The Euler characteristic of hypersurfaces in space forms and applications to isoparametric hypersurfaces
aa r X i v : . [ m a t h . DG ] F e b The Euler characteristic of hypersurfaces in space forms andapplications to isoparametric hypersurfaces
R. AlbuquerqueFebruary 18, 2021
Abstract
We revisit Allendoerfer-Weil’s formula for the Euler characteristic of embedded hypersur-faces in constant sectional curvature manifolds, first taking the opportunity to re-prove it whiledemonstrating techniques of [2] and then applying it to gain new understanding of isoparametrichypersurfaces.
Key Words:
Euler characteristic, hypersurface, mean curvatures, space form, isoparametric hy-persurface.
MSC 2010:
Primary: 53C35, 57R20; Secondary: 53C17, 58A15
1. Allendoerfer-Weil’s formula
Let (
N, g ) denote an oriented C Riemannian ( n + 1)-dimensional manifold.By a hypersurface f : M ֒ → N we refer to a closed orientable C embedded n -dimensionalRiemannian submanifold M of N . The second fundamental form A = ∇ · ~n , where ~n denotes a unitvector field normal to M , becomes a class C endomorphism of T M . The i th-mean curvature H i ,1 ≤ i ≤ n , is defined by H i = 1 (cid:0) ni (cid:1) X ≤ j < ··· Let P denote the function on M , dependent on the 2nd fundamental form, P = P( λ , . . . , λ n )= (2 l − c l + (2 l − c l − S + · · · +(2 l − p − p − c l − p S p + · · · + (2 l − S l = (2 l − l X p =0 (cid:18) lp (cid:19) c l − p H p (3) where (2 p − p − p − · · · · · and ( − . An alternative expression for P in (3) uses the trivial identity (2 p )! /p ! = 2 p (2 p − Theorem 1.1. Let f t : M → N ( c ) denote a C deformation of a hypersurface f : M → N ( c ) with empty boundary. Then, up to a scalar multiple, P is the only constant coefficients linearcombination on the S i which yields dd t (cid:12)(cid:12)(cid:12)(cid:12) t Z M t P vol t = 0 . (4) Proof. Immediate application of Reilly’s formula (a) in [16, Theorem B], valid for all t :dd t Z M t S i vol t = Z M t g ( X, ~n )(( i + 1) S i +1 − c ( n − i + 1) S i − ) vol t , (5)for any 0 ≤ i ≤ n , where X = ∂f t ∂t is the deformation vector field. (cid:4) We thus have the integral invariant χ ( M ) = 12 l π l Z M P vol = 12 l π l l X p =0 (2 l − p − p − c l − p Z M S p vol . (6)We shall see below that the expression on the right hand side is indeed the Euler-Poincar´e char-acteristic χ ( M ) of M . In other words, that l π l P vol M is the n -form found by the Theorem of . Albuquerque M with certain planes. Interesting enough,Allendoerfer-Weil’s formula is shown in the celebrated book of L. Santal´o [17, p.303] and the formulaof R. Reilly is deduced by Santal´o [18] for parallel deformations of the given hypersurface. 2. Variational calculus with the tangent sphere bundle The theory of a fundamental exterior differential system “ θ, α , . . . , α n ” of Riemannian geometrywas first presented in [2]. Applications in various fields soon emerged, some also related with theintegral of mean curvatures of hypersurfaces, cf. [2, 3, 4].Again we let ( N, g ) denote an oriented Riemannian ( n + 1)-dimensional manifold of class C .We consider the unit tangent sphere bundle π : SN −→ N with its natural Sasaki metric andinduced SO( n ) structure. The metric gives the horizontal tangent subbundle and the geodesic flowvector field; its dual is the most fundamental contact 1-form on SN , denoted by θ .We recall the mirror endomorphism B : T T N → T T N , which sends a horizontal lift of a vectorfield on N to the respective vertical lift and sends verticals to 0. B is indeed an endomorphism andan important object in the definition of the α i ∈ Ω nSN , i = 0 , . . . , n . Letting α n be the volume-formof the fibers, we recall α i = 1 i !