Affine hypersurfaces of arbitrary signature with an almost symplectic form
aa r X i v : . [ m a t h . DG ] F e b AFFINE HYPERSURFACES OF ARBITRARY SIGNATUREWITH AN ALMOST SYMPLECTIC FORM
MICHAL SZANCER AND ZUZANNA SZANCER
Abstract.
In this paper we study affine hypersurfaces with non-degeneratesecond fundamental form of arbitrary signature additionally equipped withan almost symplectic structure ω . We prove that if R p ω = 0 or ∇ p ω = 0 forsome positive integer p then the rank of the shape operator is at most one. Theresults provide complete classification of affine hypersurfaces with higher orderparallel almost symplectic forms and are generalization of recently obtainedresults for Lorentzian affine hypersurfaces. Introduction
Parallel structures are of the great interest in the classical Riemannian geometry(see [7, 19, 3]) as well as in affine differential geometry ([4, 10, 11, 15, 17, 14, 16]).Higher order parallel structures are the natural generalization of parallel structuresand are widely studied as well ([7, 8, 28, 29, 27]).In [2] O. Baues and V. Cort´es studied affine hypersurfaces equipped with analmost complex structure. They showed that there is direct relation between simplyconnected special K¨ahler manifolds ([12]) and improper affine hyperspheres. LaterV. Cort´es together with M.-A. Lawn and L. Sch¨afer proved a similar result forspecial para-K¨ahler manifolds ([5]). In both cases an important role was played bythe K¨ahlerian (resp. para-K¨ahlerian) symplectic form ω . The concept of specialaffine hyperspheres was generalized by the first author in [23]. Some other resultsrelated to affine hypersurfaces with almost complex structures can be also found inthe paper of M. Kon ([18]). In all the above cases the important role was played byan (almost) symplectic structure related in some way to the induced affine structureon a hypersurface. In particular, relation between an almost symplectic structure ω and the induced affine connection ∇ and its curvature R seemed crucial.The above results motivated the first author to study non-degenerate affine hy-persurfaces f : M → R n +1 with a transversal vector field ξ additionally equippedwith an almost symplectic structure ω in a more general setting. More preciselyaffine hypersurfaces with the property ∇ p ω = 0 or even more general R p ω = 0.In [21] it was shown that if dim M ≥ Rω = 0 implies that ∇ must beflat (what is generalization of result obtained in [18]) and the result generalizes toan arbitrary power of R under additional assumption that the second fundamentalform is positive definite and the transversal vector field ξ is locally equiaffine (i.e. dτ = 0). Namely, we have Mathematics Subject Classification.
Key words and phrases. affine hypersurface, almost symplectic structure, symplectic form.
Theorem 1.1 ([21]) . Let f : M → R n +1 be a non-degenerate affine hypersurface( dim M ≥ ) with a locally equiaffine transversal vector field ξ and an almost sym-plectic form ω . Additionally assume that the second fundamental form is positivedefinite on M . If R p ω = 0 for some positive integer p then ∇ is flat. From the above theorem it follows that
Theorem 1.2 ([21]) . Let f : M → R n +1 be a non-degenerate affine hypersurface( dim M ≥ ) with a locally equiaffine transversal vector field ξ and an almost sym-plectic form ω . Additionally assume that the second fundamental form is positivedefinite on M . If ∇ p ω = 0 for some positive integer p then ∇ is flat. In the very same paper it was shown that the condition that the second funda-mental form is positive definite cannot be relaxed since there exists affine hyper-surface with property R ω = 0 which is not flat.Later in [22] it was shown that for affine hypersurfaces with Lorentzian secondfundamental form we still have strong constrains on the shape operator S if onlydim M ≥
6. More precisely, although it cannot be shown that S = 0 (what isequivalent with ∇ being flat) one may still show that rank S ≤
1. Recently ([24])it was shown that the same constrains apply when dim M = 4. Combining resultsfrom [22] and [24] we have the following theorems: Theorem 1.3 ([22], [24]) . Let f : M → R n +1 ( dim M ≥ ) be a non-degenerateaffine hypersurface with a locally equiaffine transversal vector field ξ and an almostsymplectic form ω . If R p ω = 0 for some p ≥ and the second fundamental form isLorentzian on M (that is has signature (2 n − , ) then the shape operator S hasthe rank ≤ . Theorem 1.4 ([22],[24]) . Let f : M → R n +1 ( dim M ≥ ) be a non-degenerateaffine hypersurface with a locally equiaffine transversal vector field ξ and an almostsymplectic form ω . If ∇ p ω = 0 for some p ≥ and the second fundamental form isLorentzian on M (that is has signature (2 n − , ) then the shape operator S hasthe rank ≤ . The main purpose of the present paper is to prove that the condition that thesecond fundamental form is Lorentzian can be dropped and Theorem 1.3 and The-orem 1.4 stay true for an arbitrary non-degenerate second fundamental form. Themain difficulty of this generalization comes from the fact that due to exponentialgrow of different possible scenarios methods from [22] and [24] cannot be easilyrepeated. For this reasons we need to change approach and first focus more onparticular types of Jordan blocks rather than on all possible configurations. Re-ducing number of allowed Jordan blocks dramatically decrease number of differentconfigurations we need to consider when proving main theorems of this paper.The paper is organized as follows. In Section 2 we briefly recall the basic formulasof affine differential geometry, Jordan decomposition and some basic definitionsfrom symplectic geometry. In this section we also prove a simple but importantlemma about simultaneous decomposition of the shape operator S and the secondfundamental form h . The Section 3 is devoted to real Jordan blocks. The mainresult of this section is that the Jordan decomposition of the shape operator cannotcontain real Jordan blocks of dimension grater than or equal 3 and it contains atmost one block of dimension 2. In section 4 we study complex Jordan blocks. Itis shown that condition R p ω = 0 implies that the Jordan decomposition of the FFINE HYPERSURFACES OF ARBITRARY SIGNATURE... 3 shape operator cannot contain complex Jordan blocks. Sections 5 contains themain results of this paper. Basing on results from sections 3 and 4, we show thatif there exists an almost symplectic structure ω satisfying condition R p ω = 0 or ∇ p ω = 0 for some positive integer p then the rank of the shape operator must be ≤
1. We conclude the section with some general example.2.
Preliminaries
We briefly recall the basic formulas of affine differential geometry. For moredetails, we refer to [20]. Let f : M → R n +1 be an orientable connected differentiable n -dimensional hypersurface immersed in the affine space R n +1 equipped with itsusual flat connection D. Then for any transversal vector field ξ we have(1) D X f ∗ Y = f ∗ ( ∇ X Y ) + h ( X, Y ) ξ and(2) D X ξ = − f ∗ ( SX ) + τ ( X ) ξ, where X, Y are vector fields tangent to M . It is known that ∇ is a torsion-freeconnection, h is a symmetric bilinear form on M , called the second fundamentalform , S is a tensor of type (1 , the shape operator , and τ is a 1-form, called the transversal connection form . The vector field ξ is called equiaffine if τ = 0.When dτ = 0 the vector field ξ is called locally equiaffine .When h is non-degenerate then h defines a pseudo-Riemannian metric on M .In this case we say that the hypersurface or the hypersurface immersion is non-degenerate . In this paper we always assume that f is non-degenerate. We have thefollowing Theorem 2.1 ([20], Fundamental equations) . For an arbitrary transversal vectorfield ξ the induced connection ∇ , the second fundamental form h , the shape operator S , and the 1-form τ satisfy the following equations: R ( X, Y ) Z = h ( Y, Z ) SX − h ( X, Z ) SY, (3) ( ∇ X h )( Y, Z ) + τ ( X ) h ( Y, Z ) = ( ∇ Y h )( X, Z ) + τ ( Y ) h ( X, Z ) , (4) ( ∇ X S )( Y ) − τ ( X ) SY = ( ∇ Y S )( X ) − τ ( Y ) SX, (5) h ( X, SY ) − h ( SX, Y ) = 2 dτ ( X, Y ) . (6)The equations (3), (4), (5), and (6) are called the equations of Gauss, Codazzifor h , Codazzi for S and Ricci, respectively.Let ω be a non-degenerate 2-form on manifold M . The form ω we call an almostsymplectic structure . It is easy to see that if a manifold M admits some almostsymplectic structure then M is orientable manifold of even dimension. Structure ω is called a symplectic structure , if it is almost symplectic and additionally satisfies dω = 0. Pair ( M, ω ) we call a (an almost) symplectic manifold , if ω is a (an almost)symplectic structure on M .Recall ([1]) that an affine connection ∇ on an almost symplectic manifold ( M, ω )we call an almost symplectic connection if ∇ ω = 0. An affine connection ∇ on analmost symplectic manifold ( M, ω ) we call a symplectic connection if it is almostsymplectic and torsion-free.Now we recall a well-know theorem about Jordan normal form (See eg. Th.A.2.6 in [13]).
M. SZANCER AND Z. SZANCER
Theorem 2.2 ([13], Jordan) . If A : V → V is an endomorphism of real finitedimensional vector space V then there exists a basis of V such that the matrix ofthe endomorphism A in this basis has a form (7) L . . . L . . . ... ... . . . . . . . . . L s , where L i is the Jordan block corresponding to the eigenvalue λ i and given by theformula (8) λ i . . . λ i . . . λ i . . . ... ... ... . . . . . . . . . . . . λ i
00 0 0 . . . λ i ∈ M ( k i , k i , R ) , when λ i is a real number, or by the formula (9) B i . . . I B i . . . I B i . . . ... ... ... . . . . . . . . . . . . B i
00 0 0 . . . I B i ∈ M (2 k i , k i , R ) , where B i = (cid:20) α i β i − β i α i (cid:21) , I = (cid:20) (cid:21) ∈ M (2 , , R ) , when λ i = α i + iβ i ( β i = 0 ) is a complex number. A square matrix P of dimension n is called the sip matrix (the standard involutorypermutation) ([13]) if it has a form:(10) · · · · · · · · · · · · · · · · · · . Note that P is non-singular symmetric matrix and P = I . In particular all itseigenvalues are equal ±
1. Moreover, it is easy to verify that we have the followingformula for signature of P :(11) sig P = ( ( n , n ) , if n is even( n +12 , n − ) , if n is odd. Theorem 2.3 (Th. 6.1.5 [13]) . Let H be a real invertible and symmetric matrixof dimension n . Then for every square n dimensional and H -selfadjoint matrix A (i.e. A T H = HA ) there exists a basis { e , . . . , e n } such that (12) A = J ⊕ · · · ⊕ J t ⊕ J t +1 ⊕ · · · ⊕ J t + s , FFINE HYPERSURFACES OF ARBITRARY SIGNATURE... 5 where J , . . . , J t are Jordan blocks of type (8) and J t +1 , . . . , J t + s are Jordan blocksof type (9) . Moreover (13) H = ε P ⊕ · · · ⊕ ε t P t ⊕ P t +1 ⊕ · · · ⊕ P t + s , where P j is a sip matrix of dimension equal to dimension of matrix J j for j =1 , . . . , t + s and ε j = ± for j = 1 , . . . , t . The signs ε j are determined uniquely by ( A, H ) up to permutation of signs in the blocks of (13) corresponding to the Jordanblocks of A with the same real eigenvalue and the same size. For a tensor field T of type (0 , p ) its covariant derivation ∇ T is a tensor field oftype (0 , p + 1) given by the formula:( ∇ T )( X , X , . . . , X p +1 ) := X ( T ( X , . . . , X p +1 )) − p +1 X i =2 T ( X , . . . , ∇ X X i , . . . , X p +1 ) . Higher order covariant derivatives of T can be defined by recursion:( ∇ k +1 T ) = ∇ ( ∇ k T ) . To simplify computation it is often convenient to define ∇ T := T .If R is a curvature tensor for an affine connection ∇ , one can define a new tensor R · T of type (0 , p + 2) by the formula( R · T )( X , X , . . . , X p +2 ) := − p +2 X i =3 T ( X , . . . , R ( X , X ) X i , . . . , X p +2 ) . Analogously to the previous case, we may define a tensor R k · T of type (0 , k + p )using the following recursive formula: R k · T = R · ( R k − · T )and additionally R · T := T .In order to simplify the notation, we will be often omitting ” · ” in R k · T whenno confusion arises. Thus we will be writing often R k T instead of R k · T .We conclude this section with the following lemma: Lemma 2.4.
Let f : M → R n +1 be a non-degenerate affine hypersurface witha locally equiaffine transversal vector field ξ . Then for every point x ∈ M thereexists a basis e , . . . , e n of T x M such that the shape operator S and the secondfundamental form h can be expressed in this basis in the block matrix form (14) S = S . . . S . . . ... ... . . . . . . . . . S q + r , h = H . . . H . . . ... ... . . . . . . . . . H q + r , and S i , H i satisfy the following conditions: • For i = 1 , . . . , q + r dim S i = dim H i . • For i = 1 , . . . , q S i is a Jordan block of type (8) and dim S i ≥ dim S i +1 for i = 1 , . . . , q − . • For i = q +1 , . . . , q + r S i is a Jordan block of type (9) and dim S i ≥ dim S i +1 for i = q + 1 , . . . , q + r − . • For i = 1 , . . . , q H i is up to a sign sip matrix. M. SZANCER AND Z. SZANCER • For i = q + 1 , . . . , q + r H i is a sip matrix.Proof. Since ξ is locally equiaffine we have dτ = 0 and in consequence h ( SX, Y ) = h ( X, SY ) for all
X, Y ∈ T x M . Now thesis immediately follows from Theorem 2.3and the fact that we can rearrange Jordan blocks S i and matrices H i in desiredorder (rearranging vectors e , . . . , e n if needed). (cid:3) Real Jordan blocks
In this chapter we study properties of real Jordan blocks of the shape operator S . In all the below lemmas we assume that f : M → R n +1 is a non-degenerateaffine hypersurface with a locally equiaffine transversal vector field ξ and an almostsymplectic form ω . About objects ∇ , h , S and τ we assume that they are inducedby ξ .In the following lemmas (if not stated otherwise) we assume that S (from Lemma2.4) is a k -dimensional block of the form S = α . . . α . . . α . . . . . . . . . . . . α
00 0 0 . . . α ∈ M ( k, k, R ) , (15)where α ∈ R and H = · · · ε · · · ε · · · · · · ε · · · ε · · · ∈ M ( k, k, R ) , where ε ∈ {− , } . By { e , . . . , e n } we will be denoting basis of T x M such that { e , . . . , e k } is a basis for S . Lemma 3.1.
