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3.5 the Grassmannian . find: Theorem . Let be . Let be unique with Then is d. Proof. 3.5 The Grassmannian 3.5.1 The exterior product of two vectors Let us recall the exterior product of a finite dimensional K-vector space V . An alternating bilinear form is a map B : V × V −! K satisfying the axioms: (Bil 1) B(v; w) = −B(w; v), (Bil 2) B(λ1v1 + λ2v2; w) = λ1B(v1; w) + λ2B(v2; w), for all λ• 2 K and v•; v; w 2 V . We can add alternating bilinear forms and multiply them with scalars and get a thus the K-vector space of bilinear forms on V . This space is isomorphic to the space of skew-symmetric matrices of the appropriate size. The second exterior power V2 V of V is the dual to the space of alternating bilinear forms on V . The exterior product of v and w is (v ^ w) 2 V2 V given by (v ^ w)(B) := B(v; w): We have the obvious properties: (i) v ^ w = −w ^ v, (ii) (λ1v1 + λ2v2) ^ w = λ1v1 ^ w + λ2v2 ^ w, V2 (iii) If fv1; : : : ; vng is a basis of V , then fvi ^ vj : 1 ≤ i < j ≤ ng is a basis of V . Furthermore: v ^ w = 0 if and only if v = λw for some λ 2 K. 3i.e. curve of degree two. 4You may check Wikipedia on this, or the paper of Ciliberto, Harbourne, Miranda, and Roé :https://arxiv.org/abs/1202.0475 19 3.5.2 The relation with two-dimensional subspaces We are interested in the exterior product because every two-dimensional subspace of V corresponds to a one- dimensional subspace of V2 V . Take U = Span fv; wg ⊂ V with dim U = 2. Then every other basis fu ; u g K K 1 2 a b of U is obtained from fv; wg via an invertible matrix A = 2 GL ( ) as u = av +bw and u = cv +dw. c d 2 K 1 2 u1 ^ u2 = (av + bw) ^ (cv + dw) = ad(v ^ w) + bc(w ^ v) = det(A)(v ^ w): So, Span fv ^ wg ⊆ V2 V is independent of the choice of basis of U. Thus, to every 2-dimensional subspace K V2 of V corresponds a point in P( V ). That is why it is good to think about projective spaces as associated to n n(n−1) vector spaces, and not simply as P . In this case, just writing P 2 would have been less illuminating. The first question that comes up is: do all 1-dimensional subspaces of Λ2V correspond to a 2-dimensional V2 subspace of V ? The answer is: No, not every element of V is of the form v ^w. For instance, v1 ^v2 +v3 ^v4, where fv1; : : : ; v4g is a basis of V . (This is an easy linear algebra exercise). So, the association U 7−! v ^ w V2 gives a subset of the projective space P( V ). Is this algebraic? We will see that it is, and that this holds more generally. 3.5.3 Exterior powers To parametrize the collection of subspaces of any dimension with a projective algebraic set, we have too take k 5 a look at alternating multilinear maps M : V −! W . A multilinear map is alternating, if M(v1; : : : ; vk) = 0 n n whenever there exists i 6= j with vi = vj. The determinant, viewed as a map det : (K ) −! K is an alternating multilinear form. Also the cross product ' :(K3)2 −! K3, given by '((a; b; c); (x; y; z)) := (bz − cy; cx − az; ay − bx); is an alternating bilinear map. The k-th exterior power of V , also known as the alternating k-fold tensor product, is given by the unique Vk Vk k Vk pair ( V; τk), where V is a k-vector space and τk : V −! V is an alternating k-linear map, satisfying the following universal property: For every k-fold alternating linear map f : V k −! W , there exists a unique Vk linear map g : −! W such that f = g ◦ τk. That is such that the following diagram commutes: f V k W τ k g Vk V One can easily see that Vk V = V ⊗k=L, where L = Span fv ⊗ · · · ⊗ v : v = v for some i 6= jg. K 1 k i j Vk Remark 3.11. Assume that fe1; : : : ; eng is a basis of V . A basis of V is given by the alternating tensors: fei1 ^ ::: ^ eik : 1 ≤ i1 < ··· < ik ≤ ng: So dim Vk V = n. In particular, V0 V = , V1 V = V and Vn V = Span (e ^ · · · ^ e ). All higher exterior K k K 1 n powers vanish. 5that is linear in each argument. 