ( H ),although it does not have zero boundary values. Proposition 4.2 (Unspecifiability of boundary values at x = 0 ) Assume f is a boundedsolution to (55) on any precompact domain Ω ⊂ H with zero boundary values on ∂ Ω \ { x =0 } . Then f is zero.Proof . Suppose f is a solution with f = 0 on ∂ Ω \ { x = 0 } . The PDE (55) is uniformlyelliptic on any open half-plane 0 < δ < x < ∞ so the strong maximum principle is availableon any Ω δ = Ω ∩ { x ≥ δ } . The the uniform boundedness of f implies that given any (cid:15) thereis a δ (cid:48) so that for any δ ∈ (0 , δ (cid:48) ], then (cid:15)x − is an upper barrier for f on Ω δ . Thus for any (cid:15) > (cid:15)x − is an upper barrier for f on Ω. Sending (cid:15) to zero gives f ≤
0. Replacing f by − f gives also f ≥ (cid:3) Remark.
This proposition is used in Proposition 4.13 to help show that boundaryvalues of f at ∂ Ω \ { x = 0 } uniquely determine boundary values at { x = 0 } , under thecondition of boundedness of f . In particular, the non-uniformly elliptic equation (55), inthe form x ( f xx + f yy ) + 3 f x = 0, is not even hypoelliptic at the boundary { x = 0 } , in thatit fails H¨ormander’s criterion there [25]. Non-hypoellipticity at the boundary also holds23or (54); this also follows from H¨ormander’s criterion, but one can see this directly fromExample 4. Remark.
The hypothesis that Ω is precompact in Proposition 4.2 can be relaxed, butwe shall not explore this further. The assumption that f ∈ L ∞ (Ω) is however necessary;see Example 3 for an example where this is violated. Remark.
The linear equations (54) and (55) yield easily to separation of variables.In the following tables the λ are positive, and the separation is f ( x, y ) = u ( x ) v ( y ) whereeither/both u , v may be replaced by ˜ u , ˜ v . Equation f xx + f yy − x − f x = 0 Eigenvalue Eigenf unctionsλ u λ ( x ) = xJ ( λ x ) v λ ( y ) = sinh ( λ y )˜ u λ ( x ) = xY ( λ x ) ˜ v λ ( y ) = cosh ( λ y )0 u ( x ) = x v ( y ) = y ˜ u ( x ) = 1 ˜ v ( y ) = 1 − λ u λ ( x ) = xI ( λ x ) v λ ( y ) = sin ( λ y )˜ u λ ( y ) = xK ( λ x ) ˜ v λ ( y ) = cos ( λ y ) (57) Equation f xx + f yy + 3 x − f x = 0 Eigenvalue Eigenf unctionsλ u λ ( x ) = x − J ( λ x ) v λ ( y ) = sinh ( λ y )˜ u λ ( x ) = x − Y ( λ x ) ˜ v λ ( y ) = cosh ( λ y )0 u ( x ) = 1 v ( y ) = y ˜ u ( x ) = x − ˜ v ( y ) = 1 − λ u λ ( x ) = x − I ( λ x ) v λ ( y ) = sin ( λ y )˜ u λ ( y ) = x − K ( λ x ) ˜ v λ ( y ) = cos ( λ y ) (58)The functions J , Y , I , K are the usual Bessel functions. These eigenfunctions will beused extensively in § We shall switch freely between the ( x, y ) and ( s, t ) coordinate systems and the respectiveequations (55) and (56).We take a moment to outline the argument we are about to make. Proposition 4.3 givesuniversal gradient control on any non-negative solution, bounded at { x = 0 } or not, of (55);these bounds instantly give polynomial growth/decay in x and exponential growth/decayin y . Then Corollary 4.7 improves the exponential control in y to polynomial control—thiscritical result allows the use of Fourier transform methods in § f is “almost” decreasing in x , or increasing in s .Finally Corollary 4.8 proves that st s = 0 (which is x = ∞ ) f is actually continuous and24onstant with respect to t , and this constant is actually the global infimum of f ; this isnecessary in the proof of Lemma 4.10. Proposition 4.3 (Interior gradient bounds)
There is some universal constant c so thatif f ∈ C ∞ ( H ) is a non-negative solution to x − ( x f x ) x + f yy = 0 , then x |∇ log f | < c .Proof . If not, there exists some sequence of points ( x i , y i ) so that x i |∇ log f | ( x i ,y i ) → ∞ .We make a point-picking improvement argument to find a ball on which |∇ log f | is controlled by |∇ log f | ( x i ,y i ) on a sufficiently large ball around ( x i , y i ). Note that wecan assume the ball of radius 2 i |∇ log f | − actually remains in the half-plane, meaning2 i |∇ log f | − << x i . This is because x i |∇ log f | ( x i ,y i ) → ∞ , so we can pass to a subse-quence if necessary to ensure i << x i |∇ log f | . Now if there is any point ( x (cid:48) i , y (cid:48) i ) withinthe ball of radius i · (cid:0) |∇ log f | ( x i ,y i ) (cid:1) − with |∇ log f | ( x (cid:48) i ,y (cid:48) i ) > |∇ log f | ( x i ,y i ) , then re-choosethe point ( x i , y i ) to be the point with this larger gradient. There may be a nearby pointwith a larger gradient still, so we can repeat this process. We may repeat this as manytimes as necessary and still remain all within a ball of radius 2 i |∇ log f | − from the originalpoint, and therefore still remain in the half-plane. Thus the pointpicking process termi-nates, resulting in a point ( x i , y i ) with the following two properties: first the ball of radius i · (cid:0) |∇ log f | ( x i ,y i ) (cid:1) − is in the right half-plane, and second |∇ log f | ≤ |∇ log f | ( x i ,y i ) at allpoints in a ball of radius i · (cid:0) |∇ log f | ( x i ,y i ) (cid:1) − .Now for each i we transition from the coordinates ( x, y ) to coordinates ( ξ, υ ) by setting x = α i ξ + x i , y = α i υ + y i where α i = (cid:0) |∇ f | ( x i ,y i ) (cid:1) − . Using ∇ ( x,y ) and ∇ ( ξ,υ ) to distin-guish the gradients in the ( x, y ) and the ( ξ, υ ) coordinates systems, we have the followingproperties: i ) |∇ ( ξ,υ ) log f | (0 , = 1 and |∇ ( ξ,υ ) log f | < ξ, υ )-coordinate ball of radius i , ii ) x i α − i = x i (cid:12)(cid:12) ∇ ( x,y ) f (cid:12)(cid:12) ( x i ,y i ) −→ ∞ , iii ) d fdξ + ξ + x i α − i ddξ + d fdυ = 0 on the ball of radius i .We can scale so that f (0 ,
0) = 1, and since f and |∇ f | have at worst exponential growthby ( i ), we have at least C ,α convergence. By ( ii ) and ( iii ) f satisfies an elliptic differentialwith bounded coefficients, so the convergence is actually C ∞ . Then passing to the limit, weobtain a function f ≥ u, v )-plane so that |∇ f | = 1 at one point, butsince x i α i → ∞ also f satisfies the Laplace equation f ξξ + f υυ = 0. However a non-negativeharmonic function on R is constant by Liouville’s theorem, so we have a contradiction. (cid:3) With x ddx = − s dds and x ddy = s − ddt , the previous lemma gives polynomial controlover s (cid:55)→ f ( s, t ) and exponential control over t (cid:55)→ f ( s, t ).25 orollary 4.4 There is a constant c < ∞ such that s (cid:12)(cid:12)(cid:12) d log fds (cid:12)(cid:12)(cid:12) < c and s − / (cid:12)(cid:12)(cid:12) d log fdt (cid:12)(cid:12)(cid:12) < c .In particular, (cid:0) ss (cid:48) (cid:1) − c ≤ f ( s,t ) f ( s (cid:48) ,t ) ≤ (cid:0) ss (cid:48) (cid:1) c for < s (cid:48) < s and e − c √ s ( t − t (cid:48) ) ≤ f ( s,t ) f ( s,t (cid:48) ) < e c √ s ( t − t (cid:48) ) for < t (cid:48) < t . Next we