The Clifford algebra and the Chevalley map - a computational approach (detailed version 1 ) Darij Grinberg Version 0.6 (3 June 2016). Not proofread!

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1 The Cliord algebra and the Chevalley map - a computational approach detailed version 1 Darij Grinberg Version June Not prooread! 1. Introduction: the Cliord algebra The theory o the Cliord algebra o a vector space with a given symmetric bilinear orm is rather well-understood: One o the basic properties o the Cliord algebra gives an explicit basis or it in terms o a basis o the underlying vector space Theorem 1 below, and another one provides a canonical vector space isomorphism between the Cliord algebra and the exterior algebra o the same vector space the so-called Chevalley map, Theorem 2 below. While both o these properties appear in standard literature such as [1] and [2], sadly I have never seen them proven in the generality they deserve: irst, the bilinear orm needs not be symmetric 2. Besides, the properties still hold over arbitrary commutative rings rather than just ields o characteristic 0. The proos given in literature are usually not suicient to cover these general cases. Here we are going to present a computational proo o both o these properties, giving integral 3 recursive ormulas or the vector space isomorphism 4 between the Cliord algebra and the exterior algebra in both directions. Remark added in As I now know, most o what is done in this paper is not new. In particular, its main results already appear in 9 o Chapter IX o [7] 5 ; they also essentially appear in Chapter 2 o [8] 6 ; the main ideas also appear in 1.7 o Chapter IV o [9] 7. Moreover, the proos given in [7], in [8] and in [9] are essentially the same as ours. Moreover, similar ideas and a variant o our map α have been used or dierent purposes in [10]. The results in Sections o this paper might still be new. First, let us deine everything in maximal generality: Deinition 1. In this note, a ring will always mean a ring with 1. I k is a ring, a k-algebra will mean a not necessarily commutative k-algebra with 1. Sometimes we will use the word algebra as an abbreviation or 1 This is a version including all the proos o the results given in [0]. While it is sel-contained and detailed, I would recommend any reader to read the much shorter summary [0] and consult this detailed version only in case o unclarity. 2 although this is not a substantial generalization as long as we are working over a ield k with characteristic 2 3 in the sense o: no division by k! 4 or, respectively, module isomorphism i we are working over a commutative ring instead o a ield 5 More precisely: Our Theorem 33 is Proposition 3 in 9 o Chapter IX o [7] and thus, our Theorem 1 is a consequence o said proposition; our Theorem 2 is a particular case or L {1, 2,..., n} o Théorème 1 in 9 o Chapter IX o [7]. 6 More precisely, Theorem 2.16 in Chapter 2 o [8] includes both our Theorem 1 and our Theorem 2 in the case when the k-module L is initely generated and projective. But the proo given in [8], as ar as it concerns our Theorem 2.16, does not require the initely generated and projective condition. 7 Thanks to Rainer Schulze-Pillot or making me aware o [9]. 1

2 k-algebra. I L is a k-algebra, then a let L-module is always supposed to be a let L-module on which the unity o L acts as the identity. Whenever we use the tensor product sign without an index, we mean k. Deinition 2. Let k be a commutative ring. Let L be a k-module. A bilinear orm on L means a bilinear map : L L k. A bilinear orm on L is said to be symmetric i it satisies x, y y, x or any x L and y L. Deinition 3. Let k be a commutative ring. Let L be a k-module, and : L L k be a bilinear orm on L. For every i N, we deine the so-called i-th tensor power L i o L to be the k-module L } L {{... L }. i times The tensor algebra L o L over k is deined to be the algebra L L 0 L 1 L 2..., where the multiplication is given by the tensor product. Now, we deine the Cliord algebra Cl L, to be the actor algebra L I, where I is the two-sided ideal o the algebra L. 8 L v v v, v v L L Remark. We denote by 0 the symmetric bilinear orm on L deined by 0 x, y 0 or every x L and y L. Then, I 0 L v v 0 v, v 0 v L L L v v v L L, and thus Cl L, 0 L I 0 L L v v v L L is the exterior algebra L o the k-module L. Hence, the exterior algebra L is a particular case o the Cliord algebra - namely, it is the Cliord algebra Cl L, 0. In general, the Cliord algebra Cl L, is not isomorphic to the exterior algebra L as algebra. However, they are isomorphic as k-modules, as the ollowing theorem states: Theorem 1 Chevalley map theorem: Let k be a commutative ring. Let L be a k-module, and : L L k be a bilinear orm on L. Then, the k-modules L and Cl L, are isomorphic. We are going to prove this theorem by explicitly constructing mutually inverse homomorphisms in both directions. This proo substantially diers rom the proos given in standard literature or the particular case o k being a ield o characteristic 0 and L being a inite-dimensional k-vector space, which proceed by constructing the isomorphism in one direction and showing either its injectivity or its surjectivity, or 8 Here, whenever U is a set, and P : U L is a map not necessarily a linear map, we denote by P v v U the k-submodule o L generated by the elements P v or all v U. 2

3 proving both using the basis theorem Theorem 2 below. 9 Using Theorem 1 we will be able to construct a basis or Cl L, in the case when L has one: Theorem 2 Cliord basis theorem: Let k be a commutative ring. Let L be a ree k-module with a inite basis e 1, e 2,..., e n, and : L L k be a bilinear orm on L. Let ϕ : L Cl L, be the k-module homomorphism deined by ϕ proj inj, where inj : L L is the canonical injection o the k-module L into its tensor algebra L, and where proj : L Cl L, is the canonical projection o the tensor algebra L onto its actor algebra L I Cl L,. Then, ϕ e i is a basis o the k-module Cl L,, where I P{1,2,...,n} P {1, 2,..., n} denotes the power set o the set {1, 2,..., n}. Here, we are using the ollowing notation: Deinition 4. Let A be a ring, and let I be a inite subset o Z. Let a i be an element o A or each i I. Then, we denote by a i the element o A deined as ollows: We write the set I in the orm I {i 1, i 2,..., i l } with i 1 < i 2 <... < i l in other words, we let i 1, i 2,..., i l be the elements o I, written down in ascending order. Then, we deine a i as the product a i1 a i2...a il. This product a i is called the ascending product o the elements a i o A. One more theorem that is oten silently used and will ollow rom our considerations: Theorem 3. Let k be a commutative ring. Let L be a k-module, and : L L k be a bilinear orm on L. Let ϕ : L Cl L, be the k-module homomorphism deined by ϕ proj inj, where inj : L L is the canonical injection o the k-module L into its tensor algebra L, and where proj : L Cl L, is the canonical projection o the tensor algebra L onto its actor algebra L I Cl L,. Then, the homomorphism ϕ is injective. Theorem 2 is known in the case o k being a ield and L being a inite-dimensional k-vector space; in this case, it is oten proved using orthogonal decomposition o L into -orthogonal subspaces - a tactic not available to us in the general case o k being an arbitrary commutative ring. We will have to derive Theorem 2 rom Theorem 1 to 9 The proo o Theorem 1 in [2] where Theorem 1 appears as Theorem 1.2, albeit only in the case o k being a ield seems dierent, but I don t completely understand it; to me it seems that it has a law it states that the r-homogeneous part o ϕ is then o the orm ϕ r a i v i v i b i where deg a i +deg b i r 2 or each i, which I am not sure about, because theoretically one could imagine that the representation o ϕ in the orm ϕ a i v i v i + q v i b i involves some a i and b i o extremely huge degree which cancel out in the sum. 3