Let p ≥ . If k ≥ and X , . . . , X p ∈ span { e , . . . , e k } =: V then forevery i ∈ { , , . . . , n } R p ω ( X , e k , X , e k , . . . , X p , e k | {z } p , e i , e k ) = π ( X ) · . . . π ( X p ) ε p α p ω ( e i , e k ) , (16) where π : V → R is a projection defined as follows: π ( X ) = λ if X = λ e + . . . + λ k e k ∈ V for some λ , . . . , λ k ∈ R .Proof. First we shall prove the following formulas: R ( X, e k ) e k = − π ( X ) εαe k , (17) R ( X, e k ) e i = − h ( X, e i ) αe k , (18) π ( R ( X, e k ) Y ) = εαπ ( X ) π ( Y ) , (19) FFINE HYPERSURFACES OF ARBITRARY SIGNATURE... 7 for all
X, Y ∈ V . To prove (17) we compute R ( X, e k ) e k = h ( e k , e k ) | {z } S X − h ( X, e k ) S e k = − h ( α e + . . . + α k e k , e k ) αe k = − α h ( e , e k ) | {z } ε αe k = − π ( X ) εαe k . To prove (18) we compute R ( X, e k ) e i = h ( e k , e i ) | {z } S X − h ( X, e i ) S e k = − h ( X, e i ) αe k . To prove (19) we compute π ( R ( X, e k ) Y ) = π ( h ( e k , Y ) S X − h ( X, Y ) S e k ) = h ( e k , Y ) π ( S X ) − h ( X, Y ) π ( S e k ) | {z } = επ ( Y ) π ( X ) α. Now using formulas (17), (18), (19) we shall prove the thesis of the lemma. For p = 1 we have( Rω )( X , e k , e i , e k ) = − ω ( R ( X , e k ) e i , e k ) − ω ( e i , R ( X , e k ) e k )= ω ( h ( X , e i ) αe k , e k ) | {z } + ω ( e i , π ( X ) εαe k )= π ( X ) εαω ( e i , e k ) . M. SZANCER AND Z. SZANCER
Assume that formula (16) is true for some p ≥ p + 1 we get R p +1 ω ( X , e k , . . . , X p +1 , e k | {z } p +2 , e i , e k )= − R p ω ( R ( X , e k ) X , e k , . . . ) − R p ω ( X , R ( X , e k ) e k , . . . ) − R p ω ( X , e k , R ( X , e k ) X , . . . ) − R p ω ( X , e k , X , R ( X , e k ) e k , . . . ) · · ·− R p ω ( X , e k . . . , R ( X , e k ) e i , e k ) − R p ω ( X , e k , . . . , e i , R ( X , e k ) e k )= − π (( R ( X , e k ) X ) π ( X ) . . . π ( X p +1 ) ε p α p ω ( e i , e k )+ R p ω ( X , π ( X ) εαe k , X , . . . ) − π ( X ) π (( R ( X , e k ) X ) π ( X ) . . . π ( X p +1 ) ε p α p ω ( e i , e k )+ R p ω ( X , e k , X , π ( X ) εαe k , . . . ) · · ·− π ( X ) . . . π ( X p +1 ) π ( R ( X , e k ) e i ) | {z } ε p α p ω ( e i , e k )+ R p ω ( X , e k , . . . , e i , π ( X ) εαe k )= − εαπ ( X ) π ( X ) π ( X ) . . . π ( X p +1 ) ε p α p ω ( e i , e k )+ εαπ ( X ) π ( X ) π ( X ) . . . π ( X p +1 ) ε p α p ω ( e i , e k ) − εαπ ( X ) π ( X ) π ( X ) . . . π ( X p +1 ) ε p α p ω ( e i , e k )+ εαπ ( X ) π ( X ) π ( X ) . . . π ( X p +1 ) ε p α p ω ( e i , e k ) · · ·− εαπ ( X ) π ( X ) . . . π ( X p +1 ) ε p α p ω ( e i , e k )= ε p +1 α p +1 π ( X ) π ( X ) . . . π ( X p +1 ) ω ( e i , e k ) . Now, by the induction principle the formula (16) holds for every p ≥ (cid:3) As an immediate consequence of Lemma 3.1 (setting X = X = . . . = X p = e )we obtain: Corollary 3.2. If k ≥ then for every i ∈ { , . . . , n } we have R p ω ( e , e k , . . . , e , e k , e i , e k ) = ε p α p ω ( e i , e k ) . In the next few lemmas we shall obtain some properties of R p ω under the as-sumption that k > FFINE HYPERSURFACES OF ARBITRARY SIGNATURE... 9
Lemma 3.3. If k > then R ( e , e k − ) e = R ( e , e k − ) e k − = 0(20) R ( e , e k − ) e = εS e = εαe + εe (21) R ( e , e k − ) e k = − εS e k − = − εαe k − − εe k (22) R ( e k − , e k ) e = εS e k − = εαe k − + εe k (23) R ( e k − , e k ) e = − εS e k = − εαe k (24) R ( e k − , e k ) e k − = R ( e k − , e k ) e k = 0(25) Proof.
The proof is an immediate consequence of the Gauss equation and the factthat h ( e , e k ) = h ( e , e k − ) = ε and h ( e , e ) = h ( e , e ) = h ( e , e k − ) = h ( e k − , e k − )= h ( e k , e k − ) = h ( e k , e k ) = h ( e , e k ) = 0 , if only k > (cid:3) Lemma 3.4. If k > and p ≥ we have R p ω ( e , e k − , e , e k − , . . . , e , e k − ) = 0 , (26) R p ω ( e , e k − , . . . , e , e k − , e , e k − ) = ( − p ε p ( αω ( e , e k − ) + ω ( e , e k − )) , (27) R p ω ( e , e k − , . . . , e , e k − , e i , e k ) = ε p ( αω ( e i , e k − ) + ω ( e i , e k ))(28) for i ∈ { , . . . , n } \ { , k } .Proof. First note that the formula (26) easily follows from (20) for every p ≥ p = 1 we have Rω ( e , e k − , e , e k − ) = − ω ( R ( e , e k − ) e , e k − )) − ω ( e , R ( e , e k − ) e k − )= − ω ( εαe + εe , e k − ) = − εαω ( e , e k − ) − εω ( e , e k − )thanks to (20) and (21). Now assume that (27) is true for some p ≥
1. Using (20),(21) and (26) we compute R p +1 ω ( e , e k − , e , e k − , . . . , e , e k − | {z } p +2 , e , e k − )= ( R ( e , e k − ) · R p ω )( e , e k − , . . . , e , e k − | {z } p , e , e k − )= − R p ω ( e , e k − , . . . , e , e k − , R ( e , e k − ) e , e k − )= − R p ω ( e , e k − , . . . , e , e k − , εS e , e k − )= − εR p ω ( e , e k − , . . . , e , e k − , αe + e , e k − )= − εα R p ω ( e , e k − , . . . , e , e k − , e , e k − ) | {z } − εR p ω ( e , e k − , . . . , e , e k − , e , e k − )= − ε ( − p ε p ( αω ( e , e k − ) + ω ( e , e k − ))= ( − p +1 ε p +1 ( αω ( e , e k − ) + ω ( e , e k − )) . Thus, by the induction principle (27) is true for all p ≥ e i ⊥{ e , e k − } we have R ( e , e k − ) e i = h ( e k − , e i ) S e − h ( e , e i ) S e k − = 0 . (29)In particular the above holds for all i ∈ { , . . . , n } \ { , k } . Using (22) and (29)we get that for every i ∈ { , . . . , n } \ { , k } Rω ( e , e k − , e i , e k ) = − ω ( R ( e , e k − ) e i | {z } , e k ) − ω ( e i , R ( e , e k − ) e k )= ε ( αω ( e i , e k − ) + ω ( e i , e k ))thus (28) is true for p = 1. Now assume that (28) holds for some p ≥ i ∈ { , . . . , n } \ { , k } . We have R p +1 ω ( e , e k − , e , e k − , . . . , e , e k − | {z } p +2 , e i , e k )= ( R ( e , e k − ) · R p ω )( e , e k − , . . . , e , e k − | {z } p , e i , e k )= − R p ω ( e , e k − , . . . , e , e k − , e i , R ( e , e k − ) e k )= εαR p ω ( e , e k − , . . . , e , e k − , e i , e k − )+ εR p ω ( e , e k − , . . . , e , e k − , e i , e k )= εR p ω ( e , e k − , . . . , e , e k − , e i , e k )= ε p +1 ( αω ( e i , e k − ) + ω ( e i , e k ))since R ( e , e k − ) e = R ( e , e k − ) e k − = R ( e , e k − ) e i = 0by (20) and (29). Now, by the induction principle (28) holds for all p ≥ (cid:3) Lemma 3.5. If k > and p ≥ we have R p +1 ω ( e , e k − , . . . , e , e k − | {z } p +2 , e , e k ) = − εα ( ω ( e , e k ) − ω ( e , e k − )) , (30) R p +2 ω ( e , e k − , . . . , e , e k − | {z } p +4 , e , e k )(31) = − α (2 αω ( e , e k − ) + ω ( e , e k − ) + ω ( e , e k )) . Proof.
First note that by straightforward computations, using (20), (21) and (22),one may easily check that (30) and (31) are true for p = 0. FFINE HYPERSURFACES OF ARBITRARY SIGNATURE... 11
Let us assume that p >
0. In order to prove (30), using Lemma 3.3, we compute R p +1 ω ( e , e k − , . . . , e , e k − | {z } p +2 , e , e k )= ( R ( e , e k − ) · R p ω )( e , e k − , . . . , e , e k − | {z } p , e , e k )= − R p ω ( e , e k − , . . . , e , e k − , R ( e , e k − ) e , e k ) − R p ω ( e , e k − , . . . , e , e k − , e , R ( e , e k − ) e k )= − εR p ω ( e , e k − , . . . , e , e k − , αe + e , e k ) − εR p ω ( e , e k − , . . . , e , e k − , e , − αe k − − e k )= − εαR p ω ( e , e k − , . . . , e , e k − , e , e k ) − εR p ω ( e , e k − , . . . , e , e k − , e , e k )+ εαR p ω ( e , e k − , . . . , e , e k − , e , e k − )+ εR p ω ( e , e k − , . . . , e , e k − , e , e k )= − εαR p ω ( e , e k − , . . . , e , e k − , e , e k )+ εαR p ω ( e , e k − , . . . , e , e k − , e , e k − ) . Now using (28) (for i = 1) and (27) we obtain R p +1 ω ( e , e k − , . . . , e , e k − | {z } p +2 , e , e k )= − εαR p ω ( e , e k − , . . . , e , e k − , e , e k )+ εαR p ω ( e , e k − , . . . , e , e k − , e , e k − )= − εαε p ( αω ( e , e k − ) + ω ( e , e k )) + εα · ( − p ε p ( αω ( e , e k − ) + ω ( e , e k − ))= − εα ( ω ( e , e k ) − ω ( e , e k − ))that is (30) is true for all p ≥ R p +2 ω ( e , e k − , . . . , e , e k − | {z } p +4 , e , e k )= − R p +1 ω ( R ( e , e k − ) e , e k − , . . . , e , e k − , e , e k ) − R p +1 ω ( e , R ( e , e k − ) e k − , . . . , e , e k − , e , e k ) · · ·− R p +1 ω ( e , e k − , . . . , e , e k − , R ( e , e k − ) e , e k ) − R p +1 ω ( e , e k − , . . . , e , e k − , e , R ( e , e k − ) e k )= − R p +1 ω ( e , e k − , . . . , e , e k − , R ( e , e k − ) e , e k ) − R p +1 ω ( e , e k − , . . . , e , e k − , e , R ( e , e k − ) e k ) where the last equality follows from (20). Now using (21) and (22) we get R p +2 ω ( e , e k − , . . . , e , e k − | {z } p +4 , e , e k )= − εαR p +1 ω ( e , e k − , . . . , e , e k − , e , e k ) − εR p +1 ω ( e , e k − , . . . , e , e k − , e , e k )+ εαR p +1 ω ( e , e k − , . . . , e , e k − , e , e k − )+ εR p +1 ω ( e , e k − , . . . , e , e k − , e , e k )= − εαR p +1 ω ( e , e k − , . . . , e , e k − , e , e k )+ εαR p +1 ω ( e , e k − , . . . , e , e k − , e , e k − ) . Again using (28) (for i = 1) and (27) we obtain R p +2 ω ( e , e k − , . . . , e , e k − | {z } p +4 , e , e k )= − εαε p +1 ( αω ( e , e k − ) + ω ( e , e k ))+ εα · ( − p +1 ε p +1 ( αω ( e , e k − ) + ω ( e , e k − ))= − α (2 αω ( e , e k − ) + ω ( e , e k − ) + ω ( e , e k )) , what completes the proof of (31). (cid:3) Lemma 3.6. If k > and p ≥ we have R p ω ( e k − , e k , e , e k − , . . . , e , e k − | {z } p − , e , e )(32) = ε p α ( ω ( e , e k ) + ω ( e , e k − )) + ε p ω ( e , e k ) Proof.