improve the exponential constraint in t to a polynomial constraint. This is doneusing a choice of lower barrier; the barrier is related to the one from Lemma 4.1. Proposition 4.5 (Quadratic Growth/Decay in t ) Assume f ∈ C ∞ ( H ) is a non-negativesolution to (56). Given any point ( s , t ) in the right half-plane, and given any M ∈ R , wehave that f ( s , t ) ≥ f ( s , t ) c − (cid:16) cc − (cid:17) c (cid:18) τ − (cid:113) τ − c − c (cid:19) c − τ + c (cid:113) τ − c − c · s ( t − M ) ,where τ = (cid:114) s ( t − M ) (59) for t ∈ [ M, t ] or t ∈ [ t , M ] respectively as M < t or M > t . If some sharpness issacrificed, this can be simplified to the estimate f ( s , t ) ≥ f ( s , t )(2 + s ( t − M ) ) c ( t − M ) ( t − M ) , (60) for t ∈ [ M, t ] or t ∈ [ t , M ] respectively as M < t or M > t . Remark . Due to translation invariance, it may typically be convenient to assume t = 0. Proof . Given ( s , t ), the previous lemma implies that the function H t ( s ) = f ( s , t ) (cid:16) ss (cid:17) c s ∈ [0 , s ) f ( s , t ) (cid:16) ss (cid:17) − c s ∈ [ s , ∞ ) (61)is a lower barrier: f ( s, t ) > H t ( s ). The function ψ ( s, t ) = Ds − (cid:0) − s s − − s − s + 2 + s ( t − M ) (cid:1) (62)satisfies s ψ ss + ψ tt = 0, and we will find D = D ( M, s , t ) so that on { t = t } we have ϕ ( s, t ) < H t ( s ). Then restricting to the precompact domainΩ (cid:44) { ϕ > } ∩ { t < t } , (63)this implies that ψ < f on Ω. To simplify appearances a bit, we’ll use substitutions σ = ss , τ = (cid:112) s ( t − M ) /
2, and τ = (cid:112) s ( t − M ) /
2, and then choose the largest
D > f σ c − Ds − (cid:0) − σ − − σ + 2 τ (cid:1) ≥ , < σ < f σ − c − Ds − (cid:0) − σ − − σ + 2 τ (cid:1) ≥ , < σ < ∞ (64)26igure 1: Graphic for the barrier in Proposition 4.5, drawn using ( s , t ) = (1 , and f = f ( s , t ) = 2 . (a) Thin curve is the polynomial lowerbound on s (cid:55)→ f ( s, t ) from Corollary4.4. Thick curve is the “optimal” curvedescribed in equation (64). (b) Computed lower barrier ψ shown forboth M < t and M > t . The depic-tion here is for t = 0; see equation (67) where f (cid:44) f ( s , t ) and τ = 1 + s ( t − M ) /
2. Noticing that these two equations areequivalent after the transformation σ (cid:55)→ σ − , we need only consider one of them, say thefirst one. Given any D > G ( σ ) = f σ c − Ds − (cid:0) − σ − − σ + 2 τ (cid:1) has asingle extremum, which is a global minimum. The optimal choice for D would then solvesimultaneously G (cid:48) ( σ ) = 0, G ( σ ) = 0. We find that σ = cc − (cid:32) τ − (cid:114) τ − c − c (cid:33) ,Ds − = ( c + 1) f σ c τ − σ = 12 ( c − f (cid:16) cc − (cid:17) c (cid:18) τ − (cid:113) τ − c − c (cid:19) c − τ + c (cid:113) τ − c − c . (65)Therefore ψ ( s, t ) = 12 ( c − f (cid:16) cc − (cid:17) c (cid:18) τ − (cid:113) τ − c − c (cid:19) c − τ + c (cid:113) τ − c − c (cid:32) − (cid:18) ss (cid:19) − − (cid:18) ss (cid:19) + 2 τ (cid:33) (66)is a lower barrier: ψ ( s, t ) < f ( s, t ) on Ω. Restricting to t = t we get our polynomial lower27ecay bound on f : f ( s , t ) ≥ ψ ( s , t ) = 12 ( c − f (cid:16) cc − (cid:17) c (cid:18) τ − (cid:113) τ − c − c (cid:19) c − τ + c (cid:113) τ − c − c · s ( t − M ) ,where τ = 1 + 12 s ( t − M ) . (67)It is possible to estimate the expression for Ds − in (65) by something simpler; for instance Ds − ≥ f τ ) c ( τ −
1) (68)so we have the simplified estimates f ( s, t ) ≥ f τ ) c ( τ − (cid:18) − (cid:16) s s (cid:17) − ss + 2 + s ( t − M ) (cid:19) f ( s , t ) ≥ f (2 + s ( t − M ) ) c ( t − M ) ( t − M ) (69) (cid:3) Remark . If one works even harder at choosing an optimal barrier in Proposition 4.5,one may obtain f ( s , t ) < Cf ( s , t ) ( t/t ) − . However this will not be necessary for us.See the conjecture at the end of this subsection. Corollary 4.6 (Almost-monotonicity in s ) Assume f ∈ C ∞ ( H ) is a non-negative so-lution of (55). There is a number α ∈ (0 , with the following property. Fixing ( s , t ) , if s > s then we have f ( s, t ) > αf ( s , t ) .Proof . Again the strategy is to place a test function underneath f . We use ψ ( s, t ) = D (cid:0) s − − ( t − t ) (cid:1) − η ( s − s ) + A (70)for positive D . Similar to the previous lemma, note that the domain Ω = { ψ > }∩{ s > s } is precompact, and ψ = 0 on ∂ Ω \ { s = s } , so to check that ψ < f on Ω, it is enough tocheck that ψ ≤ f on { s = s } . Setting s = s the previous lemma gives f ( s , t ) ≥ α · f · ( t − M ) ( t − M ) , where α = ( c + 1) s ( t − M ) σ c τ − σ ,σ = cc − (cid:32) τ − (cid:114) τ − c − c (cid:33) , τ = 1 + 12 s ( t − M ) . (71)28igure 2: Graphic for the barrier in Corollary 4.6, drawn using ( s , t ) = (1 , and αf = 2 . (a) The thin curve is the quadratic lowerbound on s (cid:55)→ f ( s, t ) from Proposition4.5. The darker curve underneath is the“optimal” parabola determined by (72).The barrier ψ ( s, t ) is shown in the limitwhere η = 0. (b) This is the same barrier ψ , ex-pressed in the ( x, y )-coordinate system. We pick some N (to be chosen below), and find the optimal D so that N − D ( t − t ) ≤ αf (cid:16) t − Mt − M (cid:17) ≤ f ( s , t ). To do so, we again use the first derivative trick, and simultaneouslysolve H ( t ) = αf (cid:18) t − Mt − t (cid:19) − N + D ( t − t ) = 0 H (cid:48) ( t ) = 2 αf t − M ( t − t ) + 2 D ( t − t ) = 0 (72)for t and D . We find D = Nαf ( αf − N )( t − M ) . Therefore the optimal barrier is ψ ( s, t ) = N (cid:18) − αf ( αf − N ) s ( t − M ) (cid:19) + N αf ( αf − N ) s ( t − M ) (cid:0) s s − − s ( t − t ) (cid:1) + η ( s − s ) . (73)29ince the D given above is the solution to the system (72), we have ψ ( s , t ) = N − N αf ( t − t ) ( αf − N )( t − M ) ≤ αf (cid:18) t − Mt − M (cid:19) ≤ f ( s , t ) . (74)Now set N = (1 − µ ) αf , to obtain ψ ( s, t ) = 1 − µµ αf µs ( t − M ) − s ( t − M ) + 1 − µµ αf s ( t − M ) (cid:0) s s − − s ( t − t ) (cid:1) + η ( s − s ) . (75)Choose M = M ( s , t ) so that s ( t − M ) = 4. Then τ = 3, σ = cc − (3 − √ c − ), and α = ( c +1) σ c − σ . Finally choose µ = . Then ψ ( s, t ) = 18 αf + 18 αf (cid:32)(cid:18) ss (cid:19) − − (cid:18) t − t t − M (cid:19) (cid:33) + η ( s − s ) (76)Now we have that ψ < f on ∂ Ω, so ψ < f on Ω, and this is independent of η . Sending η → t = t , we obtain f ( s, t ) > α f ( s , t ). (cid:3) Corollary 4.7 (Polynomial Upper bounds for t ) There exists some c < ∞ so that if f ∈ C ∞ ( H ) is a non-negative solution of (56), then f ( s, t ) ≤ c +2 (1 + st ) c f ( s, t ) (77) for any t, t ∈ R . In the ( x, y ) system, this is f ( x, y ) ≤ C (cid:16) x +( y − y ) x (cid:17) c f ( x, y ) . Proof . First assume t > t in the conclusion of Corollary 4.5, and set M = t − ( t − t ) = 2 t − t so t − M = 2( t − t ) and t − M = t − t . We get f ( s , t ) ≥ f ( s , t )(1 + s ( t − t ) ) c c +2 . (78) (cid:3) Corollary 4.8 (Global minimum occurs at s = 0 ) If f ≥ solves (56) on { s ≥ } ,then f is continuous at all boundary points { s = 0 } , f | s =0 is constant, and the value of f at s = 0 is a strict minimum.Proof . For ( s , t ), again choose M = M ( s , t ) so s ( t − M ) = 4 as above, but nowassume t ∈ (cid:104) t − (cid:113) s , t + (cid:113) s (cid:105) so | t | ≤ ( M − t ) / √
