# Lie group. How is manifold defined?

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?

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 $\det.$

• 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. – RealAnalysis Mar 19 '15 at 19:38
• I have editted the answer to show that the only critical value of $\det$ is zero. – mfl Mar 19 '15 at 19:46

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: $$(\phi_{11}^{-1}\circ\phi_{12})(x,\alpha,\beta)=\left(\alpha,x,\frac{\alpha\beta-1}{x}\right) \\ (\phi_{22}^{-1}\circ\phi_{11})(x,\alpha,\beta)=\left(\frac{1-\alpha\beta}{x},\alpha,\beta\right)$$

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.