Theorem: Quadratic component classification

Let $Q$ be a nondegenerate quadratic form on $V$, with $\dim V=m\ge 3$. Suppose $\{Q>0\}\ne \varnothing$, and a connected component $F$ of $\{Q<0\}$ has pointed closure. Then $Q$ has negative index one and positive index $m-1$.

Conversely, for this signature, $\{Q<0\}$ has two components, each an open convex cone with proper closure.

Proof. By a linear change of coordinates,

The assumptions imply $k,\ell \ge 1$. The map

is a homeomorphism, with inverse

If $k\ge 2$, the negative region is connected and contains both $v$ and $-v$ for each of its vectors. Its closure is not pointed. Thus $k=1$.

Writing $u=t$, the components are

The triangle inequality proves convexity. Their closures $t\ge \Vert w{\Vert }_2$ and $t\le -\Vert w{\Vert }_2$ are pointed, closed, and have nonempty interior.