For p = 1, using (23) and (24), we directly check that Rω ( e k − , e k , e , e ) = εα ( ω ( e , e k ) + ω ( e , e k − )) + εω ( e , e k ) . FFINE HYPERSURFACES OF ARBITRARY SIGNATURE... 13
Now assume that (32) is true for some p ≥
1. First we compute that R p +1 ω ( e k − , e k , e , e k − , . . . , e , e k − | {z } p , e , e )= ( R ( e k − , e k ) · R p ω )( e , e k − , . . . , e , e k − | {z } p , e , e )= − R p ω ( R ( e k − , e k ) e , e k − , . . . , e , e k − , e , e ) − R p ω ( e , R ( e k − , e k ) e k − , . . . , e , e k − , e , e ) − . . . − R p ω ( e , e k − , . . . , R ( e k − , e k ) e , e k − , e , e ) − R p ω ( e , e k − , . . . , e , R ( e k − , e k ) e k − , e , e ) − R p ω ( e , e k − , . . . , e , e k − , R ( e k − , e k ) e , e ) − R p ω ( e , e k − , . . . , e , e k − , e , R ( e k − , e k ) e )= − εR p ω ( e k , e k − , . . . , e , e k − , e , e ) − εR p ω ( e , e k − , e k , e k − , . . . , e , e k − , e , e ) − . . . − εR p ω ( e , e k − , . . . , e k , e k − , e , e ) − R p ω ( e , e k − , . . . , e , e k − , εαe k − + εe k , e ) − R p ω ( e , e k − , . . . , e , e k − , e , − εαe k ) . Before we proceed we shall show that for p ≥ R p ω ( (1) z }| { e , e k − , . . . , ( i ) z }| { e k , e k − , . . . , ( p ) z }| { e , e k − , e , e ) = 0if only i ∈ { , . . . , p } . Indeed we have R p ω ( (1) z }| { e , e k − , . . . , ( i ) z }| { e k , e k − , . . . , ( p ) z }| { e , e k − , e , e )= − R p − ω ( R ( e , e k − ) e , e k − , . . . ) − R p − ω ( e , R ( e , e k − ) e k − , . . . ) − · · ·− R p − ω ( . . . , ( i − z }| { R ( e , e k − ) e k , e k − , . . . ) − R p − ω ( . . . , ( i − z }| { e k , R ( e , e k − ) e k − , . . . ) − · · ·− R p − ω ( . . . , R ( e , e k − ) e , e ) − R p − ω ( . . . , e , R ( e , e k − ) e )= − R p − ω ( . . . , ( i − z }| { R ( e , e k − ) e k , e k − , . . . ) − R p − ω ( . . . , e , R ( e , e k − ) e ) thanks to (20). Now formulas (21) and (22) imply that R p ω ( (1) z }| { e , e k − , . . . , ( i ) z }| { e k , e k − , . . . , ( p ) z }| { e , e k − , e , e ) = 0 . (33)Now applying (33) if needed (that is if p >
1) we conclude that R p +1 ω ( e k − , e k , e , e k − , . . . , e , e k − | {z } p , e , e )= εR p ω ( e k − , e k , e , e k − , . . . , e , e k − , e , e ) − εαR p ω ( e , e k − , . . . , e , e k − , e k − , e ) − εR p ω ( e , e k − , . . . , e , e k − , e k , e )+ εαR p ω ( e , e k − , . . . , e , e k − , e , e k )= εR p ω ( e k − , e k , e , e k − , . . . , e , e k − , e , e )+ εαR p ω ( e , e k − , . . . , e , e k − , e , e k − )+ εR p ω ( e , e k − , . . . , e , e k − , e , e k )+ εαR p ω ( e , e k − , . . . , e , e k − , e , e k ) . By the induction principle, using formula (27) and formula (28) (for i = 1) weobtain R p +1 ω ( e k − , e k , e , e k − , . . . , e , e k − | {z } p , e , e )= ε (cid:16) ε p α ( ω ( e , e k ) + ω ( e , e k − )) + ε p ω ( e , e k ) (cid:17) + εα (cid:16) ( − p ε p ( αω ( e , e k − ) + ω ( e , e k − )) (cid:17) + εR p ω ( e , e k − , . . . , e , e k − , e , e k )+ εα (cid:16) ε p ( αω ( e , e k − ) + ω ( e , e k )) (cid:17) = ε p +1 (cid:16) αω ( e , e k ) + ( α + ( − p α ) ω ( e , e k − )+ ( α + ( − p α ) ω ( e , e k − ) + ω ( e , e k ) (cid:17) + εR p ω ( e , e k − , . . . , e , e k − , e , e k ) . Finally we need to consider two cases. If p is odd, using Lemma 3.5, we obtain R p +1 ω ( e k − , e k , e , e k − , . . . , e , e k − | {z } p , e , e )= ε p +1 (cid:16) αω ( e , e k ) + ( α + ( − p α ) ω ( e , e k − )+ ( α + ( − p α ) ω ( e , e k − ) + ω ( e , e k ) (cid:17) + ε ( − εα ( ω ( e , e k ) − ω ( e , e k − )))= α ( ω ( e , e k ) + ω ( e , e k − )) + ω ( e , e k ) . FFINE HYPERSURFACES OF ARBITRARY SIGNATURE... 15 If p is even, again from Lemma 3.5 we get R p +1 ω ( e k − , e k , e , e k − , . . . , e , e k − | {z } p , e , e )= ε p +1 (cid:16) αω ( e , e k ) + ( α + ( − p α ) ω ( e , e k − )+ ( α + ( − p α ) ω ( e , e k − ) + ω ( e , e k ) (cid:17) + ε ( − α (2 αω ( e , e k − ) + ω ( e , e k − ) + ω ( e , e k )))= εα ( ω ( e , e k ) + ω ( e , e k − )) + εω ( e , e k ) . The proof is completed. (cid:3)
Lemma 3.7. If k > then for every p ≥ we have R p ω ( e , e k − , e , e k − , . . . , e , e k − , e , e ) = 0(34) Proof. If p = 1 we have Rω ( e , e k − , e , e ) = − ω ( R ( e , e k − ) e , e ) − ω ( e , R ( e , e k − ) e ) . For p > R p ω ( e , e k − , e , e k − , . . . , e , e k − , e , e )= − R p − ω ( R ( e , e k − ) e , e k − , . . . , e , e k − , e , e ) − R p − ω ( e , R ( e , e k − ) e k − , . . . , e , e k − , e , e ) − . . . − R p − ω ( e , e k − , . . . , e , R ( e , e k − ) e k − , e , e ) − R p − ω ( e , e k − , . . . , e , e k − , R ( e , e k − ) e , e ) − R p − ω ( e , e k − , . . . , e , e k − , e , R ( e , e k − ) e ) . Now the thesis follows immediately from (20) and the fact that for k > R ( e , e k − ) e = 0. (cid:3) Lemma 3.8.
Let f : M → R n +1 ( dim M ≥ ) be a non-degenerate affine hyper-surface with a locally equiaffine transversal vector field ξ and an almost symplecticform ω . If R p ω = 0 for some p ≥ and S = S . . . S . . . ... ... . . . . . . . . . S q + r (35) is the Jordan decomposition of S as stated in the Lemma 2.4 then dim S ≤ .Proof. Let us assume that S has the form (15) and k >
3. If α = 0, from Corollary3.2 we obtain ω ( e , e k ) = . . . = ω ( e n , e k ) = 0 . From Lemma 3.5, formula (30) we have ω ( e , e k ) = ω ( e , e k − ) . From Lemma 3.6, formula (32) we get ω ( e , e k ) = − ω ( e , e k − ) , since ω ( e , e k ) = 0. Therefore ω ( e i , e k ) = 0 for i ∈ { , . . . , n } what contradictsassumption that ω is non-degenerate. Thus it must be α = 0.When α = 0 from Lemma 3.4, formula (28) we get ω ( e i , e k ) = 0 for i ∈{ , , . . . , n } \ { , k } . Of course ω ( e k , e k ) = 0 as well. From Lemma 3.6, for-mula (32) we have that also ω ( e , e k ) = 0, so again we obtain that ω is degeneratethus dim S cannot exceed 3. (cid:3) Lemma 3.9. If S is a 3-dimensional real Jordan block, p ≥ then R p ω ( e , e , e , e , . . . , e , e ) = ( − p ε p p ! ω ( e , e )(36) Proof.
From the Gauss equation we have R ( e , e ) e = 0 , R ( e , e ) e = εS e = εαe + εe . (37)Now, for p = 1 we easily check that Rω ( e , e , e , e ) = − εω ( e , e ) . Let us assume that (36) is true for some p ≥
1. Using (37) we compute R p +1 ω ( e , e , e , e , . . . , e , e )= ( R ( e , e ) · R p ω )( e , e , . . . , e , e | {z } p , e , e )= − R p ω ( e , εS e , . . . , e , e , e , e ) − R p ω ( e , e , e , εS e , . . . , e , e ) . . . − R p ω ( e , e , e , e , . . . , e , εS e , e , e ) − R p ω ( e , e , e , e , . . . , e , e , e , εS e )= − ε ( p + 1) R p ω ( e , e , e , e , . . . , e , e )= − ε ( p + 1) · ( − p ε p p ! ω ( e , e ) = ( − p +1 ε p +1 ( p + 1)! ω ( e , e )Now by the induction principle (36) holds for all p ≥ (cid:3) Lemma 3.10. If S is a 3-dimensional real Jordan block, p ≥ and i, j > then R p ω ( e , e , e , e , . . . , e , e , e , e i , e , e j )(38) = ( − p ε p − ( p − h ( e i , e j )(2 αω ( e , e ) + ω ( e , e )) Proof.
Using the Gauss equetion let us note that R ω ( e , e , e , e i , e , e j ) = εh ( e i , e j )(2 αω ( e , e ) + ω ( e , e )) , (39) R ω ( e , e , e , e i , e , e j ) = 0(40)Now, for p ≥ R p +1 ω ( e , e , . . . , e , e , e , e i , e , e j ) = − ε ( p − R p ω ( e , e , . . . , e , e , e , e i , e , e j )since R ( e , e ) e = R ( e , e ) e i = R ( e , e ) e j = 0 and R ( e , e ) e = εS e = εαe + εe . In consequence, by the induction principle R p ω ( e , e , . . . , e , e , e , e i , e , e j ) = 0(41) FFINE HYPERSURFACES OF ARBITRARY SIGNATURE... 17 for all p ≥
2. We also compute R p +1 ω ( e , e , . . . , e , e , e , e i , e , e j ) = − ε ( p − R p ω ( e , e , . . . , e , e , e , e i , e , e j ) − R p ω ( e , e , . . . , e , e , R ( e , e ) e , e i , e , e j )= − εpR p ω ( e , e , . . . , e , e , e , e i , e , e j ) − εαR p ω ( e , e , . . . , e , e , e , e i , e , e j )= − εpR p ω ( e , e , . . . , e , e , e , e i , e , e j ) , where the last equality follows from (41). The induction principle implies that (38)holds for all p ≥ (cid:3) Lemma 3.11. If S is a 3-dimensional real Jordan block, p ≥ and α = 0 then R p ω ( e , e , e , e , . . . , e , e , e , e i ) = ( − p ε p p ! ω ( e , e i )(42) for any i ∈ { , . . . , n } \ { } .Proof. First notice that (42) holds trivially for i = 2. Now we assume that i ∈{ , . . . , n } \ { , } . Note that from the Gauss equation and assumption α = 0 wehave R ( e , e ) e = R ( e , e ) e i = 0 , R ( e , e ) e = εS e = εe . (43)For p = 1 we easily get Rω ( e , e , e , e i ) = − εω ( e , e i ) . Let us assume that (42) holds for some p ≥
1, we shall show that it also holds for p + 1. Indeed we obtain R p +1 ω ( e , e , . . . , e , e | {z } p +2 , e , e i )= ( R ( e , e ) · R p ω )( e , e , . . . , e , e | {z } p , e , e i )= − ε ( p + 1) R p ω ( e , e , . . . , e , e , e , e i )= − ε ( p + 1) · ( − p ε p p ! ω ( e , e i ) = ( − p +1 ε p +1 ( p + 1)! ω ( e , e i )thanks to (43). Now by the induction principle (42) holds for all p ≥ (cid:3) Lemma 3.12. If S is a 3-dimensional real Jordan block, p ≥ and α = 0 then R p ω ( e , e , . . . , e , e , e , e , e , e ) = ( − p +1 ε p ( p − ω ( e , e ) . (44) Proof.
First we notice that (44) is satisfied for p = 1. Indeed, since R ( e , e ) e = εS e = εe , R ( e , e ) e = − εS e = 0we easily obtain Rω ( e , e , e , e ) = − ω ( R ( e , e ) e , e ) − ω ( e , R ( e , e ) e ) = εω ( e , e ) . Now assume that (44) holds for some p ≥
1, we shall show that it also holds for p + 1. From the Gauss equation and our assumptions we have R ( e , e ) e = 0 , R ( e , e ) e = εS e = εe , R ( e , e ) e = − εS e = − εe . Now we obtain R p +1 ω ( e , e , . . . , e , e | {z } p , e , e , e , e )= ( R ( e , e ) · R p ω )( e , e , . . . , e , e | {z } p − , e , e , e , e )= − ε ( p + 1) R p ω ( e , e , . . . , e , e , e , e , e , e ) − R p ω ( e , e , . . . , e , e , e , R ( e , e ) e , e , e )= − εpR p ω ( e , e , . . . , e , e , e , e , e , e )= − εp · ( − p +1 ε p ( p − ω ( e , e ) = ( − p +2 ε p +1 p ! ω ( e , e ) . Now by the induction principle (44) holds for all p ≥ (cid:3) Lemma 3.13.
Let us assume that S , S (from Lemma 2.4) are 2-dimensional realJordan blocks. That is S = (cid:20) α α (cid:21) , S = (cid:20) β β (cid:21) , where α, β ∈ R . We also assume that H , H have the form H = (cid:20) εε (cid:21) , H = (cid:20) ηη (cid:21) , where ε, η ∈ {− , } .Then for every i ∈ { , . . . , n } \ { , } we have R p ω ( e , e , . . . , e , e , e i , e ) = 0 for p ≥ , (45) R p +1 ω ( e , e , . . . , e , e , e i , e )(46) = ( − p +1 η p +1 ε p ( αω ( e i , e ) + ω ( e i , e )) for p ≥ . Proof.
First note that from the Gauss equation we easily obtain that R ( e , e ) e i = 0for i ∈ { , . . . , n } \ { , } . In particular R ( e , e ) e = R ( e , e ) e = 0 . (47)Thus (45) follows immediately.In order to prove (46) first notice that (46) is satisfied for p = 0, since R ( e , e ) e i =0 and R ( e , e ) e = ηS e = ηαe + ηe . (48)Now assume that (46) holds for some p ≥
0, we shall show that it also holds for p + 1. From the Gauss equation we also have R ( e , e ) e = − εS e = − εβe − εe . (49) FFINE HYPERSURFACES OF ARBITRARY SIGNATURE... 19
Now using (47)–(49) we obtain R p +3 ω ( e , e , . . . , e , e , e i , e )= − ηα R p +2 ω ( e , e , . . . , e , e , e i , e ) | {z } − ηR p +2 ω ( e , e , . . . , e , e , e i , e )= ηR p +1 ω ( e , e , . . . , e , e , e i , R ( e , e ) e )= − ηεβR p +1 ω ( e , e , . . . , e , e , e i , e ) − ηεR p +1 ω ( e , e , . . . , e , e , e i , e )= − ηεR p +1 ω ( e , e , . . . , e , e , e i , e ) , where the last equality follows from (45). Finally we have R p +3 ω ( e , e , . . . , e , e , e i , e )= − ηεR p +1 ω ( e , e , . . . , e , e , e i , e )= − ηε ( − p +1 η p +1 ε p ( αω ( e i , e ) + ω ( e i , e ))= ( − p +2 η p +2 ε p +1 ( αω ( e i , e ) + ω ( e i , e )) . By the induction principle (46) holds for all p ≥ (cid:3) Lemma 3.14.
