MATH 210A, FALL 2017 HW 9 SOLUTIONS WRITTEN BY DAN DORE (If you find any errors, please email [email protected]) Let V be a vector space over a field F, and let ! : V × V ! F be an alternating form. An !-symplectic basis is an ordered basis a1; b1; a2; b2; : : : ; an; bn for V with the property that !(ai; bi) = 1 for all i !(ai; aj) = !(ai; bj) = !(bi; aj) = !(bi; bj) = 0 if i 6= j Question 1. Suppose that ! is a nondegenerate alternating form over an arbitrary1 field F. Prove there exists an !-symplectic basis. Solution. Note that in particular, we are showing that any vector space which admits an alternating non- degenerate form has even dimension 2n. We will show by induction on n that a vector space of dimension 2n with a non-degenerate alternating form admits a symplectic basis and that a vector space of dimension 2n + 1 does not admit a non-degenerate symplectic form. The base case is n = 0. When V has dimension 0 the claim is vacuously true. When V = F · e has dimension 1, there cannot be a non-degenerate alternating form ! on V : !(xe; ye) = xy!(e; e) = 0 for any x; y 2 F. Now, both inductive steps will rest on the following lemma: Lemma 1. Let V be a vector space with a non-degenerate alternating form ! on V . If V1 ⊆ V is a two- ? dimensional subspace with basis vectors e; f with !(e; f) = 1, then if V2 := V1 = fv 2 V j !(v; V1) = 0g is the orthogonal complement of V1 in V with respect to !, we have V = V1 ⊕ V2 and furthermore !jV2 is alternating and non-degenerate. Before we prove the lemma, let’s see why it suffices for both inductive steps. For arbitrary a 2 V , there 0 0 1 0 exists b 2 V with !(a; b ) 6= 0 by nondegeneracy. Taking b = !(a;b) b we have !(a; b) = 1. In particular a and b are linearly independent (because otherwise !(a; b) = !(a; ca) = c!(a; a) = 0). So let V1 be the space of a; b. Thus, we may apply the lemma. In particular it implies dim V2 = dim V − 2. If the dimension of V is odd, so is the dimension of V2, but the lemma shows that !jV2 is an alternating non-degenerate form on V2. By induction, we know that this is impossible. If the dimension of V is even, so is the dimension of V2, so we may find a symplectic basis a2; b2; : : : ; an; bn for V2. Then letting a1 = a; b1 = b, we can see that a1; b1; a2; b2; : : : ; an; bn is a symplectic basis since !(a1; b1) = 1, a2; b2; : : : ; an; bn is a symplectic basis for V2, and a1; b1 live in the orthogonal complement to V2. Now, we must prove the lemma: Proof. First, note that V1 \ V2 = 0. To see this, let ce + df 2 V1 with c; d 2 F. If this is in V2, then we have 0 = !(ce + df; e) = d!(f; e) = −d and 0 = !(ce + df; f) = c!(e; f) = c, so c = 0 and d = 0. Now, we need to show that V = V1 + V2. To do this, since V1 \ V2 = 0, it suffices to show that _ dim V2 = dim V − 2. Now, the bilinear form ! induces a map !e : V ! V by sending v 2 V to the map 1Note we do not need to assume anything about which elements are squares, nor anything about char F. 1 w 7! !(v; w). Non-degeneracy of ! means exactly that !e is injective: if !(v; w) = 0 for all w 2 V , then _ v = 0. Since V; V are vector spaces of the same finite dimension, this implies that !e is an isomorphism. V v !(v)j = 0 V _ (V )_ ' 'j 2 exactly consists of the such that e V1 . But the map from to 1 sending to V1 is a _ _ surjection onto the two-dimensional vector space (V1) , so its kernel has codimension 2 in V . Therefore, V2 has codimension 2 in V , so we have shown V = V1 ⊕ V2. Now, we need to show that !jV2 is alternating and non-degenerate. The fact that it is alternating is obvious: for v 2 V2, !jV2 (v; v) = !(v; v) = 0. To see that it is non-degenerate, fix some v 2 V2. We need to find some w 2 V2 with !jV2 (v; w) = !(v; w) 6= 0. Since ! is non-degenerate, we may pick some w0 2 V with !(v; w0) 6= 0. Since V = V1 ⊕ V2, we may uniquely write w0 = w1 + w2 with wi 2 Vi. Then we have !(v; w0) = !