4 prove it in this generality. Most proos o Theorem 1 rely on Theorem 2, and Theorem 3 is usually proven using either Theorem 1 or Theorem 2. The nature o our proo will be computational - we are going to deine some k- module automorphisms o the tensor algebra L by recursive ormulae. During the course o the proo, we will show a lot o ormulas, each o which has a more or less straightorward inductive proos. The inductive proos will always use one and the same tactic: a tactic I call tensor induction. Here is what it is about: Deinition 5. a Let k be a commutative ring, and L be a k-module. Let p N. An element o L p is said to be let-induced i and only i it can be written in the orm u U or some u L and some U L p 1. Then, or every p N +, the k-module L p is generated by its let-induced elements because L p L L p 1, and thereore the k-module L p is generated by its elements o the orm u U or some u L and some U L p 1 ; in other words, the k-module L p is generated by its let-induced elements. b Let k be a commutative ring, and L be a k-module. Let p N. An element o L p is said to be right-induced i and only i it can be written in the orm U u or some u L and some U L p 1. Then, or every p N +, the k-module L p is generated by its right-induced elements because L p L p 1 L, and thereore the k-module L p is generated by its elements o the orm U u or some u L and some U L p 1 ; in other words, the k-module L p is generated by its right-induced elements. The let tensor induction tactic. Let p N +. Let η and ε be two k-linear maps rom L p to some other k-module. Then, in order to prove that η T ε T or every T L p, it is enough to prove that η T ε T or every let-induced T L p because the k-module L p is generated by its let-induced elements. In words: In order to prove that all elements o L p satisy some given k-linear equation, it is enough to show that all let-induced elements o L p satisy this equation. The right tensor induction tactic. Let p N +. Let η and ε be two k-linear maps rom L p to some other k-module. Then, in order to prove that η T ε T or every T L p, it is enough to prove that η T ε T or every right-induced T L p because the k-module L p is generated by its right-induced elements. In words: In order to prove that all elements o L p satisy some given k-linear equation, it is enough to show that all right-induced elements o L p satisy this equation. 4

5 The tensor algebra induction tactic. Let η and ε be two k-linear maps rom L to some other k-module. Then, in order to prove that η T ε T or every T L, it is enough to prove that η T ε T or every p N and every T L p because the k-module L is the direct sum L 0 L 1 L 2..., and thereore is generated by its submodules L p or all p N. 2. Let interior products on the tensor algebra From now on, we ix a commutative ring k, and a k-module L. Let be some bilinear orm on L. First, we deine some operations o L on L - the so-called interior products. Our deinition will be rather dry - i you want a ormula or these operations, scroll down to Theorem 5 below. Deinition 6. Let : L L k be a bilinear orm. For every p N and every v L, we deine a k-linear map δ v,p : L p L p 1 where L 1 means 0 by induction over p: Induction base: For p 0, we deine the map δ v,p : L 0 L 1 to be the zero map. Induction step: For each p N +, we deine a k-linear map δ v,p : L p L p 1 by δ v,p u U v, u U u δ v,p 1 U or every u L and U L p 1, 1 assuming that we have already deined a k-linear map δ v,p 1 : L p 1 L p 2. This deinition is justiied, because in order to deine a k-linear map rom L p to some other k-module, it is enough to deine how it acts on tensors o the orm u U or every u L and U L p 1, as long as this action is bilinear with respect to u and U. This is because L p L L p 1. This way we have deined a k-linear map δ v,p : L p L p 1 or every p N. We can combine these maps δ v,0, δ v,1, δ v,2,... into one k-linear map δ v : L L since L L 0 L 1 L 2..., and the ormula 1 rewrites as δ v u U v, u U u δ v U or every u L and U L p 1. 2 It is easily seen by induction over p N that the map δ v,p depends linearly on the vector v L. Hence, the combination δ v o the maps δ v,0, δ v,1, δ v,2,... must also depend linearly on v L. In other words, the map L L L, v, U δ v U 5

6 is k-bilinear. Hence, this map gives rise to a k-linear map δ : L L L, v U δ v U. We are going to denote δv U by v U or each v L and U L. Thus, the equality 2 takes the orm v u U v, u U u v U or every u L and U L p 1. The tensor v U is called the let interior product o v and U with respect to the bilinear orm. Let us note that many authors omit the in the notation ; in other words, they simply write or. However, we are going to avoid this abbreviation, as we aim at considering several bilinear orms at once, and omitting the name o the bilinear orm could lead to conusion. The above inductive deinition o is not particularly vivid. Here is an explicit ormula or albeit we are mostly going to avoid using it in proos: Theorem 5. Let : L L k be a bilinear orm. a For every λ k and every v L, we have v λ b For every u L and v L, we have v u v, u. 11 c Let u 1, u 2,..., u p be p elements o L. Let v L. Then, v u 1 u 2... u p 3 p 1 i 1 v, u i u 1 u 2... û i... u p. 4 Here, the hat over the vector u i means that the vector u i is being omitted rom the tensor product; in other words, u 1 u 2... û i... u p is just another way to write u 1 u 2... u }{{ i 1 u } i+1 u i+2... u p. tensor product o the tensor product o the irst i 1 vectors u l last p i vectors u l Proo o Theorem 5. a We have λ k L 0 and thus δv λ δ v,0 λ 0 λ 0 0. Thus, v λ δv λ 0, and Theorem 5 a is proven. b Applying 3 to U 1, we see that v u 1 v, u 1 u v 1 v, u 1 u 0 v, u. 0 by Theorem 5 a Since u 1 u, this rewrites as v u v, u. Thus, Theorem 5 b is proven. 10 Here, λ k is considered as an element o L by means o the canonical inclusion k L 0 L. 11 Here, v, u k is considered as an element o L by means o the canonical inclusion k L 0 L. 6

