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!


1 Answer 1


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.

  • $\begingroup$ Thank you very much for this complete answer! I will tell you if I have some more questions. $\endgroup$
    – Alphonse
    Jun 13, 2017 at 21:27
  • $\begingroup$ Very impressive! Besides, I have revised a typo in your post~ $\endgroup$
    – Bach
    Jun 7, 2019 at 16:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .