21-801: Algebraic Methods in Combinatorics Multilinear notes 1 Multilinear and alternating functions We have already encountered multilinear polynomials in the proofs of intersection theorems of Frankl{Wilson and Ray-Chaudhuri{Wilson. We will nowextendthenotiontofunctionsonanarbitraryproductofvectorspaces. Let W ;:::;W ;V be vector spaces over some (cid:12)eld F. A function 1 k f: W (cid:2)(cid:1)(cid:1)(cid:1)(cid:2)W ! V 1 k is multilinear if it is linear in each of the k variables separately. In other words, f(:::;wi(cid:0)1; (cid:11)jwij;wi+1;:::) = (cid:11)jf(:::;wi(cid:0)1;wij;wi+1;:::); j j X X for any choice of w 2 W , w 2 W , etc. 1 1 2 2 There are special names for multilinear functions when k is small. When k = 2 we speak of a bilinear function functions, and of trilinear functions when k = 3. In general, we may call f a k-linear function. Example: Recall that a polynomial on Fn is called multilinear if it is a linear combination of monomials of the form x d=ef x . Such a I i2I i polynomial is the same as a multilinear function f: F(cid:2)(cid:1)(cid:1)(cid:1)(cid:2)F ! F. Q A special class of multilinear functions are alternating functions. If f is a multilinear function, W = ::: = W and f(w ;:::;w ) = 0 whenever 1 k 1 k w = w for some i 6= j, then we say that f is alternating. i j Example: Given a vectors w ;:::;w 2 F let [w ;:::;w ]T be the 1 n n 1 n n-by-n matrix whose rows are w ;:::;w in that order. Then the function 1 n D(w ;:::;w ) = det[w ;:::;w ]T is alternating. 1 n 1 n Lemma 1. Let f: W (cid:2)(cid:1)(cid:1)(cid:1)(cid:2)W ! V be an alternating function. Then the following hold: a) f(:::;wi(cid:0)1; (cid:11)jwj;wi+1;:::) = (cid:11)if(:::;wi(cid:0)1;wi;wi+1;:::) b) f(:::;w ;:::P;w ;:::) = (cid:0)f(:::;w ;:::;w ;:::). i j j i c) If e ;:::;e are a basis for W, then for every w ;:::;w 2 W we have 1 n 1 k f(w ;:::;w ) 2 spanff(e ;:::;e ) : j < (cid:1)(cid:1)(cid:1) < j g: 1 k j1 jk 1 k 1 21-801: Algebraic Methods in Combinatorics Multilinear notes The property (b) is the one responsible for the name ‘alternating’. If charF 6= 2, then it is easy to see that a multilinear functions satisfying property (b) is alternating in our sense. However, if charF = 2, the impli- cation fails, as witnessed by the function F (cid:2)F ! F that is non-zero only 2 2 2 when both of its arguments are non-zero. Proof of Lemma 1. a) This follows by invoking multilinearity, and noting all the terms save for one vanish. b) This is a consequence of (a) and the following computation: f(:::;w ;:::;w ;:::) = f(:::;w +w ;:::;w ;:::) i j i j j = f(:::;w +w ;:::;w (cid:0)(w +w );:::) i j j i j = (cid:0)f(:::;w +w ;:::;w ;:::) i j i = (cid:0)f(:::;w ;:::;w ;:::): j i c) Let w = a e . Then by multilinearity i j ij j P f(w ;:::;w ) = (cid:11) (cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:11) f(e ;:::;e ): 1 n 1j1 kjk j1 jk j1X;:::;jk The terms in which same basis vectors appears twice vanish, whereas the remaining terms can be arranged so that j < j < ::: < j by 1 2 k repeated application of (b). 