Let S , S be 2-dimensional real Jordan blocks like in Lemma 3.13.If α = β = 0 then R p ω ( e , e , . . . , e , e , e , e , e , e ) = ( − p ( εη ) p − · p − ω ( e , e ) , (50) R p ω ( e , e , . . . , e , e , e , e , e , e ) = ( − p +1 ( εη ) p · p − ω ( e , e )(51) for every p ≥ .Proof. First note that formulas (47), (48) and (49) from the proof of Lemma 3.13are still valid in our case. Using them and taking into account that α = β = 0 weeasily compute that R ω ( e , e , e , e , e , e ) = − ε ω ( e , e ) = − ω ( e , e )and R ω ( e , e , e , e , e , e ) = εηω ( e , e ) . That is (50) and (51) are true for p = 1. Now assume that (50) and (51) hold forsome p ≥
1. Again using (47)–(49) and the fact that α = β = 0 we obtain R p +2 ω ( e , e , . . . , e , e , e , e , e , e )= − R p +1 ω ( e , e , . . . , e , e , e , R ( e , e ) e , e , e ) − R p +1 ω ( e , e , . . . , e , e , e , e , e , R ( e , e ) e )= εR p +1 ω ( e , e , . . . , e , e , e , e , e , e )+ εR p +1 ω ( e , e , . . . , e , e , e , e , e , e )= − εR p ω ( e , e , . . . , e , e , e , R ( e , e ) e , e , e ) − εR p ω ( e , e , . . . , e , e , e , e , e , R ( e , e ) e ) − εR p ω ( e , e , . . . , e , e , e , R ( e , e ) e , e , e ) − εR p ω ( e , e , . . . , e , e , e , e , e , R ( e , e ) e )= − εηR p ω ( e , e , . . . , e , e , e , e , e , e )+ 2 ε R p ω ( e , e , . . . , e , e , e , e , e , e )= 2 ε R p ω ( e , e , . . . , e , e , e , e , e , e ) − εηR p ω ( e , e , . . . , e , e , e , e , e , e )= 2( ε ( − p +1 ( εη ) p · p − ω ( e , e ) − εη ( − p ( εη ) p − · p − ω ( e , e ))= ( − p +1 ( εη ) p · p ω ( e , e ) . In a similar way we show that R p +2 ω ( e , e , . . . , e , e , e , e , e , e )= 2 η R p ω ( e , e , . . . , e , e , e , e , e , e ) − εηR p ω ( e , e , . . . , e , e , e , e , e , e )= 2 η ( − p ( εη ) p − · p − ω ( e , e ) − εη ( − p +1 ( εη ) p · p − ω ( e , e )= ( − p +2 p ( εη ) p +1 ω ( e , e )Now by the induction principle (50) and (51) hold for all p ≥ (cid:3) Theorem 3.15.
Let f : M → R n +1 ( dim M ≥ ) be a non-degenerate affine hyper-surface with a locally equiaffine transversal vector field ξ and an almost symplecticform ω . Let S = S . . . S . . . ... ... . . . . . . . . . S q + r (52) be the Jordan decomposition of S as stated in the Lemma 2.4. If R p ω = 0 for some p ≥ then dim S ≤ and dim S i = 1 for i = 2 , . . . , q .Proof. By Lemma 3.8 dim S ≤
3. If dim S = 3, using Corollary 3.2 we obtain ε p α p ω ( e , e ) = . . . = ε p α p ω ( e n , e ) = 0 . (53) FFINE HYPERSURFACES OF ARBITRARY SIGNATURE... 21
By Lemma 3.9 ω ( e , e ) = 0. Now using Lemma 3.10 we obtain h ( e i , e j ) ω ( e , e ) = 0for i, j >
3. Since dim M ≥ h is non-degenerate there exist i, j > h ( e i , e j ) = 0 and in consequence ω ( e , e ) = 0. If α = 0 then from (53) weget ω ( e , e ) = . . . = ω ( e n , e ) = 0, so ω is degenerate. That is we must have α = 0. Now using Lemma 3.11 and Lemma 3.12 we show that ω ( e , e i ) = 0 for i = 1 , . . . , n that is ω is degenerate again. In consequence the case dim S = 3 isnot possible.Assume now that dim S = 2. If dim S = 2 then S = (cid:20) α α (cid:21) , S = (cid:20) β β (cid:21) , where α, β ∈ R . If R p ω = 0 for some p ≥ R p +1 ω = 0. Now usingLemma 3.13 we get αω ( e i , e ) + ω ( e i , e ) = 0for i ∈ { , . . . , n } \ { , } . In particular, (for i = 1) we get that ω ( e , e ) = 0. Onthe other hand by Corollary 3.2 we have ε p α p ω ( e , e ) = . . . = ε p α p ω ( e n , e ) = 0 . Note that case α = 0 is not possible since then ω ( e i , e ) = 0 for all i ∈ { , . . . , n } and ω is degenerate. Thus we must have α = 0. In this case Lemma 3.13 impliesthat ω ( e i , e ) = 0 for i ∈ { , . . . , n } \ { , } . Since, without loss of generality, wecan exchange S with S we also get that β = 0. Now by Lemma 3.14 we get that ω ( e , e ) = 0 and ω is degenerate, so this case is also not possible. Summarisingwe must have dim S ≤ S i = 1 for i = 2 , . . . , q , what completes theproof. (cid:3) Complex Jordan blocks
In this chapter we study properties of complex Jordan blocks of the shape oper-ator S . Before we proceed, to simplify proofs in this chapter, we need to do slightmodification in the notation of Lemma 2.4.Let { e , . . . , e n } be the basis of T x M from Lemma 2.4. Without loss of generalityrearranging and renaming vectors e , . . . , e n we can change order of S i and H i insuch way that S , . . . , S r will be complex blocks and S r +1 , . . . , S r + q will be realblocks. If we assume that S is a 2 k -dimensional block, k ≥ S = α β . . . − β α . . . α β . . . − β α . . . . . . . . . . . . α β . . . − β α ∈ M (2 k, k, R ) , where α, β ∈ R , β = 0 and H = · · · · · · · · · · · · · · · · · · ∈ M (2 k, k, R ) . Moreover, vectors { e , . . . , e k } will be a basis for S .In all the below lemmas (if not stated otherwise) we always assume that S and H are as above.Let us start with the following three lemmas related to 2-dimensional complexJordan blocks. Lemma 4.1. If S is a 2-dimensional complex Jordan block then for every i ∈{ , . . . , n } we have R ( e , e ) e = Se = αe − βe ,R ( e , e ) e = − Se = − βe − αe ,R ( e , e ) e i = 0 . Proof.
Proof easily follows from the Gauss equation and the fact that h ( e , e i ) = h ( e , e i ) = 0 for i > (cid:3) Lemma 4.2. If S is a 2-dimensional complex Jordan block then for every p ≥ , i ∈ { , . . . , n } we have R p ω ( e , e , e , e , . . . , e , e , e , e i ) = (det S ) p ω ( e , e i )(54) R p ω ( e , e , e , e , . . . , e , e , e , e i ) = (det S ) p ω ( e , e i )(55) Proof.
We shall prove (54). For p = 1, using Lemma 4.1 we compute R ω ( e , e ,e , e , e , e i )= − Rω ( R ( e , e ) e , e , e , e i ) − Rω ( e , R ( e , e ) e , e , e i ) − Rω ( e , e , R ( e , e ) e , e i ) − Rω ( e , e , e , R ( e , e ) e i )= − αRω ( e , e , e , e i ) + αRω ( e , e , e , e i ) − Rω ( e , e , αe − βe , e i )= − αRω ( e , e , e , e i ) + βRω ( e , e , e , e i )= − α ( − ω ( R ( e , e ) e , e i ) − ω ( e , R ( e , e ) e i ))+ β ( − ω ( R ( e , e ) e , e i ) − ω ( e , R ( e , e ) e i ))= α ω ( e , e i ) − αβω ( e , e i ) + β ω ( e , e i ) + βαω ( e , e i )= ( α + β ) ω ( e , e i ) . FFINE HYPERSURFACES OF ARBITRARY SIGNATURE... 23
Assume that (54) holds for some p ≥ p + 1.Indeed, using Lemma 4.1 we have R p +2 ω ( e , e , . . . , e , e , e , e i )= − R p +1 ω ( R ( e , e ) e , e , . . . , e , e , e , e i ) − R p +1 ω ( e , R ( e , e ) e , . . . , e , e , e , e i ) − . . . − R p +1 ω ( e , e , . . . , R ( e , e ) e , e , e , e i ) − R p +1 ω ( e , e , . . . , e , R ( e , e ) e , e , e i ) − R p +1 ω ( e , e , . . . , e , e , R ( e , e ) e , e i ) − R p +1 ω ( e , e , . . . , e , e , e , R ( e , e ) e i )= − αR p +1 ω ( e , e , . . . , e , e , e , e i ) + αR p +1 ω ( e , e , . . . , e , e , e , e i ) − . . . − αR p +1 ω ( e , e , . . . , e , e , e , e i ) + αR p +1 ω ( e , e , . . . , e , e , e , e i ) − R p +1 ω ( e , e , . . . , e , e , αe − βe , e i ) − − R p +1 ω ( e , e , . . . , e , e , αe − βe , e i )= − αR p +1 ω ( e , e , . . . , e , e , e , e i ) + βR p +1 ω ( e , e , . . . , e , e , e , e i ) . Now we compute that R p +1 ω ( e , e , . . . , e , e , e , e i )= − R p ω ( R ( e , e ) e , e , . . . , e , e , e , e i ) − R p ω ( e , R ( e , e ) e , . . . , e , e , e , e i ) . . . − R p ω ( e , e , . . . , e , e , R ( e , e ) e , e i ) − R p ω ( e , e , . . . , e , e , e , R ( e , e ) e i )= − R p ω ( e , e , . . . , e , e , R ( e , e ) e , e i ) . Similarly R p +1 ω ( e , e , . . . , e , e , e , e i ) = − R p ω ( e , e , . . . , e , e , R ( e , e ) e , e i ) . Thus we obtain R p +2 ω ( e , e , . . . , e , e , e , e i )= αR p ω ( e , e , . . . , e , e , αe − βe , e i ) − βR p ω ( e , e , . . . , e , e , − βe − αe , e i )= α R p ω ( e , e , . . . , e , e , e , e i ) + β R p ω ( e , e , . . . , e , e , e , e i )= det S · R p ω ( e , e , . . . , e , e , e , e i ) = (det S ) p +1 ω ( e , e i ) . Now by the induction principle (54) holds for all p ≥
1. The formula (55) can beshown in a similar way. (cid:3)
Lemma 4.3. If S is a 2-dimensional complex Jordan block then for every p ≥ , i, j > we have R p ω ( e , e , . . . , e , e , e , e i , e , e j ) = 2 p − β (det S ) p − h ( e i , e j ) ω ( e , e )(56) R p ω ( e , e , . . . , e , e , e , e i , e , e j ) = 2 p − β (det S ) p − h ( e i , e j ) ω ( e , e )(57) R p ω ( e , e , . . . , e , e , e , e i , e , e j ) − R p ω ( e , e , . . . , e , e , e , e i , e , e j )(58) = − p αβ (det S ) p − h ( e i , e j ) ω ( e , e ) Proof.