7 c We are going to prove Theorem 5 c by induction over p: The induction base is clear, since or p 0, Theorem 5 c trivially ollows rom Theorem 5 a 12. Now to the induction step: Let p N +. Let us prove Theorem 5 c or this p, assuming that we have already veriied Theorem 5 c applied to p 1 instead o p. In act, we have assumed that we have already shown Theorem 5 c applied to p 1 instead o p. In other words, we have already shown the equality p 1 v u 1 u 2... u p 1 1 i 1 v, u i u 1 u 2... û i... u p 1 5 or any p 1 vectors u 1, u 2,..., u p 1 in L. Now, our goal is to prove the equality 4 or any p vectors u 1, u 2,..., u p in L. In act, substituting the vectors u 2, u 3,..., u p instead o u 1, u 2,..., u p 1 into the already proven equality 5, we get p 1 v u 2 u 3... u p 1 i 1 v, u i+1 u 2 u 3... û i+1... u p p 1 i 2 v, u i u 2 u 3... û i... u p 1 i 1 i2 here, we have substituted i or i + 1 in the sum p 1 i 1 v, u i u 2 u 3... û i... u p. 6 i2 12 because or p 0, we have u 1 u 2... u p empty product 1 k and p 1 i 1 v, u i u 1 u 2... û i... u p empty sum 0 7

8 Now, applying 3 to u u 1 and U u 2 u 3... u p, we get v u 1 u 2 u 3... u p v, u 1 u 2 u 3... u p u 1 v, u 1 u 2 u 3... u p v,u 1 u 1 u 2... û 1... u p v u 2 u 3... u p p 1 i 1 v,u i u 2 u 3... û i... u p i2 by 6 p + u 1 1 i 1 v, u i u 2 u 3... û i... u p i2 p 1 i 1 v,u i u 1 u 2 u 3... û i... u p i2 p 1 i 1 v,u i u 1 u 2... û i... u p i v, u 1 u 1 u 2... û 1... u p p + 1 i 1 v, u i u 1 u 2... û i... u p i2 p 1 i 1 v, u i u 1 u 2... û i... u p. Thus, 4 is proven or our p N. In other words, Theorem 5 c is proven or our p N. This completes the induction step, and thus the proo o Theorem 5 c is complete. We are now going to prove some properties o the interior product. The most important one is the bilinearity o ; this property states that the map L L L, v, U v U is k-bilinear 13, i. e. that αv + βv U αv U + βv U and that v αu + βu αv U + βv U or any v L, v L, U L and U L. Theorem 6. I u L, U L, and v L, then v u U v, u U u v U This is because v U δ v U, and because the map L L L, v, U δ v U is k-bilinear. 8

9 Proo o 7. I U is a homogeneous tensor i. e. an element o L r or some r N, then 7 ollows directly rom 3 applied to p r + 1. Otherwise, we can write U as a k-linear combination o homogeneous tensors o various degrees, and then apply 3 to each o these tensors; summing up, we obtain 7. Thus, 7 is proven. Theorem 7. I v L and U L, then v v U 0. 8 Proo o Theorem 7. Fix some v L. First we will prove that or every p N and every U L p, the equation 8 holds. In act, we will show this by induction over p: The induction base p 0 is clear thanks to Theorem 5 a, which yields v U 0 or every U L 0 k. Now or the induction step: Fix some p N +. Let us now prove 8 or all U L p, assuming that 8 is already proven or all U L p 1. We want to prove 8 or all U L p. But in order to achieve this, it is enough to prove 8 or all let-induced U L p because o the let tensor induction tactic, since the equation 8 is linear in U. So let us prove 8 or all let-induced U L p. In act, let U L p be a let-induced tensor. Then, U can be written in the orm U u Ü or some u L and Ü L p 1 since U is let-induced. Then, v U v u Ü v, u Ü u v Ü by 7, applied to Ü instead o U yields v v U v v, u Ü u v Ü v, u v Ü v u by the bilinearity o v, u v Ü v, u v Ü u v v Ü since v u v Ü v, u v Ü u by 7, applied to v Ü instead o U u v v Ü 0 v Ü v v Ü because v v Ü 0 by 8 applied to Ü instead o U 14. Thus, we have proven that v v U 0 or all let-induced U L p. Consequently, by the let tensor induction tactic as we said above, we conclude that 8 holds or all U L p. This completes the induction step. Thereore we have now proven that or every p N, and every U L p, the equation 8 holds. This yields that the equation 8 holds or every U L since every element o L is a k-linear combination o elements o L p or various p N, and since the equation 8 is linear in U. This proves Theorem 7. Theorem 8. I v L, w L and U L, then v w U w v U In act, we are allowed to apply 8 to Ü instead o U, since Ü L p 1 and since we have assumed that 8 is already proven or all U L p 1. 9

10 First proo o Theorem 8. Theorem 7 yields v v U 0. Theorem 7, applied to w instead o v, yields w w U 0. Finally, Theorem 7, applied to v + w instead o v, yields v + w v + w U 0. Thus, 0 v + w v + w U v + w v U + w U by the bilinearity o v v U + w U + w v U + w U by the bilinearity o v v U +v w U + w v U + w w U 0 0 by the bilinearity o v w U + w v U. This yields 9, and thus Theorem 8 is proven. Second proo o Theorem 8. Fix some v L and w L. First we will prove that or every p N and every U L p, the equation 9 holds. In act, we will show this by induction over p: The induction base p 0 is clear since Theorem 5 a yields w U 0 and v U 0 in the case p 0. Now or the induction step: Fix some p N +. Let us now prove 9 or all U L p, assuming that 9 is already proven or all U L p 1. We want to prove 9 or all U L p. But in order to achieve this, it is enough to prove 9 or all let-induced U L p because o the let tensor induction tactic, since the equation 9 is linear in U. So let us prove 9 or all let-induced U L p. In act, let U L p be a let-induced tensor. Then, U can be written in the orm U u Ü or some u L and Ü L p 1 since U is let-induced. Thereore, v U v u Ü v, u Ü u v Ü by 7, applied to Ü instead o U yields w v U w v, u Ü u v Ü v, u w Ü w u v Ü by the bilinearity o v, u w Ü w, u v Ü w u v Ü since w u v Ü w, u v Ü u w v Ü by 7, applied to w and v Ü instead o v and U v, u w Ü w, u v Ü w + u v Ü. 10 Applying this equality 10 to w and v instead o v and w, we obtain v w U w, u v Ü v, u w Ü + u v w Ü. 10