2. Then using the barrier (76) withthese values of t we have ψ ( s, t ) ≥ αf + η ( s − s ) (79)30gain we can send η →
0, so therefore f ( s, t ) > αf on ( s, t ) ∈ [ s , ∞ ) × (cid:104) t − (cid:113) s , t + (cid:113) s (cid:105) (where f = f ( s , t )). This is independent of s . So let ( s i , t i ) be a sequence of pointswhere s i (cid:38) t i → t ∞ . Define numbers f i = f ( s i , t i ). First, it is impossible thatsup i f i = ∞ , because f ( s, t i ) > αf ( s i , t i ) for s > s i would then force f ( s, t i ) → ∞ for anyfixed s , which is impossible.Define the values of f on { s = 0 } by simply setting f (0 , t ) = lim inf ( s (cid:48) ,t (cid:48) ) → (0 ,t ) f ( s (cid:48) , t (cid:48) ).Also, by subtracting a constant from f if necessary, assume inf f ( s, t ) = 0 (where theinfimum is taken over the whole half-plane).Now passing to a subsequence of ( s i , t i ), we can assume f i converges to some f ∞ < ∞ ,where the subsequence can be chosen so that f ∞ = lim sup i →∞ f ( s i , t i ). Then (using theextended definition of f ) we have that f (0 , t ) ≥ αf ∞ . Then from the previous result also f ( s, t ) ≥ α (cid:0) αf ∞ (cid:1) . But inf f = 0, so necessarily f ∞ = 0. Since f ≥ f ∞ waschosen as a “lim inf,” we have that if ( s j , t j ) is any other sequence that converges to (0 , t ),this forces lim j f ( s j , t j ) to exist, and to equal zero. But t is arbitrary, so f is continuousand constant at s = 0, as claimed. (cid:3) Remark . The foundational result is Proposition 4.3, which cannot hold for supersolu-tions; see Example 1. Nevertheless we conjecture that Corollary 4.4 holds for supersolutions,whether bounded at { x = 0 } or not, and also that c can be taken to be 2. If f is a solution function and x >
0, then the function y (cid:55)→ f ( x , y ) has polynomialgrowth in y by Corollary 4.7. The main idea of this section is that the polynomial growthbound allows the use of Fourier transform methods, and allows us to construct fundamentalsolutions on strips { < x < x } as well as half-planes { x < x } . By the standard methodof convolutions of boundary data with with fundamental solutions, we are able to thenshow that necessarily y (cid:55)→ f (0 , y ) also has polynomial growth. Then we show that this isimpossible.For the moment we require the apriori assumption that y (cid:55)→ f (0 , y ) already has expo-nential growth: this technical assumption is necessary in the proof of 4.11, where we showthat, on strips { < x < (cid:15) } then a solution f is actually equal to the usual convolution witha fundamental solution. This exponential assumption is removed in § G ( x, y ) = x ( x + y ) / . (80)This solves G xx − x − G x + G yy = 0, and, restricted to { x = 0 } , y (cid:55)→ G (0 , y ) is the Dirac31elta (this is justified in Example 2 below). If ψ ( y ) are boundary values on { x = 0 } , then ϕ ( x, y ) = (cid:90) ∞−∞ G ( x, y − t ) ψ ( t ) dt (81)solves ϕ xx − x − ϕ x + ϕ yy = 0 with ϕ (0 , y ) = ψ ( y ).By contrast, in light of Proposition 4.2 there can be no fundamental solution for theequation f xx + 3 x − f x + f yy = 0 on { x ≥ } . We have the following fundamental solutionson certain right half-planes and certain strips: G (cid:15) ( x, y ) = (cid:15)x (cid:90) ∞ K ( ωx ) K ( ω(cid:15) ) cos( ωy ) dω, on { x ≥ (cid:15) } , G (cid:15) ( x, y ) = (cid:15)x (cid:90) ∞ I ( ωx ) I ( ω(cid:15) ) cos( ωy ) dω, on { ≤ x ≤ (cid:15) } , (82)where K and I are the usual modified Bessel functions of the first and second kinds.Define one-variable functions ψ (cid:15) by ψ (cid:15) ( y ) = f ( (cid:15), y ) (83)For simplicity, from here on we assume f ( x, y ) = f ( x, − y ), as we could always replace f ( x, y )with f ( x, y ) + f ( x, − y ). This allows us to use a Fourier cosine representation for ψ (cid:15) , whichthe author likes better than the exponential representation. Lemma 4.9
The Fourier transform (cid:99) ψ (cid:15) of ψ (cid:15) exists in the sense of distributions, and (cid:99) ψ (cid:15) ( ω ) is smooth except at ω = 0 .Proof . Since ψ (cid:15) is an even function, it has a Fourier cosine transform (cid:99) ψ (cid:15) ( ω ) = (cid:90) ∞−∞ ψ (cid:15) ( t ) cos( ωt ) dt. (84)The function ψ (cid:15) ( y ) has polynomial growth at worst by Corollary 4.7, so this transformexists in the distributional sense. Because ψ (cid:15) is non-negative and therefore has no largeoscillations, the transform must be smooth except possibly at ω = 0. (cid:3) Lemma 4.10
We have that f ( x, y ) = ( G (cid:15) ∗ y ψ (cid:15) ) ( x, y ) on the half-plane { x ≥ (cid:15) } .Proof . The assertion is that if we set F (cid:15) ( x, y ) (cid:44) (cid:90) ∞−∞ G (cid:15) ( x, y − t ) ψ (cid:15) ( t ) dt, (85)then actually f ( x, y ) = F (cid:15) ( x, y ) on { x ≥ (cid:15) } . To prove this, we first look at the strip { (cid:15) < x < (cid:15) (cid:48) } and prove that fundamental solution methods let us recover the solutionexactly, and then we let (cid:15) (cid:48) → ∞ and see that we are left with precisely (85).32or certain values of α ω , β ω to be determined below, define G (cid:15)(cid:15) (cid:48) ( x, y ) = 1 x (cid:90) ∞ ( α ω K ( ωx ) + β ω I ( ωx )) dω (86)Then on whatever the domain of convergence happens to be, we have that G (cid:15)(cid:15) (cid:48) indeed solves(55). Choosing α ω = (cid:15)I ( ω(cid:15) (cid:48) ) K ( ω(cid:15) ) I ( ω(cid:15) (cid:48) ) − K ( ω(cid:15) (cid:48) ) I ( ω(cid:15) ) and β ω = − (cid:15) (cid:48) K ( ω(cid:15) (cid:48) ) K ( ω(cid:15) ) I ( ω(cid:15) (cid:48) ) − K ( ω(cid:15) (cid:48) ) I ( ω(cid:15) ) we havethat G (cid:15)(cid:15) (cid:48) ( (cid:15), y ) = δ y , G (cid:15)(cid:15) (cid:48) ( (cid:15) (cid:48) , y ) = 0 . (87)Similarly, if we set (cid:101) G (cid:15)(cid:15) (cid:48) ( x, y ) = 1 x (cid:90) ∞ (cid:16) ˜ α ω K ( ωx ) + ˜ β ω I ( ωx ) (cid:17) dω (88)with ˜ α ω = − (cid:15)I ( ω(cid:15) ) K ( ω(cid:15) ) I ( ω(cid:15) (cid:48) ) − K ( ω(cid:15) (cid:48) ) I ( ω(cid:15) ) and ˜ β ω = (cid:15) (cid:48) K ( ω(cid:15) ) K ( ω(cid:15) ) I ( ω(cid:15) (cid:48) ) − K ( ω(cid:15) (cid:48) ) I ( ω(cid:15) ) we obtain (cid:101) G (cid:15)(cid:15) (cid:48) ( (cid:15), y ) = 0 , (cid:101) G (cid:15)(cid:15) (cid:48) ( (cid:15) (cid:48) , y ) = δ y . (89)Then define F (cid:15)(cid:15) (cid:48) ( x, y ) = (cid:90) ∞−∞ G (cid:15)(cid:15) (cid:48) ( x, y − t ) ψ (cid:15) ( t ) dt + (cid:90) ∞−∞ G (cid:15)(cid:15) (cid:48) ( x, y − t ) ψ (cid:15) (cid:48) ( t ) dt. (90)Because of the polynomial growth of ψ (cid:15) , ψ (cid:15) (cid:48) , the function F (cid:15)(cid:15) (cid:48) is well-defined on the strip { (cid:15) < x < (cid:15) (cid:48) } , and agrees with f ( x, y ) on the boundary { x = (cid:15) } ∪ { x = (cid:15) (cid:48) } . Standard Fouriertheory implies that F (cid:15)(cid:15) (cid:48) ( x, y ) also has polynomial growth in y for each fixed x ∈ ( (cid:15), (cid:15) (cid:48) ).Next we show that F (cid:15)(cid:15) (cid:48) = f on the interior of the strip { (cid:15) < x < (cid:15) (cid:48) } . Note that F (cid:15)(cid:15) (cid:48) − f has zero boundary conditions and at worst polynomial growth; we’ll use a barrier argumentto show it is exactly zero. Consider the function η ( x, y ) = 1 x J ( x/ (2 (cid:15) (cid:48) α )) cosh( y/ (2 (cid:15) (cid:48) α )) . (91)where α is the second zero of J ( x ). As noted in Table 58, η solves the PDE (55). Further, η ( x, y ) > x ∈ [0 , (cid:15) (cid:48) ), and indeed we have an exponential lower growth estimate: η ( x, y ) ≥ (cid:15) (cid:48) J (1 / (2 α )) cosh( y/ (2 (cid:15) (cid:48) α )) . (92)Due to the polynomial growth of F (cid:15)(cid:15) (cid:48) − f , we see that give any (cid:15) , (cid:15) (cid:48) and any positive constant C we have that Cη ( x, y ) > F (cid:15)(cid:15) (cid:48) ( x, y ) − f ( x, y ) for sufficiently large y . But then the maximumprinciple says indeed Cη ( x, y ) > F (cid:15)(cid:15) (cid:48) ( x, y ) − f ( x, y ) on the entire strip. But C is arbitrary,we can let C → F (cid:15)(cid:15) (cid:48) ( x, y ) − f ( x, y ) ≤
0. Similarly using − Cη ( x, y ) we obtain F (cid:15)(cid:15) (cid:48) ( x, y ) − f ( x, y ) ≥
0, so indeed we have proven that F (cid:15)(cid:15) (cid:48) ( x, y ) = f ( x, y ) on the strip.Next we fix (cid:15) and let (cid:15) (cid:48) →
0. Considering the coefficients α ω , β ω , ˜ α ω , ˜ β ω , we have α ω → (cid:15)K ( ω(cid:15) ) , while β ω , ˜ α ω , and ˜ β ω all converge to zero. Convergence of the second integral33n (90) is not an issue, as the polynomial growth bounds on ψ (cid:15) (cid:48) ( y ) only improve as (cid:15) (cid:48) → ∞ .This proves the second equality of f ( x, y ) = lim (cid:15) (cid:48) →∞ F (cid:15)(cid:15) (cid:48) ( x, y ) = F (cid:15) ( x, y ) (93)and finishes the lemma. (cid:3) Lemma 4.11
Assume y (cid:55)→ f (0 , y ) has exponential growth bounds, meaning there is some C where f (0 , y ) ≤ e C | y | . Set ψ (cid:15) ( y ) = f ( (cid:15), y ) . If (cid:15) = (cid:15) ( C ) > is sufficiently small, then actually f ( x, y ) = ( G (cid:15) ∗ y ψ (cid:15) )( x, y ) on { ≤ x ≤ (cid:15) } . Further, y (cid:55)→ f (0 , y ) is actually polynomiallybounded: f (0 , y ) ≤ C + C | y | c for some universal constants c , C , and C .Proof . We would like to imitate the proof above, setting F (cid:15)(cid:15) (cid:48) ( x, y ) = (cid:90) ∞−∞ G (cid:15)(cid:15) (cid:48) ( x, y − t ) ψ (cid:15) ( t ) dt + (cid:90) ∞−∞ (cid:101) G (cid:15)(cid:15) (cid:48) ( x, y − t ) ψ (cid:15) (cid:48) ( t ) dt (94)and this time letting (cid:15) →
0. However the polynomial growth bounds from Corollary 4.7actually deteriorate as (cid:15) → (cid:15) (cid:48) → ∞ ), so even though thecoefficients converge: α ω → β ω →
0, ˜ α ω →
0, ˜ β ω → (cid:15) (cid:48) I ( ω(cid:15) (cid:48) ) , it is not clear that the firstintegral in (94) actually vanishes in the limit.So we argue differently. Simply define F (cid:15) ( x, y ) (cid:44) (cid:90) ∞−∞ G (cid:15) ( x, y − t ) ψ (cid:15) ( t ) dt = (cid:15)x (cid:90) ∞ I ( ωx ) I ( ω(cid:15) ) (cid:18)(cid:90) ∞−∞ cos( ω ( y − t )) ψ (cid:15) ( t ) dt (cid:19) dω = (cid:15)x (cid:90) ∞ I ( ωx ) I ( ω(cid:15) ) cos( ωy ) (cid:99) ψ (cid:15) ( ω ) dω (95)Then F (cid:15) ( x, y ) has polynomial growth bounds everywhere, including at x = 0. To see this,note that F (cid:15) (0 , y ) = (cid:15) (cid:90) ∞ ω I ( ω(cid:15) ) cos( ωy ) (cid:99) ψ (cid:15) ( ω ) dω (96)which is certainly convergent. To explain why, notice this is simply the Fourier inverse of (cid:15)ω I ( ω(cid:15) ) (cid:99) ψ (cid:15) ( ω ), and notice that indeed ω(cid:15) I ( ω(cid:15) ) (cid:99) ψ (cid:15) ( ω ) − (cid:99) ψ (cid:15) ( ω ) is actual a smooth function of ω .This is seen by noticing that lim ω → ω(cid:15) I ( ω(cid:15) ) = 1, and that by Lemma 4.9 the non-smoothpart of (cid:99) ψ (cid:15) ( ω ) occurs only at ω = 0. Then, because only the non-smooth part of (cid:99) ψ (cid:15) controlsthe growth of F (cid:15) (0 , y ), we see that F (cid:15) (0 , y ) and F (cid:15) ( (cid:15), y ) have the same polynomial growth (eg.a Dirac δ -function implies O (1) growth, the k th derivative δ ( k ) of a delta implies polynomialgrowth of order k , and so on). 34ow consider the function f ( x, y ) − F (cid:15) ( x, y ). This is zero on { x = (cid:15) } , and by assumptionhas, at worst, exponential growth in y of order C at { x = 0 } . But then if (cid:15) is small enough,the function η ( x, y ) = A x J (2 Cx ) cosh(2 Cy ) (97)exists and is positive on { ≤ x ≤ (cid:15) } , has exponential growth in y of order 2 C at { x = 0 } ,and solves the PDE (55). Thus, no matter what A > η ( x, y ) >f ( x, y ) − F (cid:15) ( x, y ) for y sufficiently large, and therefore η ( x, y ) > f ( x, y ) − F (cid:15) ( x, y ) everywhereon { ≤ x ≤ (cid:15) } . Sending A → f ( x, y ) ≤ F (cid:15) ( x, y ). Similarly using − η ( x, y ) we obtainthe opposite inequality, and so f ( x, y ) = F (cid:15) ( x, y ) on { ≤ x ≤ (cid:15) } . (cid:3) Proposition 4.12
Assume f ∈ C ( H ) ∩ C ∞ ( H ) is a non-negative solution to (55) onthe right half-plane, and assume y (cid:55)→ f (0 , y ) has exponential growth bounds. Then f ( x, y ) is constant.Proof . By subtracting a constant, we can assume inf x> f ( x, y ) = 0. Lemma 4.11 impliespolynomial bounds on the boundary: f (0 , y ) < C + C y c . By Lemma 4.10 we have that f ( x, y ) on { x ≥ (cid:15) } is f ( x, y ) = 1 x (cid:90) ∞−∞ (cid:90) ∞ K ( ωx ) K ( ω(cid:15) ) cos( ω ( y − t )) ψ (cid:15) ( t ) dω dt = 1 x (cid:90) ∞ K ( ωx ) K ( ω(cid:15) ) (cid:99) ψ (cid:15) ( ω ) cos( ωy ) dω (98)Sending (cid:15) → f ( x, y ) ≡
0. To see this, note that x (cid:55)→ K ( ωx ) x exists and is finiteeverywhere including { x = 0 } , and that K ( ω(cid:15) ) → (cid:15) →
0. Using the uniform polynomialbounds on ψ (cid:15) ( y ) as (cid:15) →
0, we have that lim (cid:15) → (cid:98) ψ (cid:15) ( x ) converges to a distribution in thedistributional sense. Therefore the function ( x, ω ) (cid:55)→ lim (cid:15) → K ( ωx ) xK ( ω(cid:15) ) (cid:99) ψ (cid:15) ( ω ) converges to 0 inthe distributional sense. (cid:3) Remark.
