SOME MULTILINEAR ALGEBRA OVER FIELDS WHICH I UNDERSTAND

SOME MULTILINEAR ALGEBRA OER FIELDS WHICH I UNDERSTAND Most of what s dscussed n ths handout extends verbatm to all felds wth the excepton of the descrpton of the Exteror and Symmetrc Algebras, whch requres more care n non-zero characterstc. These dfferences can be easly accounted for except for the propertes of the Exteror Algebra n characterstc 2, whch requre a sgnfcant amount of work to understand. I wll, however, assume that the vector feld s R. 1 1. The Tensor Product Defnton 1. Let, U and W be vector spaces. A blnear map φ from U to W s a map whch satsfes the followng condtons: φ(a 1 v 1 + a 2 v 2, u) = a 1 φ(v 1, u) + a 2 φ(v 2, u) for all v 1, v 2, u U, a 1, a 2 R. φ(v, b 1 u 1 + b 2 u 2 ) = b 1 φ(v, u 1 ) + b 2 φ(v, u 2 ) for all v, u 1, u 2 U, b 1, b 2 R. Here are some examples of blnear maps: (1) If U = R and s any vector space, we may defne a blnear map wth range W = usng the formula φ(v, c) = cv. (2) If U s any vector space, and s the dual vector space U, we may defne a blnear map wth range R usng the formula φ(v, α) = α(v). (3) If A s an algebra over R, then the product operaton defnes a blnear map φ : A A A: φ(a, b) = ab. (4) More generally, f A s an algebra over R, and M s a module over the algebra A, then the acton of A on M defnes a blnear map φ : A M M: φ(a, m) = am. (5) Gong back to realty, we can pck M = R n, and A = M(n, n), the set of n n matrces wth coeffcents n R. Here, the standard rules for multplyng vectors by matrces make R n nto a module over n n matrces. Now that you re convnced that blnear maps are lurkng everywhere, let s try to understand all of them at once: Defnton 2. Let and U be two vector spaces. The tensor product of and U s a vector space, denoted U, for whch there exsts blnear map α : U U whch satsfes the followng property: Whenever φ : U W s any blnear map, there exsts a unque lnear map(lnear transformaton) ˆφ : U W such that ˆφ α = φ. In other words, there s a map of vector spaces ˆφ such that the followng dagram commutes: U α U W φ ˆφ 1 The opnons expressed theren are solely the responsblty of the author, and then, only partally. Regardless, no other entty bears ANY responsblty for them. In partcular, they should not be taken to represent opnons (voced or otherwse) of the WOMP organzng commttee, the Unversty of Chcago Mathematcs Department, or (Heaven Forbd!) some supposed consensus of the Mathematcal Communty. 1

2 SOME MULTILINEAR ALGEBRA OER FIELDS WHICH I UNDERSTAND Of course, the above defnton may be that of an object whch doesn t exst, and whch, moreover, may not be unque f t were to exst. The ssue of unqueness does not actually arse, snce the requrement that the map ˆφ be unque elmnates any possble ambguty. Ths s a common feature of such defntons featurng unversal dagrams. As to the problem of exstence, we may construct the tensor product by hand n the followng exercse: Exercse 1. Work out the detals of ths vague descrpton: Consder the vector space freely generated by all symbols of the form v u for v and u U. The tensor product wll be the quotent of ths vector space by the subspace generated by elements of the form (cv) u c(v u), (v 1 + v 2 ) u v 1 u v 2 u, and, snce U and should be treated on equal footng, the correspondng elements f they swtch roles. The map α wll then map the par (v, u) to the mage of v u n the quotent; ths map wll be blnear because we took the quotent by the correct subspace. Smlarly, any map φ tell us where the generatng set v u should be mapped to. Ths map s well defned on the quotent snce φ s blnear. If we re wllng to jump n the mud hole and wrestle wth our vector spaces, we get another descrpton of the tensor product. Exercse 2. If has bass {e } n =1 and U has bass {f j} m j=1, then U has bass e f j. To show that ths constructon works, we may ether show that t s somorphc to the prevous constructon (n fact, t embeds n the bg ugly vector spaces we frst constructed, the projecton defnes a map to the quotent, so t suffces to show that ths s an somorphsm), or we may show that t satsfes the unversal dagram. For all my skeptcsm about categorcal propaganda, the second opton s easer. As another applcaton of the unversal property, we get the followng theorem: Theorem 3. The tensor product s commutatve, assocatve, and respects dualty of vector spaces. In other words, f are vector spaces, then 1 2 = 2 1 ( 1 2 ) 3 = 1 ( 2 3 ) ( 1 2 ) = 1 2 Proof: The drect sum ( W ) of vector spaces satsfes the same condtons. The unversal property allows one to pass from drect sums to tensor products. I should remark that the above theorem s true for all vector spaces at once. In categorcal words, the above somorphsms are all natural. Note further that the second property above means that we can unambguously wrte: 1 k for any fnte collecton of vector spaces wthout havng to worry about parentheses. In fact, ths object could have been defned wthout usng nducton: Exercse 3. Defne a k-multlnear map from a product of vector space 1 k n analogy wth the defnton of a blnear map (For a specal case, see Defnton 6). Use ths to defne the tensor product of k vector spaces followng Defnton 2. From Exercse 2, t s clear that the dmenson of U s the product of the dmenson of the two vector spaces. Here s another basc result about tensor products: Exercse 4. If M and N are m 1 m 2 and n 1 n 2 matrces, then they defne a map R m 1 R n 1 R m 2 R n 2 whch we may compose wth the defnng map for R m 2 R n 2 to get a blnear map. The defnton of the tensor product says that ths determnes a unque map: R m 1 R n 1 R m 2 R n 2. What s the matrx of ths map n the bass gven above?