11 Adding this equality to 10, we obtain w v U + v w U v, u w Ü w, u v Ü + u w v Ü + w, u v Ü v, u w Ü + u v w Ü u w v Ü + u v w Ü u w v Ü + v w Ü 0 because w v Ü + v w Ü to Ü instead o U 15 0, since v, and thereore v w Ü w w U w v Ü by 9 applied. Thus, we have v U proven that 9 holds or all let-induced U L p. Consequently, by the let tensor induction tactic as we said above, we conclude that 9 holds or all U L p. This completes the induction step. Thereore we have now proven that or every p N, and every U L p, the equation 9 holds. This yields that the equation 9 holds or every U L since every element o L is a k-linear combination o elements o L p or various p N, and since the equation 9 is linear in U. This proves Theorem 8. Theorem 9. I p N, u L, U L p, and v L, then v U u 1 p v, u U + v U u. 11 Instead o proving this directly, we show something more general: Theorem 10. I p N, v L, U L p, and V L, then v U V 1 p U v V + v U V. 12 Proo o Theorem 10. We are going to prove 12 by induction over p: The induction base p 0 is obvious 16. Now let us come to the induction step: Fix some p N + and some V L. Let us now prove 12 or all U L p, assuming that 12 is already proven or all U L p In act, we are allowed to apply 9 to Ü instead o U, since Ü L p 1 and since we have assumed that 9 is already proven or all U L p In act, in the case p 0, we have U L p L 0 k and thus 1 p v U V U v V and UV U v V + v U V 1U v V + 0 V U v V, } {{ } 0 by Theorem 5 a, since U k U v V since U k and thereore 12 is valid in the case p 0. 11

12 We want to prove 12 or all U L p. But in order to achieve this, it is enough to prove 12 or all let-induced U L p by the let tensor induction tactic, because the equation 12 is linear in U. Thus, let us prove 12 or all let-induced U L p. In other words, let us prove 12 or all tensors U L p o the orm U u Ü with u L and Ü L p 1 because every let-induced tensor U L p has the orm U u Ü or some u L and Ü L p 1. In other words, let us prove that v u Ü V 1 p u Ü or every u L, Ü L p 1, v L and V L. In act, u Ü V u Ü V yields v V + v u Ü V 13 v u Ü V v u Ü V v, u Ü V u v Ü V. by 7, applied to Ü V instead o U. But since v Ü V 1 p 1 Ü v V + v Ü V this ollows rom applying 12 to p 1 and Ü instead o p and U 17, this becomes v u Ü V v, u Ü V u 1 p 1 Ü v V + v Ü V v, u Ü V 1 p 1 u Ü v V u v Ü V v, u 1 p Ü V + 1 p u Ü v V u Comparing this to 1 p u Ü v V + v u Ü V 1 p u Ü v V + v, u Ü u v Ü V because 7, applied to Ü instead o U, yields v u Ü v, u Ü u v Ü 1 p u Ü v V + v, u Ü V u v Ü V v, u Ü V + 1p u Ü v V u v Ü V v, u Ü V + 1 p u Ü v V u v Ü V, v Ü V. we obtain 13. Hence, we have proven 13. As already explained above, this completes the induction step. Thus, 12 is proven or all p N. In other words, the proo o Theorem 10 is complete. 17 In act, we are allowed to apply 12 to p 1 and Ü instead o p and U, because we have assumed that 12 is already proven or all U L p 1. 12

13 Proo o Theorem 9. Applying Theorem 10 to V u, we obtain v U u 1 p U v u + v U u. Since v u v, u by Theorem 5 b, this becomes v U u 1 p U v, u + v U u 1 p v, u U + v U u, v,uu and thereore Theorem 9 is proven. Note that we will oten use a trivial generalization o Theorem 9 rather than Theorem 9 itsel: Theorem I p N, u L, U 2 i N; v U u 1 p v, u U + Proo o Theorem Since U i N; i N; L i, and v L, then v U u. 14 L i, we can write U in the orm U U i, where U i L i or every i N satisying i p mod 2. Now, 11 applied to i and U i instead o p and U yields v U i u 1 i v, u U i + v U i u or every i N satisying i p mod 2. Since 1 i 1 p because i p mod 2, this becomes v U i u 1 p v, u U i + v U i u

14 Now, U i N; v U u v i N; U i yields i N; 1 p v, u U i u 1 p v, u U i + i N; U i + i N; 1 p v, u U i + v i N; U 1 p v, u U + v U u. v U i i N; v U i u u v U i v U i i N; by the bilinearity o U i i N; } {{ } U u u by the bilinearity o by 15 This proves Theorem Finally, another straightorward act: Theorem Let : L L k and g : L L k be two bilinear 4 orms. I w L and U L, then w U + w g U w +g U. 16 This theorem is immediately trivial using Theorem 5 c, but as we want to avoid using Theorem 5 c, here is a straightorward proo o Theorem 10 3 using tensor 4 induction: Proo o Theorem Fix some w L. We will irst show that or every p N, 4 the equation 16 holds or every U L p. In act, we will prove this by induction over p: The induction base case p 0 is obvious because in this case, U L p L 0 k and thus Theorem 5 a yields w U 0, w U g 0 and w +g U 0, making the equation 16 trivially true. So let us now come to the induction step: Let p N +. We must prove 16 or every U L p, assuming that 16 has already been proven or every U L p 1. 14

15 We want to prove that 16 holds or every U L p. In order to do this, it is enough to prove that 16 holds or every let-induced U L p by the let tensor induction tactic, because the equation 16 is linear in U. So, let us prove this. Let U L p be a let-induced tensor. Then, we can write U in the orm U u Ü or some u L and Ü L p 1 because U is let-induced. Thus, w U w u Ü w, u Ü u w Ü by 7, applied to w and Ü instead o v and U. Also, w U g g w, u w Ü u Ü g by 17, applied to g instead o and w +g U + g w, u Ü u w +g Ü by 17, applied to + g instead o. Hence, w U + w U g w, u Ü u w Ü + g w, u Ü u w Ü g w, u Ü + g w, u Ü u w Ü + u w Ü g w,u+gw,uü u w Ü+wg Ü w, u + g w, u w Ü u Ü + w Ü g w, u + g w, u Ü u w +g Ü +gw,u since 16 applied to Ü instead o U yields w Ü + w Ü g w+g Ü in act, we are allowed to apply 16 to Ü instead o U, since Ü L p 1 and since 16 has already been proven or every U L p 1 + g w, u Ü u w +g Ü w +g U. Hence, the equality 16 is proven or every let-induced tensor U L p. As we already said above, this entails that 16 must also hold or every tensor U L p, and thus the induction step is complete. Hence, 16 is proven or every p N and every U L p. Consequently, the equation 16 holds or every U L since every U L is a k-linear combination o elements o L p or various p N, and since the equation 16 is k-linear. In other words, Theorem is proven. 3. Right interior products on the tensor algebra We have proven a number o properties o the interior product. We are now going to introduce a very analogous construction which works rom the right almost the same way as works rom the let : 15 17

