Let $f : A\subset \mathbb{R}^{n+1} \to \mathbb{R}$, what does mean that $f$ is a submersion?

Question

I am trying to answer the following question:

Let $M_a := \{ (x^1,\ldots,x^n,x^{n+1}) \in \mathbb{R}^{n+1} : (x^1)^2 + \cdots +(x^n)^2 - (x^{n+1})^2 = a\}$. For which values of $a$, $M_a$ is a submanifold of $\mathbb{R}^{n+1}?$

I have to use the regular value theorem, but I think I don't know in fact how to do it... I thought the following:

Let $f: \mathbb{R}^{n+1} \to \mathbb{R}$ defined as: $f(x^1,\ldots,x^{n+1}) = (x^1)^2 + \cdots+ (x^n)^2 - (x^{n+1})^2.$ Then if $f$ has constant rank, then every set level is a manifold... How to relate this with $a$? I am not seeing the connection...

Answer 1

If $f$ has constant rank, you have a submersion and $a$ does not matter. But it is not the case (as noticed by Tsemo Aristide): $$\nabla f(x^1,...,x^{n+1})=(2x^1,...,2x^n,-2x^{n+1}).$$ Therefore, $(0,...,0)$ is not a regular point.

In the theorem, the dependence of the point $a$ appears when you need to check regularity (in this case): $M_a$ is a submanifold if $\nabla f(x)$ is non-zero for all $x\in M_a$. So, if you change $a$, you change the points where you have to check the condition on the gradient/differential.

Note that this argument guarantee that $M_a$ is a submanifold for $a\neq 0$.

Answer 2

You have to compute $df =2x_1dx_1+...2x_ndx_n -2x_{n+1}dx_{n+1}$. Let $a$ be a non zero real. for every $x=(x_1,...,x_n)\in f^{-1}(a)$, there exists $i$ such that $x_i\neq 0$. Let $e_i$ be the vector of $R^{n+1}$ whose $i$-coordinate is $1$ and its other coordinates are $0$. $df_x(0,..,0,1,0..)=x_i\neq 0$. Thus $df_x$ is surjective, and $f^{-1}(a)$ is a manifold.

If $a=0$, $0_{n+1}$ the vector of $R^{n+1}$ whose coordinates are $0$ is in $f^{-1}(0)$. $df_{o_{n+1}}=0$. Thus the tangent space of $f^{-1}(0)$ is not defined at $0_{n+1}$. Thus $f^{-1}(0)$ is not a submanifold.