Connected components of a subset of E

Question

Let $E$ be a real vector space of dimension n+1 with a symmetric bilinear form B of signature (n,1). Let $H=\{x \in E : B(x,x) <0\}$. Somewhere I saw that it has two connected components. Can anybody please tell me what are its connected components?

Answer 1

Let $E=\mathbb{R}^{n+1}$, the set of all $x=(x_0,x_1,\ldots,x_n)$ with $x_0,x_1,\ldots,x_n\in\mathbb{R}$, and $B(x,y)=x_1y_1+\cdots+x_ny_n-x_0y_0$. (For any real vector space $E$ of dimension $n+1$ and any symmetric bilinear form $B$ on $E$ of signature $(n,1)$ we can choose a basis of $E$ such that $B$ will have the diagonal form given above in the cooordinates relative to the basis.) Then the two connected components of $H$ are $H_-=\{x\in H\mid x_0<0\}$ and $H_+=\{x\in H\mid x_0>0\}$. The sets $H_-$ and $H_+$ are connected because they are convex (and so they are path connected since in either one of the two sets any two points are connected by a line segment contained in the set). It suffices to prove that $H_+$ is convex, since $H_-=-H_+$. Suppose that $x,y\in H_+$, which means that $x_1^2+\cdots+x_n^2<x_0^2$ and $y_1^2+\cdots+y_n^2<y_0^2$, where $x_0>0$ and $y_0>0$. Let $0\leq\lambda\leq 1$ and set $z:=(1-\lambda)x+\lambda y$; we claim that $z\in H_+$. This boils down to proving that $x_1y_1+\cdots+x_ny_n< x_0y_0$, which we can do using the Schwarz inequality: $x_1y_1+\cdots+x_ny_n\leq\sqrt{x_1^2+\cdots+x_n^2}\sqrt{y_1^2+\cdots+y_n^2} < x_0y_0$.

Geometric explanation: the sets $H_-$ and $H_+$ are the interiors of the two halves of a double cone.
To see this, draw the figures of the sets $H_-$ and $H_+$ for $n=2$ and $n=3$.