( n − i )! α n ◦ ( B n − i ∧ i ) , (7)with ◦ denoting a natural alternating operator. It is also useful to define α − = α n +1 = 0 and torecall the identity d θ ∧ α i = 0, for all i = 0 , . . . , n .Now, let f : M → N denote an oriented hypersurface. On M we have the canonical liftˆ f : M ֒ → SN , defined by ˆ f ( x ) = ~n f ( x ) . We have the following formulas, cf. [2, Proposition 3.2]:ˆ f ∗ θ = 0 , ˆ f ∗ α i = S i vol M . (8)As in [2, 4, 9], we now consider a Lagrangian Λ ∈ Ω nSN . This is just any given n -form on SN .Then we may develop the variations of the following functional on M F Λ ( M t ) := Z M ˆ f t ∗ Λ . (9) Remark. Let X be any vector field on the base space N and let { φ t : N → N } denote its flow.Then the flow of the horizontal lift π ∗ X of X on the unit tangent sphere bundle is rather difficultto find. A rather natural object to consider is the complete lift or extension of X onto T N .Nevertheless we shall observe some results where horizontal lifts are just helpful enough.The vector field X t = ∂f t ∂t is related with the derivative of (9). We shall look for variationsof M with the boundary fixed. Hence X vanishes on ∂M . The derivative requires us to find the . Albuquerque θ ∧ Ψ associated to dΛ, this is, the forms θ ∧ Ψ ∈ Ω n +1 and β ∈ Ω n − such that dΛ = θ ∧ Ψ + d( θ ∧ β ) — a decomposition assured to exist by a Theorem of Lepage,[9]. We then find δ Xt F Λ ( M t ) = dd t (cid:12)(cid:12)(cid:12)(cid:12) t Z M ˆ f t ∗ (Λ − θ ∧ β ) = Z M ˆ f t ∗ L X t (Λ − θ ∧ β )= Z M ˆ f t ∗ (cid:0) d( X t y (Λ − θ ∧ β )) + X t y ( θ ∧ Ψ ) (cid:1) = Z ∂M ˆ f t ∗ X t y (Λ − θ ∧ β ) + Z M ˆ f t ∗ ( θ ( X t ) Ψ ) . (10)Notice we must take X t = ∂ ˆ f t ∂t for the Lie derivative. But the horizontal part X ht is the horizontallift of X t . Easy enough, π ∗ ( X t ) = ∂∂t π ◦ ˆ f t = ∂f t ∂t = X t . Since, by definition, θ u ( Z ) = g π ( u ) ( π ∗ Z, u )for all u ∈ SN and Z ∈ T SN , we findˆ f t ∗ ( θ ( X t )) = ( θ ( X t )) ˆ f t = g ( X t , ~n ) . Proposition 2.1. For a hypersurface M ⊂ N to be a stationary submanifold of the functional F Λ ,under all variations with fixed boundary ∂M and fixed unit normal to the boundary, it is necessaryand sufficient that ˆ f ∗ Ψ = 0 . Notice the hypothesis X t | ∂M = 0. These are deformations which leave ∂M strongly fixed in thesense that both X t and its gradient vanish on ∂M (as in [16, p.467]). Of course, if the boundary isempty, then there is no restriction on the deformation. Remark. On the assumptions of Proposition 2.1, we may compute the second derivative: δ Xt F Λ ( M t ) = Z M ( dd t g ( X t , ~n )) ˆ f t ∗ Ψ + g ( X t , ~n ) ˆ f t ∗ L X t Ψ . (11)If θ ∧ Ψ is the new Poincar´e-Cartan form, dΨ ≡ θ ∧ Ψ + d( θ ∧ β ), then on the stationary M = M we obtain δ X F Λ ( M ) | t =0 = Z M g ( X, ~n ) ˆ f ∗ (cid:0) d( X y Ψ ) + X y d( θ ∧ β ) (cid:1) + ( g ( X, ~n )) ˆ f ∗ Ψ . (12) 3. Alternative proof of Theorem 1.1 Let us return to the case of an oriented Riemannian ( n + 1)-dimensional manifold of constantsectional curvature N = N ( c ) and let us study the functional F α i . We have the following magic formula from [2, formula 2.26]:d α i = θ ∧ (cid:0) ( i + 1) α i +1 − c ( n − i + 1) α i − (cid:1) . (13)These are already in the Poincar´e-Cartan form. Combining with (8,10), yields immediately formula(a) of R. Reilly in [16, Theorem B] which we have used in (5). . Albuquerque α i = θ ∧ Ψ and thus dΨ = θ ∧ Ψ whereΨ = ( i + 1)( i + 2) α i +2 − c ( i ( n − i + 1) + ( i + 1)( n − i )) α i + c ( n − i + 1)( n − i + 2) α i − . (14)Hence the following result. Proposition 3.2. Under all variations with fixed boundary ∂M and fixed unit normal to the bound-ary, at a stationary submanifold of F α i we have δ X F α