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Bilinear Forms

Tim Campion

January 28, 2011

1 TerminologyLet V be a finite-dimensional vector space over a field k of characteristic = 2and Q V V a bilinear form. We say that V is a Q-module.

Nondegeneracy Q induces a map Q : VV by v Q(v, ). V iscalled nondegenerate if Q is an isomorphism. The largest-dimension of asubspace ofV on which V is nondegenerate is called its rank. Note that asubspace of a nondegenerate space need not be nondegenerate.

Symmetry (Q,V) is called braided if there is k such that Q(v, w) =

Q(w, v) for all v, w V. The case = 1 is called symmetric; the case = 1is called alternating; the case = 0 is the trivial module we will generallysilently exclude this case unless we explicitly mention it.

If char k = 2, then the notions of symmetric and alternating coincide. Ifchar k = 2, then V is alternating if and only if Q(v, v) = 0 for all v V.For Q(v, v) = Q(v, v) implies 2Q(v, v) = 0; since char k = 2, this impliesQ(v, v) = 0. Conversely, we have

0 = Q(v+w, v+w) = Q(v, v)+Q(v, w)+Q(w, v)+Q(w, w) = Q(v, w)+Q(w, v)

Any subspace of an -braided space is -braided, in contrast to the fact

that a subspace of a nondegenerate space need not be nondegenerate. A1-dimensional alternating space in char k = 2 is degenerate; we will shortlysee that this is true of all odd-dimensional alternating spaces.

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Orthogonality If Q(v, w) = 0 then we write v w and say v is left

orthogonal to w. The left orthogonal complement W

of a subspace W V is the set of all vectors left orthogonal to every element of W. If v isleft orthogonal to w for all w V, then v is called left isotropic. Q isnondegenerate iff none of its elements are isotropic. 1

If V is nondegenerate and W V, then dimW = dimV dimWbecause Q is an isomorphism; the image ofW has the same dimension asW, and so ker Q|W = W

has the complementary dimension in V. 2

All of these notions can, of course be defined on the right, too. IfV isbraided, then left orthogonal is the same as right orthogonal; left isotropic isthe same as right isotropic. For ifQ(v, w) = Q(w, v), then Q(v, w) = 0 ifand only ifQ(w, v) = 0. A space where left and right orthogonality coincide

is called ambidextrous.

Bases A basis {e1, . . . , en} for V is called sparse if for each ei there is atmost one eji with Q(ei, eji) = 0 and moreover Q(ek, ei) = 0 for k = ji.A basis {e1, . . . , en} for V is called an orthogonal basis if Q(ei, ej) = 0 fori = j. A basis {e1, . . . , en, f1, . . . , f n} for V is called symplectic ifQ(ei, fj) =Q(fj , ei) = ij . Orthogonal and symplectic bases are sparse.

1It suffices to consider left isotropy alone, because Q is nondegenerate iff its matrix isinvertible; but inverses of square matrices can have only 2-sided inverses.

2Linear algebra fact: Let V be a finite dimensional vector space. Then W V is

uniquely determined by ker W, by which I mean the intersection of the kernels of allelements of W, and they are complementary under any identification V = V sendingker W to a comlementary subspace of W, of which there exists at least one. This is aspecial case of the Nullstellensatz... Heres an algorithmic way to do it (I would like tohave a more conceptual way, but perhaps thats not in the cards after all... No, I still thinkit might be.): choose a basis for ker W; we will extend it to one on V. Let 0 = 0 V, andinductively choose vi ker0 and i W such that i(vi) = 0, setting i = i1 {i}.This continues until ker = ker W. Then we can apply Gram-Schmidt: normalize so thati(vi) = 1. Set 1 = 1; we have 1(vj) = 1j. Inductively suppose that ivj = ij fori < k. Then let k = k

i

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2 Decomposition of Bilinear Forms

Lemma 1. Let (Q,V) be a nondegenerate bilinear module and e V. Ifv e is left-isotropic in e, thenv is a scalar multiple of e.

Proof. Since V is nondegenerate, e is of codimension 1. Pick f V\e, acomplement. Since V is nondegenerate, we can normalize f so that Q(e, f) =1. Ife e is left-isotropic, then by the nondegeneracy ofV we must haveQ(e, f) = c = 0. Then W(ece,f) = 0. Since ece is also left-isotropic one, we see that ece is in fact left-isotropic on V. Since V is nondegenerate,in fact e = ce.

