Does $\mathrm{SL}(2, \mathbb{R})$ have a non-empty boundary?

Question

Does $\mathrm{SL}(2, \mathbb{R})$ have a non-empty boundary?

Edit:

I've shown that $1$ is a regular value, and hence $\mathrm{SL}(2, \mathbb{R})$ is a three-dimensional manifold as Ted's hint. But I am not sure about what to do next. One more step please?

score 5 · Accepted Answer · answered Aug 15 '13 at 03:28

5

Hint: Prove it is a $3$-dimensional manifold.

answered Aug 15 '13 at 03:28

Ted Shifrin

115,160

er....... yes, I got that.... but i can't get conclusion about boundary... – WishingFish Aug 15 '13 at 23:22
If every point has a neighborhood diffeo to an open set in $\mathbb R^3$, what do you conclude? – Ted Shifrin Aug 16 '13 at 00:39
Hi Ted, it is not diffeomorphic to $\mathbb{H}^3$ hence does not have boundary. I thought "manifold" means "manifold with or without boundary", but now I realized "manifold" means "manifold without boundary". thank you :) – WishingFish Aug 16 '13 at 00:59
1

Yup! Terminology is important :) – Ted Shifrin Aug 16 '13 at 01:21
1

You might want to accept an answer if this question is settled to your satisfaction. – Ted Shifrin Aug 16 '13 at 01:26
Oh, yes I should have accepted! Sorry Ted. :-) Once again: thank you very much. – WishingFish Aug 16 '13 at 01:28

Neal · Answer 2 · 2013-08-15T12:31:10.410

1

Since $1$ is a regular value of the determinant function, $\det^{-1}\{1\} = \mathrm{SL}(2;\mathbb{R})$ is a submanifold of the two-by-two matrices. In particular, since it is a manifold, it has no boundary.

edited Aug 15 '13 at 12:31

answered Aug 15 '13 at 03:28

Neal

32,659

5

This is not quite right. Where did $-1$ come from? – Ted Shifrin Aug 15 '13 at 03:29
My fevered imagination. I'm used to working with $\mathrm{PSL}(2;\mathbb{R})$, where you have to ignore $\pm I$, so I jumped to the conclusion that "determinant one" really means "determinant plus/minus one." Thanks for the catch! – Neal Aug 15 '13 at 12:32

Does $\mathrm{SL}(2, \mathbb{R})$ have a non-empty boundary?

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