Prove that $v \in T_mSL(n) \Leftrightarrow trv = 0$. Prove that $v \in T_mSL(n) \Leftrightarrow trv = 0$.
I haven't idea how to prove that. I understand next steps:
1) $SL(n)$ - is matrices $n^2$ that have $\det = 1$
2)$T_mSL(n) = ^{Curv_m(SL(n))}/_{\underset{m}{\sim}}$
Where $Curv_m(SL(n)) = \underset{\alpha \in [\small{-}\infty,0),\beta \in (0,\small{+}\infty]}{\cup} \{\gamma \in C^{\infty}((\alpha,\beta), SL(n)) |\gamma(0) = m\}$
And $\gamma_1 \underset{m}{\sim} \gamma_2 \Leftrightarrow \exists$ coordinates $\xi$ that $\gamma_1^{'}(0)^{\xi} = \gamma_2^{'}(0)^{\xi}$
Next i can't understand how to do tasks like this
 A: Straight approach. Direct computation from the definition.
By definition, one has:
$$\textrm{SL}_n(\mathbb{R})={\det}^{-1}(\{1\}).$$
Furthermore, for all $X\in\textrm{SL}_n(\mathbb{R})$, $M\mapsto XM$ is a diffeomorphism of $\mathcal{M}_n(\mathbb{R})$, hence it suffices to show that $\textrm{SL}_n(\mathbb{R})$ is a manifold in a neighbourhood of $I_n$ and one will have:
$$T_X\textrm{SL}_n(\mathbb{R})=XT_{I_n}\textrm{SL}_n(\mathbb{R}).$$
Let us show that $\det\colon\mathcal{M}_n(\mathbb{R})\rightarrow\mathbb{R}$ is a submersion at $I_n$. It is a well-known result that:
$$\mathrm{d}_{I_n}\det\cdot H=\textrm{tr}(H)\tag{1}.$$
Hence, $\mathrm{d}_{I_n}\det$ is a submersion since it is a nonzero linear map with codomain of dimension $1$, $\textrm{tr}(I_n)=n$. Finally, $\mathrm{SL}_n(\mathbb{R})$ is a manifold of dimension $n^2-1$ and one has:
$$T_X\textrm{SL}_n(\mathbb{R})=X\ker(\mathrm{d}_{I_n}\det)=\{M\in\mathcal{M}_n(\mathbb{R})\textrm{ s.t. }\textrm{tr}(X^{-1}H)=0\}.$$
If you do not recall how to establish $(1)$ here is a sketch of a proof. For all $(i,j)\in\{1,\ldots,n\}^2$, let $E_{i,j}$ be the matrix with zero entries everywhere except the entry $(i,j)$ which is a $1$. Then, notice that:
$$\frac{\partial\det}{\partial E_{i,j}}(I_n):=\lim_{t\to 0}\frac{\det(I_n+tE_{i,j})-1}{t}=\delta_{i,j}.$$
Indeed, if $i\neq j$, then $\det(I_n+tE_{i,j})=1$ and $\det(I_n+tE_{i,i})=1+t$. Finally, one has:
$$\mathrm{d}_{I_n}\det\cdot H=\sum_{i=1}^n\sum_{j=1}^nh_{i,j}\frac{\partial\det}{\partial E_{i,j}}(I_n)=\sum_{i=1}^nh_{i,i}=\textrm{tr}(H).$$

Addendum. Submersions and tangent spaces.
I am going to expand a bit on the link between the tangent space and the kernel of the differential of a submersion, maybe it will shed some light.
Let $M$ be a manifold of dimension $n$, let $v$ be a point of $M$ and let $(U,\Phi)$ be a chart of $M$ centered at $v$. The map $\ell\colon T_vM\rightarrow\mathbb{R}^n$ defined by:
$$\ell([c])=(\Phi\circ c)'(0),$$
where $[c]$ denotes the tangency class of a curve $c\colon]-\varepsilon,\varepsilon[\rightarrow M$ going through $v$, is well-defined, namely the value $\ell([c])$ only depends of the tangency class of $c$ and is bijective. Indeed, it is injective by definition of the tangency relation and is surjective. If $x\in\mathbb{R}^n$, then $t\mapsto\Phi^{-1}(tx)$ is a curve of $M$ going through $v$ such that:
$$\ell([c])=x.$$
Therefore, one can endow $T_vM$ with a structure of a $n$-dimensional vector space.
Now, let $f\colon X\rightarrow Y$ be a submersion at $v$, with $M\subseteq X$ and $n=\dim(X)-\dim(Y)$ such that:
$$M=f^{-1}(\{y\})$$
Now, notice that if $c\colon]-\varepsilon,\varepsilon[\rightarrow M$ is a curve going through $v$, then:
$$f\circ c=y.$$
In particular, one has $\Phi\circ f\circ c=\Phi(y)$ and by differentiation, one gets:
$$\mathrm{d}_vf([c]):=(\Phi\circ f\circ c)'(0)=0.$$
namely $[c]\in\ker(\mathrm{d}_vf)$. Finally, we have establish that:
$$\ker(\mathrm{d}_vf)\subseteq T_vM.$$
Whence $T_vM=\ker(\mathrm{d}_vf)$, using the equality of dimension between the two vector spaces.

Another approach. The Cartan-Von Neumann theorem.
If you want to involve some machinery, you can always use the following:

Theorem. Let $G$ be a closed subgroup of $\textrm{GL}_n(\mathbb{R})$, then $G$ is a manifold whose tangent space at $I_n$ is:
  $$\mathfrak{g}:=\{X\in\mathcal{M}_n(\mathbb{R})\textrm{ s.t. }\forall t\in\mathbb{R},\exp(tX)\in G\}.$$

This is not a too hard result, the key point is observing that $\mathfrak{g}$ is a vector space which follows from $G$ being closed and the Lie-Trotter formula.
In our case, $\textrm{SL}_n(\mathbb{R})$ is indeed a closed subgroup of $\textrm{GL}_n(\mathbb{R})$, therefore:
$$T_{I_n}\textrm{SL}_n(\mathbb{R})=\{X\in\mathcal{M}_n(\mathbb{R})\textrm{ s.t. }\forall t\in\mathbb{R},\det(\exp(tX))=1\}.$$
Using trigonalization over $\mathbb{C}$, one has:
$$\det\circ\exp=\exp\circ\textrm{tr}.$$
Finally, one gets:
$$\begin{align}T_{I_n}\textrm{SL}_n(\mathbb{R})&=\{X\in\mathcal{M}_n(\mathbb{R})\textrm{ s.t. }\forall t\in\mathbb{R},\exp(t\textrm{tr}(X))=1\}\\&=\{X\in\mathcal{M}_n(\mathbb{R})\textrm{ s.t. }\textrm{tr}(X)=0\}.\end{align}$$