2 Exterior powers In this section we shall work with vector space W = Fn exclusively. The part (c) of Lemma 1 tells us that if f is an alternating function on Wk, then f(Wk)spansanatmost n -dimensionalspace. Intuitively,dimspanf(Wk) = k n holds whenever there are no relations between values of f other than k (cid:0) (cid:1) those implied by the alternating property of f. We shall make this intuition (cid:0) (cid:1) rigorous below, but for now we exhibit an alternating function such that f(W ) spans a n -dimensional space. k k Theorem 2. F(cid:0)or(cid:1)each k, there exists a nk -dimensional space Tk and an alternating function f : Fn (cid:2)(cid:1)(cid:1)(cid:1)(cid:2)Fn ! T , and, for each k and l, there k k (cid:0) (cid:1) exists a bilinear function g : T (cid:2)T ! T that satisfy the following two kl k l k+l properties: 2 21-801: Algebraic Methods in Combinatorics Multilinear notes a) dimspanff (w ;:::;w ) : w ;:::;w 2 Fng = n , k 1 k 1 k k b) g f (w ;:::;w );f (u ;:::;u ) = f (w ;::(cid:0):;(cid:1)w ;u ;:::;u ). kl k 1 k l 1 l k+l 1 k 1 l Proof. W(cid:0)e de(cid:12)ne Tk d=ef F([nk]). In othe(cid:1)r words, Tk a nk -tuples of elements from F indexed by k-element subsets of [n]. (cid:0) (cid:1) Given vectors w ;:::;w 2 Fn, form a k-by-n matrix A whose rows are 1 k w ;:::;w (in that order). Then for a set I 2 [n] let A be the k-by-k 1 k k I submatrix of A that is made of columns indexed by the set I. Then de(cid:12)ne (cid:0) (cid:1) f : W (cid:2)(cid:1)(cid:1)(cid:1)(cid:2)W ! T by k k f (w ;:::;w ) = detA : k 1 k I I The function f is multinear because det is multilinear, and it is alternating because det is alternating. To see that f satis(cid:12)es part (a), consider any k k distinct basis vectors e ;:::;e in Fn. Then i1 ik (cid:6)1 if I = fi ;:::;i g; 1 k f(e ;:::;e ) = i1 ik I (0 otherwise; where the sign is determined by the order of i ;:::;i . So, each element of 1 k the standard basis of T is in the image of f, and so spanf(Wk) = T . k k De(cid:12)ning g is simple but cumbersome: we (cid:12)rst let (cid:6)(I) d=ef i, then kl i2I for each I 2 [n] we put k+l P (cid:0) g(cid:1)(x;y) = ((cid:0)1)(cid:6)([k]) ((cid:0)1)(cid:6)(I1)x y : kl I I1 I2 I1[XI2=I I12([nk]) I22([nl]) The part (b) of the lemma amounts to the identity detC = ((cid:0)1)(cid:6)([k]) ((cid:0)1)(cid:6)(I1)detA (cid:1)detB I1 I2 I1[XI2=I I12([nk]) I22([nl]) where C is a matrix of the form C = A where A;B are the k-by-n, and B l-by-n matrices respectively. To see the identity, note that both sides are (cid:2) (cid:3) polynomials made of the same monomials, and we just need to check that the signs match. The latter is veri(cid:12)ed by noting that the signs match for I = [k], and that swapping an integer from I with one in I changes the 1 1 2 sign correctly. 3 21-801: Algebraic Methods in Combinatorics Multilinear notes The vector spaces T are called exterior powers of Fn, and are tradi- k tionally denoted by kFn. The functions f and g are called alternating k kl product anddenotedbythesamewedgesymbol^. Inparticular,theidentity V g (f ;f ) = f from part (b) of the preceding lemma is written concisely kl k l k+l as (w ^(cid:1)(cid:1)(cid:1)^w )^(v ^(cid:1)(cid:1)(cid:1)^v ) = w ^(cid:1)(cid:1)(cid:1)^w ^v ^(cid:1)(cid:1)(cid:1)^v ; 1 k 1 l 1 k 1 l and expresses associativity of the alternating product. Of course, the alter- 0 0 nating product is not commutative since w^w = (cid:0)w ^w. 3 From subspaces to vectors The k-dimensional subspaces in Fn can be naturally regarded