To simplify computations let us denote B p ( X, Y, i, j ) := R p ω ( e , e , e , e , . . . , e , e , X, e i , Y, e j )(59)where X, Y ∈ span { e , e } , i, j ∈ { , . . . , n } and p ≥ p = 1. Indeed, using Lemma 4.1,we obtain R ω ( e , e , e , e i , e , e j )= − Rω ( Se , e i , e , e j ) + Rω ( e , e i , Se , e j )= − αRω ( e , e i , e , e j ) + βRω ( e , e i , e , e j )+ βRω ( e , e i , e , e j ) + αRω ( e , e i , e , e j )= β ( Rω ( e , e i , e , e j ) + Rω ( e , e i , e , e j ))= β ( − ω ( R ( e , e i ) e , e j ) − ω ( e , R ( e , e i ) e j ) − ω ( R ( e , e i ) e , e j ) − ω ( e , R ( e , e i ) e j ))= − β ( ω ( e , R ( e , e i ) e j ) + ω ( e , R ( e , e i ) e j ))= − βh ( e i , e j ) (cid:16) ω ( e , Se ) + ω ( e , Se ) (cid:17) = 2 β h ( e i , e j ) ω ( e , e ) . that is B ( e , e , i, j ) = 2 β h ( e i , e j ) ω ( e , e ) . (60)Exactly in the same way we show that B ( e , e , i, j ) = 2 β h ( e i , e j ) ω ( e , e ) . (61)For (58) we compute B ( e , e , i, j ) − B ( e , e , i, j ) = − B ( R ( e , e ) e , e , i, j ) − B ( e , R ( e , e ) e , i, j )+ B ( R ( e , e ) e , e , i, j ) + B ( e , R ( e , e ) e , i, j )= − B ( Se , e , i, j ) − B ( e , Se , i, j ) − B ( Se , e , i, j ) − B ( e , Se , i, j )= − α ( B ( e , e , i, j ) + B ( e , e , i, j ))= − α ( − h ( e i , e j ) ω ( e , Se ) − h ( e i , e j ) ω ( e , Se ))= 2 αh ( e i , e j )( − βω ( e , e ) + βω ( e , e ))= − αβh ( e i , e j ) ω ( e , e ) . FFINE HYPERSURFACES OF ARBITRARY SIGNATURE... 25
Now assume that (56)–(58) are all true for some p ≥
1. We compute B p +2 ( e , e , i, j ) = R p +2 ω ( e , e , . . . , e , e , e , e i , e , e j )= − R p +1 ω ( Se , e , . . . , e , e , e , e i , e , e j )+ R p +1 ω ( e , Se , . . . , e , e , e , e i , e , e j ) · · ·− R p +1 ω ( e , e , . . . , Se , e , e , e i , e , e j )+ R p +1 ω ( e , e , . . . , e , Se , e , e i , e , e j ) − R p +1 ω ( e , e , . . . , e , e , Se , e i , e , e j )+ R p +1 ω ( e , e , . . . , e , e , e , e i , Se , e j )= − R p +1 ω ( e , e , . . . , e , e , Se , e i , e , e j )+ R p +1 ω ( e , e , . . . , e , e , e , e i , Se , e j )= − B p +1 ( Se , e , i, j ) + B p +1 ( e , Se , i, j ) , since for l = 1 , . . . , p , terms 2 l − l cancel each other. Now we easily computethat B p +1 ( Se , e , i, j ) = − B p ( R ( e , e ) Se , e , i, j ) + B p ( Se , Se , i, j )and B p +1 ( e , Se , i, j ) = − B p ( Se , Se , i, j ) − B p ( e , R ( e , e ) Se , i, j )In consequence B p +2 ( e , e , i, j ) = − B p ( Se , Se , i, j ) + B p ( R ( e , e ) Se , e , i, j ) − B p ( e , R ( e , e ) Se , i, j )= − αβB p ( e , e , i, j ) + 2 αβB p ( e , e , i, j ) − α B p ( e , e , i, j ) + 2 β B p ( e , e , i, j )+ ( α + β ) B p ( e , e , i, j ) + ( α + β ) B p ( e , e , i, j )= − αβ ( B p ( e , e , i, j ) − B p ( e , e , i, j ))+ 2 β ( B p ( e , e , i, j ) + B p ( e , e , i, j )) , since R ( e , e ) Se = ( α + β ) e and R ( e , e ) Se = − ( α + β ) e . Now usingassumptions (56)–(58) we obtain B p +2 ( e , e , i, j ) = − αβ ( − p αβ (det S ) p − h ( e i , e j ) ω ( e , e ))+ 4 β (2 p − β (det S ) p − h ( e i , e j ) ω ( e , e ))= 2 p +1 ( α β + β )(det S ) p − h ( e i , e j ) ω ( e , e )= 2 p +1 β (det S ) p h ( e i , e j ) ω ( e , e )In a similar way we show that B p +2 ( e , e , i, j ) = 2 p +1 β (det S ) p h ( e i , e j ) ω ( e , e ) . Eventually B p +2 ( e , e , i, j ) − B p +2 ( e , e , i, j )= − B p +1 ( R ( e , e ) e , e , i, j ) − B p +1 ( e , R ( e , e ) e , i, j )+ B p +1 ( R ( e , e ) e , e , i, j ) + B p +1 ( e , R ( e , e ) e , i, j )= − B p +1 ( Se , e , i, j ) − B p +1 ( e , Se , i, j ) − B p +1 ( Se , e , i, j ) − B p +1 ( e , Se , i, j )= − α ( B p +1 ( e , e , i, j ) + B p +1 ( e , e , i, j ))= − α ( − B p ( Se , e , i, j ) − B p ( e , Se , i, j )+ B p ( Se , e , i, j ) + B p ( e , Se , i, j ))= − α ( − α ( B p ( e , e , i, j ) − B p ( e , e , i, j )) + 4 βB p ( e , e , i, j ))= 4 α ( B p ( e , e , i, j ) − B p ( e , e , i, j )) − αβB p ( e , e , i, j ) . Using (56)–(58) we get B p +2 ( e , e , i, j ) − B p +2 ( e , e , i, j )= 4 α ( B p ( e , e , i, j ) − B p ( e , e , i, j )) − αβB p ( e , e , i, j )= 4 α ( − p αβ (det S ) p − h ( e i , e j ) ω ( e , e )) − αβ · p − β (det S ) p − h ( e i , e j ) ω ( e , e )= − p +2 ( α β + αβ )(det S ) p − h ( e i , e j ) ω ( e , e )= − p +2 αβ (det S ) p h ( e i , e j ) ω ( e , e ) . Now by the induction principle (56)–(58) hold for all p ≥ (cid:3) In the next three lemmas we study properties of complex Jordan block of di-mension greater than 2 in relation to other Jordan blocks from the decomposition.Thus in these lemmas, we implicitly assume that the Jordan decomposition containsmore than one (not necessarily complex) block.
Lemma 4.4.
Let S be a k -dimensional complex Jordan block, k ≥ , i ∈{ , . . . , k − } \ { } , s ∈ { , . . . , k } , j > k , p ≥ . Then R p ω ( e , e k − , . . . , e , e k − , e i , e s , e , e j )(62) = ( , for s < k − ω ( S e i , e j ) for s = 2 k .Proof. First let us notice that R ( e i , e s ) e = ( , for s < kS e i , for s = 2 k and R ( e i , e s ) e j = 0 . FFINE HYPERSURFACES OF ARBITRARY SIGNATURE... 27
From the above properties we obtain Rω ( e i , e s , e , e j ) = − ω ( R ( e i , e s ) e , e j ) − ω ( e , R ( e i , e s ) e j )= ( , for s < k − ω ( S e i , e j ) , for s = 2 k thus (62) is true for p = 1. Moreover we have that R ( e , e k − ) e s = , for s ∈ { , . . . , k − } \ { } S e , for s = 3 − S e k − , for s = 2 k ,and R ( e , e k − ) e = R ( e , e k − ) e k − = R ( e , e k − ) e i = R ( e , e k − ) e j = 0 . Now, assume that (62) is true for some p ≥
1. Using the above formulas we easilyget R p +1 ω ( e , e k − , . . . , e , e k − , e i , e s , e , e j )= ( R ( e , e k − ) · R p ω )( e , e k − , . . . , e , e k − , e i , e s , e , e j )= − R p ω ( e , e k − , . . . , e , e k − , e i , R ( e , e k − ) e s , e , e j )= , for s ∈ { , . . . , k − } \ { }− R p ω ( e , e k − , . . . , e , e k − , e i , S e , e , e j ) , for s = 3 R p ω ( e , e k − , . . . , e , e k − , e i , S e k − , e , e j ) , for s = 2 k = , for s ∈ { , . . . , k − } \ { }− R p ω ( e , e k − , . . . , e , e k − , e i , αe − βe + e , e , e j ) , for s = 3 R p ω ( e , e k − , . . . , e , e k − , e i , βe k − + αe k − + e k , e , e j ) , for s = 2 k = , for s ∈ { , . . . , k − } \ { } , for s = 3 R p ω ( e , e k − , . . . , e , e k − , e i , e k , e , e j ) , for s = 2 k. = , for s ∈ { , . . . , k − } \ { } , for s = 3 − ω ( S e i , e j ) , for s = 2 k. Now by the induction principle (62) holds for all p ≥ (cid:3) Lemma 4.5.
Let S be a k -dimensional complex Jordan block, k ≥ , j > k , p ≥ . Then R p ω ( e , e k − , . . . , e , e k − , e , e k , e , e j ) = − ( − β ) p − ω ( S e , e j ) . (63) Proof.
For p = 1, by direct computations, we obtain Rω ( e , e k , e , e j ) = − ω ( R ( e , e k ) e , e j ) − ω ( e , R ( e , e k ) e j )= − ω ( S e , e j ) . Now let us assume that (63) holds for some p ≥
1. Since R ( e , e k − ) e = R ( e , e k − ) e = R ( e , e k − ) e k − = R ( e , e k − ) e j = 0 ,R ( e , e k − ) e k = − S e k − = − αe k − + βe k we compute that( R p +1 ω )( e , e k − , . . . , e , e k − , e , e k , e , e j )= ( R ( e , e k − ) · R p ω )( e , e k − , . . . , e , e k − , e , e k , e , e j )= − R p ω ( e , e k − , . . . , e , e k − , e , − αe k − + βe k , e , e j )= αR p ω ( e , e k − , . . . , e , e k − , e , e k − , e , e j ) − βR p ω ( e , e k − , . . . , e , e k − , e , e k , e , e j ) . Since R ( e , e k − ) e = R ( e , e k − ) e j = 0one may easily deduce that Rω ( e , e k − , e , e j ) = 0 and more general R p ω ( e , e k − , . . . , e , e k − , e , e k − , e , e j ) = 0for all p ≥
1. Thus we have( R p +1 ω )( e , e k − , . . . , e , e k − , e , e k , e , e j )= − βR p ω ( e , e k − , . . . , e , e k − , e , e k , e , e j )= − β · ( − ( − β ) p − ) ω ( S e , e j ) = − ( − β ) p ω ( S e , e j ) , where the last equality follows from (63). Now by the induction principle (63) holdsfor all p ≥ (cid:3) Lemma 4.6.
Let S be a k -dimensional complex Jordan block, k ≥ , i ∈{ , . . . , k } , j > k . Then for every p ≥ we have R p ω ( e , e k , . . . , e , e k , e i , e k , e k , e j )(64) = ( , for i ∈ { , . . . , k } ( − β ) p − ω ( S e k , e j ) for i = 1 . Proof.
First notice that R ( e i , e k ) e k = ( , for i ∈ { , . . . , k }− S e k , for i = 1 R ( e i , e k ) e j = 0 . From the above we obtain Rω ( e i , e k , e k , e j ) = − ω ( R ( e i , e k ) e k , e j ) − ω ( e k , R ( e i , e k ) e j )= ( , for i ∈ { , . . . , k } ω ( S e k , e j ) , for i = 1 , thus (64) is true for p = 1. Moreover, we have R ( e , e k ) e i = ( , for i ∈ { , . . . , k } S e , for i = 1 FFINE HYPERSURFACES OF ARBITRARY SIGNATURE... 29
Now let us assume that (64) holds for some p ≥
1. Using the above formulas weobtain( R p +1 ω )( e , e k , . . . , e , e k , e i , e k , e k , e j )= − R p ω ( e , e k , . . . , e , e k , R ( e , e k ) e i , e k , e k , e j )= ( , for i ∈ { , . . . , k }− R p ω ( e , e k , . . . , e , e k , S e , e k , e k , e j ) , for i = 1= ( , for i ∈ { , . . . , k }− R p ω ( e , e k , . . . , e , e k , βe + αe + e , e k , e k , e j ) , for i = 1= ( , for i ∈ { , . . . , k }− βR p ω ( e , e k , . . . , e , e k , e , e k , e k , e j ) , for i = 1= ( , for i ∈ { , . . . , k }− β · ( − β ) p − ω ( S e k , e j ) = ( − β ) p ω ( S e k , e j ) , for i = 1Now by the induction principle (64) holds for all p ≥ (cid:3) Now we can prove
Corollary 4.7.
Let S be a k -dimensional complex Jordan block, k ≥ . If R p ω =0 for some p ≥ then for every X ∈ span { e , . . . , e k } := V and j > kω ( X, e j ) = 0 . (65) Proof.
For k = 1 Corollary 4.7 is an immediate consequence of Lemma 4.2. For k ≥ ω ( S e i , e j ) = 0 for i ∈ { , . . . , k } . Since det S = 0 and S : V → V is an isomorphism (so { S e , . . . , S e k } generate V ) we obtain that ω ( SX, e j ) = 0for all X ∈ V . (cid:3) In the next few lemmas we study intrinsic properties of 2 k -dimensional complexJordan block for k ≥ Lemma 4.8.
Let S be a k -dimensional complex Jordan block, k ≥ , i ∈{ , . . . , k − } \ { } . Then for every p ≥ R p ω ( e , e k − , . . . , e , e k − , e i , e k − ) = 0 , (66) R p ω ( e , e k − , . . . , e , e k − , e i , e k ) = ( − β ) p − ω ( e i , S e k − ) . (67) Proof.
The Gauss equation implies that R ( e , e k − ) e = R ( e , e k − ) e k − = R ( e , e k − ) e i = 0 . By straightforward computations we get (66) for every p ≥
1. In order to prove(67) again by the Gauss equation we have R ( e , e k − ) e k = − S e k − = − αe k − + βe k . In particular, for p = 1, we get Rω ( e , e k − , e i , e k ) = ω ( e i , S e k − ) . Now, assume that the formula (67) is true for some p ≥
1. Then, for p + 1, we get( R p +1 ω )( e , e k − , . . . , e , e k − , e i , e k )= ( R ( e , e k − ) · R p ω )( e , e k − , . . . , e , e k − , e i , e k )= − R p ω ( e , e k − , . . . , e , e k − , e i , − S e k − )= R p ω ( e , e k − , . . . , e , e k − , e i , αe k − − βe k )= αR p ω ( e , e k − , . . . , e , e k − , e i , e k − ) − βR p ω ( e , e k − , . . . , e , e k − , e i , e k )Now, by (66) and the induction principle, the formula (67) holds for every p ≥ (cid:3) Lemma 4.9.
Let S be a k -dimensional complex Jordan block, k ≥ , p ≥ , i ∈ { , . . . , k } \ { , k − } . Then R p ω ( e , e k , . . . , e , e k , e i , e k ) = 0 , (68) R p ω ( e , e k , . . . , e , e k , e i , e k − ) = β p − ω ( e i , S e k ) . (69) Proof.
The Gauss equation implies that R ( e , e k ) e = R ( e , e k ) e k = R ( e , e k ) e i = 0 . By straightforward computations we get (68) for every p ≥
1. In order to prove(69) again by the Gauss equation we have R ( e , e k ) e k − = − S e k = − βe k − − αe k . In particular, for p = 1, we get Rω ( e , e k , e i , e k − ) = ω ( e , S e k ) . Now, assume that the formula (69) is true for some p ≥
1. Then, for p + 1, we get( R p +1 ω )( e , e k , . . . , e , e k , e i , e k − )= ( R ( e , e k ) · R p ω )( e , e k , . . . , e , e k , e i , e k − )= − R p ω ( e , e k , . . . , e , e k , e i , − S e k )= R p ω ( e , e k , . . . , e , e k , e i , βe k − + αe k )= αR p ω ( e , e k , . . . , e , e k , e i , e k ) + βR p ω ( e , e k , . . . , e , e k , e i , e k − )Now, by (68) and the induction principle, the formula (69) holds for every p ≥ (cid:3) Lemma 4.10.
