Let consider $\mathcal{Q} := \{(x,y) \in \mathbb{R}^n \times \mathbb{R}^p \ / \ \vert x \vert^2 -\vert y\vert ^2=1 \}$ where $\vert . \vert$ represents the euclidean norm on $\mathbb{R}^n$ and $\mathbb{R}^p$. Is it a submanifold of $\mathbb{R}^{n\times p}$ ?
Let $f(x,y)=\vert x \vert^2 -\vert y\vert ^2-1= (x\mid x)-(y\mid y)-1$ where $(.\mid .)$ represents the scalar product associated to $\vert . \vert$.
The map is differentiable and for $(x_0,y_0)$ we have :
$\mathrm{d}f_{(x_0,y_0)}(h,k)=2(x_0 \mid h)-2(y_0 \mid k)$ where $h$ is a vector from $\mathbb{R}^n$ and $k$ a vector from $\mathbb{R}^p$.
Do I have to prove that the rank is equal to $1$ to conclude ?
Thanks in advance !
Hint 1: All you have to see is that the differential (or derivative as you call it) is not zero at any $(x_0,y_0)$ of $\mathcal{Q}$. If you show that, $d_{(x_0,y_0)}f$ will have rank 1 and therefore it will be a submersion, hence the result.
Hint 2: Notice that if $x=(x_1,\dots,x_n)$ and $y=(y_1,\dots,y_p)$, the differential at pont $(x,y)\in\mathcal{Q}$ is $$d_{(x,y)}f=2(x_1,\dots,x_n,-y_1,\dots,-y_p)$$ And all you have to do is take a look at the previous hint and see that it cannot be possible that all the components of the differential are zero.