Math 120c: Algebra HW 5

Let F be a field. Let V be a finite dimensional vector space over F and B be a nondegenerate bilinear form on V . Note that when V is finite dimensional, V is nondegenerate if and only if it is nonsingular, i.e. LB is an isomorphism.

1) Let (V,B) be a nondegenerate bilinear space.

a) Assume B is symmetric. An orthogonal basis is a basis e1, e2, ..., en such that ei ⊥ ej for i 6= j. Show that if characteristic is not 2, V has an orthogonal basis. (We note that if charF 6= 2 then B is completely determined by its value on B(v, v) for v ∈ V .) b) Assume B is alternating and V has dimension 2m. A symplectic basis is a basis e1, f1, e2, f2, ..., em, fm such that B(ei, fi) = 1 and the planes Ui = F ei +F fi are mutually perpendicular. Show that V has a symplectic basis. 2) a) Assume (V,B) is an alternating space. Show V is necessarily even dimen- sional. (Hint: Induction on the dimension. For each v 6= 0 in V , take a vector w ∈ V such that B(v, w) = 1. Show that F v + F w is a nondegen- erate subspace the make use of nondegeneracy of V to apply induction.)

b) Show that any two nondegenerate alternating bilinear spaces with same dimension are equivalent. Namely, up to a change of bases the matrices  0 I  associated to the bilinear forms are the same. (This matrix is r −Ir 0 relative to a sympletic basis with suitable ordering.) 3) Assume charF 6= 2 and B is nondegenerate symmetric. A quadratic form on V is a function q : V → F such that q(cv) = c2q(v) for all c ∈ F and v ∈ V 1 and the pairing Bq(v, w) := 2 (q(v + w) − q(v) − q(w)) is a bilinear form on V . a) Show that Q(v) := B(v, v) is a quadratic form and show if there exists v ∈ V nonzero such that Q(v) = 0 then Q takes on all values, i.e. Q(V ) = F . We say a quadratic form q such that q(V ) = F is universal. b) Show that if dim V ≥ 3 and F is a finite field then there exists v ∈ V nonzero such that Q(v) = 0. Deduce that Q is universal if dim V ≥ 2 and F is a finite field.

1 c) Show that if there exists v ∈ V nonzero such that Q(v) = 0, then V ' H ⊥ W for some nondegenerate subspace W , where H is a plane with quadratic form equals xy under certain basis. d) Assume V has dimension 2m or 2m + 1 and B = 0 on an m-dimensional subspace. Show that V has a nondegenerate subspace isomorphic to H⊥m, m copies of the hyperbolic plane H. In this case we say the quadratic space (V,Q) split. As a consequence the matrix associated to a split quadratic form of discriminant 1 has 1 on anti-diagonal and 0 elsewhere relative to some basis.

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