16 Deinition 7. Let : L L k be a bilinear orm. For every p N and every v L, we deine a k-linear map ρ v,p : L p L p 1 where L 1 means 0 by induction over p: Induction base: For p 0, we deine the map δ v,p : L 0 L 1 to be the zero map. Induction step: For each p N +, we deine a k-linear map δ v,p : L p L p 1 by ρ v,p U u u, v U ρ v,p 1 U u or every u L and U L p 1, 18 assuming that we have already deined a k-linear map ρ v,p 1 : L p 1 L p 2. This deinition is justiied, because in order to deine a k-linear map rom L p to some other k-module, it is enough to deine how it acts on tensors o the orm U u or every u L and U L p 1, as long as this action is bilinear with respect to u and U. This is because L p L p 1 L. This way we have deined a k-linear map ρ v,p : L p L p 1 or every p N. We can combine these maps ρ v,0, ρ v,1, ρ v,2,... into one k-linear map ρ v : L L since L L 0 L 1 L 2..., and the ormula 1 rewrites as ρ v u U u, v U ρ v U u or every u L and U L p It is easily seen by induction over p N that the map ρ v,p depends linearly on the vector v L. Hence, the combination ρ v o the maps ρ v,0, ρ v,1, ρ v,2,... must also depend linearly on v L. In other words, the map L L L, U, v ρ v U is k-bilinear. Hence, this map gives rise to a k-linear map ρ : L L L, U v ρ v U. We are going to denote ρ v U by U v or each v L and U L. Thus, the equality 2 takes the orm U u v u, v U U v u or every u L and U L p The tensor U v is called the right interior product o v and U with respect to the bilinear orm. Again, many authors omit the in the notation ; in other words, they simply write or. However, we are going to avoid this abbreviation, as we aim at considering several bilinear orms at once, and omitting the name o the bilinear orm could lead to conusion. 16

17 Everything that we have proven or has an analogue or. In act, we can take any identity concerning, and read it rom right to let to obtain an analogous property o 18 For instance, reading the property 3 o rom right to let, we obtain 20, because reading the term v u U rom right to let means replacing it by U u v; reading the term v, u U rom right to let means replacing it by U u, v u, v U since u, v k is a scalar; reading the term u v U rom right to let means replacing it by U v u. I we take a theorem about the let interior product or example, one o the Theorems 5-10, and read it rom right to let, we obtain a new theorem about the right interior product, and this new theorem is valid because we can read not only the theorem, but also its proo rom right to let. This way, we get the ollowing new theorems: Theorem 11. Let : L L k be a bilinear orm. a For every λ k and every v L, we have λ v b For every u L and v L, we have u v u, v. 20 c Let u 1, u 2,..., u p be p elements o L. Let v L. Then, u 1 u 2... u p v p 1 p i u i, v u 1 u 2... û i... u p. 21 Here, the hat over the vector u i means that the vector u i is being omitted rom the tensor product; in other words, u 1 u 2... û i... u p is just another way to write u 1 u 2... u }{{ i 1 u } i+1 u i+2... u p. tensor product o the tensor product o the irst i 1 vectors u l last p i vectors u l 18 Reading rom right to let means replacing every term o the orm v U by U v where v L and U L, and vice versa; reversing the order in every tensor product; replacing every u, v by v, u. However, some care must be taken here: when our identity is o the orm sum o some terms involving vectors, tensors and and signs another sum o terms o this kind, then we should not read each o the sums rom right to let, but we should read every o their terms rom right to let. Thus, reading a term like a b c d rom right to let, we get b a d c, and not d c b a. 19 Here, λ k is considered as an element o L by means o the canonical inclusion k L 0 L. 20 Here, u, v k is considered as an element o L by means o the canonical inclusion k L 0 L. 17

18 Theorem 12. I u L, U L, and v L, then U u v u, v U U v u. 22 Theorem 13. I v L and U L, then U v v Theorem 14. I v L, w L and U L, then U w v U v w. 24 Theorem 15. I p N, u L, U L p, and v L, then u U v 1 p u, v U + u U v. 25 Theorem 16. I p N, v L, U L p, and V L, then V U v 1 p V v U + V U v. 26 Theorem I p N, u L, U 2 i N; u U v 1 p u, v U + u L i, and v L, then U v. 27 These Theorems are simply the results o reading Theorems 5-10 rom right to let, so as we said, we don t really need to give proos or them because one can simply read the proos o Theorems 5-10 rom right to let, and thus obtain proos o Theorems Yet, we are going to present the proo o Theorem 11 explicitly 21, and we will later reprove Theorems in a dierent way. Proo o Theorem 11. a We have λ k L 0 and thus ρ v λ ρ v,0 0 0 λ 0. Thus, λ v δv λ 0, and Theorem 11 a is proven. b Applying 20 to U 1, we see that 1 u v u, v 1 1 v u u, v 1 0 u u, v. 0 by Theorem 11 a Since 1 u u, this rewrites as u v u, v. Thus, Theorem 11 b is proven. c We are going to prove Theorem 11 c by induction over p: λ 21 This is mainly because Theorem 11 does not result verbatim rom reading Theorem 5 rom right to let, but instead requires some more changes such as renaming u 1 u 2... u p by u p u p 1... u 1, and renaming i 1 by p i. 18

19 The induction base is clear, since or p 0, Theorem 11 c trivially ollows rom Theorem 11 a 22. Now to the induction step: Let p N +. Let us prove Theorem 11 c or this p, assuming that we have already shown Theorem 11 c applied to p 1 instead o p. In act, we have assumed that we have already shown Theorem 11 c applied to p 1 instead o p. In other words, we have already shown the equality p 1 u 1 u 2... u p 1 v 1 p 1 i u i, v u 1 u 2... û i... u p or any p 1 vectors u 1, u 2,..., u p 1 in L. Now, our goal is to prove the equality 21 or any p vectors u 1, u 2,..., u p in L. 22 because or p 0, we have u 1 u 2... u p empty product 1 k and p 1 p i v, u i u 1 u 2... û i... u p empty sum 0 19