Let S be a k -dimensional complex Jordan block, k ≥ , p ≥ . If ω ( e j , e k − ) = ω ( e j , e k ) = 0 for j ∈ { , . . . , k } then R p ω ( e , e k − , . . . , ( i ) z }| { e k − , e k . . . , e , e k − , e , e ) = 0(70) for i ∈ { , . . . , p } FFINE HYPERSURFACES OF ARBITRARY SIGNATURE... 31
Proof.
For p = 1 we have Rω ( e k − , e k , e , e ) = − ω ( R ( e k − , e k ) e , e ) − ω ( e , R ( e k − , e k ) e )= − ω ( R ( e k − , e k ) e , e ) = − ω ( S e k − , e ) = 0 , since ω ( e , e k − ) = ω ( e , e k ) = 0 by assumption.Assume that (70) holds for some p ≥ i ∈ { , . . . , p } . Let i ∈{ , . . . , p + 1 } . If i > R p +1 ω ( e , e k − , . . . , ( i ) z }| { e k − , e k . . . , e , e k − , e , e )= − R p ω ( e , e k − , . . . , ( i − z }| { e k − , R ( e , e k − ) e k . . . , e , e k − , e , e )= R p ω ( e , e k − , . . . , ( i − z }| { e k − , S e k − . . . , e , e k − , e , e )= R p ω ( e , e k − , . . . , ( i − z }| { e k − , αe k − − βe k . . . , e , e k − , e , e )= − βR p ω ( e , e k − , . . . , ( i − z }| { e k − , e k . . . , e , e k − , e , e )since R ( e , e k − ) e = R ( e , e k − ) e k − = R ( e , e k − ) e = 0 and R ( e , e k − ) e k = − S e k − = − αe k − + βe k . Now by (70) we obtain that for i > R p +1 ω ( e , e k − , . . . , ( i ) z }| { e k − , e k . . . , e , e k − , e , e ) = 0 . If i = 1 we compute R p +1 ω ( (1) z }| { e k − , e k , e , e k − , . . . , . . . , e , e k − , e , e )= − R p ω ( R ( e k − , e k ) e , e k − , e , e k − , . . . , e , e k − , e , e ) − R p ω ( e , e k − , R ( e k − , e k ) e , e k − , . . . , e , e k − , e , e ) · · ·− R p ω ( e , e k − , e , e k − , . . . , R ( e k − , e k ) e , e k − , e , e ) − R p ω ( e , e k − , e , e k − , . . . , e , e k − , R ( e k − , e k ) e , e )= βR p ω ( (1) z }| { e k , e k − , e , e k − , . . . , e , e k − , e , e )+ βR p ω ( e , e k − , (2) z }| { e k , e k − , . . . , e , e k − , e , e ) · · · + βR p ω ( e , e k − , e , e k − , . . . , ( p ) z }| { e k , e k − , e , e ) − R p ω ( e , e k − , e , e k − , . . . , e , e k − , S e k − , e )= − R p ω ( e , e k − , e , e k − , . . . , e , e k − , S e k − , e ) , since all terms but last are equal 0 thanks to (70). Now it is enough to show that R p ω ( e , e k − , e , e k − , . . . , e , e k − , S e k − , e ) = 0 . Indeed, by Lemma 4.8 we have R p ω ( e , e k − , e , e k − , . . . , e , e k − , S e k − , e )= R p ω ( e , e k − , e , e k − , . . . , e , e k − , αe k − − βe k , e )= − αR p ω ( e , e k − , e , e k − , . . . , e , e k − , e , e k − )+ βR p ω ( e , e k − , e , e k − , . . . , e , e k − , e , e k )= β · ( − β ) p − ω ( e , S e k − ) = 0 , where the last equality follows from the assumption that ω ( e , e k − ) = ω ( e , e k ) =0. Summarising, we have shown that R p +1 ω ( e , e k − , . . . , ( i ) z }| { e k − , e k . . . , e , e k − , e , e ) = 0for all i ∈ { , . . . , p +1 } . Now by the induction principle (70) holds for all p ≥ (cid:3) Lemma 4.11.
Let S be a k -dimensional complex Jordan block, k ≥ , p ≥ . If ω ( e , e k − ) = ω ( e , e k ) = 0 then R p ω ( e , e k − , . . . , e , e k − , e , e k − ) = β p − ( − αω ( e , e k − ) + βω ( e , e k − ))(71) R p ω ( e , e k − , . . . , e , e k − , e , e k ) = αβ p − ( − αω ( e , e k − ) + βω ( e , e k − ))(72) − α ( − β ) p − ω ( e , e k − ) − α ( − β ) p − ω ( e , e k ) Proof.
For p = 1 we compute Rω ( e , e k − , e , e k − ) = − ω ( R ( e , e k − ) e , e k − ) − ω ( e , R ( e , e k − ) e k − )= − ω ( αe − βe + e , e k − )= − αω ( e , e k − ) + βω ( e , e k − ) , since by assumption ω ( e , e k − ) = 0. Now, assume that the formula (71) is truefor some p ≥
1. Then, for p + 1, we get( R p +1 ω )( e , e k − , . . . , e , e k − , e , e k − )= ( R ( e , e k − ) · R p ω )( e , e k − , . . . , e , e k − , e , e k − )= − R p ω ( e , e k − , . . . , e , e k − , R ( e , e k − ) e , e k − )= − R p ω ( e , e k − , . . . , e , e k − , S e , e k − )= − R p ω ( e , e k − , . . . , e , e k − , αe − βe + e , e k − )= − αR p ω ( e , e k − , . . . , e , e k − , e , e k − )+ βR p ω ( e , e k − , . . . , e , e k − , e , e k − ) − R p ω ( e , e k − , . . . , e , e k − , e , e k − )= βR p ω ( e , e k − , . . . , e , e k − , e , e k − ) FFINE HYPERSURFACES OF ARBITRARY SIGNATURE... 33 where the last equality follows from Lemma 4.8 (formula (66), i = 1 , R p +1 ω )( e , e k − , . . . , e , e k − , e , e k − )= βR p ω ( e , e k − , . . . , e , e k − , e , e k − )= β ( β p − ( − αω ( e , e k − ) + βω ( e , e k − )))= β p ( − αω ( e , e k − ) + βω ( e , e k − )) . By the induction principle (71) holds for all p ≥ (cid:3) From Lemma 4.11 we immediately get
Corollary 4.12.
Let S be a k -dimensional complex Jordan block, k ≥ , p ≥ .If ω ( e , e k − ) = ω ( e , e k ) = 0 then R p ω ( e , e k − , . . . , e , e k − , e , S e k − ) = ( − p β p − α ( αω ( e , e k − ) − βω ( e , e k ))(73) Lemma 4.13.
Let S be a k -dimensional complex Jordan block, k ≥ , p ≥ . If ω ( e , e k − ) = ω ( e , e k ) = 0 then for i ∈ { , . . . , p } R p ω ( e , e k − , . . . , ( i ) z }| { e k − , e k . . . , e , e k − , e , e )(74) = ( ( − β ) p − ( − ω ( S e k − , e ) + ω ( e , S e k )) , if i = 10 , if i > . Proof.
We directly check that Rω ( e k − , e k , e , e ) = − ω ( R ( e k − , e k ) e , e ) − ω ( e , R ( e k − , e k ) e )= − ω ( S e k − , e ) + ω ( e , S e k ) , so (74) is true for p = 1. Now let us assume that (74) is true for some p ≥ i ∈ { , . . . , p } . Let i ∈ { , . . . , p + 1 } . When i > R ( e , e k − ) e = R ( e , e k − ) e k − = 0 we get R p +1 ω ( e , e k − , . . . , ( i ) z }| { e k − , e k . . . , e , e k − , e , e )= − R p ω ( e , e k − , . . . , ( i − z }| { e k − , R ( e , e k − ) e k . . . , e , e k − , e , e ) − R p ω ( e , e k − , . . . , ( i − z }| { e k − , e k . . . , e , e k − , e , R ( e , e k − ) e )= R p ω ( e , e k − , . . . , ( i − z }| { e k − , S e k − . . . , e , e k − , e , e ) − R p ω ( e , e k − , . . . , ( i − z }| { e k − , e k . . . , e , e k − , e , Se )= − βR p ω ( e , e k − , . . . , ( i − z }| { e k − , e k . . . , e , e k − , e , e )+ βR p ω ( e , e k − , . . . , ( i − z }| { e k − , e k . . . , e , e k − , e , e ) − R p ω ( e , e k − , . . . , ( i − z }| { e k − , e k . . . , e , e k − , e , e )= − R p ω ( e , e k − , . . . , ( i − z }| { e k − , e k . . . , e , e k − , e , e ) = 0where the last equality follows from Lemma 4.10.If i = 1, using the fact that R ( e k − , e k ) e k − = 0 we obtain R p +1 ω ( (1) z }| { e k − , e k , e , e k − , . . . , e , e k − , e , e )= − R p ω ( R ( e k − , e k ) e , e k − , e , e k − , . . . , e , e k − , e , e ) − R p ω ( e , e k − , R ( e k − , e k ) e , e k − , . . . , e , e k − , e , e ) · · ·− R p ω ( e , e k − , e , e k − , . . . , R ( e k − , e k ) e , e k − , e , e ) − R p ω ( e , e k − , e , e k − , . . . , e , e k − , R ( e k − , e k ) e , e ) − R p ω ( e , e k − , e , e k − , . . . , e , e k − , e , R ( e k − , e k ) e )= βR p ω ( (1) z }| { e k , e k − , e , e k − , . . . , e , e k − , e , e )+ βR p ω ( e , e k − , (2) z }| { e k , e k − , . . . , e , e k − , e , e ) · · · + βR p ω ( e , e k − , e , e k − , . . . , ( p ) z }| { e k , e k − , e , e ) − R p ω ( e , e k − , e , e k − , . . . , e , e k − , S e k − , e )+ R p ω ( e , e k − , e , e k − , . . . , e , e k − , e , S e k ) FFINE HYPERSURFACES OF ARBITRARY SIGNATURE... 35 = − βR p ω ( (1) z }| { e k − , e k , e , e k − , . . . , e , e k − , e , e )+ R p ω ( e , e k − , e , e k − , . . . , e , e k − , e , S e k − )+ R p ω ( e , e k − , e , e k − , . . . , e , e k − , e , S e k ) , since all but first and the last two terms are equal to 0 by (74). Now Lemma 4.8and Corollary 4.12 imply that R p ω ( e , e k − , e , e k − , . . . , e , e k − , e , S e k )+ R p ω ( e , e k − , e , e k − , . . . , e , e k − , e , S e k − )= α ( − β ) p − ω ( e , S e k − ) + ( − p β p − α ( αω ( e , e k − ) − βω ( e , e k ))= 0 . In consequence, using (74) we get R p +1 ω ( (1) z }| { e k − , e k , e , e k − , . . . , e , e k − , e , e )= − βR p ω ( (1) z }| { e k − , e k , e , e k − , . . . , e , e k − , e , e )= ( − β ) p ( − ω ( S e k − , e ) + ω ( e , S e k )) . Now the thesis follows from the induction principle. (cid:3)
To simplify further notation let us denote: A pij := R p ω ( e , e k − , e , e k − , . . . , e , e k − , e i , e k − , e j , e k − )(75)where i, j ∈ { , , } and p ≥
1. We have the following lemma:
Lemma 4.14.
Let S be a k -dimensional complex Jordan block, k ≥ , p ≥ .Then A p = A p = A p = A p = 0 , (76) A p +112 = βA p , (77) A p +121 = βA p , (78) A p +123 = βA p , (79) A p +132 = βA p , (80) A p +122 = − α ( A p + A p ) + 2 βA p − ( A p + A p ) . (81) Proof.
In order to prove (76) first note that A p = A p = 0immediately follows from Lemma 4.8, formula (66). By the Gauss equation we have R ( e , e k − ) e = R ( e , e k − ) e k − = R ( e , e k − ) e = R ( e , e k − ) e = R ( e , e k − ) e k − = R ( e , e k − ) e = 0 . Using the above equalities we easy obtain that A p = A p = 0 . To prove (77) we compute A p +112 = R p +1 ω ( e , e k − , . . . , e , e k − , e , e k − , e , e k − )= − R p ω ( e , e k − , . . . , e , e k − , e , e k − , R ( e , e k − ) e , e k − )= − R p ω ( e , e k − , . . . , e , e k − , e , e k − , S e , e k − )= − R p ω ( e , e k − , . . . , e , e k − , e , e k − , αe − βe + e , e k − ) , = − αA p + βA p − A p = βA p , where the last equality follows from (76). The formulas (78)–(80) we prove in asimilar way like (77). To prove (81) we compute A p +122 = R p +1 ω ( e , e k − , . . . , e , e k − , e , e k − , e , e k − )= − R p ω ( e , e k − , . . . , e , e k − , R ( e , e k − ) e , e k − , e , e k − ) − R p ω ( e , e k − , . . . , e , e k − , R ( e , e , e k − , R ( e , e k − ) e , e k − )= − R p ω ( e , e k − , . . . , e , e k − , S e , e k − , e , e k − ) − R p ω ( e , e k − , . . . , e , e k − , e , e k − , S e , e k − )= − R p ω ( e , e k − , . . . , e , e k − , αe − βe + e , e k − , e , e k − ) − R p ω ( e , e k − , . . . , e , e k − , e , e k − , αe − βe + e , e k − , e , e k − )= − αA p + βA p − A p − αA p + βA p − A p = − α ( A p + A p ) + 2 βA p − ( A p + A p ) . (cid:3) From the above lemma we obtain:
Corollary 4.15.
Let S be a k -dimensional complex Jordan block, k ≥ , p ≥ .If ω ( e j , e k − ) = ω ( e j , e k ) = 0(82) for j ∈ { , . . . , k } then A p = β p − ( − αω ( e , e k − ) + βω ( e , e k − )) , (83) A p = β p − ( αω ( e , e k − ) − βω ( e , e k )) , (84) A p +122 = − αβ p ( ω ( e , e k − ) − ω ( e , e k )) + 2 βA p . (85) Proof.