i ( M t ) | t =0 = d d t (cid:12)(cid:12)(cid:12)(cid:12) t =0 Z M S i vol t = Z M g ( X, ~n ) ˆ f ∗ d( X y Ψ ) + g ( X, ~n ) ˆ f ∗ Ψ . (15)Now any non-trivial Lagrangian Λ which satisfies dΛ = 0 gives rise to a conservation law : here,an integral invariant of hypersurfaces under C deformations. Thus we consider a formΛ = b α + · · · + b n α n (16)with constant coefficients b j . Such a form is an invariant of the geodesic flow vector field S = θ ♯ . Proposition 3.3. If dΛ = θ ∧ Λ , then Λ = L S Λ . Proposition 3.4. dΛ = 0 if and only if • in case c = 0 , we have b = b = . . . = b n − = 0 • in case c = 0 and n is odd, we have b = b = . . . = b n − = b n = 0 , • in case c = 0 and n is even, we have b = b = b = . . . = b n − = 0 and b p = 1 · · · · · (2 p − b c p ( n − n − · · · ( n − p + 1) , ∀ p = 1 , . . . , n . (17) Proof. It is clear that dΛ = θ ∧ n X j =0 ( jb j − − c ( n − j ) b j +1 ) α j . (18)The α , . . . , α n are linearly independent, so from the cases j = 0 , n we obtain cb = 0 , b n − = 0,respectively. We may immediately assume c = 0. Then b = 0; and then b = 0; and so on and soforth. If also n is odd, then it follows b n = 0. Moreover, in the descending order, b n − = 0 impliesall the b even = 0. Finally, if n is even, then the last remaining condition must hold: b odd = 0 andan induction on jb j − − c ( n − j ) b j +1 = 0 yields the formula. (cid:4) In case c = 0, we may have b n = 0 and thus find a variational trivial functional. S n is theso-called Gauss-Kronecker curvature.Let n = 2 l and let us take b = c l (2 l − f t ∗ Λ = n X i =0 b i S i vol M = l X p =0 (2 l − p − p − c l − p S p vol M (19) . Albuquerque χ is the Euler characteristic To the best of our knowledge, the generalized Theorem of Gauss-Bonnet-Chern is first surveyedin its most comprehensive and insightful form in [14], and such is the form we wish to followhere. The theorem establishes the identity χ ( M ) = R M Pf(Ω / π ), where Ω = [Ω ij ] i,j =1 ,...,n is thecurvature 2-form matrix and Pf(Ω) is the Pfaffian of Ω, a closed n -form and a non-trivial object ineven dimensions in general.On a constant sectional curvature ambient manifold N ( c ), for a hypersurface f : M → N ( c )and a positively oriented orthonormal frame e , . . . , e n of principal curvatures, we have by Gauss-Codazzi equations, cf.[2]: R Mpkij = ( c + λ i λ j )( δ pi δ kj − δ pj δ ki ) . (20)Hence the matrix of 2-forms corresponds to Ω ij = ( c + λ i λ j ) e ij . The Pfaffian n -form is the Pfaffianof the following skew-symmetric matrix c + λ λ ) e ( c + λ λ ) e ( c + λ λ n ) e n ( c + λ λ ) e c + λ λ ) e · · · ( c + λ λ n ) e n ( c + λ λ ) e ( c + λ λ ) e c + λ λ n ) e n · · · . . .( c + λ λ n ) e n ( c + λ λ n ) e n . (21)Let us denote such matrix by Ω( c ; λ , . . . , λ n ).For n = 2, clearly Pf(Ω( c ; λ , λ )) = ( c + λ λ ) e , proving (6). As well as with n = 4, where awell-known formula is appliedPf(Ω( c ; λ , λ , λ , λ )) = Ω ∧ Ω − Ω ∧ Ω + Ω ∧ Ω = (cid:0) c + c ( λ λ + λ λ + λ λ + λ λ + λ λ + λ λ ) + 3 λ λ λ λ (cid:1) e . (22)Now, we recall a ‘Laplace rule’ for the Pfaffian of an n × n skew-symmetric matrix X , cf. [20,p.27]: Pf( X ) = n X j =2 ( − j x j Pf( X j ) (23)where X ij denotes the skew-symmetric matrix with the i th and j th rows and columns removed.Since Λ even T ∗ M is commutative, the formula is valid for the curvature form, i.e. giving the preferredsquare-root of the determinant of Ω.For n = 2 l , it follows that Pf(Ω( c ; 0 , . . . , n − c l vol. Therefore,Pf(Ω(0; λ , . . . , λ n )) = ( n − λ · · · λ n vol . (24)The next result finally establishes Allendoerfer-Weyl’s formula (6). . Albuquerque Proposition 4.5. We have that Pf(Ω) = P vol M , where P is given in (3) .Proof. Let