20 Applying 20 to u u p and U u 1 u 2... u p 1, we get u 1 u 2... u p 1 u p v u p, v u 1 u 2... u p 1 u p, v 1 p p u p,v p 1 u 1 u 2... u p 1 u 1 u 2... u p 1 v p 1 1 p 1 i u i,v u 1 u 2... û i... u p 1 by 28 1 p 1 i 1 p i 1 1 p i u i, v u 1 u 2... û i... u p 1 1 p p u p, v u 1 u 2... u p 1 p 1 1 p i u i, v u 1 u 2... û i... u p 1 u p 1 p p u p, v u 1 u 2... u p 1 + p 1 u 1 u 2... û p... u p 1 p i u i, v u 1 u 2... û i... u p 1 u p p 1 1 p i u i,v u 1 u 2... û i... u p 1 u p p 1 1 p i u i,v u 1 u 2... û i... u p 1 p p u p, v u 1 u 2... û p... u p p p i u i, v u 1 u 2... û i... u p p 1 p i u i, v u 1 u 2... û i... u p. u p u p Thus, 21 is proven or our p N. In other words, we have proven Theorem 11 c or our p N. This completes the induction step, and thus the proo o Theorem 11 c is complete. Let us notice another property o : the bilinearity o. This property states that the map L L L, U, v U v 20

21 is k-bilinear 23, i. e. that U αv + βv αu v + βu v and that αu + βu v αu v + βu v or any v L, v L, U L and U L. As we said above, Theorems don t need to be proven in details, because simply reading the proos o Theorems 6-10 rom right to let yields proos o Theorems However, there also is a dierent way to prove these theorems, namely by deining an automorphism o the k-module L: Deinition 8. For every p N, we deine an endomorphism t p : L p L p o the k-module L p by t p u 1 u 2... u p u p u p 1... u 1 or any vectors u 1, u 2,..., u p in L These endomorphisms t 0, t 1, t 2,... can be combined together to an endomorphism t : L L o the k-module L since L L 0 L 1 L 2... This map t satisies t u 1 u 2... u p u p u p 1... u 1 30 or any p N and any vectors u 1, u 2,..., u p in L. due to 29. This obviously yields t 2 id bijective. Besides, 25. Hence, the map t : L L is t U V t V t U or every U L and V L Also, obviously, t u u or every u L. Our use or the map t is now to reduce the right interior product to the let interior product. For this we need yet another deinition: 23 This is because U v ρ v U, and because the map L L L, U, v ρ v U is k-bilinear. 24 This deinition is legitimate, because the map L L... L L p given by p times u 1, u 2,..., u p u p u p 1... u 1 or any vectors u 1, u 2,..., u p in L is k-multilinear, and thus yields a map t p : L p L p satisying In act, t 2 u 1 u 2... u p t t u 1 u 2... u p t u p u p 1... u 1 u 1 u 2... u p u p u p 1... u 1 due to 30 due to 30, applied to u p, u p 1,..., u 1 instead o u 1, u 2,..., u p or any p N and any vectors u 1, u 2,..., u p in L. Thus, t 2 U U or every U L because every U L is a k-linear combination o tensors o the orm u 1 u 2... u p or p N and vectors u 1, u 2,..., u p in L, and because the equation t 2 U U is linear in U. In other words, t 2 id, qed. 26 Proo o 31. We WLOG assume that U u 1 u 2... u q or some q N and some vectors 21

22 Deinition 9. Let : L L k be a bilinear orm. Then, we deine a new bilinear orm t : L L k by t u, v v, u or every u L and v L. This bilinear orm t is called the transpose o the bilinear orm. It is clear that t t or any bilinear orm, and that a bilinear orm is symmetric i and only i t. Now, here is a way to write in terms o : t Theorem 17. Let v L and U L. Then, t U v v t t U 32 and t v U t t U v. 33 Proo o Theorem 17. Fix some v L. We are irst going to prove that t U v v t t U or every p N and every U L p. In act, we will prove this by induction over p: The induction base case p 0 is trivial 27. So let us come to the induction step: Let p N +. Assume that we have already proven t U v v t t U or every U L p Now we must prove t U v v t t U or every U L p. In order to do this, it is clearly enough to prove t U v v t t U or every right-induced U L p by the right tensor induction tactic, because the equation t U v v t t U is linear in U. u 1, u 2,..., u q in L. In act, this assumption is legitimate, since every U L can be written as a k-linear combination o tensors o the orm u 1 u 2... u q or q N and vectors u 1, u 2,..., u q in L, and since the equation 31 is linear in U. We also WLOG assume that V u q+1 u q+2... u p or some p N and or some vectors u q+1, u q+2,..., u p in L. In act, this assumption is legitimate, since every V L can be written as a k-linear combination o tensors o the orm u q+1 u q+2... u p or p N and vectors u q+1, u q+2,..., u p in L, and since the equation 31 is linear in V. Then, U V u 1 u 2... u q u q+1 u q+2... u p u 1 u 2... u p, so that t U V t u 1 u 2... u p u p u p 1... u 1 u p u p 1... u q+1 u q u q 1... u 1 t V t U, tu q+1 u q+2... u ptv tu 1 u 2... u qtu so that 31 is proven. 27 In act, in the case p 0, we have U L p L 0 k and thus U v 0 by Theorem 11 a and v t t U 0 by Theorem 5 a, applied to t instead o, which makes the assertion t U v v t t U obvious. 22

23 But this is easy: I U L p is a right-induced tensor, then U Ü u or some u L and Ü L p 1, and thus t U v v t Ü u t u, v Ü Ü v u v since Ü u u, v Ü Ü v u by 20 applied to Ü instead o U u, v t Ü t Ü v u u, v t Ü u, v t Ü u t tu t Ü v by 31, applied to Ü v and u instead o U and V t u t u since u L Ü v Ü v since t is k-linear u, v t Ü u v t Ü t since Ü L p 1 yields t Ü v v t Ü t, according to 34 applied to Ü instead o U and t U v t v t t u u since u L t Ü since U Ü Ü u yields t U t u t u t Ü by 31 v t u t Ü applied to Ü and u instead o U and V t v, u t Ü u v Ü t u,v by 3, applied to t and t Ü u, v t Ü u v Ü t instead o and U lead to t U v v t t U. This completes our induction step, and thus we have proven that t U v v t t U or every p N and every U L p. This immediately yields that t U v v t t U or every U L because every U L is a k- linear combination o elements o L p or dierent p N, and since the equation t U v v t t U is linear in U. Thus, 32 is proven. In order to prove Theorem 17, it now only remains to prove 33. In act, applying 32 to t U instead o U, we obtain t 23 t U v v t t t U t 2 UU since t 2 id

24 v U. t Thus, t t t U v t v U t. Since t t U v since t 2 id, this becomes t U v t t t U v t t 2 U v, and thus 33 is proven. v t U So we have now proven both 32 and 33. Thus, Theorem 17 is proven. Now, Theorem 17 enables us to prove Theorems quickly: Proo o Theorem 12. We have t U v u t u u since u L t U v by 31, applied to U v and u instead o U and V u t U v u v t t U by and t U u v v t t U u by 32, applied to U u instead o U v t t u u since u L t U since 31 applied to V u yields t U u t u t U v t u t U t v, u t U u v t t U u,v by 7, applied to t and t U instead o and U u, v t U u v t t U t u, v U t U v u by 35 U v u u, v t U t U v u since the map t is k-linear. Since t is injective because t is bijective, this yields U u v u, v U U v u. This proves Theorem 12. Proo o Theorem 13. The equation 32, applied to U v instead o U, yields t U v v v t t U v v t v t t U by 32 0 by 8, applied to t instead o. This proves Theorem