It may be helpful to explain the proof of Proposition 4.12 in naturallanguage. We found earlier that, on the half-planes { x ≥ (cid:15) } , f equals the convolutionof f ( (cid:15), y ) with the fundamental solutions G (cid:15) ( x, y ) produced above. We also shows that y (cid:55)→ f (0 , y ) indeed must have polynomial bounds, and therefore we can send (cid:15) → G (cid:15) actu-ally converges to 0 everywhere on { x > } . In the distributional sense, one has actuallylim (cid:15) → G (cid:15) ( x, y ) = δ ( x, y ). 35igure 3: The thinning of the fundamental solution on half-planes. (a) Fundamental solution on { x ≥ } . (b) Fundamental solution on { x ≥ } . Here we remove the exponential growth condition on y (cid:55)→ f (0 , y ). To do this, we first showthat if f ∈ C ( H ) ∩ C ∞ ( H ), then f ∈ C ∞ ( H ), and then we use a blow-up argumentto obtain a non-constant, non-negative function with exponential growth bounds on theboundary, leading to a contradiction. Proposition 4.13 (Regularity at the boundary)
Assume Ω is a pre-compact domainwith Ω ∩ { x = 0 } non-empty. Assume f satisfies (55) on Ω , and f ∈ C (Ω) ∩ C ∞ (Ω) . Then f is C ∞ on Ω ∪ ( ∂ Ω ∩ { x = 0 } ) .Proof . We have f ∈ C ∞ (Ω). If p is a point in the interior of ∂ Ω ∩ { x = 0 } , we can find anopen quadrilateral Q ⊂ Ω with p ∈ ∂Q . By translation invariance with respect to y and bysimultaneous-scale invariance in x and y , we may assume Q = { < x < } ∩ {− < y < } . This shows a fundamental difference between equations (54) and (55); see Example 4. f on Q . We get f ( x, y ) = A ( x, y ) + B ( x, y ) + C ( x, y ) (99)where A , B , and C are given by A ( x, y ) = ∞ (cid:88) n =0 a n x I ( πnx ) I ( πn ) cos( πny ) + ∞ (cid:88) n =0 b n x I ( πnx ) I ( πn ) sin( πny ) (100)where a n , b n are the usual Fourier cosine and sine coefficients for y (cid:55)→ f (1 , y ); B ( x, y ) = ∞ (cid:88) n =0 c n xα n J ( λ n x ) sinh( λ n (1 + y ))2 cosh( λ n ) sinh( λ n ) (101)where λ n is the n th zero of J , α n = − J ( λ n ) J ( λ n ) is a standard normalizing constant,and the c n are the Bessel series coefficients for the function x (cid:55)→ f ( x, − A ( x, C ( x, y ) = ∞ (cid:88) n =0 d n xα n J ( λ n x ) sinh( λ n (1 − y ))2 cosh( λ n ) sinh( λ n ) (102)where λ n is the n th zero of J , α n is the normalizing constant from above, and the d n arethe Bessel series coefficients for the function x (cid:55)→ f ( x, − − A ( x, − x → x − I ( πnx ) → πn and x − J ( λ n x ) → λ n , we have A (0 , y ) = ∞ (cid:88) n =0 πna n / I ( πn ) cos( πny ) + ∞ (cid:88) n =0 b n πnb n / I ( πn ) sin( πny ) B (0 , y ) = ∞ (cid:88) n =0 λ n c n α n sinh( λ n (1 + y ))2 cosh( λ n ) sinh( λ n ) C (0 , y ) = ∞ (cid:88) n =0 λ n d n α n sinh( λ n (1 − y ))2 cosh( λ n ) sinh( λ n ) (103)Of course the sequences { a n } , { b n } , { c n } , { d n } are square-summable (the Plancherel the-orem), and since I ( nπ ) is exponentially decreasing with n , we easily see that the seriesfor A (0 , y ) and for every derivative ( ∂/∂y ) k A (0 , y ) is summable. For B and C , note that α n ≈ √ λ n , and that for each y ∈ ( − ,
1) the coefficients sinh( λ n (1+ y ))2 cosh( λ n ) sinh( λ n ) , sinh( λ n (1 − y ))2 cosh( λ n ) sinh( λ n ) are exponentially decreasing with respect to n . Therefore for y ∈ ( − ,
1) each series in (103)converges absolutely. Taking derivatives with respect to y in any of these series only createscoefficients that are polynomial in λ n (which grows like n ), and so the series in (103) allremain absolutely convergent no matter how many derivatives with respect to y are taken.The fact that f ( x, y ) is identically equal to A ( x, y ) + B ( x, y ) + C ( x, y ) on Q followsfrom the fact that both expressions are finite and equal to one another on ∂Q \ { x = 0 } (byconstruction), and then by Proposition 4.2. Thus f is C ∞ on Q , including at { x = 0 } . (cid:3) emark . Proposition 4.13 helps justify the colloquialism that points on the “singularboundary,” namely ∂ Ω ∩ ∂H , are “interior” boundary points. Not only are the values of ϕ uniquely determined there by its values on ∂ Ω \ ∂H , but C ∞ regularity holds. Remark . Our regularity proof in Proposition 4.13 was a simple spin-off of the Fourieranalysis explored above, but other proofs exist. In the literature, the generalized Hestonequation x (cid:0) σ ϕ xx + 2 ρσϕ xy + ϕ yy (cid:1) + (2 c − q − x ) ϕ y + 2 κ ( θ − x ) ϕ x + c ϕ = 0 (104)has seen substantial study in recent years, although it appears that our Liouville-type theo-rem was not proved, and the methods are substantially different. Our equation x ( ϕ xx + ϕ yy )+ νϕ x = 0 does not quite have the form (104), but does have the same behavior at the sin-gular boundary. See [15] for existence/uniqueness of the Dirichlet problem, and [10], [16],[17] (and references therein) for C ∞ regularity on non-singular boundary components. In[10] the parabolic version of our equation ϕ t = x ( ϕ xx + ϕ yy ) + νϕ x , specifically, was studiedfor ν >
0. Notice that our conclusion in Proposition 4.13 specifically fails to deal with the“corner points” of the domain. This is dealt with at length in [16].
Lemma 4.14 (Reduction to the exponential case)
Assume f ≥ is C ( H ) ∩ C ∞ ( H ) solves f xx + f yy + 3 x − f x = 0 on the right half-plane. Then there is another function ˜ f > on H with ˜ f ∈ C ( H ) ∩ C ∞ ( H ) so that the function y (cid:55)→ ˜ f (0 , y ) has at worst exponentialgrowth. If f is non-constant, then ˜ f is non-constant.Proof . So assume there is such an f ≥ f ∈ C ∞ ( H ), and Corollary 4.6 shows that actually f > H . We can thusconsider the function y (cid:55)→ ddy (cid:12)(cid:12)(cid:12) (0 ,y ) log f . If this is bounded, then obviously y (cid:55)→ f (0 , y ) hasexponential growth and putting ˜ f = f , we are done.Otherwise, there exists some sequence { y i } such that lim i →∞ (cid:12)(cid:12)(cid:12)(cid:12) ddy (cid:12)(cid:12)(cid:12) (0 ,y i ) log f (cid:12)(cid:12)(cid:12)(cid:12) = ∞ . Foreach i , we can re-choose the point y i so that (cid:12)(cid:12)(cid:12) ddy log f (cid:12)(cid:12)(cid:12) is “almost largest” in an appropriateneighborhood. Specifically, set suppose M i = (cid:12)(cid:12)(cid:12)(cid:12) ddy (cid:12)(cid:12)(cid:12) (0 ,y i ) log f (cid:12)(cid:12)(cid:12)(cid:12) , and suppose there is some y ∈ ( y i − i M i , y i + 2 i M i ) with (cid:12)(cid:12)(cid:12)(cid:12) ddy (cid:12)(cid:12)(cid:12) (0 ,y ) log f (cid:12)(cid:12)(cid:12)(cid:12) > (cid:12)(cid:12)(cid:12)(cid:12) ddy (cid:12)(cid:12)(cid:12) (0 ,y i ) log f (cid:12)(cid:12)(cid:12)(cid:12) , then re-choose y i to bethis new y . Continuing, we eventually obtain a sequence y i so that i ) lim i →∞ (cid:12)(cid:12)(cid:12)(cid:12) ddy (cid:12)(cid:12)(cid:12) (0 ,y ) log f (cid:12)(cid:12)(cid:12)(cid:12) → ∞ ii ) (cid:12)(cid:12)(cid:12)(cid:12) ddy (cid:12)(cid:12)(cid:12) (0 ,y ) log f (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12) ddy (cid:12)(cid:12)(cid:12) (0 ,y i ) log f (cid:12)(cid:12)(cid:12)(cid:12) on ( y i − i M i , y i +2 i M i ), where M i = (cid:12)(cid:12)(cid:12)(cid:12) ddy (cid:12)(cid:12)(cid:12) (0 ,y ) log f (cid:12)(cid:12)(cid:12)(cid:12) .38inally, consider the sequence of functions f i ( x, y ) = f ( x/M i , ( y − y i ) /M i ) / f (0 , y i ) . (105)These have the properties i ) Each f i is non-negative and solves the PDE on H . ii ) f i (0 ,
0) = 1 and (cid:12)(cid:12)(cid:12)(cid:12) ddy (cid:12)(cid:12)(cid:12) (0 , log f i (cid:12)(cid:12)(cid:12)(cid:12) = 1. iii ) We have (cid:12)(cid:12)(cid:12)(cid:12) ddy (cid:12)(cid:12)(cid:12) (0 ,y ) log f i (cid:12)(cid:12)(cid:12)(cid:12) ≤ − i , i )Combining ( ii ) with Corollary 4.6 shows that f i is uniformly bounded at least at the point(1 , f i does not converge to ∞ anywhere, and therefore f i convergesto a C ∞ ( H ) solution ˜ f . From ( iii ) we have that ˜ f (0 , y ) is at worst exponential. Of course( ii ) implies ˜ f is non-constant. Proposition 4.12 now provides the contradiction. (cid:3) Finally we write down our primary analytical theorems.
