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$\forall M \in \mathrm{SL}(2, \mathbb{R})$ has a neighborhood diffeomorphic to $\mathbb{R}^3$?

Known that $\mathrm{SL}(2, \mathbb{R})$ is a three-dimensional manifold.

WishingFish
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    isn't every three dimensional manifold locally diffeomorphic to $\mathbb{R}^3$? – James S. Cook Aug 15 '13 at 23:50
  • Thank you James!! Now I got it!!! 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 01:00
  • ah ha, yes, no edges by default. Which is a bit funny when you think about all the integration we do in calculus three, always a manifold with boundary. Well, that's even inaccurate, a square is not a manifold with boundary as I've seen it defined. The corners mess it up. Honestly, I've yet to see an abstract manifold-type definition which covers the all the constructions we make in ordinary multivariate calculus. – James S. Cook Aug 16 '13 at 01:40
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    @JamesS.Cook: Lee's Introduction to Smooth Manifolds has a definition of a manifold with corners. – Michael Albanese Aug 16 '13 at 01:56
  • @MichaelAlbanese thanks! Sadly I do not have that one yet... however, you just gave me an idea for the library purchase at my school this year. – James S. Cook Aug 16 '13 at 02:09

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(Just so this question has an answer.)

The answer to your question is yes. Any point in a smooth $n$-dimensional manifold has a neighbourhood diffeomorphic to $\mathbb{R}^n$ (this is part of the definition). In particular, any $M \in \operatorname{SL}(2, \mathbb{R})$ has a neighbourhood diffeomorphic to $\mathbb{R}^3$ as $\operatorname{SL}(2, \mathbb{R})$ is a smooth three-dimensional manifold.

Michael Albanese
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