From formulas (77), (78) and assumption (82) we obtain A p = β p − · A = β p − · Rω ( e , e k − , e , e k − )= − β p − ω ( R ( e , e k − ) e , e k − ) = − β p − ω ( S e , e k − )= − αβ p − ω ( e , e k − ) + β p ω ( e , e k − ) .A p = β p − · A = β p − · Rω ( e , e k − , e , e k − )= − β p − ω ( e , R ( e , e k − ) e k − ) = β p − ω ( e , S e k − )= αβ p − ω ( e , e k − ) − β p ω ( e , e k ) , what proves (83) and (84). Using (83) and (84) we obtain A p + A p = β p ( ω ( e , e k − ) − ω ( e , e k )) . FFINE HYPERSURFACES OF ARBITRARY SIGNATURE... 37
Moreover from formulas (79), (80) and (82) we get A p = β p − A = β p − Rω ( e , e k − , e , e k − )= − β p − ω ( R ( e , e k − ) e , e k − ) = − β p − ω ( S e , e k − ) = 0 ,A p = β p − · A = 0 . Now (85) immediately follows from (81). (cid:3)
Lemma 4.16.
Let S be a k -dimensional complex Jordan block, k ≥ . If R p ω = 0 for some p ≥ then for i ∈ { , . . . , k } ω ( e i , e k − ) = ω ( e i , e k ) = 0 . Proof.
From Lemma 4.8, for i = 3 , . . . , k − ω ( e i , S e k − ) = 0 . (86)In particular for i = 2 k − ω ( e k − , S e k − ) = ω ( e k − , αe k − − βe k ) = − βω ( e k − , e k ) . Thus ω ( e k − , e k ) = 0 , since β = 0. Now we have ω ( e k , S e k − ) = ω ( e k , αe k − − βe k ) = αω ( e k , e k − ) = 0 . That is (86) is also true for i = 2 k . From Lemma 4.9 for i ∈ { , . . . , k } \ { k − } we have ω ( e i , S e k ) = 0 . (87)Since ω ( e k − , e k ) = 0 we also get that ω ( e k − , S e k ) = ω ( e k − , βe k − + αe k ) = αω ( e k − , e k ) = 0 . That is (87) also valid for i = 2 k −
1. Now from (86) and (87), for i ∈ { , . . . , k } we get 0 = αω ( e i , S e k − ) + βω ( e i , S e k )= ω ( e i , α · ( αe k − − βe k ) + β · ( βe k − + αe k ))= ( α + β ) ω ( e i , e k − ) . In a similar way we compute that0 = − βω ( e i , S e k − ) + αω ( e i , S e k )= ( α + β ) ω ( e i , e k ) . Since α + β = 0 the above equations imply that ω ( e i , e k − ) = ω ( e i , e k ) = 0for i ∈ { , . . . , k } . (cid:3) Now we are ready to prove that the decomposition from the Lemma 2.4 cannotcontain complex Jordan blocks. Namely we have the following:
Theorem 4.17.
Let f : M → R n +1 ( dim M ≥ ) be a non-degenerate affine hyper-surface with a locally equiaffine transversal vector field ξ and an almost symplecticform ω . If R p ω = 0 for some p ≥ and S = S . . . S . . . ... ... . . . . . . . . . S q + r (88) is the Jordan decomposition of S as stated in the Lemma 2.4 then (88) does notcontain complex Jordan blocks (that is r = 0) .Proof. Let { e , . . . , e n } be the basis of T x M from Lemma 2.4. Without loss ofgenerality, as described at the beginning of this section, we can change order of S i and H i in such way that S , . . . , S r will be complex blocks and S r +1 , . . . , S r + q willbe real blocks. Moreover, we can assume that dim S ≥ dim S i for i = 2 , . . . , r .First assume that S is a complex block of dimension 2 k and k ≥
2. By Lemma4.16 we have that ω ( e i , e k − ) = ω ( e i , e k ) = 0 for i ∈ { , . . . , k } . Now fromCorollary 4.15 (formulas (83) and (84)) we get ω ( e , e k − ) = αβ ω ( e , e k − )(89)and ω ( e , e k ) = αβ ω ( e , e k − ) , (90)since R p ω = 0 and β = 0. In particular ω ( e , e k − ) = ω ( e , e k ) and (85) simplifyto the form A p +122 = 2 βA p . Now one can easily find explicit formula for A p . Namely we have A p = 2 p − β p − A = 2 p − β p − Rω ( e , e k − , e , e k − )= 2 p − β p − ( − ω ( S e , e k − ) + ω ( e , S e k − ))= 2 p − β p − ( − βω ( e , e k − ) − αω ( e , e k − )+ αω ( e , e k − ) − βω ( e , e k ))= − p − β p ( ω ( e , e k − ) + ω ( e , e k )) . On the other hand we have A p = 0 (since R p ω = 0) and in consequence ω ( e , e k ) = − ω ( e , e k − ) . (91)From Lemma 4.13 (formula (74)) we obtain ω ( S e k − , e ) − ω ( e , S e k ) = 0that is αω ( e k − , e ) − βω ( e k , e ) − βω ( e , e k − ) − αω ( e , e k ) = 0 . Now using (89)–(91) the above implies that − α · αβ ω ( e , e k − ) − βω ( e , e k − ) − βω ( e , e k − ) − α · αβ ω ( e , e k − )= − α β + β ) ω ( e , e k − ) = − β ( α + β ) ω ( e , e k − ) = 0 . FFINE HYPERSURFACES OF ARBITRARY SIGNATURE... 39
In this way we have shown that ω ( e , e k − ) = 0 and in consequence also ω ( e , e k − ) =0. Hence ω ( e i , e k − ) = 0for i ∈ { , . . . , k } . From Corollary 4.7 we also have that ω ( e i , e k − ) = 0for i > k , that is ω is degenerate, what leads to contradiction and we must have k <
2. In this way we have shown that if Jordan decomposition of S contains somecomplex Jordan blocks they all must be 2-dimensional.It remained to show that also 2-dimensional complex Jordan blocks are notpossible. In order to prove it let us assume that S is a 2-dimensional complexblock. Since R p ω = 0 then also R p ω = 0 and Lemma 4.2 implies that ω ( e , e l ) = 0for l = 3 , . . . , n . Now observe that since dim M ≥ i , j > h ( e i , e j ) = 0 (otherwise h would be degenerate). Now from Lemma 4.3we have 2 p − β (det S ) p − h ( e i , e j ) ω ( e , e ) = 0 . That is ω ( e , e ) = 0 , since h ( e i , e j ) = 0. In this way we have shown that ω is degenerate since ω ( e , e l ) = 0 for l = 1 , . . . , n , what leads us again to contradiction. (cid:3) Main Results
Before we proceed with main results of this paper we need to recall the followingtwo lemmas:
Lemma 5.1 ([21]) . Let f : M → R n +1 be a non-degenerate affine hypersurface( dim M ≥ ) with a transversal vector field ξ and an almost symplectic form ω .Let x ∈ M . If there exist a natural number ≤ s ≤ n and a basis { e , . . . , e n } of T x M such that h ( e i , e j ) = ε i δ ij , ε i = ± for i, j = 1 , . . . , s , h ( e i , e j ) = 0 for i = 1 , . . . , s , j = s + 1 , . . . , n and Se i = λ i e i for i = 1 , . . . , s , λ i ∈ R . Then forevery k, j = 1 , , . . . , s , k = j and for every i = 1 , . . . , n , i = j , i = k we have (92) R l ω ( e k , e j , e k , e j , . . . , e k , e j , e k , e i ) = ( − l ε lk ε lj λ lk λ lj ω ( e k , e i ) for every l ∈ N . Lemma 5.2 ([21]) . Let f : M → R n +1 be a non-degenerate affine hypersurface( dim M ≥ ) with a transversal vector field ξ and an almost symplectic form ω .Let x ∈ M and let X, Y, Z , Z be vector fields from T x M such that SX = λX , SZ = 0 , SZ = 0 , h ( Z , Z ) = 0 and h ( Y, Z ) = 0 . Then, for every l ≥ we have R l ω ( X, Z , Z , Z , Z , . . . , Z , Z | {z } l , Y )(93) = ( − l λ l h ( Z , Z ) l h ( Z , Z ) l ω ( X, Y ) . We will need also below three lemmas.
Lemma 5.3.
Let { e , e , e } be vectors such that Se = αe + e , Se = αe , Se = λe and h ( e , e ) = h ( e , e ) = h ( e , e ) = h ( e , e ) = 0 , h ( e , e ) = η , h ( e , e ) = ε , η = 0 , ε = 0 . Then for every p ≥ we have (94) R p ω ( e , e , e , e , . . . , e , e , e , e , e , e ) = ( − p · (2 ηα ) p − εω ( e , e ) . Proof.
By straightforward computations we obtain Rω ( e , e , e , e ) = − εω ( e , e )(95) Rω ( e , e , e , e ) = εαω ( e , e )(96) Rω ( e , e , e , e ) = − εαω ( e , e )(97) Rω ( e , e , e , e ) = 0 . (98)Now for p ≥ i, j ∈ { , } we get R p +1 ω ( e , e , e , e , . . . , e , e , e i , e , e j , e )= − R p ω ( e , e , e , e , . . . , e , e , R ( e , e ) e i , e , e j , e ) − R p ω ( e , e , e , e , . . . , e , e , e i , e , R ( e , e ) e j , e )= − αh ( e , e i ) R p ω ( e , e , e , e , . . . , e , e , e , e , e j , e )+ ( αh ( e , e i ) − h ( e , e i )) R p ω ( e , e , e , e , . . . , e , e , e , e , e j , e ) − αh ( e , e j ) R p ω ( e , e , e , e , . . . , e , e , e i , e , e , e )+ ( αh ( e , e j ) − h ( e , e j )) R p ω ( e , e , e , e , . . . , e , e , e i , e , e , e ) . Using different configurations of i and j we obtain the following four relations: R p +1 ω ( e , e , e , e , . . . , e , e , e , e , e , e )(99) = 2 αh ( e , e ) R p ω ( e , e , e , e , . . . , e , e , e , e , e , e ) R p +1 ω ( e , e , e , e , . . . , e , e , e , e , e , e )(100) = − h ( e , e ) R p ω ( e , e , e , e , . . . , e , e , e , e , e , e ) R p +1 ω ( e , e , e , e , . . . , e , e , e , e , e , e )(101) = − h ( e , e ) R p ω ( e , e , e , e , . . . , e , e , e , e , e , e ) R p +1 ω ( e , e , e , e , . . . , e , e , e , e , e , e )(102) = − αh ( e , e ) R p ω ( e , e , e , e , . . . , e , e , e , e , e , e ) − h ( e , e ) R p ω ( e , e , e , e , . . . , e , e , e , e , e , e ) − h ( e , e ) R p ω ( e , e , e , e , . . . , e , e , e , e , e , e ) . Now from (98) and (99), by the induction principle we easily obtain that R p ω ( e , e , e , e , . . . , e , e , e , e , e , e ) = 0for all p ≥
1. In consequence from (100) and (101) we get that also R p ω ( e , e , e , e , . . . , e , e , e , e , e , e ) = 0and R p ω ( e , e , e , e , . . . , e , e , e , e , e , e ) = 0for all p ≥
2. Now (102) simplify to the form R p +1 ω ( e , e , e , e , . . . , e , e , e , e , e , e )(103) = − αh ( e , e ) R p ω ( e , e , e , e , . . . , e , e , e , e , e , e )for p ≥
2. Note that using (95)–(97) one may easily show that (103) is also true for p = 1. From (95) we see that (94) holds for p = 1. Assume now that (94) is true FFINE HYPERSURFACES OF ARBITRARY SIGNATURE... 41 for some p ≥
1. From (103) we compute that R p +1 ω ( e , e , e , e , . . . , e , e , e , e , e , e )= − αh ( e , e ) · ( − p · (2 ηα ) p − εω ( e , e )= ( − p +1 · (2 ηα ) p εω ( e , e ) . Now by the induction principle (94) is true for all p ≥ (cid:3) Lemma 5.4.
Let { e , e , . . . , e n } be vectors such that Se = e , Se = 0 , h ( e , e ) = h ( e , e ) = 0 , h ( e , e ) = η and Se i = λ i − e i , h ( e , e i ) = h ( e , e i ) = 0 , h ( e i , e i ) = ε i − for i = 3 , . . . , n where η = 0 , ε = 0 . Then for every p ≥ and for every i = 3 , . . . , n we have R p ω ( e , e , e , e , . . . , e , e , e i , e ) = − η p λ p ω ( e i , e ) . (104) Proof.
