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Topic: Orthogonal complement
Replies: 15   Last Post: Mar 5, 2013 6:48 PM

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 David C. Ullrich Posts: 21,553 Registered: 12/6/04
Re: Orthogonal complement
Posted: Mar 4, 2013 8:34 AM
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On Sun, 03 Mar 2013 20:16:52 -0500, Shmuel (Seymour J.) Metz
<spamtrap@library.lspace.org.invalid> wrote:

>In <9s07j89ui5l98kpkfnmv1l2oq5ocdt14tg@4ax.com>, on 03/03/2013
> at 11:11 AM, David C. Ullrich <ullrich@math.okstate.edu> said:
>

>>Maybe the "+" was supposed to mean _direct_ sum? It seems likely
>>that it's easy to show that f is non-degenerate if and only if S
>>intersect C(S) = 0 for all S.

>
>Let f be a Lorentz metric, s nonzero with f(s,s)=0 and S = {s}. Then S
>intersect C(S) is nonnull.

The two of us are evidently interpreting the word "non-degenerate"
differently. I was just sort of guessing what that meant...

Hmm. According to the definition at

http://en.wikipedia.org/wiki/Bilinear_form

your f is _not_ non-degenerate. In the notation used there
we have f_1(s)(s) = 0, hence f_1 is not an isomorphism.

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