Proof that such a set is a smooth submanifold. Let $M = \left \{ (x,y) \in \mathbb R^3 \times \mathbb R^3: \|x\|=\|y\|=1, \, x \perp y \right \}$. I'm trying to solving the following questions:


*

*Show a smooth structure as submanifold in $M$ of $\mathbb R^6$.

*Is $M$ compact?

*Describe its tangent space in $(e_1,e_2)$ where $\{ e_1, e_2, e_3\}$ is the canonic basis of $\mathbb R^3$.


I'm stuck in the first question. How can we construct a smooth atlases in $M?$
 A: Define $f : \mathbb R^6 \to \mathbb R^3, f(x_1\dots,x_6)= (x_1^2+x_2^2+x_3^2,x_4^2+x_5^2+x_6^2,x_1x_4 + x_2x_5+x_3x_6)$. Then you have $M = f^{-1}(1,1,0)$. This map is smooth with Jacobian
$$Jf(x_1,\dots,x_6) = \left( \begin{array}{rrrrrr}
2x_1 & 2x_2 & 2x_3 & 0 & 0 & 0 \\
0 & 0 & 0 & 2x_4 & 2x_5 & 2x_6 \\
x_4 & x_5 & x_6 & x_1 & x_2 & x_3\\
\end{array} \right) $$
For $(x_1,\dots,x_6)\in M$ we have $(x_1,x_2,x_3) \ne 0$ and $(x_1,x_2,x_3) \ne 0$, thus the first two rows are linearly independent. If the third row would not be linearly independent from the first two, we would get $(x_4,x_5,x_6) = \lambda \cdot 2(x_1,x_2,x_3)$, $(x_1,x_2,x_3) = \mu \cdot 2(x_4,x_5,x_6)$ for some pair $(\lambda,\mu) \ne (0,0)$. We conclude that $(x_4,x_5,x_6) = \rho(x_1,x_2,x_3)$ for some $\rho \ne 0$ which yields $x_1x_4 + x_2x_5+x_3x_6 = \rho(x_1^2 +x_2^2+x_3^2) \ne 0$. This is impossible for points of $M$.
Hence the Jacobian has rank $3$ in all points of $M$ which means that $(1,1,0)$ is a regular value of $f$. Therefore $M$ is a smooth $3$-dimensional submanifold of $\mathbb R^6$.
Clearly $M$ is a closed and bounded subset of $\mathbb R^6$, hence compact.
The tangent space $T_\epsilon M$ of $M$ in the point $\epsilon = (e_1,e_2) = (1,0,0,0,1,0) \in M$ can be regarded as a $3$-dimensional linear subspace of $\mathbb R^6$. Alternatively you may also view it as the affine subspace $\epsilon + T_\epsilon M$.
It is well-known that the tangent space $T_pM$ at a point $p$ of a manifold $M$ which the preimage of some regular value $\eta$ of a smooth map $f$ is nothing else than the kernel of the differential $df(p)$. In the present case this kernel is the set of solutions of $Jf(1,0,0,0,1,0) \cdot \xi = 0$, i.e. of the three equations
$2\xi_1 = 0, 2\xi_5 = 0, \xi_2 + \xi_4 = 0$. Therefore
$$T_\epsilon M = \{ (0,\xi_2,\xi_3,-\xi_2, 0, \xi_6) \mid \xi_2,\xi_3, \xi_ \in \mathbb  R \} .$$
