8
$\begingroup$

If I have the Lie group $SL(2,\mathbb{R})$. Then how is the manifold structure on this algebraic group defined, could anybody explain this to me? I mean this is the group of matrices that determinant one, but where comes the manifold structure from?

$\endgroup$
11
$\begingroup$

You can identify $M_2(\mathbb{R})$ with $\mathbb{R}^4.$ Now, the function

$$\det : \mathbb{R}^4 \to \mathbb{R}$$ is differentiable. Since $1$ is a regular value of $\det$ one has that $SL(2,\mathbb{R})=\det^{-1}(1)$ is a submanifold.

Edit

Note that the only critical point of $\det$ is $(0,0,0,0).$ This can be shown by computing the gradient of $\det(x,y,z,t)=xt-yz:$

$$\nabla \det(x,y,z,t)=(t,-z,-y,x)=0 \iff (x,y,z,t)=(0,0,0,0).$$ Thus, any value different from zero is a regular value of $\det.$

$\endgroup$
  • $\begingroup$ could you explain why $1$ is a regular value? We need to show that the derivative of the determinant for any matrix in $SL(2,\mathbb{R})$ is surjective. $\endgroup$ – RealAnalysis Mar 19 '15 at 19:38
  • $\begingroup$ I have editted the answer to show that the only critical value of $\det$ is zero. $\endgroup$ – mfl Mar 19 '15 at 19:46
3
$\begingroup$

If you want concrete charts, you can consider the open sets $$U_{ij}=\left\lbrace\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\mathrm{SL}_2(\mathbb{R})\text{ s.t. the }(i,j)\text{-th coordinate is nonzero}\right\rbrace$$ These are obvioulsy in bijection with $\mathbb{R}^*\times\mathbb{R}^2$ by obvious maps $\phi_{ij}:\mathbb{R}^*\times\mathbb{R}^2\to U_{ij}$, and you can easily check by hand that the transition maps are smooth in coordinates, actually rational functions. This describes an explicit atlas on $\mathrm{SL}_2(\mathbb{R})$, since the four open sets $U_{ij}$ form a cover.


Here are the four maps:

$$\phi_{11}(x,\alpha,\beta)=\begin{pmatrix}x &\alpha\\\beta & \frac{1+\alpha\beta}{x} \end{pmatrix}, \quad\phi_{12}(x,\alpha,\beta)=\begin{pmatrix} \alpha & x\\ \frac{\alpha\beta-1}{x} & \beta\end{pmatrix} \\ \phi_{21}(x,\alpha,\beta)=\begin{pmatrix} \alpha & \frac{\alpha\beta-1}{x} \\ x& \beta \end{pmatrix}, \quad\phi_{22}(x,\alpha,\beta)=\begin{pmatrix} \frac{1+\alpha\beta}{x}& \alpha \\ \beta & x\end{pmatrix}$$ and here are two of the twelve (equally simple) transition functions: $$(\phi_{11}^{-1}\circ\phi_{12})(x,\alpha,\beta)=\left(\alpha,x,\frac{\alpha\beta-1}{x}\right) \\ (\phi_{22}^{-1}\circ\phi_{11})(x,\alpha,\beta)=\left(\frac{1-\alpha\beta}{x},\alpha,\beta\right)$$

$\endgroup$
2
$\begingroup$

A slightly different take: you can show that $SL_2\mathbb{R}$ is diffeomorphic to $S^1 \times \mathbb{H}$, where $\mathbb{H}$ is the upper half-plane.

The idea is as follows. $SL_2\mathbb{R}$ acts on $\mathbb{C}$ via fractional linear transformations. That is, via $$ \begin{pmatrix}a & b \\ c & d \end{pmatrix} \cdot z = \frac{az + b}{cz + d} $$ Since all entries of the matrix are real, it in particular preserves the real line in $\mathbb{C}$. Moreover, since the determinant is 1 (and not -1), it preserves the upper half-plane. So we have an action (actually---this is an action of $PSL_2\mathbb{R}$, and we are really showing that $PSL_2\mathbb{R} \cong S^1 \times \mathbb{H}$---but since $SL_2\mathbb{R}$ is a double cover...).

Essentially, we now pick our favorite point in the upper half plane (say, $\sqrt{-1}$), and note that its stabilizer is just rotations around that point i.e. a copy of $S^1$. Since the group acts transitively on the upper half-plane, it follows then that $PSL_2\mathbb{R}\cong S^1 \times \mathbb{H}$ as desired.

Anyhow, this is a super sketchy argument, but I always think it's a nice one, and it brings the group action nicely into the picture.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.