# Show isomorphism $W_1 \hookrightarrow V \twoheadrightarrow W_2$

Let $\langle , \rangle$ be a non-degenerate bilinear form with the signature $(p,q)$ on a real vectorspace $V$ and $W_1, W_2$ subspaces, such that the restriction $\langle , \rangle |_{W_i}$ is non-degenerate with signature $(p,0)$.

Show that for the orthogonal direct sum decomposition $V = W_2 \oplus W_2^\perp$ and the induced projection $V \to W_2$ the composition $$W_1 \hookrightarrow V \twoheadrightarrow W_2$$ is an isomorphism.

• From the data you have, what do you know about the dimensions of $W_1,W_2$ and $V$? – Daniel Fischer Apr 18 '14 at 8:50
• Well, to be an isomorphism the dimensions of $W_1$ and $W_2$ have to be equal, and since $V$ is the direct sum of $W_2 \oplus W_2$ it should have the same dimension as $W_1$ and $W_2$. To be honest I'm confused of the data, I don't see how it all connects... – sj134 Apr 18 '14 at 16:53
• Point 1: What does the signature of the bilinear form signify? Point 2: What does it mean that the form is nondegenerate (on $V$, resp. on $W_i$)? – Daniel Fischer Apr 18 '14 at 17:03
• The signature gives the rank and dimension of the bilinear form with $p+q$ and being non-degenerate means that the determinant of the gramian matrix is non zero and that the map $V \to V^*,v \mapsto \langle v,\cdot \rangle$ is injective. – sj134 Apr 19 '14 at 8:10
• Sorry for the delay. That's right, but only in part what I was aiming for (but, how were you to know what I was aiming for). Anyway, since the form is non-degenerate on both $W_i$ and has the same signature on both, it follows that $\dim W_1 = \dim W_2$. Now, for $W_1 \hookrightarrow V \twoheadrightarrow W_2$ to be an isomorphism, what still is needed is that $W_1 \cap W_2^\perp = \{0\}$. That follows from the signatures. But to see that, you need to know what the signature means, what the parts stand for. – Daniel Fischer Apr 20 '14 at 23:09

## 1 Answer

The signature $(p,q)$ of the bilinear form gives the maximal dimensions of the subspaces on which $\langle\cdot,\cdot\rangle$ is positive respectively negative definite.

Hence the premise that the restriction of the bilinear form to both $W_i$ is nondegenerate with signature $(p,0)$ means

• $\dim W_1 = \dim W_2 = p$,
• $\langle\cdot,\cdot\rangle\lvert_{W_i}$ is positive definite, and
• $\langle\cdot,\cdot\rangle$ is not positive definite on any subspace properly containing one of the $W_i$.

Now, with the injection $\iota \colon W_1 \hookrightarrow V$ and the projection $\pi \colon V \twoheadrightarrow W_2$ with kernel $W_2^\perp$, since the dimensions of $W_1$ and $W_2$ are the same, the composition $\pi \circ \iota$ is an isomorphism if and only if it is injective. We have

$$\ker (\pi\circ \iota) = \iota^{-1}(\ker \pi) = W_1\cap \ker\pi = W_1 \cap W_2^\perp,$$

so we must prove that $W_1\cap W_2^\perp = \{0\}$.

Consider the subspace $U = W_2 \oplus (W_1 \cap W_2^\perp)$ of $V$. Deduce that $\langle\cdot,\cdot\rangle\lvert_U$ is positive definite, and hence $\pi\circ\iota$ is an isomorphism.