4
$\begingroup$

Assume I have a real bilinear form $b$ of signature $(p,q)$ on $E=\Bbb R^n$, and a subspace $F$ of $E$ of dimension $d$. What are the possible signatures $(\alpha, \beta)$ of the restriction of $b$ to $F \times F$?

I saw the question possible signatures of bilinear form on subspaces, but it is unanswered.

For instance, if $q=1, p=n-1$, apparently the only possibilities for $b\vert_{F \times F}$ are $(\alpha, \beta) = (d,0), (d-1,1), (d-1,0)$. I was able to show that if $\beta \geq 1$, then we must have $(\alpha, \beta) = (d-1,1)$, but how to do the case $\beta=0$? What about more general cases for $p$ and $q$? (For instance, do we have $\alpha ≤ p, \beta ≤ q$ ?).

Thank you!

$\endgroup$
4
$\begingroup$

I'm assuming from what you wrote that you want $E$ to be a non-degenerate space, i.e. $p+q = n$. The arguments below can also be extended to the case $p+q<n$ if needed, but are nicer to see in the non-degenerate case.

Take an orthogonal basis of $E$, of the form $e_1,e_2,\ldots e_p,f_1,f_2,\ldots f_q$ such that $\left<e_i,e_i\right> = 1$ and $\left<f_i,f_i\right> = -1$ (and everything else zero). Then given a desired signature $(a,b)$ and a dimenson $d$, we can construct a space $F$ of this dimension and signature as follows:

  1. Take $e_1,e_2,\ldots, e_a$.
  2. Take $f_1,f_2,\ldots, f_b$.
  3. Fill up with pairs $e_i+f_i$ from the remaining vectors until reaching dimension $d$.

Now let's look when this is possible. We need $a \leq p$, we need $b \leq q$ and we need $d-a-b \leq \min(p-a,q-b)$. If we have this, we can construct a space of desired dimension and signature.

Now all that is left is to show that these are all the signatures we can construct. For that let $F \leq E$ be a subspace of signature $(a,b)$ and dimension $d$ and let $F^{\perp}$ be the orthogonal complement of $F$ in $E$. Set $F_1 := F \cap F^{\perp}$. Then there exists $F_2 \leq F$ such that $$F = F_1 \oplus F_2$$ and $F_2$ is non-degenerate, having signature $(a,b)$. Hence, we can extend an orthogonal basis of $F_2$ to an orthogonal basis of $E$ and get $a \leq p$ and $b \leq q$. For the last condition, look at the space $$V := E/F_2.$$ This space is non-degenerate of signature $(p-a,q-b)$. and under the standard projection $\pi$, we get that $\pi(F_1)$ has dimension $$\dim(\pi(F_1)) = \dim(F_1) = d-a-b.$$ This is due to the fact that $F_1 \cap F_2 = \{0\}$.

Now we are almost done: $\pi(F_1)$ is an isotropic subspacie (i.e. $\langle x,y\rangle = 0$ for all $x,y \in \pi(F_1)$) of dimension $d-a-b$ of a space of signature $(p-a,p-b)$. This is only possible if $d-a-b \leq \min(p-a,p-b)$.

There are quite some results I used here, e.g. the last one is due to the definition of the Witt index, so if something is unclear feel free to ask.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you very much for this complete answer! I will tell you if I have some more questions. $\endgroup$ – Alphonse Jun 13 '17 at 21:27
  • $\begingroup$ Very impressive! Besides, I have revised a typo in your post~ $\endgroup$ – Bach Jun 7 '19 at 16:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.