I've recently taught about manifolds and some of their properties, and I'm working on problems to help me get a grasp of them. So far I've been trying to show that certain sets are manifolds, but I'm stuck on showing that the following is a $\mathcal{C}^\infty$ manifold of dimension $n-1$:
Let $M=\{x=(x_1,\dots,x_n)\in\mathbb{R}^n :\sum^n_{i,j=1} a_{ij}x_i x_j = 1\}$, where $A=(a_{ij})$ is an invertible symmetric matrix of rank $n$.
I'm trying to show that the gradient of the function $f(x)=\left(\sum^n_{i,j=1}a_{ij}x_ix_j\right)-1$ has rank $n-1$, but it's getting really complicated and messy. I'm just wondering if there is a better way of doing it, and how one would go about doing it.
Thanks!
You can rewrite $M=\{x \in \mathbb R^n \mid x^T A x = 1\}$. Therefore $f(x)=x^TAx-1$.
The Fréchet derivative of $f$ if $f^\prime(x) = 2x^TA$ which is a non-zero linear form for all $x \neq 0$. Hence the kernel of $f^\prime(x)$ is of dimension $n-1$. Therefore $M$ is a manifold of dimension $n-1$, and is $\mathcal C^\infty$ as $f$ is $\mathcal C^\infty$.