First we shall show that for every p ≥ i = 3 , . . . , n thefollowing formula holds: R p ω ( e , e , e , e , . . . , e , e , e i , e ) = 0 . (105)For p = 1 we have Rω ( e , e , e i , e ) = − ω ( R ( e , e ) e i , e ) − ω ( e i , R ( e , e ) e ) = 0 , since R ( e , e ) e i = 0 , R ( e , e ) e = 0 . (106)Now assume that (105) holds for some p ≥
1, we shall show that it also holds for p + 1. From the Gauss equation we have R ( e , e ) e = ηSe = ηλ e . (107)Now, using (106) and (107) we obtain R p +1 ω ( e , e , e , e , . . . , e , e , e i , e )= − R p ω ( ηλ e , e , e , e , . . . , e , e , e i , e ) − R p ω ( e , e , ηλ e , e , e , e , . . . , e , e , e i , e ) − . . . − R p ω ( e , e , e , e , . . . , ηλ e , e , e i , e )= − ηλ R p ω ( z }| { e , e , e , e , . . . , e , e , e i , e ) − ηλ R p ω ( e , e , z }| { e , e , e , e , . . . , e , e , e i , e ) − . . . − ηλ R p ω ( e , e , e , e , . . . , p z }| { e , e , e i , e )= − ηλ R p ω ( e , e , e , e , . . . , e , e , e i , e ) , since all terms but first are equal 0 due to the following identities R ( e , e ) e = R ( e , e ) e i = R ( e , e ) e = 0 ,R ( e , e ) e = ηSe = ηe . Now by the induction principle (105) is true for all p ≥ Now we can prove (104). First we check that (104) is satisfied for p = 1 and p = 2. Indeed, since R ( e , e ) e i = − h ( e , e i ) e , R ( e , e ) e = ηλ e , R ( e , e ) e = 0 , (108) R ( e , e ) e = R ( e , e ) e = R ( e , e ) e i = 0 , R ( e , e ) e = ηe (109)by straightforward computations we easily obtain that Rω ( e , e , e i , e ) = − ω ( R ( e , e ) e i , e ) − ω ( e i , R ( e , e ) e )= h ( e , e i ) ω ( e , e ) − ηλ ω ( e i , e ) = − ηλ ω ( e i , e )and R ω ( e , e , e , e , e i , e )= − Rω ( R ( e , e ) e , e , e i , e ) − Rω ( e , R ( e , e ) e , e i , e ) − Rω ( e , e , R ( e , e ) e i , e ) − Rω ( e , e , e i , R ( e , e ) e )= − Rω ( e , ηλ e , e i , e ) − Rω ( e , e , e i , ηλ e )= ηλ ( − ηλ ω ( e i , e )) − − η λ ω ( e i , e ) . Let us assume that (104) holds for some p ≥
2. Then using (108) we compute R p +1 ω ( e , e , e , e , . . . , e , e , e i , e )= − R p ω ( e , ηλ e , e , e , e , . . . , e , e , e i , e ) − R p ω ( e , e , e , ηλ e , e , e , . . . , e , e , e i , e ) . . . − R p ω ( e , e , e , e , . . . , e , ηλ e , e i , e ) − R p ω ( e , e , e , e , . . . , e , e , − h ( e , e i ) e , e ) − R p ω ( e , e , e , e , . . . , e , e , e i , ηλ e )= − ηλ R p ω ( z }| { e , e , e , e , . . . , e , e , e i , e ) − ηλ R p ω ( e , e , z }| { e , e , e , e , . . . , e , e , e i , e ) . . . − ηλ R p ω ( e , e , . . . , e , e , p z }| { e , e , e i , e ) − ηλ R p ω ( e , e , . . . , e , e , e i , e ) . From (109) it follows that for j = 2 , . . . , p and i ∈ { , . . . , n } \ { } R p ω ( e , e , . . . , j z }| { e , e , . . . , e , e , e i , e )(110) = − R p − ω ( e , e , . . . , j − z }| { ηe , e , . . . , e , e , e i , e ) . We also have R p ω ( e , e , . . . , e , e , e i , e ) = 0 FFINE HYPERSURFACES OF ARBITRARY SIGNATURE... 43 for i = 3 , , . . . , n thanks to (109) and since p ≥
2. Now we obtain R p +1 ω ( e , e , e , e , . . . , e , e , e i , e )= ηλ R p ω ( e , e , e , e , . . . , e , e , e i , e )+ ηλ R p − ω ( z }| { ηe , e , e , e , . . . , e , e , e i , e ) . . . + ηλ R p − ω ( e , e , . . . , e , e , p − z }| { ηe , e , e i , e )= ηλ R p ω ( e , e , e , e , . . . , e , e , e i , e )+ η λ R p − ω ( e , e , e , e , . . . , e , e , e i , e )= ηλ R p ω ( e , e , e , e , . . . , e , e , e i , e ) , where the last two equalities follow from (109) and (105) respectively. By theinduction principle (104) is true for all p ≥ (cid:3) Lemma 5.5.
Let { e , e , . . . , e n } be vectors with properties like in the Lemma 5.4.Then for every p ≥ we have R p ω ( e , e , e , e , . . . , e , e ) = ( − p η p λ p ω ( e , e )(111) R p ω ( e , e , e , e , . . . , e , e ) = − η p λ p ω ( e , e )(112) + 12 (cid:16) ( − p + 1 (cid:17) η p λ p − ω ( e , e ) . Proof.
Since basis { e , . . . , e n } satisfy conditions of Lemma 5.4. in particular wehave (109) and (110). We also have R ( e , e ) e = ηλ e , R ( e , e ) e = 0 . (113)By direct computations we check that Rω ( e , e , e , e ) = − ηλ ω ( e , e ) ,Rω ( e , e , e , e ) = − ηλ ω ( e , e ) ,R ω ( e , e , e , e , e , e ) = − η λ ω ( e , e ) + η λ ω ( e , e ) . It means that (111) is true for p = 1 and (112) is true for p = 1 , p ≥
1, using (113) we compute that R p +1 ω ( e , e , e , e , . . . , e , e )= − R p ω ( ηλ e , e , e , e , . . . , e , e ) − R p ω ( e , e , ηλ e , e , . . . , e , e ) . . . − R p ω ( e , e , . . . , e , e , ηλ e , e )= − ηλ R p ω ( e , e , e , e , . . . , e , e ) , since all terms but first are equal to 0 due to (109). Now by the induction principle(111) is true for all p ≥ To prove (112) let us assume that (112) is true for some p ≥ R p +1 ω ( e , e , e , e , . . . , e , e )= − R p ω ( e , ηλ e , e , e , . . . , e , e ) − R p ω ( e , e , e , ηλ e , . . . , e , e ) . . . − R p ω ( e , e , . . . , e , e , e , ηλ e )= − ηλ R p ω ( e , e , e , e , . . . , e , e ) − ηλ R p ω ( e , e , e , e , . . . , e , e ) . . . − ηλ R p ω ( e , e , . . . , e , e , e , e ) . Now using (110) (for i = 1) we obtain R p +1 ω ( e , e , e , e , . . . , e , e )= − ηλ R p ω ( e , e , e , e , . . . , e , e )+ η λ R p − ω ( e , e , e , e , . . . , e , e )+ η λ R p − ω ( e , e , e , e , . . . , e , e ) . . . + η λ R p − ω ( e , e , . . . , e , e , e , e ) − ηλ R p ω ( e , e , . . . , e , e , e , e )= − ηλ R p ω ( e , e , e , e , . . . , e , e )+ η λ R p − ω ( e , e , e , e , . . . , e , e )= ηλ (cid:16) − η p λ p ω ( e , e ) + 12 (( − p + 1) η p λ p − ω ( e , e ) (cid:17) − η λ ( − p − η p − λ p − ω ( e , e )= − η p +1 λ p +11 ω ( e , e ) + 12 (( − p + 1 + 2 · ( − p − ) η p +1 λ p ω ( e , e )= − η p +1 λ p +11 ω ( e , e ) + 12 (( − p +1 + 1) η p +1 λ p ω ( e , e )where the last equalities are consequence of (112) and (111) (for p − p ≥ (cid:3) Theorem 5.6.
Let f : M → R n +1 ( dim M ≥ ) be a non-degenerate affine hyper-surface with a locally equiaffine transversal vector field ξ and an almost symplecticform ω . If R p ω = 0 for some p ≥ then for every point x ∈ M either S = 0 in x or there exists a basis e , . . . , e n of T x M such that S in this basis has the form S = . . .
01 0 0 . . .
00 0 0 . . . ... ... . . . . . . . . . (114) FFINE HYPERSURFACES OF ARBITRARY SIGNATURE... 45
Proof.
From Theorem 4.17 we have that Jordan block decomposition of S do notcontain complex blocks. From Theorem 3.15 we also know that S contains at mostone real Jordan block of dimension 2 and remaining blocks are all of dimension 1.Now (rearranging vectors e , . . . , e n if needed) S and h can be represented inone of the following two forms: S = λ . . . λ . . .
00 0 λ . . . . . . λ n , h = ε . . . ε . . .
00 0 ε . . . . . . ε n (115)or S = α . . . α . . .
00 0 λ . . . . . . λ n − , h = η . . . η . . .
00 0 ε . . . . . . ε n − (116)where ε i ∈ {− , } for i = 1 , . . . , n , η ∈ {− , } and | λ | ≥ · · · ≥ | λ n | . (117)We need to show that λ = · · · = λ n = 0 (respectively α = 0 and λ = · · · = λ n − = 0).First assume that (115) holds. Since ω is non-degenerate there exists i suchthat ω ( e , e i ) = 0. If i > s = 2 n , k = 1, j = 2, i = i )we get R p ω ( e , e , e , e , . . . , e , e , e , e i ) = ( − p ε p ε p λ p λ p ω ( e , e i ) . If i = 2 then from Lemma 5.1 ( s = 2 n , k = 1, j = 3, i = i ) we get R p ω ( e , e , e , e , . . . , e , e , e , e i ) = ( − p ε p ε p λ p λ p ω ( e , e i ) . Since R p ω = 0 then also R p ω = 0 and the above implies that( − p ε p ε p λ p λ p ω ( e , e i ) = ( − p ε p ε p λ p λ p ω ( e , e i ) = 0 . Taking into account (117) we deduce that λ = λ = 0 if i > λ = λ = 0if i = 2. Now, if i = 4 using Lemma 5.2 ( X = e , Y = e i , Z = e , Z = e ) weget R p ω ( e , e , e , e , e , . . . , e , e | {z } p , e i ) = ( − p λ p ε p ε p ω ( e , e i ) . If i = 4 then we have that also λ = 0 and in this case from Lemma 5.2 ( X = e , Y = e i , Z = e , Z = e ) we get R p ω ( e , e , e , e , e , . . . , e , e | {z } p , e i ) = ( − p λ p ε p ε p ω ( e , e i ) . The above implies that λ = 0 and thanks to (117) we get that S = 0.Now assume that S and h have the form (116). First we shall show that α = 0.For this purpose note that from Corollary 3.2 ( k = 2) we have R p ω ( e , e , . . . , e , e , e i , e ) = η p α p ω ( e i , e ) for i ∈ { , . . . , n } . From Lemma 5.3 ( ε = ε ) we have R p ω ( e , e , e , e , . . . , e , e , e , e , e , e ) = ( − p · (2 ηα ) p − ε ω ( e , e ) . Since R p ω = 0 and η, ε = 0 we obtain α p ω ( e , e ) = · · · = α p ω ( e n , e ) = α p − ω ( e , e ) = 0and in consequence α = 0 (since ω is non-degenerate).Now we are able to show that λ = · · · = λ n − = 0. Indeed, from (117) itfollows that it is enough to show that λ = 0. Since α = 0, the basis { e , . . . , e n } satisfy conditions of Lemma 5.4 and Lemma 5.5. Thus using formulas (104), (111)and (112) and taking into account that R p ω = 0 we obtain − η p λ p ω ( e i , e ) = 0 for i = 3 , . . . , n, (118) ( − p η p λ p ω ( e , e ) = 0 , (119) − η p λ p ω ( e , e ) + 12 (cid:16) ( − p + 1 (cid:17) η p λ p − ω ( e , e ) = 0 . (120)Since ω is non-degenerate there must exist i ∈ { , . . . , n }\{ } such that ω ( e i , e ) =0. If i > λ = 0. If i = 2 then (119)implies that λ = 0. Eventually, if i = 1 and ω ( e , e ) = 0 we obtain that λ = 0from (120). The proof of the theorem is completed. (cid:3) As a consequence of Theorem 5.6 we obtain
Theorem 5.7.
Let f : M → R n +1 ( dim M ≥ ) be a non-degenerate affine hyper-surface with a locally equiaffine transversal vector field ξ and an almost symplecticform ω . If R k ω = 0 for some k ≥ then the shape operator S has the rank ≤ .Proof. From Theorem 5.6 it follows that for every x ∈ M either S x = 0 or S x is ofthe form (114) thus rank S x ≤ x ∈ M . (cid:3) Recall that we have the following lemma ([21]).
Lemma 5.8 ([21]) . Let T be a tensor of type (0 , p ) and let ∇ be an affine torsion-free connection. Then for every k ≥ and for any k + p vector fields X ± , . . . , X k ± , Y , . . . , Y p the following identity holds: ( R k · T )( X , X − , . . . , X k , X k − , Y , . . . , Y p )(121) = X a ∈J sgn a ( ∇ k T )( X a (1) , X − a (1) , . . . , X ka ( k ) , X k − a ( k ) , Y , . . . , Y p ) , where J = { a : I k → {− , }} and sgn a := a (1) · . . . · a ( k ) . From Theorem 5.7 and Lemma 5.8 we have the following
Theorem 5.9.
Let f : M → R n +1 ( dim M ≥ ) be a non-degenerate affine hyper-surface with a locally equiaffine transversal vector field ξ and an almost symplecticform ω . If ∇ p ω = 0 for some p ≥ then the shape operator S has the rank ≤ . We conclude this section with the following example
Example 5.10.
Let n ≥ γ, α i : R → R n +1 be curves given as follows: γ ( t ) := (cos t, sin t, , . . . , T ; α i ( t ) := (0 , . . . , , ( i +3) z}|{ cos t, ( i +4) z}|{ sin t, , . . . , T FFINE HYPERSURFACES OF ARBITRARY SIGNATURE... 47 for i = 0 , . . . , n −
3. Let ε i ∈ {− , } for i = 1 , . . . , n − f : R \ { } × (0 , ∞ ) × (0 , π ) n − → R n +1 givenby the formula: f ( x, y, z , . . . , z n − ) := yγ ′ ( x ) + xα ( z ) + n − X i =1 ε i α i ( z i )together with the transversal vector field ξ : R \ { } × (0 , ∞ ) × (0 , π n − ∋ ( x, y, z , . . . , z n − )
7→ − γ ( x ) ∈ R n +1 . By straightforward computations we get h = · · ·
01 0 0 0 · · ·
00 0 xy · · ·
00 0 0 ε y sin( z )cos( z ) · · · · · · ε n − y sin( z )cos( z n − ) Q n − i =1 tan z i and S = · · ·
01 0 0 · · ·
00 0 0 · · · · · · , τ = 0 . Thus f is a non-degenerate equiaffine hypersurface with the second fundamentalform of signature (1 , − , ε , ε , . . . , ε n − ) where ε = 1 for x > ε = − x <
0. Note also that R = 0 since S = 0.Now let us define Ω = [ ω i,j ] i,j =1 ... n such that ω i,j = − ω j,i and det Ω = 0. That is Ω is a symplectic form. We easilycheck that R Ω (cid:16) ∂∂x , ∂∂z , ∂∂x , ∂∂z (cid:17) = − xyω , and R Ω (cid:16) ∂∂x , ∂∂z , ∂∂x , ∂∂z , ∂∂x , ∂∂z (cid:17) = xyω , thus R Ω = 0 and R Ω = 0 if only ω , = 0 , ω , = 0. On the other hand one mayshow that R Ω = 0.
This Research was financed by the Ministry of Science and Higher Education ofthe Republic of Poland.
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