So ife e, then e is nondegenerate, while ife e in an ambidextrous

space, then any complementary subspace to e in e is nondegenerate (since ife were isotropic in W, then it would also be isotropic in e, as Q(e, e) = 0).

Lemma 2. Let(Q,V) be a nondegenerate module. Lete, f V andW e.If W is nondegenerate and there is a sparse basis for W, then there existsf W with f f e.

Proof. Let w1, . . . , wn be a sparse basis for W, and let wi be the partnerof wi. Since W is nondegenerate, we have Q(wi, wi) = 0 for all i. Letf =

i iwi for scalars i we are about to set. Certainly f f

e.

Then Q(wi, f) = Q(wi, f) +

i iQ(wi, wi). So let i =

Q(wi,f)Q(wi,wi)

; then

f W.

Theorem 3. An ambidextrous module (Q,V) has a sparse basis.

Proof. Let (Q,V) be an -braided module, and let e V. First considerthe case that e e. By Lemma 1, e is nondegenerate. By induction (thetheorem holds if dimV = 0, 1) there is a sparse basis B for e. Adjoining eto this basis provides a sparse basis for V, with e being its own partner.

Now consider the case that e e. By Lemma 1, ifW e is com-plementary to e, then W is nondegenerate. By induction, there is a sparsebasis B for W. Since V is nondegenerate, there is f V with Q(e, f) = 1;

since f e, we have V = e f (abstractly, not orthogonally). By Lemma2, there is f W with f f e. Then Q(e, f) = Q(e, f) = 1 whileQ(b, f) = 0 for any b B. Since V is ambidextrous, we also have Q(f, b) = 0,so adjoining e, f to B provides a sparse basis for V.

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Corollary 4. A symmetric module has an orthogonal basis. An alternating

module has a symplectic basis.Proof. The alternating case is clear: take a sparse basis, the partner of abasis vector is itself only if that vector is isotropic. Otherwise, we have apartition of the basis into {e1, . . . , en, f1, . . . , f n} where ei, fi are partners; anypartners with Q(ei, fi) = 0 are isotropic and can be reassigned to themselves.Normalize fi so that Q(ei, fi) = 1; since V is alternating then Q(fi, ei) = 1.So this is a symplectic basis.

In the symmetric case, we can again separate out the isotropic vectors. Ofthe remaining vectors, some are self-partnered, and hence already orthogonal.The others can be normalized as in the alternating case, except now we have

Q(ei, fi) = Q(fi, ei) = 1. Then the vectors vi = ei + fi, wi = ei fi haveQ(vi, wi) = 0 and Q(vi, vi) = 2, Q(wi, wi) = 2, and span the same spaceas ei, fi (recall char k = 2). So these along with the self-partners provide anorthogonal basis.

Theorem 5. Any bilinear module can be decomposed as orthogonal the di-

rect sum of a nondegenerate symmetric module, a nondegenerate alternating

module, and uniformly isotropic module.

Proof. IfV is alternating, we are done. Otherwise, there is an element e Vwhich is not self-orthogonal. Then e is nondegenerate by Lemma 1, andthere is a sparse basis for e by Corollary 4. By Lemma 2, there is e

orthogonal to e with e e e. Then adjoining e to the sparse basison e continues to provide a sparse basis, whether Q(e, e) = 0 (and weexpand the uniformly zero part) or whether Q(e, e) = 0 (and we expand thesymmetric part).

3 Invariants of Bilinear Forms

The alternating and symmetric and trivial parts of a bilinear form are in-variant, because Q|W is -braided iffQ(v, w) =

12(Q|W(v, w) + Q|W(w, v))

for all v, w W (including the = 1, 0).

Trivial modules have only one invariant: dimension.It is immediate from the above discussion that, up to isomorphism, there

is a unique nondegenerate alternating form of rank 2n for each n N, and itis the direct sum ofn copies of the 2-dimensional alternating form; degeneratealternating forms simply tack on direct sums of the trivial form.

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Symmetric modules are more complicated. Weve seen that they can

be diagonalized; dimensions can be stretched so that the diagonal entriesare defined up to isomorphism only mod k2; in particular if k

2= k

(for instance, ifk is algebraically closed), then all nondegenerate symmetricmodules of the same rank are isomorphic. But change of basis multiplies thedeterminant by the square of the basis matrix, so the determinant in k/k

2

is an invariant. Over the reals, the signature is another invariant. In otherfields, other invariants are used.

4 Automorphisms of Bilinear Modules

This is the great can of worms! This is essentially equivalent to the study ofsemisimple Lie groups (or algebraic groups)!

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