# Signature of restriction of a bilinear form

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!

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 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 $$\left = 1$$ and $$\left = -1$$ (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.

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