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Topic: Bilinear forms and reflexivity
Replies: 24   Last Post: Jan 1, 2012 5:21 AM

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Paul Sperry

Posts: 1,392
Registered: 12/6/04
Re: Bilinear forms and reflexivity
Posted: Dec 26, 2011 12:11 AM
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In article <MPG.2961f642dd362374989a02@news.cc.tut.fi>, Kaba
<kaba@nowhere.com> wrote:

> Paul Sperry wrote:
> > > But how about the other direction, that reflexive implies symmetric or
> > > alternating?

> >
> > Reflexive only implies symmetric if the field has characteristic 2 - so
> > your assumption that the field is the reals means you would show
> > reflexive implies anti-symmetric.

>
> It think this is not true. See below.


I think you are correct. I have no idea what I had in mind.

> > A far as I know this is not trivial. A Google search turns up
> > <www.maths.qmul.ac.uk/~pjc/class_gps/ch3.pdf>

>
> The Theorem 3.3 of that paper gives ideas for a proof, although I think
> the proof itself is misleading if not flawed.
>
> Claim
> -----
>
> Let B : V times V --> F be a bilinear form, where V is a vector space
> over the field F. If B is reflexive _and_ non-degenerate, then B is
> symmetric. (Note that these are the same conditions as in the paper)
>
> Proof
> -----
>
> Denote the bilinear form B(u, v) by u * v. It trivially holds forall u,
> v in V that
>
> (u * u)(u * v) - (u * u)(u * v) = 0.
>
> This can be modified by bilinearity to
>
> u * (u * v)u - u * (u * u)v = 0
> <=>
> u * ((u * v)u - (u * u)v) = 0.
>
> By reflexivity
>
> ((u * v)u - (u * u)v) * u = 0.
>
> By bilinearity again
>
> (u * u)((u * v) - (v * u)) = 0.
>
> Since B is non-degenerate, u * u != 0, forall u != 0. Thus
>
> forall u != 0, v: u * v = v * u.


For _some_ v u*v != 0.

The cited article now proceeds like this:

Call u good if there is a v such that u*v = v*u != 0.

Use the above equation to show that if u is good then u*w = w*u for all
w. This commutivity implies that if u is good and u*w != 0 then w is
also good.

Let u be good and w any non-zero vector. Use non-degeneracy to show
that there is a v such that u*v != 0 and w*v != 0. Since u is good v is
good; since v is good w is good; so, if there is at least one good
vector, _all_ vectors are good and B is symmetric.

If there are _no_ good vectors set w = u in the equation and get
u*u = 0 for all u and B is alternating.

[...]

> This still leaves open the case of degenerate reflexive bilinear forms.
> Let us assume we are away from 2. The only bilinear form that is both
> symmetric and skew-symmetric is the zero form:


I suspect the author of the wiki article just forgot to add
"non-degenerate". I can't recall seeing the result elsewhere without
"non-degenerate".

> u * v = -v * u = -u * v
> ==>
> 2 (u * v) = 0
> ==>
> u * v = 0
>
> Thus a skew-symmetric bilinear form must be degenerate (since skew-
> symmetry implies reflexivity and the zero form is degenerate):
>
> B is skew-symmetric ==> B is reflexive and degenerate.


It is true that skew-symmetric implies reflexive if char(F) != 2 but
reflexive does _not_ imply both skew-symmetric _and_ alternating. There
is at least one well known skew-symmetric non-degenerate form.

> Does someone see a way forward?

--
Paul Sperry
Columbia, SC (USA)



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