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Re: Bilinear forms and reflexivity
Posted:
Dec 27, 2011 10:33 PM
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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 non-degeneracy. 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 non-zero 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)
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