Theorem 4.15 If f ∈ C ( H ) ∩ C ∞ ( H ) , f ≥ , and f solves f xx + f yy + 3 x − f x = 0 ,then f is constant.Proof . By Lemma 4.1 and Proposition 4.13 we have that f > f ∈ C ∞ ( H ). ByLemma 4.14, we may assume that y (cid:55)→ f (0 , y ) has exponential growth at worst. ThenProposition 4.12 implies f is constant. (cid:3) Theorem 4.16
Assume ϕ ∈ C ( H ) ∩ C ∞ ( H ) , ϕ ≥ , ϕ = 0 at { x = 0 } , and ϕ solves ϕ xx + ϕ yy − x − ϕ x = 0 . Then ϕ ( x, y ) = A x for some constant A ≥ .Proof . We have that f ( x, y ) = ϕ ( x, y ) · x − solves f xx + f yy + 3 x − f x = 0. By Lemma4.1 f is actually C ( H ) ∩ C ∞ ( H ). Therefore Theorem 4.15 implies f is constant, and so ϕ ( x, y ) = A · x for some constant A ≥ (cid:3) Here we apply our analytic results to the geometric situation.
Theorem 4.17 (Potentials on half-plane polytopes)
Assume the polytope (Σ , g Σ ) isa closed half-plane and obeys (A)-(F). Possibly after affine recombination of ϕ , ϕ , we have ϕ = 12 x , ϕ = y. (106)39 roof . After a possible affine re-combination of { ϕ , ϕ } , the image of the moment mapΦ = ( ϕ , ϕ ) is precisely the set H = { ϕ ≥ } . Then x = √V is zero on the boundary,and Proposition 2.8 guarantees the complex coordinate z = x + √− y is 1-1, onto, C ∞ , andhas C ∞ inverse. In real terms, the map ( x, y ) is actually a coordinate system on Σ , sendingthe half-plane { φ ≥ } to the half-plane { x ≥ } . Because the map ( x, y ) : Σ → { x ≥ } has C ∞ inverse, the functions ϕ , ϕ are C ( H ) ∩ C ∞ ( H ) where H = { x > } .By Proposition 2.11, we have ϕ , ϕ ∈ C ( H ) ∩ C ∞ ( H ). Since ϕ (0 , y ) = 0, Theorem4.16 guarantees ϕ = α x .As for ϕ , by ( ix ) of § ϕ = y + C ( y ) x + C ( y ) x + . . . . Thus the function ϕ − y is zero on { x = 0 } so that Theorem 4.16 gives ϕ − y = β x . After possiblyrecombination of ϕ , ϕ , we therefore have ϕ = x , ϕ = y .We have the transition matrix A = ( ∂ϕ i ∂x j ) = (cid:18) x
00 1 (cid:19) , so det( A ) x = 1 and from (23)and (29) we have g Σ = dx + dy and K Σ = 0. In momentum coordinates, from (23) wehave g Σ = ϕ ( dϕ ) + ( dϕ ) = (cid:16) d (cid:112) ϕ (cid:17) + ( dϕ ) .Using (7) this gives the metric g on M : g = (cid:16) d (cid:112) ϕ (cid:17) + 4 ϕ (cid:0) dθ (cid:1) + (cid:0) dϕ (cid:1) + (cid:0) dθ (cid:1) . (107)Substituting, say, ϕ = r , we easily see this is the flat metric on R × R , where the Killingfield on the first factor is a rotational field and on the second factor is a translation field. (cid:3) Corollary 4.18 (Half-plane polytopes are flat)
Assume ( M , J, ω, X , X ) has polytope (Σ , g Σ ) where Σ is a closed half-plane, and assume s = 0 on M . Then, possibly afteraffine recombination of ϕ , ϕ , we have ϕ = 12 x , ϕ = y. (108) Further, the metric g on M is flat.Proof . The K¨ahler reduction satisfies the hypotheses of Theorem 4.17. (cid:3) In this, our first of two “examples” section, we focus on analytical examples chosen to demon-strate the themes and the limitations of our analytical lemmata. See § Example 1 . Superharmonic solutions. ϕ ( x, y ) = √ − (cid:113) − ( x + y ) + (cid:112) (1 − ( x + y )) + 4 y (109)satisfies ϕ (0 , y ) = 0, and solves ϕ xx − x − ϕ x + ϕ yy = 0 away from a singular set. The point(1 ,
0) can be seen to be a branch point, and the singular ray is a branch cut. The function f ( x, y ) = ϕ ( x, y ) · x − is smooth at { x = 0 } , and has the same singular locus.Figure 4: Graphs for Example 1: Supersolutions. (a) Graph of ϕ , solving ϕ xx + ϕ yy − x − ϕ x ≤ f = ϕ · x − solving f xx + f yy + 3 x − f x ≤ Neither ϕ nor f is Lipschitz, but are Holder continuous with Holder exponent . Thegradient does not exists along the ray { ( x, (cid:12)(cid:12) x ≥ } so Proposition 4.3 is obviously mean-ingless, but we note that the conclusions of Corollary 4.4 and Corollary 4.7 actually holdwith c = 2.In the general superharmonic case, provided some version of Corollary 4.4 actuallyholds, then the proofs of Corollary 4.6, Corollary 4.7, and Corollary 4.8 actually go through.In this example, we see the conclusions of all of these results indeed hold. Example 2 . Solutions that are Step Functions and δ -functions on { x = 0 } .Consider the functions ϕ ( x, y ) = 12 (cid:32) y (cid:112) x + y (cid:33) ψ ( x, y ) = 12 x ( x + y ) . (110)These both solve the PDE (54) for ν = − ψ = ∂ϕ∂y . Also,we have y (cid:55)→ ϕ (0 , y ) is the unit step function, and therefore y (cid:55)→ ψ (0 , y ) is the unit Dirac41elta function. This justifies the assertion, from the beginning of § G ( x, y ) = ψ ( x, y )is a fundamental solution on the right half-plane.Figure 5: Graphs for Example 2: Steps and Deltas on { x = 0 } . (a) Graph of non-negative solution ϕ where y (cid:55)→ ϕ (0 , y ) is a step. (b) Graph of non-negative solution ψ where y (cid:55)→ ψ (0 , y ) is a Dirac- δ . Setting f = ϕ · x − and g = ψ · x − , we have non-negative functions on the half-planethat satisfy (55), but which are not C ( H ). In the case of f , we have a solution with finiteboundary values on x < x ≥
0. In the case of g , we have a solutionwith finite boundary values on x (cid:54) = 0, and g (0 ,
0) = ∞ . Example 3 . Solution that is a pulse on { x = 0 } .Solutions are translation-invariant in y , so using translations of (110) we construct ϕ ( x, y ) = 12 (cid:32) y + 1 / (cid:112) x + ( y + 1 / − y − / (cid:112) x + ( y − / (cid:33) (111)which solves ϕ xx − x − ϕ x + ϕ yy = 0. Restricted to the boundary { x = 0 } , this is a unitpulse. The associated function f = ϕ · x − = ( x + y ) − / solves f xx + f yy + 3 x − f x = 0.Then f has finite values on x / ∈ [ − , ]. If one considers, say, the quadrilateral42igure 6: Graphs for Example 3: A pulse, and a solution of (55) not determined by boundaryvalues. (a) Graph of non-negative ϕ where y (cid:55)→ ϕ (0 , y ) is a pulse function. (b) Graph of non-negative, non-constant f with finite and infinitevalues on the boundary. Q = { < x < } ∩ {− < y < } , then we see that f has finite values on ∂Q , and so usingthe Fourier and Bessel series from (103) we can construct an L ∞ solution ˜ f to the PDEthat has ˜ f = f on ∂Q . In particular, the function f − ˜ f is zero on ∂Q but is non-constant.This illustrates the necessity of the hypothesis of boundedness in Lemma 4.2. Example 4 . A solution that is only C ,α ( H ) ∩ C ∞ ( H ).Consider the function ϕ = y (cid:112) x + y + x log (cid:32) y + (cid:112) x + y x (cid:33) , (112)which solves ϕ xx + ϕ yy − x − ϕ x = 0. Note that ϕ ∈ C ,α ( H ) ∩ C ∞ ( H ) for all α ∈ [0 , ϕ / ∈ C , ( H ).The fact that y (cid:55)→ ϕ (0 , y ) is piecewise quadratic rather than piecewise-linear provides animportant counterpoint to the crucial assertion in Proposition 2.11 where piecewise linearityis asserted when the solution ϕ is a momentum function in a geometric situation.This example shows the particular need for Propostion 2.11 where we show that thesolutions that arise from our geometrical situations are in fact C up to the boundary, exceptat corners where they are Lipschitz. 