25 Proo o Theorem 14. The equation 32, applied to U v and w instead o U and v, yields t U v w w t t U v w t v t t U by 32. Similarly, t w t v t U t U w U w v v t v v t w t t U. Together with the equality v t w U t which ollows rom 9, applied to t instead o, this yields w t t U w t v U t } {{ } t U v w t U v w t U v w. Since t is injective because t is bijective, this results in U w v U v w, which proves Theorem 14. Proo o Theorem 16. Applying 32 to V U instead o U, we get t V U v v t t V U v t t U t V due to 31, applied to V and U instead o U and V 1 p t U v t t V + v t t U t V 36 by 12, applied to t, t U and t V instead o, U and V. On the other hand, 31 applied to V v and U instead o U and V leads to t V v U t U t V v t U v t tv by 32, applied to V instead o U v t t V. Also, 31 applied to V and U v instead o U and V leads to t V U v t U v t V v t t U t V. Thus, v t tu by 32 t 1 p V v U + V U v 1 p t V v U + t V U v tu 1 p t U v t tv v t t V + v t tu tv v t t U t V t V U v 25

26 by 36. Since t is injective because t is bijective, this entails 1 p V v U + V U v V U v. Thus, Theorem 16 is proven. Proo o Theorem 15. Applying Theorem 16 to V u, we obtain u U v 1 p u v U + u U v. Since u v u, v by Theorem 11 b, this becomes u U v 1 p u, v U +u U v 1 p u, v U + u u,vu and thereore Theorem 15 is proven. Proo o Theorem Applying 32 to u U instead o U, we get 2 t u U v v t t u U v t t U t u U v, due to 31, applied to u and U instead o U and V v t t U u since t u u, because u L 1 p t v, u t U + v t t U u 37 by 14 applied to t and t U instead o and U, since U t U i N; i N; L i yields L i because the map t is composed o the maps t i : L i L i or all i N, and thus maps L i into L i or all i N. On the other hand, 31 applied to u and U v instead o U and V leads to t u U v t U v v t tu by 32 t u u since u L v t t U u. Since the map t is k-linear, we now have t 1 p u, v U + u U v 1 p u, v t U + t t v,u 1 p t v, u t U + t u U v u U v v t tu v t t U u u 26

27 by 37. Since t is injective because t is bijective, this entails 1 p u, v U + u U v u U v. Thus, Theorem is proven. This proo o Theorem yields a new proo o Theorem 15. O course, Theorem has its right counterpart as well: Theorem Let : L L k and g : L L k be two bilinear 4 orms. I w L and U L, then U w + U g w U +g w. We won t prove this theorem, since we won t ever use it and since it should now be absolutely clear how to derive it rom Theorem prove it analogously to Theorem The two operations commute with the help o 32 or how to Now that we know quite a lot about each o the operations and, let us show a relation between them: Theorem 18. Let v L, w L and U L. Then v U w v U w. 38 More generally, i : L L k and g : L L k are two bilinear orms, then v U w g v U g w. 39 Proo o Theorem 18. In order to prove Theorem 18, it is enough to prove the equality 39 only because the equality 38 directly ollows rom the equality 39, applied to g. So let us prove the equality 39. First, let us prove that or every p N, every U L p satisies the equation 39. In act, we are going to prove this by induction: The base case o our induction - the case p 0 - is evident 28. Now the induction step: Let p N +. Assume that we have proven that every U L p 1 satisies the equation 39. Now, in order to complete the induction step, we have to show that every U L p satisies the equation 39. In order to achieve this goal, it will be enough to show that every let-induced U L p 28 In act, in the case p 0, every U L p is a scalar since L p L 0 k, and thus U g w 0 by Theorem 11 a, applied to g instead o and v U 0 by Theorem 5 a, so that the equation 39 trivially holds. 27

28 satisies the equation So let us prove this: For every let-induced U L p, we can write U in the orm U u Ü or some u L and some Ü L p 1, and thereore U satisies v U w g v g w u Ü since U u Ü v 1 p 1 g u, w Ü Ü + u g w since 25 applied to p 1, g, Ü and w instead o p,, U and v yields g w u Ü 1 p 1 g u, w Ü Ü + u g w because Ü L p 1 g u Ü w 1 p 1 g u, w v Ü + v 1 p 1 g u, w v Ü + v, u Ü w g u v Ü g w since 7, applied to Ü w g instead o U, yields v g u Ü w v, u Ü w g u v Ü g w 1 p 1 g u, w v Ü + v, u Ü w g u v Ü g w because v Ü g w v Ü g w, which ollows rom applying the equality 39 to Ü instead o U 30 and v U g w v g w u Ü since U u Ü v, u Ü u v Ü g w since 7, applied to Ü instead o U, yields v u Ü v, u Ü u v Ü v, u Ü w g u v Ü g w v, u Ü w g 1 p 2 g u, w v Ü + u v Ü g w since 25 applied to p 2, g, v Ü and w instead o p,, U and v yields u v Ü g w 1 p 2 g u, w v Ü + u v Ü g w 1 p 2 1 p 1 g u, w v Ü + v, u Ü w g u 1 p 1 g u, w v Ü + v, u Ü g w u because v Ü L p 2 v Ü g w v Ü g w v U w g 29 This ollows rom the let tensor induction tactic, because the equation 39 is linear in U. 30 In act, we are allowed to apply the equality 39 to Ü instead o U, since Ü L p 1 and since we have proven that every U L p 1 satisies the equation 39 by our assumption

