
Re: Bilinear forms and reflexivity
Posted:
Dec 27, 2011 10:33 PM


In article <MPG.2963e62ad4a55a17989a08@news.cc.tut.fi>, Kaba <kaba@nowhere.com> wrote:
> Paul Sperry wrote: > > In the book "Groups and Characters" (Prop 1.7.6 page 24) Larry C Grove > > claims to have proved the result without assuming nondegeneracy. If > > the proof is correct, I don't follow it. The book is searchable at > > Amazon. (Use "reflexive" for the search.) > > Thanks for the link. The proof seems valid. Here's a more detailed > proof: > > As already shown in this thread, reflexivity is equivalent to > > forall u, v, w in V: > B(u, v)B(w, u) = B(v, u)B(u, w) (1) > > This is the assumption of Proposition 1.7.6. Taking u = v, it follows > that > > forall v, w in V: > B(v, v)[B(w, v)  B(v, w)] = 0. (2) > > We would like to to show that this implies symmetry or alternation: > > [forall v in V: B(v, v) = 0] or > [forall v, w in V: B(w, v)  B(v, w) = 0]. (3) > > Assume (3) does not hold. Then > > [exists y in V: B(y, y) != 0] and > [exists x, z in V: B(x, z)  B(z, x) != 0]. (4) > > The (2) holds in particular for x and z: > > B(x, x)[B(z, x)  B(x, z)] = 0 > ==> > B(x, x)[B(x, z)  B(z, x)] = 0 > ==> > B(x, x) = 0. > > On the other hand, by (2) again > > B(z, z)[B(x, z)  B(z, x)] = 0 > ==> > B(z, z) = 0. > > Third time by (2) gives > > B(y, y)[B(x, y)  B(y, x)] = 0 > ==> > B(x, y) = B(y, x), (5) > > since B(y, y) != 0 by (4). Similarly > > B(y, y)[(B(z, y)  B(y, z)] = 0 > ==> > B(z, y) = B(y, z). (6) > > Substitute u = x, v = y, and w = z in (1). Then > > B(x, y)B(z, x) = B(y, x)B(x, z) > ==> > B(x, y)[B(z, x)  B(x, z)] = 0 > ==> > B(x, y) = 0 (= B(y, x)), > > since B(x, y) = B(y, x), by (5), and B(z, x)  B(x, z) != 0, by (4). > > Substitute u = z, v = y, and w = x in (1). Then > > B(z, y)B(x, z) = B(y, z)B(z, x) > ==> > B(z, y)[B(x, z)  B(z, x)] = 0 > ==> > B(z, y) = 0 (= B(y, z)), > > since B(z, y) = B(y, z), by (6), and B(z, x)  B(x, z) != 0, by (4). > > Summarizing our results thus far, we have found out that for the x, y, > and z of (4), it holds that > > B(x, x) = 0 (7) > B(y, y) != 0 (4) > B(z, z) = 0 (8) > B(x, y) = B(y, x) = 0 (9) > B(y, z) = B(z, y) = 0 (10) > B(x, z) != B(z, x) (4) > > Now > > B(x, y + z) = B(x, y) + B(x, z) = B(x, z) > > and > > B(y + z, x) = B(y, x) + B(z, x) = B(z, x). > > By (4), > > B(y + z, x) != B(x, y + z). > > Choose v = y + z and w = x in (2). Then > > B(y + z, y + z)[B(x, y + z)  B(y + z, x)] = 0 > ==> > B(y + z, y + z) = 0. > > On the other hand, > > B(y + z, y + z) = B(y, y) + B(y, z) + B(z, y) + B(z, z) > = B(y, y) != 0, > > a contradiction. Thus (3) holds, and B must be either symmetric or > alternating. QED.
I'm convinced. I think I missed some of the nonzero things so some cancellations bothered me  read in haste, repent at leisure I guess. > I think I'll add a proof sketch along with this reference to Wikipedia > at some time.
 Paul Sperry Columbia, SC (USA)