SOME MULTILINEAR ALGEBRA OER FIELDS WHICH I UNDERSTAND 3 You already know another object (Hom) whose dmenson s multplcatve. Ths s no concdence: Theorem 4. If and W are two vector spaces, then Hom(, W ) = W, Proof: We defne the map n one drecton by mappng w φ to the map whch takes v to φ(v)w. Once t s shown that ths map s njectve, a dmenson count proves the requred somorphsm. 2. The Tensor Algebra We now consder a sngle vector space. Snce the tensor product s commutatve and assocatve, we may defne: = }{{} tmes and usng the fact we have somorphsms j = +j, these vector spaces ft together n a (graded) algebra: T ( ) =, where 0 = R. As usual, there s a categorcal defnton of the Tensor Algebra (stated here as a theorem): Theorem 5. There s a lnear map : T ( ) such that whenever φ : A s a lnear map from to an algebra A, there exsts a unque map of algebras ˆφ : T ( ) A such that φ = ˆφ. In other words, there exsts a unque ˆφ such that the followng dagram commutes: A φ ˆφ =0 T ( ). Proof: s the ncluson of = 1 nto T ( ). Snce T ( ) s generated as an algebra by 1, the rest follows. Exercse 5. Construct T ( ) as the quotent of a huge vector space n analogy wth the constructon of the tensor product. One advantage of the categorcal defnton, s that t gve us a lot of nformaton about maps from the Tensor Algebra. For example, f f : W s a map of vector spaces, we can use the unversal property of T ( ) to conclude that there s a unque map of algebras T (f) whch makes the followng dagram commute: T ( ) W f T (f) T (W ). Further, f g : W U s another map, then the maps T (g) T (f) and T (g f) agree. Ths s a consequence of the unversal property appled to the followng dagram: f T ( ) T (f) W T (W ) U g T (g) T (U).

4 SOME MULTILINEAR ALGEBRA OER FIELDS WHICH I UNDERSTAND Ths property s what allows us to thnk of the tensor algebra as a functor. It takes a vector space and returns a (non-commutatve) algebra, but t does so n some coherent sense, takng maps of vector spaces to maps of algebras. There s an alternatve way of lookng at the tensor product: Defnton 6. Let W be a vector space. A map f : k R s a k-lnear form f t satsfes the followng condtons: f(v 1,..., cv,..., v k ) = cf(v 1,..., v,..., v k ) for all c R f(v 1,..., v + v,..., v k) = f(v 1,..., v,..., v k ) + f(v 1,..., v,..., v k) We wll denote the space of k-lnear forms on by L k ( ). In Dfferental Geometry the term multlnear form s reserved for what s called here skewsymmetrc multlnear form (See Defnton 10). Multplcaton defnes a map L k ( ) L m ( ) L k+m ( ) (φ, ψ) (φ ψ)(x 1,..., x k, x k+1,..., x k+m ) = φ(x 1,..., x k )ψ(x k+1,..., x k+m ) Ths allows us to turn the space of all multlnear maps nto an algebra: L( ) = L k. Theorem 7. k=0 L( ) = T ( ) Proof: We wll do ths at the level of graded vector spaces, and leave checkng the compatblty of the algebra structures to the dlgent reader. Frst, we observe that by Theorem 3, we have an somorphsm ( r ) = ( ) r, so t suffces to show: ( r ) = L r ( ). Ths s a good example of the use of the unversal property of the tensor product. Keep the followng dagram n mnd: R An element of ( r ) s just a lnear map r R. Composng wth the multlnear map r we obtan an r-lnear form on. To go n the other drecton, we use Exercse 3, to see that every r-lnear form on nduces a unque lnear map from r to R. r 3. The Symmetrc Algebra If we were nterested only n commutatve algebras, we would be studyng the Symmetrc Algebra over : Defnton 8. The Symmetrc Algebra Sym( ) s defned by the followng unversal property: There exsts a lnear map j : Sym( ), such that whenever φ : A s a lnear map nto a commutatve algebra A, there exsts a unque map of algebras ˆφ : Sym( ) A such that φ = ˆφ. In other words, the followng dagram commutes: A φ ˆφ Sym( ). Exercse 6. Sym( ) s generated as an algebra by. (Hnt: Consder the subalgebra generated by ; apply the unversal property).

