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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 !

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    Yes, thanks to the regular value theorem. https://people.math.gatech.edu/~ghomi/LectureNotes/LectureNotes9G.pdf – 2017-02-18
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    @GSF The derivative is a map which belongs to $\mathcal{L}(\mathbb{R}^{n\times p}, \mathbb{R}$) so if I find a value such that the derivative has a rank equal to $0$ it will be ok ? – 2017-02-18

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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.

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    Thank you so with your differential : if the rank was equal to $0$ it would mean that $x=0$ and $y=0$. But the point $(0,0) \notin \mathcal{Q}$. So the rank is necessary equal to $1$ . But how do you go from my differential to your differential ? How do go from the scalar product to your form ? – 2017-02-18
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    Remember that the scalar product associated to the euclidean norm on $\mathbb{R}^n$ is defined, for $x,y\in\mathbb{R}^n$ as $\langle x,y\rangle:=\sum_{i=1}^nx_iy_i$, from that follows the differential! – 2017-02-18
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    For instance, if you fix a certain $0\leq i\leq n$, $\frac{\partial}{\partial x_i}(\sum_{i=1}^n x_i^2-\sum_{j=1}^py_j^2-1)=2x_i$. – 2017-02-18
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    That is why your differential is equal to mine, $d_{(x_0,y_0)}f(x,y)=\langle\nabla f(x_0,y_0),(x,y)\rangle=2\langle x_0, x\rangle-2\langle y_0,y\rangle$. – 2017-02-18
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    there is the map and the map applied to a vector so $d_{(x_0,y_0)}f=2\langle x_0, .\rangle-2\langle y_0,. \rangle$ ? – 2017-02-18
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    You could see it that way, yes. @Maman – 2017-02-18
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    Why could we write $d_{(x_0,y_0)}f=2\langle x_0,.\rangle-2\langle y_0,.\rangle=2x_0 -2y_0$ ? This equality is confusing. – 2017-02-18
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    $$d_{(x,y)}f=2(x_1,\dots,x_n,-y_1,\dots,-y_p)$$ therefore $$d_{(x,y)}f(h,k)=2(x_1,\dots,x_n,-y_1,\dots,-y_p)\cdot\begin{pmatrix} h_1\\ h_2\\ \vdots \\ k_p \\ \end{pmatrix}=\langle d_{(x,y)}f, (h,k)\rangle$$ – 2017-02-18