43igure 7: Graph for Example 4: Piecewise quadratic growth at { x = 0 } . Finally, this example show that although the PDE (54) has fundamental solutions (80)and has the ability to specify boundary values, but it is not hypoelliptic at the boundary { x = 0 } . We have already seen that (55) is not hypoelliptic at the boundary. First we consider the case that the polytope has just one corner. Take the example ϕ = ay + b (cid:112) x + y + α x ,ϕ = cy + d (cid:112) x + y + β x . (113)Setting Φ = ( ϕ , ϕ ) T we have Jacobian and Jacobian determinant D Φ = (cid:32) bx (cid:0) x + y (cid:1) − + αx a + by (cid:0) x + y (cid:1) − dx (cid:0) x + y (cid:1) − + βx c + dy (cid:0) x + y (cid:1) − (cid:33) , det ( D Φ) = − x (cid:0) x + y (cid:1) − (cid:16) ( ad − cb ) + ( aβ − cα ) (cid:0) x + y (cid:1) + y ( bβ − αd ) (cid:17) . (114)It is impossible that det( D Φ) be zero away from { x = 0 } or undefined away from ( x, y ) =(0 , dϕ and dϕ would be colinear which would imply the polytope has someadditional edge or corner somewhere. Therefore r = (cid:112) x + y and y must obey the linearinequalities det( M ) + det( N ) r + det( K ) y (cid:54) = 0 , ≤ | y | ≤ r (115)44here M = (cid:18) a bc d (cid:19) , N = (cid:18) a αc β (cid:19) , K = (cid:18) b αd β (cid:19) . (116)In order for (115) to hold, it is necessary and sufficient that (cid:12)(cid:12)(cid:12) det( K )det( N ) (cid:12)(cid:12)(cid:12) ≤ det( K )det( M ) ≥ M ) (cid:54) = 0. Lemma 5.1
Consider the map
Φ : H → R where Φ = ( ϕ , ϕ ) T , and each ϕ i satisfies ( ϕ i ) xx − x − ( ϕ i ) x + ( ϕ i ) yy = 0 . Assume Φ is C except at (0 , , and that Φ is Lipschitzeverywhere. Then either the image is either the half-plane and (possibly after GL (2 , R ) recombination of ϕ , ϕ ) we have ϕ = x ϕ = y (117) or else (possibly after a GL (2 , R ) recombination) the polytope is the first quadrant with ϕ = 1 √ (cid:16) − y + (cid:112) x + y (cid:17) + α x ,ϕ = 1 √ (cid:16) y + (cid:112) x + y (cid:17) + β x (118) for some α , β where α, β ≥ .Proof . As indicated set Φ( x, y ) = ( ϕ ( x, y ) , ϕ ( x, y )) T . At any smooth point on { x = 0 } ,the fact that ( ϕ i ) xx − x − ( ϕ i ) x + ( ϕ i ) yy = 0 implies that ( ϕ i ) x = 0 to first order. But thenlim x → x − ( ϕ i ) x = ( ϕ i ) xx , so the first two terms cancel and we have ( ϕ i ) yy = 0 at { x = 0 } .Therefore, away from (0 , y (cid:55)→ Φ(0 , y ) is actually linear. Including (0 , . After possiblyan affine recombination of ϕ , ϕ , we can therefore assume that y (cid:55)→ Φ(0 , y ) is the map y (cid:55)→ (cid:16) √ ( − y + | y | ) , √ ( y + | y | ) (cid:17) .Consider the function ψ = ϕ − √ (cid:16) − y + (cid:112) x + y (cid:17) . On { x = 0 } we now have that ψ (0 , y ) = 0. We wish to use Theorem 4.16, but we must show that ψ ≥
0. First consider˜ ψ = ψ + x , which is still zero on { x = 0 } . This may have negative components, but thenegative part is within the wedge below the line y = 1 − x (in fact, it is below the parabola y = (1 − x ). The function η (cid:15) = − (cid:15) ( x − y − x ) is zero on x = 0 and on y = 1 − x ;therefore η (cid:15) < y = 1 − x . Further, the growth of η (cid:15) is quadratic, whereas the growthof the negative part ˜ ψ − is linear. For an example of how this can fail if the “Lipschitz” hypothesis is removed, see Example 2. For anexample of how this can fail if the “ C except at (0 , (cid:15) > η (cid:15) < − ˜ ψ − for any (cid:15) . Sending (cid:15) →
0, we see that ˜ ψ − = 0.Thus ˜ ψ ≥
0. Now we can use Theorem 4.16 to see there exists some α ∈ R so that˜ ψ ( x, y ) = ˜ α x and therefore necessarily ϕ ( x, y ) = √ (cid:16) − y + (cid:112) x + y (cid:17) + α x for someconstant α = ˜ α −
1. Similarly we have ϕ = √ (cid:16) y + (cid:112) x + y (cid:17) + β x . The fact that wemust have both α, β ≥ (cid:3) In [3] it was noticed that there is a 2-parameter family of metrics on any open polytopeΣ without parallel lines that produces scalar flat metrics on ( M , J, ω ). The next theoremshows that up to homothety these are precisely all metrics when the polytope has a singlecorner. This is the geometric version of Theorem 5.1. Theorem 5.2 (cf. Theorem 1.7)
Assume (Σ , g Σ ) satisfies (A)-(F)—for instance it maybe the reduction of some scalar flat ( M , J, ω, X , X ) —and assume the associated metricpolytope (Σ , g Σ ) is closed and has a single corner. Then (up to homothethy and SL (2 , R ) re-combination) the momentum functions ϕ , ϕ necessarily have the form ϕ = 1 √ (cid:16) − y + (cid:112) x + y (cid:17) + α x ,ϕ = 1 √ (cid:16) y + (cid:112) x + y (cid:17) + β x (119) where α, β ≥ . Therefore metric g Σ belongs precisely to a 2-parameter family of possibilities,parametrized by α and β .Proof . The “closure” assumption is that the polytope contains both of its rays: geometri-cally, neither ray can be “infinitely far away”. By Proposition 2.11, the potential functions ϕ , ϕ , when considered as functions of ( x, y ) and as solutions to the PDE (54), are in-deed C on edges and are Lipschitz everywhere. But Lemma 5.1 then implies ϕ , ϕ (afterpossible GL (2 , R )-recombination) necessarily have the form given above. (cid:3) Corollary 5.3
If the K¨ahler manifold ( M , J, ω, X , X ) is scalar-flat and has a closed poly-tope with a single corner, then the metric is necessarily in the 2-parameter family given bythe momentum functions (119) and the recipe in § . Combine Theorem 5.2 with the description in § (cid:3) In the case where the polytope is unbounded but has more than two edges, the situation isa bit more complex. Homothetic GL (2 , R ) transformations of ϕ , ϕ are able to stabilize46recisely two vertices, but unlike the one-vertex case, the speed of parametrization of thetwo ray edges cannot also be fixed with such transformations.Define the “outline” of the polytope to be the image of the map y (cid:55)→ (cid:0) ϕ (0 , y ) , ϕ (0 , y ) (cid:1) T .In the case that the polytope contains all of its boundaries, then Proposition 2.11 asserts thatthis map is Lipschitz and piecewise linear. After fixing two vertices in place via GL (2 , R )-transformation, the speed at which each linear segment is traversed is an independent vari-able. Therefore each of the N edges contributes a degree of freedom to the space of polytopemetrics. Two additional degrees of freedom express the possible addition of x -terms to eachmomentum function. Theorem 5.4 (Generic Polytopes)
