Show isomorphism $W_1 \hookrightarrow V \twoheadrightarrow W_2$

Question

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.

Answer 1

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.