# (n - 1) dimensional submanifold of the manifold R^n

• Dec 19th 2012, 07:15 AM
vercammen
(n - 1) dimensional submanifold of the manifold R^n
Let $A$ be a symmetric $n \times n$ matrix over $\mathbb{R}$.
Let $0 \neq b \in \mathbb{R}$.

Show that the surface $M = \{x\in \mathbb{R}^n \mid x^T A x = b\}$ is an $(n - 1)$ dimensional submanifold of the manifold $\mathbb{R}^n$.

The differential of the smooth function $f: R^n \rightarrow R, f: x \mapsto x^TAx-b$ is
$f_*: R^n \rightarrow R, f_*: v \mapsto 2x^TAv$
Since $b \ne 0$, $x^tA$ never vanishes on the locus $f=0$ so $f_*$ has full rank.