A l,p = (2 l − p − p − A l, = A l,l = (2 l − A l,p = (2 p − A l − ,p − = (2 l − p − A l − ,p . Now we shall prove the result by induction. Since itis true for n = 2, we next assume l > l X j =2 ( − j ( c + λ λ j ) e j ∧ Pf(Ω j )= l X j =2 ( c + λ λ j )P( λ , . . . , λ j − , λ j +1 , . . . , λ l )= l X j =2 l − X p =0 ( c + λ λ j ) c l − − p A l − ,p S p ( λ , . . . , λ j − , λ j +1 , . . . , λ l )= l X j =2 l − X p =0 c l − p A l − ,p S p ( λ , . . . , b λ j , . . . , λ l ) + l X j =2 l X q =1 c l − q A l − ,q − λ λ j S q − ( λ , . . . , b λ j , . . . , λ l )= (2 l − c l + λ · · · λ l ) + l − X q =1 c l − q l X j =2 (cid:18) A l − ,q S q ( λ , . . . , b λ j , . . . , λ l )+ A l − ,q − λ λ j S q − ( λ , . . . , b λ j , . . . , λ l ) (cid:19) = A l, c l + A l,l S l + l − X q =1 c l − q A l,q l X j =2 (cid:18) S q ( λ , . . . , b λ j , . . . , λ l )2 l − q − λ λ j S q − ( λ , . . . , b λ j , . . . , λ l )2 q − (cid:19) . The result follows once we prove, for all 1 ≤ q ≤ l − l X j =2 (cid:18) S q ( λ , . . . , b λ j , . . . , λ l )2 l − q − λ λ j S q − ( λ , . . . , b λ j , . . . , λ l )2 q − (cid:19) = S q ( λ , . . . , λ l ) . ( ⋆ )Firstly, arguing by symmetry and with each monomial, we find S q ( λ , λ , . . . , λ l ) + S q ( λ , λ , . . . , λ l ) + · · · + S q ( λ , λ , . . . , λ l − , λ l ) + S q ( λ , λ , . . . , λ l − )= (2 l − q − S q ( λ , λ , λ , . . . , λ l − , λ l ) . Secondly, another careful inspection gives λ S q − ( λ , λ , . . . , λ l ) + λ S q − ( λ , λ , . . . , λ l ) + · · · + λ l S q − ( λ , λ , . . . , λ l − )= (2 q − S q − ( λ , λ , λ , . . . , λ l − , λ l ) . Hence the left hand side of ( ⋆ ) becomes= S q ( λ , . . . , λ l ) + λ S q − ( λ , . . . , λ l ) = S q ( λ , . . . , λ l )as we wished. (cid:4) . Albuquerque λ, µ of multiplicities respectively1 , n − c ; λ, µ, . . . , µ )) = ( c + λµ ) l X j =2 ( − j e j ∧ Pf(Ω( c + µ ; 0 , . . . , l − c + λµ )( c + µ ) l − vol . (25)This can be checked either through (21) or (3).Example: We consider the canonical immersion of hypersurfaces S p,qr := S pr × S q √ − r ֒ → S n +11 for any integers 0 ≤ p, q ≤ n = 2 l = p + q and r ∈ ]0 , ω m of an m -sphere, m ∈ N , of radius 1, which is ω m := 2 π ( m +1) / / Γ(( m + 1) / χ ( S p,qr ) = 0, for p, q odd; it is equal to 2, for p = 0 or q = 0; equal to 4,for p, q = 0 even. Consulting [5, Example 2], cf. [4], we find the principal curvatures: λ = · · · = λ p = − √ − r r , λ p +1 = · · · = λ n = r √ − r . (26)The two values, say λ, µ , respectively, satisfy 1 + λµ = 0. By (25) it follows that χ ( S n − , r ) = 0 asexpected. The following results are much easier to find directly from the Pfaffian of (21). In fact,since c + λµ = 0, the multiplicative nature of χ on S p,qr is immediately seen through the matrixcomputation. Yet the following is still interesting to see using P vol, that is, by (25): χ ( S nr ) = 12 l π l Z Pf(Ω) = (2 l − l π l Z S nr (cid:0) − r r (cid:1) l vol r = (2 l − l π l vol( S n ) = 2 . (27)In the same way, one finds χ ( S , )) = 4 = ( χ ( S )) = χ ( S , )) . (28)By (27) there remains no doubt that dividing the invariant R M P vol by (2 π ) l yields always aninteger in the equivalence class of the 2 l -sphere. 5. Chern’s proof of Gauss-Bonnet and a theorem of H. Hopf Let us try to follow S.S. Chern’s proof in [11] of the celebrated result bearing his name, in thecase of N = N ( c ), of course of even dimension. Thus now n is odd.A striking passage recurs to Hopf’s index Theorem and therefore to the choice of a unit vectorfield X with isolated zeros. Indeed, the unit sphere bundle plays a fundamental role in [11]. Since π is a Riemannian submersion, the integral of any form ˜Ω over N is the same as the integral of thepullback form over the ( n + 1)-submanifold or cycle V = X ( N \{ singular points } ) of SN . Moreover, − ∂V equals the n -cycle S n χ ( N ), where S n represents the standard fiber or fundamental class, cf.