as vectors in kFn. Indeed, given a subspace U with a basis u ;:::;u we can consider 1 k the vector V Û d=ef u ^(cid:1)(cid:1)(cid:1)û : 1 k The vector is non-zero since the matrix A = [u ;:::;u ]T has rank k, and 1 k so detA 6= 0 for some k-by-k minor. I As de(cid:12)ned, Û depends on the choice of the basis in U, but it does so 0 0 only slightly. If u ;:::;u is any other basis, then by part (c) of Lemma 1 1 k 0 0 it follows that u ^(cid:1)(cid:1)(cid:1)û = (cid:21)u ^(cid:1)(cid:1)(cid:1)û for some scalar (cid:21). Hence, we 1 k 1 k can regard Û as a vector in the projective space (^kFn)=F(cid:3). Below we will abuse notation and pretend that Û is an element of ^kFn. Note that (Û)^v = 0 if and only if v 2 Û. In particular, this implies thatÛ completelydeterminesthesetofvectorsinU,i.e.,themapU 7! Û is injective. However, not every vector in kFn is of the form Û. Indeed, the space of all k-dimensional subspaces of Fn is k(n (cid:0) k)-dimensional1, whereas dim Fn = n is much larger. V k A special case important in the computer graphics is the case k = 2 and V (cid:0) (cid:1) n = 4. Thinking projectively, a 2-dimensional subspace of R4 corresponds to a line in R3. Thus, the map U 7! Û associates to each line in R3 a sixtuple of numbers. This sixtuple is known as Plu(cid:127)cker coordinates of the line. 1The collection of all k-dimensional subspaces forms an algebraic variety, called the Grassmannian. Aswehavenotyetde(cid:12)nedthenotionofdimensionforalgebraicvarieties, the statement is informal: to specify a k-dimensional subspace of Fn we ‘need’ k(n(cid:0)k) scalars. 4 21-801: Algebraic Methods in Combinatorics Multilinear notes 4 Bollob(cid:19)as-type theorems We start by giving another proof of the skew version of Bolloba(cid:19)s’s theorem. The proof is due to Lova(cid:19)sz. Theorem 3. Suppose (A ;B );:::;(A ;B ) are pairs of sets that satisfy 1 1 m m a) A is an r-element set, and B is an s-element set, for all i, i i b) A \B = ;, for all i, i i c) A \B 6= ;, for all i < j. i j Then the number of pairs is m (cid:20) r+s . s Proof. Let X be a (cid:12)nite set contai(cid:0)ning(cid:1)all the A ’s and B ’s. Let W = Rr+s i i and associate to each x 2 X a vector w such that the vectors fw : x 2 Xg x x are in general position, i.e., no r+s of them are linearly dependent. To each subset S (cid:26) X associate a vector w d=ef w . With this S x2S s association we have, for any sets A;B (cid:26) X satisfying jAj+jBj (cid:20) r+s, V w \w = 0 () A\B 6= ;: A B Indeed, if A \ B 6= ;, then the alternating product w ^ w contains a A B repeated element. Conversely, if A\B = ;, then A[B is a linearly independent set, and so w ^w is non-zero. A B In particular, 6= 0 if i = j; w ^w Ai Bi (= 0 if i < j; and so the vectors w ;:::;w 2 rRr+s are linearly independent. Since A1 Am dim kRr+s = r+s , it follows that m (cid:20) r+s . r V r IVn Bolloba(cid:19)s’(cid:0)s the(cid:1)orem the pairs are cla(cid:0)ssi(cid:12)(cid:1)ed according to whether they intersect or not. In the following extension of the result, due to Fu(cid:127)redi, the pairs are classi(cid:12)ed by the size of the intersection. Theorem4. Lettbeanonnegativeinteger. Suppose(A ;B );:::;(A ;B ) 1 1 m m are pairs of sets that satisfy a) A is an r-element set, and B is an