$SL_2(\mathbb{R}) \cong S^1 \times D^2$

Question

I’m trying to prove $SL_2(\mathbb{R}) \cong S^1 \times D^2$, but I don’t know how to construct.

Where $S^1$ is the circle, and $D^2$ is the disk.

Using the implicit function theorem, I could prove that $SL_2(\mathbb{R})$ is $3$-dimensions manifold.

Please tell me the construction of $SL_2(\mathbb{R}) \cong S^1 \times D^2$.

$SL_2(\Bbb{R})$ can not be homeomorphic to $S^1 \times D^2$, since the latter is compact but the former isn't. But $SL_2(\Bbb{R}) \cong S^1 \times \Bbb{R}^2$, see this thread. — Zerox, May 31 '22 at 12:21
Does this answer your question? $SL(n,\mathbb R)$ diffeomorphic to $SO(n) \times \mathbb R^{n(n+1)/2-1}$? — Dietrich Burde, May 31 '22 at 13:34

score 1 · Answer 1 · answered May 31 '22 at 14:09

It can be easily seen that $SL_2(\Bbb R)$ is not compact. Consider elements in this space such that the non diagonal entries are $0$. Take the diagonal entries to be $n$ and $\frac{1}{n}$, where $n$ are natural numbers. We can make $n$ as large as possible and thus $SL_2(\Bbb R)$ is not bounded. Since $SL_2(\Bbb)$ is contained in $\Bbb R^{n^2}$, it is compact iff it is closed and bounded. The other space is compact since it is the product of compact spaces $S^1$ and $D^2$. Hence, these spaces cannot be homeomorphic.

$SL_2(\mathbb{R}) \cong S^1 \times D^2$

1 Answers1