(v; w1) + !(v; w2) = !(v; w2), since w1 2 V1 and V2 is !-orthogonal to V1. Thus, !(v; w2) 6= 0, so we are done. Question 2. Let V be a 2n-dimensional vector space over F. Recall that V _ denotes the dual vector space _ V = HomF(V; F). Let ! : V × V ! F be an alternating form. We can view ! as an element of V2(V _). (make sure you understand how this correspondence works) Is it true that ! is nondegenerate as a bilinear form if and only if ! ^ · · · ^ ! 2 V2n(V _) is nonzero? Solution. First, let’s make the correspondence between the space of alternating forms on V and V2(V _) precise. 2n V2 _ Proposition 2. For a vector space V ' F , there is a natural linear isomorphism ! 7! ev! between V and the vector space of alternating bilinear forms on V . Proof. Note that the space of skew-symmetric bilinear forms on V is canonically isomorphic to (V2V )_: the universal property of exterior powers says exactly that a linear map from V2V to R is the same thing as a skew-symmetric bilinear form on V . So we are defining a map from V2(V _) to (V2V )_. We define this by mapping ' ^ to the bilinear form ev'^ : (v; w) 7! '(v) (w) − (v)'(w). Since this map is clearly _ _ _ linear in each of v; w; '; , this at least defines a map from V ⊗ V to (V ⊗ V ) . Since ev'^ = −ev ^', it factors through the canonical projection V _ ⊗ V _ ! V2(V _), so it gives us a map V2(V _) to (V ⊗ V )_. Finally, since ev'^ (v; v) = '(v) (v) − '(v) (v) = 0, the image lands inside the subspace of alternating _ _ forms (V ^ V ) ⊆ (V ⊗ V ) . Note that this definition makes it clear that ev• is functorial in V , i.e. that if T : V ! W is a linear map, then evT ∗('^ )(v1; v2) = ev(T ∗'^T ∗ )(v1; v2) = '(T (v1)) (T (v2)) − (T (v2))'(T (v1)) = ev'; (T (v1);T (v2)) ∗ = (T ev'; )(v1; v2) We can compute what this is explicitly in a basis (note that the above construction was basis-independent!) 1 2n _ i i V2 _ v1; : : : ; v2n of V , with associated dual basis v ; : : : ; v of V (defined by v (vj) = δj). For ! 2 (V ), P i j P P we may write ! = i<j aijv ^ v , v = i bivi, and w = j cjvj. Then we have X ev!(v; w) = aij(bicj − bjci) (1) i<j 2 P We can check explicitly that this is alternating: if bi = ci, this is i<j aij(bibj − bjbi) = 0. To see that the map is an isomorphism, note that for i < j, ev!(vi; vj) = aij. Thus, if ev! = 0, we have aij = 0 for all V2 V2 _ _ 2n i < j, so ! = 0. This shows that ev• is injective, and since V and ( V ) both have dimension 2 , we conclude that ev• is an isomorphism. V2n _ We can show one direction right away: if ! ^ · · · ^ ! 6= 0 in (V ), then ev! is non-degenerate. Indeed, assume that ev! is degenerate so that there exists some v 2 V with ev!(v; w) = 0 for all w 2 V . Let V1 = F · v, and let W = V=V1. Then also ev!(w; v) = −ev!(v; w) = 0, so ev! descends to an alternating form on W . By functoriality of the map ! 7! ev!, this means that ! is in the image of the natural inclusion V2(W _) ,−! V2(V _). Thus, ! ^ · · · ^ ! 2 V2n(V _) is in the image of the natural inclusion of V2n(W _). But this space is 0 since W has dimension 2n − 1. We can also work explicitly in a basis: assume that ev!(v; w) = 0 for all w 2 V . Then we can choose 1 2n _ i i a basis v1; : : : ; v2n of V with v1 = v. Let ' ;:::;' be the dual basis of V , i.e. ' (vj) = δj. Write P i j 1 ! = i<j aij' ^ ' . Then for all j = 1;:::; 2n, we have 0 = ev!(v1; vj) = a1j. Thus, ' does not occur in the expression for !, so ! is in V2V 0, where V 0 is the span of '2;:::;'2n. Thus, ^n(!) is in V2nV 0 = 0, so ^n(!) = 0. Now, assume that ev! is non-degenerate as a bilinear form. By Question 1, we may pick a symplectic basis a1; b1; : : : ; an; bn for V , i.e. ev!(ai; bi) = 1 and ev!(ai; aj) = ev!(bi; bj) = ev!(ai; bj) = 0 for all i 6= j. If we let a1; b1; : : : ; an; bn be the dual basis for V _ and express ! in terms of this basis, Equation (1) shows that ! = a1 ^ b1 + a2 ^ b2 + ··· + an ^ bn.