|
|
Re: Orthogonal complement
Posted:
Mar 4, 2013 8:34 AM
|
|
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.
|
|
