I am looking for the tangent space of $SL(n,\mathbb{R}) = \{A\in\mathbb{R}^{n\times n} | \det{A}=1 \}$ (actually $n=3$) at an arbitrary $M\in SL(n,\mathbb{R})$. From now on I will omit the $\mathbb{R}$ for convenience.
It is well known and easy to prove that the tangent space at the identity matrix $\mathbb{1}$, $T_1SL(n)$, is the set of all traceless matrices of the same size.
However I am not able to generalize this result. These are my thoughts:
Let $\gamma:\mathbb{R}\to SL(n)$ be a curve with known $\gamma(0) = M$ and unknown $\gamma^\prime(0)=X$. Let's differentiate the $SL$-characteristic equation: $$\frac{d}{dt}|_{t=0}\det{\gamma}(t) = D(\det)|_{\gamma(0)}\,\cdot\,\gamma^\prime(0) = \frac{d}{dt}1 = 0$$
Here $A\cdot B = A_{ij}B_{ij}$. The trick for $M=\mathbb{1}$ is
$$D(\det)|_1 = \mathbb{1},$$
so the above differentiation equation reads ${\rm tr}(\gamma^\prime(0))=0$. However in the general case
$$D(\det)|_M = {\rm Cof}(M)$$
the cofactor matrix. As far as I see ${\rm Cof}(M) \cdot X = 0$ is of no use. Any ideas out there? Thanks.
Multiplying by a matrix $M$ is a linear transformation from $\Bbb{R}^{n\times n}$ to itself. It is also continuous, differentiable,..., whatever. Furthermore, if $M\in SL_n$ multiplication by $M$ maps any element of the group to another element. So if $T$ is the well-known tangent space of $SL_n$ at the identity, then surely $$MT=\{MX\mid X\in T\}$$ is the tangent space at $M$.
Observe that because $MTM^{-1}=T$ we have $MT=TM$. IOW you can multiply from the left or from the right as you see fit.
If $\gamma(t)$ is a path thru the identity, then $M\gamma(t)$ is a path thru $M$. Also $\dfrac d{dt}(M\gamma(t))=M\dfrac d{dt}\gamma(t)$ etc.