Assume (Σ , g Σ ) is a closed polytope that satisfies(A)-(F) and has n ≥ edges. Then the pair ( ϕ , ϕ ) , as a function of ( x, y ) , belongsto an ( n + 2) -parameter family of possibilities. As a result, the metric g Σ belongs to an ( n + 2) -parameter family of possible polytope metrics.Proof . The proof is to give a recipe for reconstructing the metric on the polytope from thebehavior of ( ϕ , ϕ ) on the edges, and noting the degrees of freedom that appear. As notedabove, the outline map y (cid:55)→ Φ(0 , y ) = (cid:18) ϕ (0 , y ) ϕ (0 , y ) (cid:19) is Lipschitz and piecewise linear, so canbe expressed as sums of terms of the formΨ i ( y ) = (cid:18) α i ( y − y i ) + β i | y − y i | + c i γ i ( y − y ) + δ i | y − y | + d i (cid:19) , Ψ(0 , y ) = n − (cid:88) i =1 Ψ i ( y ) . (120)Since y is required to satisfy only dy = − J Σ dx , we may add a constant to y to makesure that y = 0. Given an outline (the image of y (cid:55)→ Φ(0 , y )) along with the fact thatthe outline map is piecewise linear, the only freedom we have is choosing the speed ofthe parametrizations of each segment; this is n degrees of freedom. With the first corneroccurring at ( ϕ , ϕ ) = (0 ,
1) we have c = 0, d = 1, and from the speed of each segment,we can determine all of the rest of the constants.Then we simply replace | y − y i | with (cid:112) x + ( y − y i ) to obtain (cid:101) Φ i ( x, y ) = (cid:18) (cid:101) ϕ i ( x, y ) (cid:101) ϕ i ( x, y ) (cid:19) = (cid:18) α i ( y − y i ) + β i (cid:112) x + ( y − y i ) + c i γ i ( y − y i ) + δ i (cid:112) x + ( y − y i ) + d i (cid:19) . (121)We have that each (cid:101) ϕ i , (cid:101) ϕ i satisfies the differential equation (54). Since (cid:80) i (cid:101) Φ i (0 , y ) = Φ(0 , y )by construction, by Theorem 4.16 there must be constants α , β such that ϕ ( x, y ) = α (cid:80) i ϕ i and ϕ = β + (cid:80) i ϕ i . Selection of α , β give the final two degrees of freedom. (cid:3) Corollary 5.5
If the scalar-flat K¨ahler manifold ( M , J, ω, X , X ) has a closed polytopewith n ≥ edges, then the momentum map Φ = ( ϕ , ϕ ) T must be a member of an ( n + 2) -parameter family of possibilities. Thus the metric g on M lies within an ( n + 2) -parameterfamily of possible scalar flat metrics. roof . Combine Theorem 5.4 with the description in § (cid:3) Remark . To express Theorem 5.5 more intrinsically, note that if the polytope of( M , J, ω, X , X ) has n many edges, then b ( M ) = n −
2. Thus if b ( M ) > b ( M ) + 4 degrees of freedom, up to homothety, in choosing scalar-flat metrics oncomplete K¨ahler 4-manifolds with two commuting holomorphic symmetries. Remark . For the polytope of 3 edges (where b ( M ) = 1), the recipe of Theorem 5.5is carried out in Example 6 below. Remark . Although we do not discuss the case that the polytope is not closed (thatit has “edges” that are infinitely far away), it is not difficult to imagine how they couldbe built up, namely by adding together potentials of the form found in Example 2, alongwith the potentials used in this section. However we run into difficulties in the proof ofuniqueness. The issue is that, after subtracting copies of such potentials, it is still possiblethat the resulting functions are not zero everywhere on the boundary. If one attempts tomultiply such potentials ϕ by x − to get f = x − · ϕ , there may be singularities on theboundary. Thus the analysis of Section 4 fails, as it relies on f ∈ C ( H ) ∩ C ∞ ( H ). Example 5 . Construction of all doubly-invariant scalar-flat metrics on C .These are precisely the Taub-NUTs, the achiral (or twisted) Taub-NUTs, and the flatmetric. All of these are ALF metrics, except the flat metric which is ALE. According toour theorem, up to linear combinations of ϕ , ϕ , all quarter-plane examples have the formgiven in (118), and using (23) we have polytope metric g Σ = 1 + α + β √ (cid:112) x + y + α − β √ y (cid:112) x + y ( dx ⊗ dx + dy ⊗ dy ) . (122)Setting M = α + β √ and k = α − βα + β (so we are free to choose any M > k ∈ [ − , g Σ = 1 + M r (1 + k sin θ ) r (cid:0) dr + r dθ (cid:1) , ≤ θ ≤ π ,K Σ = − M − M k r ( k + sin θ )(1 + M r (1 + k sin θ )) . (123)When k = 0, these are the Taub-NUT metrics. Also using (23), we can write down themetric in ( ϕ , ϕ )-coordinates, which allows us to easily write down the metric on M aswell. However for generic α, β >
0, the epression for G = x ( det ( A )) AA T is rather complex. Example 6 . Construction of all doubly-invariant scalar-flat metrics on T C P .48hese include the Eguchi-Hanson metric. These metrics are represented by polytopeswith three edges: two rays and an adjoining segment. Under GL (2 , R ) recombinations of ϕ , ϕ , this polytope is the standard polytope with vertices at (0 ,
1) and (1 , Plot for Example 6: Polytope outline for toric metrics on T C P , with directionof parametrization indicated. two degrees of freedom arising from the possible addition of x terms to ϕ , ϕ . Thus upto homothety all such metrics are given by ϕ = v √ (cid:16) y + (cid:112) x + y (cid:17) + (cid:18) v − v √ (cid:19) (cid:32) y − √ v (cid:33) + (cid:118)(cid:117)(cid:117)(cid:116) x + (cid:32) y − √ v (cid:33) + α x ϕ = 1 + v (cid:16) − y + (cid:112) x + y (cid:17) − v √ (cid:16) y + (cid:112) x + y (cid:17) + v √ (cid:32) y − √ v (cid:33) + (cid:118)(cid:117)(cid:117)(cid:116) x + (cid:32) y − √ v (cid:33) + β x (124)The numbers v , v , v > y (cid:55)→ Φ(0 , y ) maps onto the three linearpieces of the outline. We remark that there are certainly other choices of parametrization.
Example 7 . A polytope with an “edge at infinity.”
We begin with the 4-manifold M = S × H , where H is the pseudosphere: g = dr ⊗ dr + sin ( r ) dθ ⊗ dθ + dr ⊗ dr + e r dθ ⊗ dθ . (125)The moment functions are ϕ = − cos( r ) and ϕ = e r . Because r ∈ [0 , π ] and r ∈ ( −∞ , ∞ ), we have that the range of ( ϕ , ϕ ) is precisely [0 , × (0 , ∞ ).49e have the metricsFigure 9: Plots for Example 7: A polytope with an “edge at infinity.” (a) The dashed segment is not part ofthe image of the moment map, and isinfinitely far away as measured in g Σ . (b) Graph of √ V = x as a function of ϕ , ϕ . g = 11 − ( ϕ ) (cid:0) dϕ (cid:1) + (cid:16) − (cid:0) ϕ (cid:1) (cid:17) (cid:0) dθ (cid:1) + 1( ϕ ) (cid:0) dϕ (cid:1) + (cid:0) ϕ (cid:1) (cid:0) dθ (cid:1) ,g Σ = 11 − ( ϕ ) (cid:0) dϕ (cid:1) + 1( ϕ ) (cid:0) dϕ (cid:1) (126)This produces x = √V = |∇ ϕ ||∇ ϕ | = (cid:113) − ( ϕ ) · ϕ , and y = ϕ ϕ ,ϕ = yx + y , and ϕ = (cid:112) x + y . (127)Figure 5a in Example 2 is actually the graph of ϕ as a function of x , y . Example 8 . A Polytope with a disconnected outline .This also furnishes an example of a polytope Σ so that x = √V with y gives a complexfunction z = x + √− y with a critical point, so z is not one-to-one; see Proposition 2.8.Using the scalar flat metric on S × H , where this time the hyperbolic piece has a“two-ended trumpet” metric, we have g = dr ⊗ dr + sin ( r ) dθ ⊗ dθ + dr ⊗ dr + cosh ( r ) dθ ⊗ dθ . (128)50his gives ϕ = − cos( r ) and ϕ = sinh( r ), with range Σ = [ − , × R . The polytopemetric g = 11 − ( ϕ ) dϕ ⊗ dϕ + 11 + ( ϕ ) dϕ ⊗ dϕ (129)and we have x = (cid:114)(cid:16) − ( ϕ ) (cid:17) (cid:16) ϕ ) (cid:17) , y = ϕ ϕ ,ϕ = 1 √ (cid:115) − ( x + y ) + (cid:114)(cid:16) − ( x + y ) (cid:17) + 4 y ϕ = 1 √ (cid:115) − x + y ) + (cid:114)(cid:16) − ( x + y ) (cid:17) + 4 y (130)Figure 10: Plots for Example 8: A polytope with disconnected outline. (a) The polytope is the strip − ≤ ϕ ≤
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