[11, 14]. In our understanding, it is only what later became known as Thom duality, brought inwith Stokes identity, which sustains Chern’s swift and ingenious breakthrough.The proof requires an intrinsic or metric-invariant global n -form Π, which we search here forwithin the exterior differential system of the { α i } and which satisfies both dΠ equal to π ∗ ˜Ω when . Albuquerque V , for some ˜Ω, and that the restriction of Π on the fibers coincides with that of b n α n ,with b n = − /ω n and ω n = vol( S n ). We recall α n is the volume form on the fibers. Hence thedesired identities Z N ˜Ω = Z ∂V Π = χ ( N ) . (29)The only invariant n + 1-form we have as a pullback is θ ∧ α = π ∗ vol N , so we are bound to find c from dΠ = θ ∧ P c j α j where Π = P b j α j . Proceeding as in Proposition 3.4, we clearly have theequations jb j − − c ( n − j ) b j +1 = c j , of which there are plenty of solutions in the setting and in case c = 0. A simple one is c j = 0, j > 0. Thus b j = 0 , b j +1 = − ( n − j − j )!! c n − j − ( n − ω n . (30)Hence b = − ( n − c n − ( n − ω n , c = n !! c n +12 ( n − ω n . (31)With c > k , n = 2 k + 1, and for the sphere N = S k +2 r , r = √ c , that2 = χ ( N ) = R N c θ ∧ α = c r k +2 ω k +2 . Hence the trivial identity ω k +2 ω k +1 = 2 (2 k )!!(2 k +1)!! .Returning to (31), we may write c ω k +2 = 2 c k +1 . In other words, Gauss-Bonnet-Chern’sformula reads 2 c n +12 vol( N ) = χ ( N ) ω n +1 for c = 0. Of course N must be of finite volume.In particular we recover a classical theorem of H. Hopf, as it is well known, with impor-tant implications to hyperbolic geometry. Let N be a hyperbolic space of even dimension m ,constant sectional curvature − N ) = ( − m/ ω m χ ( N ), a cornerstone in the arithmetic theory of hyperbolic geometry, cf.[8, 13]. 6. Other formulas from Allendoerfer’s article We continue with the previous setting. Let n be odd and ( N, g ) denote an oriented C Rieman-nian ( n + 1)-dimensional manifold of constant sectional curvature c .Let Q be a compact ( n + 1)-submanifold with C boundary ∂Q . We assume the existence of a C vector field X on Q such that: • X has only isolated singularities and these lie in the interior of Q ; • X coincides with the outward unit-normal ~n over ∂Q .We may conjecture this is true for all Q . Certainly, an open, relatively compact and one-pointcontractible subset Q ⊂ N with C boundary admits such a vector field.The questions raised in [6] regarding a new formula of Steiner type, i.e. one which looks forthe enclosed volume and area of hypersurfaces enlarged in geodesic outward directions, also referto the case when n is odd and the submanifold Q is closed and bounding, of class C . Moreover, acomplete metric on N is assumed. . Albuquerque χ ( Q ), [6, formula 20] must be the same as the one in the next theorem, forwhich we find a quick proof using the already seen argument from [11]. Again the result may alsobe seen in [19], with a complete proof. Theorem 6.2. Let n = 2 k + 1 and c = 0 . In the above conditions, we have: c k +1 vol( Q ) + 1 n !! k X j =0 (2 j )!!