20 V Ln Vi The exterior algebra of the n-dimensional K-vector spaces V is V := i=0 V , with product operation induced by Vi+j (v1 ^ ::: ^ vi) ^ (w1 ^ : : : wj) := v1 ^ ::: ^ vi ^ w1 ^ ::: ^ wj 2 V; Vi Vj 8 v1 ^ ::: ^ vi 2 V and w1 ^ ::: ^ wj 2 V: n Remark 3.12. Let V = K and vi = (ai;1; : : : ; ai;n) for i = 1; : : : ; k. Then, we can express the coordinates of v1 ^ · · · ^ vk with respect to the canonical basis in terms of the k-minors of (aij): X v1 ^ · · · ^ vl = a1;j1 : : : ak;jk · ej1 ^ · · · ^ ejk j1;:::;jk ! X X = sign(σ)a1,σ(j1) : : : ak,σ(jk) · ej1 ^ · · · ^ ejk 1≤j1<···<jk≤n σ2Sk X = det(Aj1;:::;jk ) · ej1 ^ · · · ^ ejk ; 1≤j1<···<jk≤n where Aj1;:::;jk is the k × k submatrix of (aij) with the columns j1; : : : ; jk. For example, if k = 2; n = 3, then a11 a12 a13 (aij) = a21 a22 a23 and v1 ^ v2 = (a11a22 − a12a21)e1 ^ e2 + (a11a23 − a13a21)e1 ^ e3 + (a12a23 − a13a22)e2 ^ e3: So the exterior product encodes all the maximal minors of a matrix in one object. As a consequence we have: v1 ^ · · · ^ vk = 0 , v1; : : : ; vk are linearly dependent. Furthermore, if the two sets fv1; : : : ; vkg ⊆ V and fw1; : : : ; wkg ⊆ V are both linearly independent, then Span (v ; : : : ; v ) = Span (w ; : : : ; w ) () v ^ · · · ^ v = λ · w ^ · · · ^ w for some λ 2 : (3.1) K 1 k K 1 k 1 k 1 k K This last equivalence needs some proof, but it’s all basic linear algebra. 3.5.4 The Plücker Embedding Definition 3.13. Let n 2 N>0 and k 2 N with 0 ≤ k ≤ n. The Grassmanian of k-planes in the n-dimensional K-vector space V is the set of all k-dimensional linear subspaces of V . We denote this by Gr(k; V ). When V = Kn, we simply write n Gr(k; n)= fW ⊆ K : dimK W = kg: For k = 1, we obtain Pn−1 (as a set for the moment). For any k ≥ 1, by the correspondence between k- dimensional subspaces of V and k − 1 linear subspaces of P(V ), we have a bijection between the collection of k-vector subspaces of Kn and k − 1-dimensional linear subspaces of Pn−1. For this reason, one may find the notation Gr(k − 1; n − 1) for the same object. In order to avoid confusion, we will denote by n−1 G(k − 1; n − 1) := fL ⊆ P : L is a linear subspace of dimension k − 1g: Vk The Plücker Embedding is the map P` : Gr(k; V ) P( (V )) given by k ^ P`(Span (v ; : : : ; v )) := [v ^ · · · ^ v ] 2 ( (V )): K 1 k 1 k P By (3.1) this is well defined and injective. For a k-dimensional subspace U ⊆ V , the homogeneous coordinates of Vk P`(U) in P( V ) are called the Plücker coordinates of U. Once a basis of V is chosen (thus an isomorphism n Vk with K and a corresponding basis of V as well), the Plücker coordinates are just the maximal (i.e. k−) minors of the matrix with the coordinates of v1; : : : ; vk as rows. 21 Example 3.14. (a) For k = 1, and U = Span (a e + ··· + a e ) we get P`(U) = (a : ::: : a ) 2 n−1. K 1 1 n n 1 n P (b) Now take U = Span (e + e ; e + e ) 2 Gr(2; 3). We have K 1 2 2 3 (e1 + e2) ^ (e1 + e3) = −e1 ^ e2 + e1 ^ e3 + e2 ^ e3; so P`(U) = (−1 : 1 : 1). To show that the image of the Plücker embedding is a projective algebraic set we need to express being a pure Vk tensor, that is being of the form v1 ^ · · · ^ vk 2 V , as a polynomial condition. For this we need the following lemma. Vk Vk+1 Lemma 3.15. Let 0 6= w 2 V , with k < n, and define fw : V −! V by fw(v) := v ^ w: Then rank fw ≥ n − k, and, most importantly, rank fw = n − k , 9 v1; : : : ; vk 2 V with w = v1 ^ · · · ^ vk: Proof. Let r = dim Ker fw = n − rank fw. Choose a basis fv1; : : : ; vrg of fw, extend it to a basis fv1; : : : ; vng of V , and express w = P p v ^ · · · ^ v . Then use the fact that for i = 1; : : : ; r we have v ^ w = 0. i1<···<ik i1:::ik i1 ik i This implies that only the pi1:::ik with f1; : : : ; rg ⊆ fi1; : : : ; ikg may be nonzero. This gives the inequality r ≤ k, which is equivalent to rank fw ≥ n − k. For the second part, clearly if r = k, then w = λ · v1 ^ · · · ^ vr. In the other direction, if w = w1 ^ ::: ^ wk, as w 6= 0 the wi are linearly independent, and they belong to Ker fw. By r ≤ k, we are forced to have the required equality.
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