SOME MULTILINEAR ALGEBRA OER FIELDS WHICH I UNDERSTAND 5 Theorem 5 gves a map T ( ) Sym( ). By the above exercse, ths map s surjectve (snce the generatng set s n the mage). Exercse 7. Wrte Sym( ) explctly (?) as a quotent of the huge vector space you used n Exercse 5. In fact, the gradng on T ( ) descends to a gradng of Sym( ), so we may wrte: Sym( ) = Sym ( ). =0 Just as fndng an explct descrpton of the tensor product allows us to gve a bass for t n the case of fnte-dmensonal vector spaces, we can use the mage of the explct bass produced n Exercse 2 to prove the followng result: Exercse 8. If has bass {e } n =1, prove that Symj ( ) has bass consstng of all elements of the form: e a 1 1 ea 2 2... ean n, Σ n =1a = j. Use ths to compute the dmenson of Sym j ( ). If ths remnds you of polynomals n several varables, you re on the rght track. The analogue of polynomals for abstract vector spaces are symmetrc multlnear forms: Defnton 9. A map f : r R s a symmetrc r-lnear form f t s r-lnear and s nvarant under the acton of the symmetrc group. Here, the symmetrc group S r acts on r by permutng components. Symmetrc multlnear forms are an algebra f we defne φ ψ as follows: 1 (φ ψ)(v 1,..., v r+s ) = (r + s)! Σ σ S r+s φ(v σ(1),..., v σ(r) )ψ(v σ(r+1),..., v σ(r+s) ) In other words, we force the product of two multlnear forms to be symmetrc by averagng over all possble permutatons. Ths s precsely the pont that doesn t work n the same way for vector spaces of non-zero characterstc, snce we are dvdng by a number that may not be nvertble n the feld. The more general thng to do requres one to not take the average, but smply take the sum over all permutatons. We can now prove the analogue of Theorem 7 Exercse 9. Prove that the algebra of symmetrc multlnear forms on s somorphc to Sym( ). Ths shows that f = R n, then Sym( ) s, ndeed, the space of polynomals n n varables. 4. The Exteror Algebra One of the fundamental tools n Dfferental Geometry s the space of k-lnear forms. If we elmnate the manfold, we are left wth the Exteror Algebra of a vector space. As for the Tensor and Symmetrc Algebras, there are dfferent ways of approachng ths object (Warnng: Ths s also called the Alternatng Algebra, and, n older and physcs texts, the Grassmann Algebra. Usng the theory of graded vector spaces, one can treat the Exteror and Symmetrc Algebras as manfestatons of the same thngs: Free Algebras on a graded vector space). Defnton 10. A skew-symmetrc k-lnear form s a k-lnear form φ whch satsfes the followng addtonal property: φ(v 1,..., v, v +1,..., v n ) = φ(v 1,..., v +1, v,..., v n ) Ths s where the trouble begns n characterstc 2. Snce 1 = 1 n that stuaton, skew-symmetry and symmetry are equvalent. The stronger condton (whch s equvalent to skew-symmetry away from the prme 2.) s that of a k- alternatng map: φ(v 1,..., v n ) = 0 f there exsts j such that v = v j. The reason most people prefer to use the skew-symmetrc defnton s that t s ncely compatble wth the sgn homomorphsm of the symmetrc group:

6 SOME MULTILINEAR ALGEBRA OER FIELDS WHICH I UNDERSTAND Exercse 10. Use the fact that transpostons generate the symmetrc group to prove that a k-lnear map φ s skew-symmetrc f and only f: φ(v 1,, v n ) = sgn(σ)φ(v σ(1),, v σ(n) ) for all permutatons σ. Use ths to conclude the equvalence of the alternatng and skew-symmetry condtons away from 2 (.e: whenever the only number that satsfes x = x s 0). One advantage of the alternatng pont of vew s that t makes the followng result clearer: Exercse 11. Prove that the zero map s the only k-lnear map on a one dmensonal vector space f k > 1. Agan, skew-symmetrc k-lnear maps form an algebra under an approprately chosen average: (φ ψ)(v 1,..., v r+s ) = 1 r! + s! Σ σ S n sgn(σ)φ(v σ(1),..., v σ(r) )ψ(v σ(r+1),..., v σ(r+s) ). Ths s, once agan, one of those stuatons where one has to make choces. Ths choce works only n characterstc 0. But even n characterstc 0, authors often have dfferent conventons n order to make some formula or the other look nce (Another reasonable choce s to replace r! + s! by (r + s)!). Just be aware of these dfferences, but don t worry too much about them, snce they rarely lead to anythng worse than beng a factoral sgn away from the rght answer. I have chosen ths conventon because t elmnates multplcatve constants n the determnant. As should be clear by now, one may proceed from here n dfferent drectons n order to defne the Exteror Algebra. We wll opt for the one encountered n most dfferental geometry textbooks: Defnton 11. The Exteror Algebra of, denoted Λ( ) s the algebra of skewsymmetrc k-lnear maps from r to R. We wrte: Λ( ) = Λ ( ) =0 Exercse 12. Defne the Exteror Algebra of a vector space wthout passng through ts dual, takng as a model Defnton 8. Explctly, one fnds that f has bass {e } n =1, then Λk ( ) has bass: e 1 e k, 1 < < k. Note that ths mples that the dmenson of Λ k s ( n k). In partcular, the 0 th and the n th exteror powers are 1 dmensonal, and Λ k s 0 f k > n. Another way of provng ths s explaned n the followng exercse: Exercse 13. Prove that: Λ k ( W ) = p+q=k Λ p Λ q W. Use nducton and Exercse 10 to prove the dmenson formula above. So far, we have more or less descrbed what our constructons do to vector spaces, gnorng maps. If we are to go any further, we have to address ths problem. In fact, the Symmetrc and Exteror Algebras are also functors (See the dscusson after Exercse 5). Exercse 14. If f : W s a map of vector spaces, then the Unversal Property of the Exteror Algebra defnes maps of (skew-commutatve) algebras Λ(f) : Λ( ) Λ(W ). Show that ths makes the exteror algebra nto a functor from the category of vector spaces to that of (skew-commutatve) algebras.

SOME MULTILINEAR ALGEBRA OER FIELDS WHICH I UNDERSTAND 7 Tracng all the dagrams n the above exercse should yeld the followng formula: Λ(φ)(v 1 v k ) = φ(v 1 ) φ(v k ). In partcular, Λ(φ) decomposes as Λ k (φ) for 0 k n. Here s a (non-standard) defnton that shows some of the use of ths machnery: Defnton 12. Let A be a lnear transformaton on a vector space of dmenson n. The determnant of A s the egenvalue of the map nduced by Λ n (A) on Λ n. In order for ths defnton to make sense, one has to remember that Λ n s 1-dmensonal. To see that ths s equvalent to the elementary (but unenlghtenng) defnton of the determnant, we pck a bass {e } n =1, wrte A as a matrx (a j ) 1,j n n that bass, and compute: Λ(A)(e 1 e n ) = A(e 1 ) A(e n ) = (Σ n j 1 =1a 1j1 e j1 ) (Σ n j n =1a njn e jn ) = Σ 1 j1,...,j n n(a 1j1 a njn )e j1 e jn At ths pont, one observe that whenever some j k = j l, the correspondng term does not contrbute to the sum, snce e j1 e jn = 0. So the only terms that matter are those n whch none of the j k are equal. We can then thnk of {j k } n k=1 as smply a reorderng of the ndexng set {1,..., n}, so we wrte j k = σ(k) for σ S n. We are now n frm, famlar ground, and we can contnue the computaton: Λ(A)(e 1 e n ) = Σ σ Sn (a 1σ(1) a nσ(n) )e σ(1) e σ(n) In partcular, the egenvalue of Λ(A) s whch s hopefully the rght answer. = Σ σ Sn (a 1σ(1) a nσ(n) )sgn(σ)e 1 e n Σ σ Sn sgn(σ)(a 1σ(1) a nσ(n) ), Exercse 15. Use the functoralty of the exteror algebra to prove that det(ab) = det(a) det(b). References [1] Spvak, Mchael. A Comprehensve Introducton to Dfferental Geometry, ol 1. Publsh Or Persh, Houston, TX, 1999. Despte the scary ttle and the fascnatng cover, ths book ncludes most of what one needs to know about basc dfferental geometry (ncludng all the multlnear algebra), wth long-wnded, sometmes qurky explanatons, and lots of exercses. Check out the yellow pg. [2] Hoffman, Kenneth, and Kunze, Ray. Lnear Algebra. Prentce-Hall, Saddle Rver, NJ, 1971. A standard ntroductory Lnear Algebra Text. [3] Harrs, Joe. Algebrac Geometry, A Frst Course. Sprnger-erlag, New York, 1992. See Lecture 6 for how to thnk geometrcally about Λ.