(2 k − j − c k − j Z ∂Q S j +1 vol ∂Q = χ ( Q )2 ω k +2 . (32) Proof. First we notice that, as a cycle in SN , the subspace V = X ( Q \{ singular points } ) has asboundary the n -cycle ∂V = − χ ( Q ) S n ∪ X ( ∂Q ) . Again we may find over the tangent sphere bundle an invariant n -form Π such that dΠ = θ ∧ α .We have seen the solution in the previous Section. Since θ ∧ α = π ∗ vol N , then X ∗ dΠ = vol Q .Applying Stokes identity and (8) the result follows. (cid:4) Clearly, the result is coherent with the formulas from the last Section where Q = N . Here weare simply not chiefly focused on χ . We remark case n = 1 is due to W. Blaschke.Example 1: It is interesting to check the case n = 1 with a sphere N = S r where r − = √ c . Letus take the usual spherical coordinates ( x, y, z ) = ( r sin φ cos θ, r sin φ sin θ, r cos φ ) with 0 ≤ θ ≤ π and 0 ≤ φ ≤ π the angle with the positive vertical axis. Let Q be the spherical cylinder − r < r cos φ ≤ z ≤ r cos φ < r. (33) χ ( Q ) = 0, as it is easy to see. One computes, area( Q ) = R φ φ R π r sin φ d θ d φ = 2 πr (cos φ − cos φ ). One may recall the horizontal projection from S r to the cylinder x + y = r is equiareal.Also the two rays r , r on the boundary components are given by r i = r sin φ i . Recalling (26)and noticing the outward normal to Q over the boundary, we find R ∂Q S vol ∂Q = − √ r − r rr πr + √ r − r rr πr = 2 π (cos φ − cos φ ). Hence, as expected, c area( Q ) + Z ∂Q S vol ∂Q = 0 . (34)Example 2: Let us otherwise take a disc cap Q ⊂ S s . Say we let φ = 0 in the same settingabove. Then χ ( Q ) = 1 and, since the boundary has just one connected component, the identity ofTheorem 6.2 is verified, like (34) but with χ ( Q )2 ω = 2 π on the right hand side.Taking the limit in c → Theorem 6.3. (i) Let n = 2 k + 1 be odd and let Q be any relatively-compact domain with C boundary in Euclidean space of dimension n + 1 = 2 k + 2 . Then χ ( Q ) = k !2 π k +1 Z ∂Q S k +1 vol ∂Q . (35) (ii) Let n = 2 l be even and let M be any compact hypersurface of class C in Euclidean space ofdimension n + 1 . Then χ ( M ) = (2 l − l π l Z M S l vol M . (36) . Albuquerque M = ∂Q isa boundary, then χ ( ∂Q ) = 2 χ ( Q ). 7. On isoparametric hypersurfaces A smooth hypersurface M isometrically immersed in an ( n +1)-dimensional manifold of constantsectional curvature c is isoparametric if and only if it has constant principal curvatures. RegardingEuler-Poincar´e characteristic, the formula of Allendoerfer yields some results which might add totheir further knowledge. We refer to [1, 12, 21, 22, 23] for the important sources of historical and upto date information on isoparametric hypersurfaces. We shall take a short detour by that theory,in so as much as to explain the leading ideas. No source we have found has been occupied with χ .The celebrated work of ´E. Cartan on isoparametric hypersurfaces starts with the fundamentalresult that X j = i m j c + λ j λ i λ i − λ j = 0 , (37)valid for each distinct principal curvature λ i , with multiplicity m i and 1 ≤ i ≤ g , where g is thestandard for the number of such λ i .If c ≤ 0, then Cartan finds there are at most two such curvatures. If g = 1, then by (25) and(27) we find χ ( M ) = 2( c + λ ) l ω n vol( M ) . (38)If g = 2, then the fundamental formula yields c + λ λ = 0 and so follows easily from (21) thatthe Euler characteristic χ ( M ) = m is odd c + λ ) m / ( c + λ ) m / ω m ω m vol( M ) if m is even . (39)For g = 1 , 2, Cartan gives the classification of the complete hypersurfaces, in [10]: embeddedhyperspheres or canonical products of hyperspheres, with χ = 2 or 4, as those from Example inSection 4. For c < 0, we see the complete hypersurfaces are non compact.We shall now assume c = 1, for it is then clear how to recover the general case. Also, by Cartan,each and every λ i = 0.In case the multiplicities are all equal, denoting λ = cotg ǫ for some 0 < ǫ < π , not a multipleof π/g , then the distinct principal curvatures are given by λ i = cotg (cid:16) ǫ + ( i − πg (cid:17) , i = 1 , . . . , g. (40)For g = 3, Cartan proves in a second article that the associated multiplicities must all