Curvature of $\operatorname{SL}(2)$ (manifolds) If we view Lie groups as manifolds, we can pose this question:

What is the curvature of the unit element of $\operatorname{SL}(2)$?

What I thought:
I know that $\operatorname{SL}(2)=\{M\in L(\mathbb{R}^2,\mathbb{R}^2):\det M=1\}$ had unit element $I_2$.
Further if $N:\mathbb{C}\rightarrow S^3$ is a unit normal field, then $\det DN(I_2)$ is the curvature.
So what we are actually looking for is such a unit normal field. What is a good example?
Edit (context of the question):

A Lie group is a manifold $G$ that is also a group such that the multiplication map and the inversion map are differentiable. Suppose that $G$ is compact, connected and is a $2n$-manifold in a $2n+1$ dimensional vector space.

That is the introduction to the question.
 A: First lets give a local coordinate patch around the identity so $\alpha \; \beta \; \gamma$ will be coordinates on the embedded $SL(2,R)$ and $a,b,c,d$ on the $\mathbb{R}^4$. These coordinates are defined by computing the formula below. You can make another choice if you want.
\begin{eqnarray*}
X (\alpha, \beta , \gamma ) &=& exp \begin{pmatrix}\alpha&\beta\\
\gamma&-\alpha\\
\end{pmatrix} = \begin{pmatrix}X^a&X^b\\
X^c&X^d\\
\end{pmatrix}
\end{eqnarray*}
So the induced metric is now $g_{mn} = \sum_{\mu\nu} \partial_m X^\mu \partial_n X^\nu g_{\mu \nu}$. mn run through a,b,c,d and $\mu,\nu$ through $\alpha \; \beta \; \gamma$. So now compute the $X^{a,b,c,d}$ in terms of $\alpha \; \beta \; \gamma$, take their derivatives and compute the induced metric.
Now you can compute all your Riemann/Ricci curvature tensors in terms of this metric and this coordinate patch and never have to touch the $\mathbb{R}^4$ again. Just plug in this $g_{mn}$.
A: If you view the space of $2 \times 2$ real matrices as
$$
\left[\begin{array}{@{}cc@{}}
    a & b \\
    c & d \\
  \end{array}\right]
\leftrightarrow (a, b, c, d),
$$
then $\operatorname{SL}(2)$ is the level set $\det(a, b, c, d) = ad - bc = 1$, the identity matrix corresponds to the point $(1, 0, 0, 1)$, and the gradient of the determinant,
$$
\nabla \det(a, b, c, d) = (d, -c, -b, a),
$$
furnishes a non-vanishing normal field along $\operatorname{SL}(2)$, so you may as well take
$$
N = \frac{(d, -c, -b, a)}{\sqrt{a^{2} + b^{2} + c^{2} + d^{2}}}.
$$
Alternatively, you could solve for $d$ locally, $d = \frac{1}{a}(1 + bc)$, and proceed using whatever tools you have for working with the curvature of a graph.

This addresses the literal question, but may not be substantively helpful, since the notation $N:\mathbf{C} \to S^{3}$ makes one wonder if some surface slice of $\operatorname{SL}(2)$ is to be taken. (If so, that raises other suspicions, since for surfaces in $\mathbf{R}^{4}$, the Gaussian curvature is not the determinant of the Gauss map.)
Separately, the explicit mention of "compact" in the problem introduction is at least mildly perverse, since $\operatorname{SL}(2)$ is a non-compact hyperboloid.