s-element set, for all i, i i b) jA \B j (cid:20) t, for all i, i i c) jA \B j (cid:21) t+1, for all i < j. i j 5 21-801: Algebraic Methods in Combinatorics Multilinear notes Then the number of pairs is m (cid:20) r+s(cid:0)2t . r(cid:0)t The result is tight. To see that(cid:0), set th(cid:1)e ground set to be [r+s(cid:0)2t], and let A be a union of [t] and any (r(cid:0)t)-element subset of [r+s(cid:0)t]n[t], and i B be the union of [t] and remaining elements from [r+s(cid:0)t]n[t]. i We will deduce 4 from a related result for subspace intersections. Theorem 5. Suppose (U ;V );:::;(U ;V ) are pairs of subspaces of a vec- 1 1 m m tors space over a (cid:12)eld F. Assume that a) U is r-dimensional, and B is s-dimensional, for all i, i i b) dim(A \B ) (cid:20) t, for all i, i i c) dim(A \B ) (cid:21) t+1, for all i < j. i j Then m (cid:20) r+s(cid:0)2t . r(cid:0)t Lemma 6.(cid:0)Suppo(cid:1)se U1;:::;Ur are subspaces of a (cid:12)nite-dimensional vector space W over an in(cid:12)nite (cid:12)eld F, and let d be a nonnegative integer not exceeding dimW. Then the following hold: a) There exists a linear mapping (cid:30): W ! Fd such that dim((cid:30)(U ) = min(dimU ;d): i i b) There exists a codimension d subspace L such that dim(L\U ) = max(dimU (cid:0)d;0): i i Wepostponetheproofoftheseassertionuntilourdiscussionofalgebraic varieties later in the course. Intuitively, since F is in(cid:12)nite, ‘almost all’ maps (cid:30) (resp. subspaces L) satisfy any one of these conditions, and there are only (cid:12)nitely many of these conditions. One can also prove this assertion directly working in coordinates (exercise!). An easier exercise is to show that the two assertions are equivalent. Proof of 5. We may assume that the (cid:12)eld F is in(cid:12)nite; if F is (cid:12)nite, we can replace F by its algebraic closure. By part (b) of Lemma 6 there exists a subspace L of codimension t that satis(cid:12)es dim(U \L) = r(cid:0)t; i dim(V \L) = s(cid:0)t; i dim(U \V \L) = 0; i i dim(U \V \L) (cid:21) 1; for i < j: i j 6 21-801: Algebraic Methods in Combinatorics Multilinear notes Let U0 d=ef U \L and V0 d=ef V \L. We have thus reduced the problem to i i i i the case t = 0 for the pairs (U0;V0);:::;(U0 ;V0 ) 2 Fr(cid:0)t (cid:2)Fs(cid:0)t. The next 1 1 m m step is to reduce the dimension of the ambient space. By part (a) of Lemma 6 there exists a linear map (cid:30): W ! Fr+s(cid:0)2t such that the spaces U~ d=ef (cid:30)(U0) and V~ d=ef (cid:30)(V0) have the same dimensions as i i j j U0 and V0 respectively, and dimspanU~ [V~ = dimspanU0[V0. The latter i j i j i j condition implies U0\V0 = f0g if and only if U~ \V~ = f0g. i j i j Let u = Û~ and v = ^V~ It follows that i i j j 6= 0 if i = j; u ^v i j (= 0 if i < j; r and so the vectors u ;:::;u 2 W are linearly independent. Since these 1 m are vectors in rFr+s(cid:0)2t, we conclude that m (cid:20) r+s(cid:0)2t . V r(cid:0)t Proof of TheorVem 4. Withoutlossthegroundseti(cid:0)s[n]. L(cid:1)ete ;:::;e bethe 1 n standardbasisvectorsforFn. AssociatetoeachsetS thevectorsspaceV = S spanfe : i 2 Sg. Then the vector space pairs (V ;V );:::;(V ;V ) i A1 B1 Am Bm satis(cid:12)es the assumption of Theorem 5. 7

Multilinear notes [expository notes]

Similar Multilinear notes [expository notes]