be equal: m = m = m = m ; moreover they can only assume the values m = 1 , , , 8. He then determinesthe associated 3 m -manifolds M (3 ,m ) , which were later, in celebrated works of H.F. M¨unzner, furtherunderstood as m -sphere tubes of standard Veronese embeddings of projective planes, over thecanonical division algebras or the Cayley ring, into 3 m + 1-dimensional spheres. . Albuquerque Theorem 7.4. For g = 3 and m = 2 , , , we have χ ( M (3 ,m,λ ) ) = P ( λ )(2 π ) m (1 − λ ) m vol( M (3 ,m,λ ) ) (41) with P ( λ ) ∈ Z [ λ ] an integer polynomial in λ = λ of degree m . In particular, for m = 2 , we have χ ( M (3 , ,λ ) ) = 3(1 + λ ) π (1 − λ ) vol( M (3 , ,λ ) ) > . (42) Proof. Using basic trigonometry identities we deduce λ = λ − √ √ λ + 1 , λ = λ + √ − √ λ . Then we define A = λ + λ = 8 λ − λ , B = λ λ = λ − − λ , and notice every symmetric polynomial S p in the Pfaffian polynomial from (6) in the 3 m variables λ, . . . , λ, λ , . . . , λ , λ , . . . , λ becomes an integer polynomial in the three variables λ, A, B . Indeed,we have S p ( λ, . . . , λ ) = P m ′ j =0 (cid:0) mj (cid:1) λ j S p − j ( λ , . . . , λ ), where m ′ = min { m, p } . Now each of thosecoefficients, indeed symmetric polynomial S p − j , is a polynomial in A, B , a result which is easilyproved by induction. So the claim follows with the numerator in (41) in Z [ λ ]. For instance, in case m = 2, we have the formulas S = 1 , S = λ + λ + λ + 4( λ λ + λ λ + λ λ ) = λ + 4 λA + A + 2 B,S = λ λ + λ λ + λ λ + 4( λ λ λ + λ λ λ + λ λ λ ) = λ ( A + 2 B ) + 4 λAB + B and S = λ B , which imply the result in (42) by not so long computations. (cid:4) Of course, the simple form of P found for m = 2 raises the question if a more structural approachshould be searched. And if P is always a polynomial in λ . Notice we have not had the input fromany preferred principal curvature. Due to both sides of (41), we must conclude P ( λ ) / (1 − λ ) m is an invariant of the λ i . If this is quite clear in case m = 2, it may help in determining P for m = 4 , λ . Indeed, for the division algebras or theoctonians F = C , H , O , it is known that χ ( FP ) = 3, so in view of the bundle structure the integerinvariant is always 6.In case m = 2, we obtain a unique absolute minimum value, 1, of the function (1+ λ ) / (1 − λ ) ,at points 0 , ±√ 3, which are not admissible λ for the construction. It is interesting thatlim λ → , ±√ vol( M (3 , ,λ ) ) = 2 π (43)is greater than ω = π / . Albuquerque g are 1 , , , , 6. And, regarding multiplicities, he deduced the general iden-tity m i = m i +2 mod g , for all i . Hence, for g = 4 , 6, there are at most two distinct multiplicities,say m and m .Letting g = 4, we may formally compute the Euler characteristic with the above techniquewhen all the multiplicities are equal to 1. It gives χ = 3 S + S + 3 S = 0 because S = S = 1and S = − λ i in the proof below). However, a further classification by U. Abresch,in [1], has excluded the existence of isoparametric hypersurfaces of this kind.A problem we must leave here is the case of m = m or equal m i but > Theorem 7.5. In case g = 4 and m = m = 2 , we find χ ( M ) = 3( λ − λ + 316 λ − λ + 1)4 π λ (1 − λ ) vol( M ) . (44) In particular χ ( M ) = 0 .Proof. As above, we denote λ by λ . By simple trigonometry, λ = λ − λ + 1 , λ = − λ , λ = − λ = λ + 11 − λ Therefore the symmetric polynomials are simplified, first by the two last helpful identities, andsecond by substituting λ . Computations thus yield S = λ + λ + 1 λ + 1 λ + 4 (cid:18) λ λ − − λ λ − λ λ − λ λ (cid:19) = 1 λ (1 − λ ) ( λ − λ + 62 λ − λ + 1) S = λ λ + 1 + λ λ + λ λ + 1 + 1 λ λ + 4 (cid:18) λ λ + λ λ − λ − λ − λ λ − λ λ − λ λ − λ λ − λ − λ + λ λ + λ λ (cid:19) + 