Computing tangent space Let
$$Sl_2(\mathbb{R})=\{M\in\mathcal{M}\mid\det M=1\}$$
after showing that this is a 3-smooth manifold, I have been told to compute the tangent space to this manifold at the point
$$p=\left(\begin{matrix}1&1\\0&1\end{matrix}\right)$$
I am not sure what to do here, I don't know what does it mean to "compute the tangent space", it's a 3 dimensional real vector space so it's $\mathbb{R}^3$, but I don't think this is what I have to do.
 A: Take a curve $\gamma(t)$ in $SL(2,\mathbb{R})$ that passes through $p$ at time $t=0$, then $\dot{\gamma}(0)$ is a matrix in the tangent space at $p$ (and by definition, all tangent vectors arise in this manner). So what conditions do we have on $\dot{\gamma}(0)$? Well, we know that $\det \gamma(t) = 1$ at all times $t$; we can differentiate this expression at $t=0$ using the Jacobi formula to get that
$$
0 = \frac{d}{dt} \det \gamma(t) \bigg|_{t=0} = \textrm{trace}\left( \dot{\gamma}(0) \gamma(0)^{-1} \right).
$$
Therefore, the matrices in the tangent space at $p$ are those of the form $\dot{\gamma}(0)$ such that $\dot{\gamma}(0) \gamma(0)^{-1}$ is traceless. Since $\gamma(0)^{-1} = p^{-1} = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$, this can be rephrased as
\begin{align*}
\dot{\gamma}(0) = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in T_p SL(2,\mathbb{R}) \iff \textrm{trace}\left( \dot{\gamma}(0) \gamma(0)^{-1} \right) = \textrm{trace} \begin{pmatrix} a + b & b \\ c + d & d \end{pmatrix}  = a + b + d = 0.
\end{align*}
A quick remark: I am thinking of $SL(2,\mathbb{R})$ as a submanifold of $GL(2,\mathbb{R})$, whose tangent space at any point can be identified with the space $M(2,\mathbb{R})$ of $2 \times 2$ real matrices. This is why I am writing tangent vectors to $SL(2,\mathbb{R})$ as matrices as well, since $T_p SL(2,\mathbb{R})$ can be identified with a vector subspace of $M_2(\mathbb{R})$.
