How to compute $T_{\left(u_{1}, u_{2}\right)} S^{3}$ and $T_{(z, t)} S^{2}$? The Euclidean space $\mathbb{R}^{4}$ can be  identified with the $\mathbb{R}$-vector space $\mathbb{C}^{2}$ provided with its canonical Euclidean structure given by:
$$
\left\langle\left(u_{1}, u_{2}\right);\left(v_{1}, v_{2}\right)\right\rangle=\Re\left(u_{ 1} \overline{v_{1}}+u_{2} \overline{v_{2}}\right),
$$
with $u_{1}, u_{2}, v_{1}, v_{2} \in \mathbb{C} .$ Similarly $\mathbb{R}^{3}$ identifies with space euclidean $\mathbb{C} \times \mathbb{R}$ equipped with its scalar product:
$$
\left\langle(z, t);\left(z^{\prime}, t^{\prime}\right)\right\rangle=\Re\left(z \overline{z^{\prime}}\right)+tt^{\prime},
$$
with $z, z^{\prime} \in \mathbb{C}$ and $t, t^{\prime} \in \mathbb{R}$, we can thus write:
$$
S^{3}=\left\{\left(u_{1}, u_{2}\right) \in \mathbb{C}^{2} /\left|u_{1}\right|^{2 }+\left|u_{2}\right|^{2}=1\right\} \text { and } S^{2}=\left\{(z, t) \in \mathbb{C} \times \mathbb{R} /|z|^{2}+t^{2}=1\right\} .
$$

*

*Give the equations of the tangent spaces $T_{\left(u_{1}, u_{2}\right)} S^{3}$ and $T_{(z, t)} S^{2}$ .


$T_{\left(u_{1}, u_{2}\right)} S^{3}= \{(u,v) \in S^3\times \mathbb C^2 : \langle u,v\rangle=0\}=\{(u,v) \in S^3\times \mathbb C^2 : \Re(u_1 \bar v_1+u_2 \bar v_2)=0 \}$
and
$T_{\left(z,t\right)} S^{2}= \{(z,t) \in S^2\times \mathbb C : \langle z,t\rangle=0\}=\{(z,t) \in S^2\times \mathbb C^2 : \Re\left(z \overline{z^{\prime}}\right)+tt^{\prime}=0 \}$
I don't know is that true or not ? What I know is that if $M$ is an $n$-manifold, so we define the tangent space in a point $x$ as $T_xM=\{(x,v)\in M\times R^n :  \exists  \beta : \mathbb R \to M  : \beta(0)=x, \beta'(0)=v\}$ !I dont know how to use it here ! so any help is really appreciated !
 A: It seems that you only consider smooth $n$-dimesional submanifolds $M \subset \mathbb R^N$ of an Euclidean space. Note that in general $n < N$; if $n = N$ you would get open subsets $M \subset \mathbb R^N$. In your example you consider $S^n \subset \mathbb R^{n+1}$.
You define $T_xM = \{x\} \times \{ v \in \mathbb R^N \mid \exists \beta : \mathbb  R \to M, \beta(0) = x, \beta'(0) = v \}$. Here of course $\beta$ is required to be smooth and $\beta'(0)$ is the "usual" derivative of $\beta$ when regarded as a function with range $\mathbb R^N$.
Let us prove that $T_xS^n = \{x\} \times \{x\}^\bot$, where $\{x\}^\bot = \{v \in \mathbb R^{n+1} \mid \langle x,v \rangle = 0 \}$ is the set of vectors which are orthogonal to $x$. It is well-known that this is an $n$-dimensional linear subspace of $\mathbb R^{n+1}$.
Since we know that $T_xS^n$ is $n$-dimensional, it therefore suffices to show that $\{x\} \times \{x\}^\bot \subset T_xS^n$. That is, for each $v \in \{x\}^\bot$ there exists $\beta$ such that $\beta(0) = x$ and $\beta'(0) = v$.
Define
$$\beta(t) = \dfrac{x+tv}{\lVert x+tv \rVert} .$$
Since $x$ and $v$ are orthogonal, they are linearly independent and therefore $x+tv \ne 0$ for all $t$. Hence $\beta$ is well-defined and smooth. Note that $\lVert x+tv \rVert = \sqrt{\langle x +tv, x+tv \rangle} = \sqrt{\lVert x \rVert^2 + t^2 \lVert v \rVert^2}$ because $\langle x, v \rangle = 0$. Hence the derivative of $u(t) = \lVert x+tv \rVert$ is $$u'(t) = \frac 1 2 (\lVert x \rVert^2 + t^2 \lVert v \rVert^2)^{-1/2}2t\lVert v \rVert^2 = \lVert x+tv \rVert ^{-1}t\lVert v \rVert^2.$$
Clearly $\lVert \beta(t) \rVert = 1$ for all $t$, thus $\beta$ is a curve in $S^n$. By definition $\beta(0) =  \frac{x}{\lVert x \rVert} = x$. Moreover the quotient rule gives for the $i$-th coordinate function $\beta_i(t) = \frac{x_i +t v_i}{\lVert x+tv \rVert}$
$$\beta_i'(t) = \dfrac{v_i\lVert x+tv \rVert - (x_i+tv_i)\lVert x+tv \rVert ^{-1}t\lVert v \rVert^2}{\lVert x+tv \rVert^2} .$$
Thus $\beta_i'(0) = \frac{v_i\lVert x \rVert}{\lVert x \rVert^2} = v_i$, i.e. $\beta'(0) = v$.