16= 1 λ (1 − λ ) ( − λ + 62 λ − λ + 62 λ − . Clearly S = S = 1, and it is easy to see S = S because in each summand of S there are alwaystwo pairs of principal curvatures λ i , − λ i − , which cancel, leaving an intact S . Finally we recur to(3), which yields P = 210 + 90 S + 9 S , and recall (6).The conclusion that χ = 0 follows by contradiction. Suppose an isoparametric hypersurface isgiven, with λ a zero of (44), isolated. We may always continuously produce a 1-parameter familyof homeomorphic so called parallel hypersurfaces, which are still isoparametric of the same type,just by varying λ in a sufficiently small interval. (cid:4) Notice we may virtually have χ of any sign. The case considered in the Theorem is classified, cf.[12, 21, 22], and we know there exists precisely one such hypersurface M up to parallel homotopy,which is homogeneous, and must correspond to one of the ‘chambers’ of the λ i designed by the 8zeros of the right hand side of (44). . Albuquerque ± ±∞ , where M has volume as small aspossible, it should be interesting to understand the extreme cases within each bounded end.As Z. Tang explained to the author, and indeed consulting [22], the class of the manifold M from Theorem 7.5 is constructed as a 2-sphere bundle over CP . Henceforth we have justvol( M ) = 32 π λ (1 − λ ) λ − λ + 316 λ − λ + 1) . (45)Finally we consider the case g = 6. Abresch also discovered that m = m and that this number ≤ g = 6 and multiplicities m = m = 1 have vanishing χ .This assertion can be proved by recalling the circle tube structure. Or we may recall the principalcurvatures differ by multiples of π/ λ i +3 = − λ i with i mod 6. It is then easy tosee S = − S and S = − S = − Theorem 7.6. In case g = 6 and m = m = 2 , we find χ ( M ) = 90(1 + λ ) π λ (1 − λ ) (3 − λ ) vol( M ) > . (46) Proof. Let λ = λ as usual. Due to the step increase by π/ 6, we have λ = √ λ − λ + √ , λ = λ − √ √ λ + 1 , λ λ = λ λ = λ λ = − . In order to deal with λ and λ we define s = λ − , t = 3 λ − , p = √ λ − λ + √ , p = √ λ + 4 λ + √ λ = ps , λ = pt , pp = st = 3 λ − λ + 3 , p − p = − λ. These enable a more simple treatment of the symmetric polynomials. Indeed, up to a commondenominator λ s t , we may rewrite S j as polynomials in λ, pp, p − p, s + t, s + t , s + t , s + t .The result is: S = 1 λ s t (cid:0) λ + 1)) − λ + λ ) + 4095( λ + λ ) − λ (cid:1) S = 1 λ s t (cid:0) − λ + 1) + 4095( λ + λ ) − λ + λ ) + 57210 λ (cid:1) S = 2 λ s t (cid:0) λ + 1) − λ + λ ) + 28605( λ + λ ) − λ (cid:1) . With S we notice that 2 pairs of principal curvatures with indices i and i + 3 mod 6 must lie withineach summand. Then a simple cancellation yields an S . The same happens with 4 and 6 pairs for . Albuquerque S = S and S = S = 1, respectively. Therefore we haveP = 11!!( S + S ) + 9!!( S + S ) + 7!!3!!( S + S ) + 5!!5!! S = 10395 · · S + 315 · S + 225 S = 2 λ s t (cid:18) λ + λ ) − λ + λ ) + 118 λ )++ (945 · − · 60 + 225 · λ + 1)+ ( − · 540 + 315 · − · λ + λ )+ (945 · − · · λ + λ )+ ( − · · − · λ (cid:19) = 2 λ s t · λ ) . Of course we have recurred to a computer program for these very last computations and simplifi-cation. (cid:4) Due to the surprising reduction of the polynomial, as we had seen before with (42), a quest isclearly set to understand more deeply the theorem above.The author acknowledges the fruitful conversations with Jos´e Navarro, from the U. Extremadura,in Portugal and Spain, and with Zizhou Tang at CIM, which helped to improve some of the con-clusions above. He also thanks deeply an anonymous Reviewer.Parts of this article were written while the author was a visitor at Chern Institute of Mathe-matics, Tianjin, China. 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