Curvature of $\operatorname{SL}(2)$ (manifolds)

Question

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.

Answer 1

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 $g_{mn}$.

Answer 2

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.

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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.

Answer 3

I will treat $M_{2x2}(\mathbb{R})$ like $\mathbb{R}^{2,2}$ in order to obtain an inner product which agrees with the determinant.

Suppose $A, B\in M_{2x2}(\mathbb{R})$ with columns $\{a_1,a_2\}$ and $\{b_1,b_2\}$ respectively.

Let $(A,B)=\frac{1}{2}(\det(a_1,b_2)+\det(b_1,a_2))$

Then $(A,A)=|A|^2:= \det(A)$

and

$(A,B)=(B,A)$

Let $\alpha=\begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}\in SL(2,\mathbb{R})$.

Then $1=\det(\alpha):=|\alpha|^2=(\alpha,\alpha) = \alpha\delta - \beta\gamma$

But $\alpha \delta-\beta \gamma=\frac{1}{4}((\alpha+\delta)^2-(\alpha-\delta)^2-(\beta+\gamma)^2+(\beta-\gamma)^2)=1$

Therefore

$\frac{1}{4}((\alpha+\delta)^2+(\beta-\gamma)^2)=\cosh^2(t)$

$\frac{1}{4}((\alpha-\delta)^2+(\beta+\gamma)^2)=\sinh^2(t)$

So we can write

$\frac{1}{2}(\alpha+\delta)=\cosh(t)\cos(\theta)$

$\frac{1}{2}(\beta-\gamma)=\cosh(t)\sin(\theta)$

$\frac{1}{2}(\alpha-\delta)=\sinh(t)\cos(\phi)$

$\frac{1}{2}(\beta+\gamma)=\sinh(t)\sin(\phi)$

Writing $X(\alpha)=\begin{pmatrix}\alpha\\\delta\\\beta\\\gamma\end{pmatrix}$ Then

$X(t,\theta,\phi)=\begin{pmatrix}\cosh(t)\cos\theta+\sinh(t)\cos\phi\\\cosh(t)\cos\theta-\sinh(t)\cos\phi\\\cosh(t)\sin\theta+\sinh(t)\sin\phi\\-\cosh(t)\sin\theta+\sinh(t)\sin\phi \end{pmatrix}$.

Notice, we need to take dot products to compute stuff so lets figure out what that is. Notice that we can use polarization. (Of course you can compute the metric by expanding dot products in terms of basis vectors and translating back to determinants, but I thought this way was nice)

Let $(\begin{pmatrix}\alpha\\\delta\\\beta\\\gamma\end{pmatrix},\begin{pmatrix}\alpha\\\delta\\\beta\\\gamma\end{pmatrix})=\alpha\delta-\gamma\beta$

By definition $(a+b,a+b)=|a|^2+|b|^2+2(a,b)$

$=(\begin{pmatrix}a^1+b^1\\a^2+b^2\\a^3+b^3\\a^4+b^4\end{pmatrix},\begin{pmatrix}a^1+b^1\\a^2+b^2\\a^3+b^3\\a^4+b^4\end{pmatrix})=a^1a^2-a^3a^4+b^1b^2-b^3b^4+a^1b^2+a^2b^1-a^3b^4-a^4b^3$

Therefore $(a,b)=\frac{1}{2}(a^1b^2+a^2b^1-a^3b^4-a^4b^3)$

I believe you can do the rest on your own. However, I recommend rewriting $X(t,\theta,\phi)$ in terms of the vectors.

$a(\theta)=\begin{pmatrix} \cos\theta\\\cos\theta\\\sin\theta\\-\sin\theta\end{pmatrix}$ and $b(\phi)=\begin{pmatrix}\cos\phi\\-\cos\phi\\\sin\phi\\\sin\phi\end{pmatrix}$

compute $a',b',a'',b''$ and show

$(a,a)=|a|^2=1,|b|^2=-1,|a'|^2=1,|b'|^2=-1 $

and

$(a,b)=0,(a,a')=0,(a,b')=0,(b,a')=0,(b,b')=0, (a',b')=0$

Rewriting $X$ in terms of $a,b$ we get

$X(t,\theta,\phi)=\cosh(t)a(\theta)+\sinh(t)b(\phi)$

We have the coordinate frame

$X_t=\sinh(t)a(\theta)+\cosh(t)b(\phi)$

$X_\theta=\cosh(t)a'(\theta)$

$X_\phi=\sinh(t)b'(\phi)$

Show that $N=X(t,\theta,\phi)$ by taking inner products. From this, we conclude $II=g$

Now you can use the characterization of curvature in terms of the second fundamental form. $R(X,Y,Z,W)= II(X,W)II(Y,Z)-II(X,Z)II(W,Y)$