29 by 40. Thus, we have proven the equality 39. Hence, every let-induced U L p satisies the equation 39. As we said above, this yields that every U L p satisies the equation 39. This completes our induction step, and thus we have shown that or every p N, every U L p satisies the equation 39. Consequently, every U L satisies the equation 39 because every U L is a k-linear combination o elements o L p or various p N, and because the equation 39 is linear in U. This proves Theorem The endomorphism α We are now going to deine an endomorphism α : L L which depends on the bilinear orm : Deinition 10. Let : L L k be a bilinear orm. For every p N, we deine a k-linear map α p : L p L by induction over p: Induction base: For p 0, we deine the map α p : L 0 L to be the canonical inclusion o L 0 into the tensor algebra L L 0 L 1 L 2... In other words, we deine the map α 0 : k L by α p λ λ or every λ k L 0. Induction step: For each p N +, we deine a k-linear map α p : L p L by α p u U u α p 1 U u α p 1 U or every u L and U L p 1, 41 assuming that we have already deined a k-linear map α p 1 : L p 1 L. This deinition is justiied, because in order to deine a k-linear map rom L p to some other k-module, it is enough to deine how it acts on tensors o the orm u U or every u L and U L p 1, as long as this action is bilinear with respect to u and U. This is because L p L L p 1. This way we have deined a k-linear map α p : L p L or every p N. We can combine these maps α 0, α 1, α 2,... into one k-linear map α : L L since L L 0 L 1 L 2..., and the ormula 41 rewrites as α u U u α U u α U or every u L and U L p 1. We note that, in contrast to the map δv which maps every homogeneous tensor rom L p to L p 1, the map α can map homogeneous tensors to inhomogeneous tensors. This endomorphism α now turns out to have plenty o properties. But irst let us 31 irst evaluate it on pure tensors o low rank 0, 1, 2, 3, 4: Action o α on tensors o rank 0: For any λ k, we have α λ λ, where we consider λ as an element o L through the canonical injection k L 0 L. In act, λ k L 0 yields α λ α 0 λ λ by the deinition o α0. 31 Note that these evaluations will never be used in uture except o 43 and 44, which are pretty much trivial, so there is no need to read them. But I think they provide a good intuition or what the map α does to tensors o low degrees

30 Action o α on tensors o rank 1: For any u L, we have α u α u 1 u α 1 u α 1 1 since 1 k 0 by Theorem 5 a, since α 11 k by 42, applied to U 1 u 1 0 u 1 u. 43 Action o α on tensors o rank 2: For any u L and v L, we have α u v u α v u α v v by 43, applied to v by 43, applied to v instead o u v instead o u by 42, applied to U v u v u v u v u, v. 44 u,v according to Theorem 5 b, applied to v and u instead o u and v Action o α on tensors o rank 3: For any u L, v L and w L, we have α u v w u α v w u α v w by 42, applied to U v w u v w v, w u v w v, w since 44 applied to v and w instead o u and v yields α v w v w v, w u v w u v, w v,wu u v w v, w u u, v w v u v w u,vw v u w by 7, applied to u, v and w instead o v, u and U u v w v, w u u, v w v u, w 0 u,wv u v w v, w u u, v w u, w v 0 u v, w u w u,w by Theorem 5 b, 0 by Theorem 5 a, since v,w k applied to u and w instead o v and u 0 u v w v, w u + u, w v u, v w. 45 Action o α on tensors o rank 4: For any u L, v L, w L and x L, we 30

31 have α u v w x u α v w x u α v w x by 42, applied to U v w x u v w x w, x v + v, x w v, w x u v w x w, x v + v, x w v, w x since 45 applied to v, w and x instead o u, v and w yields α v w x v w x w, x v + v, x w v, w x u v w x w, x u v + v, x u w v, w u x u v w x u,vw x v u w x by 7, applied to u, v and w x instead o v, u and U w, x + v, x u w u,w by Theorem 5 b, applied to u and w instead o v and u v, w u v u,v by Theorem 5 b, applied to u and v instead o v and u u x u,x by Theorem 5 b, applied to u and x instead o v and u 31

32 u v w x w, x u v + v, x u w v, w u x u, v w x v u w x u,wx w u x by 7, applied to u, w and x instead o v, u and U w, x u, v + v, x u, w v, w u, x u v w x w, x u v + v, x u w v, w u x u, v w x v u, w x w u x w, x u, v + v, x u, w v, w u, x u v w x w, x u v + v, x u w v, w u x u, v w x + v u, w x w + w, x u, v v, x u, w + v, w u, x u x u,x by Theorem 5 b, applied to u and x instead o v and u u v w x w, x u v + v, x u w v, w u x u, v w x + v u, w x w u, x + w, x u, v v, x u, w + v, w u, x u v w x w, x u v + v, x u w v, w u x u, v w x + v u, w x v w u, x u,wv x u,xv w + w, x u, v v, x u, w + v, w u, x. u v w x w, x u v + v, x u w v, w u x u, v w x + u, w v x u, x v w + w, x u, v v, x u, w + v, w u, x. These computations can be generalized to α u 1 u 2... u p or general p N. As a result, we get the ormula α u 1 u 2... u p 1 number o all bad pairs u i1, u j1 u i2, u j2... u ik, u jk u r1 u r2... u rp 2k or any p vectors u 1, u 2,..., u p in L, where the sum is over all partitions o the set {1, 2,..., p} into three subsets {i 1, i 2,..., i k }, {j 1, j 2,..., j k } and {r 1, r 2,..., r p 2k } or various k which satisy i 1 < i 2 <... < i k, j 1 < j 2 <... < j k, r 1 < r 2 <... < r p 2k and i l < j l or every l {1, 2,..., k}. Here, a bad pair means a pair l, l {1, 2,..., k} 2 satisying l l and i l < j l so, in particular, or every l {1, 2,..., k}, 32

33 the pair l, l is bad, since i l < j l. 32 Thus we have an explicit ormula or α u 1 u 2... u p, but it is extremely hard to deal with; this is the reason why I deined α by induction rather than by a direct ormula. We remark that the ormula 42 can be slightly generalized, in the sense that U doesn t have to be a homogeneous tensor: Theorem 19. Let u L and U L. Then, α u U u α U u α U. 46 Proo o Theorem 19. Fix u L. We have to prove the equation 46 or every U L. We can WLOG assume that U L p or some p N since every tensor U L is a k-linear combination o elements o L p or various p N, and since the equation 46 is linear in U. But then, 46 ollows rom 42 applied to p + 1 instead o p. Thus, the equation 46 is proven or every U L, and thereore Theorem 19 is proven. Another act is, while α is not necessarily homogeneous, the degrees o all the terms it spits out have the same parity as that o the original tensor: Theorem 20. Let U L p or some p N. Then, α U Even a stronger assertion holds: α U i N; i {0,1,...,p}; L i. 47 L i. 48 Proo o Theorem 20. We are going to prove 48 by induction over p. The induction base case p 0 is obvious 33. So let us pass on to the induction step: Let p N +. Assume that we have proven 48 or p 1 instead o p; that is, we have shown that α U L i or every U L p i {0,1,...,p 1}; i p 1 mod 2 Now we have to establish 48 or our value o p as well, i. e. we have to prove that α U L i or every U L p. 50 i {0,1,...,p}; 32 I hope I haven t made a mistake in the ormula. 33 In act, in this case, U L p L 0 k, and thus α U U i {0,1,...,p}; L i. 33