Orbit closures of real symmetric bilinear form

Question

Let $\alpha$ and $\beta$ be two real symmetric bilinear forms in $\operatorname{sym}(\mathbb{R}^n)$, with signatures $(p_{\alpha},n_{\alpha},z_{\alpha})$ and $(p_{\beta},n_{\beta},z_{\beta})$.

I would please like some help to (dis)prove:

$\beta \in \overline{GL_{n}(\mathbb{R})\cdot \alpha}$ IF AND ONLY IF $p_{\alpha} \geq p_{\beta}$ AND $n_{\alpha}\geq n_{\beta}$.

Here, $GL_{n}(\mathbb{R})\cdot \alpha :=\{\alpha(g^{-1}\cdot,g^{-1}\cdot) : g\in GL_{n}(\mathbb{R})\}$ and $\overline{GL_{n}(\mathbb{R})\cdot \alpha}$ is the closure of $GL_{n}(\mathbb{R})\cdot \alpha$ with respect to the usual topology of $\operatorname{sym}(\mathbb{R}^n)$.

Answer 1

Hint. It's true. The backward implication is easy. Let's consider only the forward implication (the "only if" part). By Sylvester's law of inertia, the set $GL_n(\mathbb R)\cdot\alpha$ in your question is just the set of all real symmetric bilinear forms with identical inertia as $\alpha$. Let $\gamma$ be any real symmetric bilinear form. Use Cauchy's interlacing theorem for Hermitian matrices to show that $(1)\Rightarrow(2)$ in the following:

1. $\gamma$ is positive when evaluated at some $m$ orthonormal vectors (w.r.t. the usual inner product of $\mathbb R^n$).
2. $p_\gamma$ is greater than or equal to some nonnegative number $m$;

Now, if $\beta$ is the limit of a sequence of bilinear forms $\alpha_1,\alpha_2,\ldots\in \left(GL_n(\mathbb R)\cdot\alpha\right)$ and $v$ is an eigenvector of $\beta$ corresponding to a positive eigenvalue, consider $\alpha_j(v,v)$ when $j$ is large. Use the previous result to assert that $p_\beta\le p_{\alpha_j}=p_\alpha$.