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I'm watching the following series of video lectures on Lie groups. In the last couple of minutes of the first lecture, he states his strategy to show that $\textrm{SL}_n(\mathbb{R})$ is a Lie subgroup of $\textrm{GL}_n(\mathbb{R})$:

1 . Show that $\textrm{SL}_n(\mathbb{R}) = \textrm{det}^{-1}\{1\}$ is a submanifold of $\textrm{GL}_n(\mathbb{R})$ at $I_n$, by showing that the map on tangent spaces at $I_n$ is surjective, and then using the "submersion principle."

2 . Use homogeneity to show that $\textrm{SL}_n(\mathbb{R})$ is a submanifold everywhere.

The second principle is clear to me, as well as the fact that the tangent space map at the identity is surjective. But I don't understand what is the "submersion principle" or how it is used. I tried googling submersion principle but nothing useful came up. This seems to have something to do with a smooth map having constant rank in a neighborhood of a point.

Edit: Not a duplicate of the previous question of why SL is a submanifold, because I am asking about a specific approach to showing it is a submanifold.

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D_S
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1 Answers1

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You're right. By note that $SL_n(R) = det^{-1}(1)$ with $det : GL_n(\mathbb{R}) \rightarrow \mathbb{R}\smallsetminus\{0\}$ is a constant rank map, $SL_n(R)$ is a embedded submanifold of $GL_n(R)$. The determinant function above is a constant rank map because its a smooth map which is also a group homomorphism. By a little work we can show every smooth map between Lie group that is also group homomorphism (called Lie group homomorphism) has constant rank.

Kelvin Lois
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  • But where does the structure of a manifold on $\textrm{SL}_n(\mathbb{R})$ come from? What general principle are you using to conclude that? – D_S Nov 25 '17 at 02:01
  • The structure of SL comes from the fact that its a level set of identity in $R\smallsetminus 0$. We can conclude that by constant rank level set theorem Th.5.12. – Kelvin Lois Nov 25 '17 at 02:05
  • Thanks for your answer, I understand why SL is a submanifold now. What you are mentioning seems different from the approach I mentioned in my question though. You would show first that the rank is everywhere constant using homogeneity then conclude the level set is a submanifold. In the video lecture, he says first show it is a submanifold at the identity, then use homegeneity. – D_S Nov 25 '17 at 04:02
  • Do you know if it is possible to make the approach I mentioned in my question work? Or is it fundamentally incorrect? – D_S Nov 25 '17 at 04:03
  • Your welcome. You can look at another approach here. But i think thats not the aprroach that you looking for too. I believe the proof that you look for is mentioned in the proof of Th.7.5 in Lee's Smooth Manifold. Only Lee's proof's is more general. After you read this proof i think you'll agree that what i mention is same as you meant in the beginning. – Kelvin Lois Nov 25 '17 at 04:16
  • https://math.stackexchange.com/questions/1712667/show-that-sln-mathbbr-is-a-n2-1-smooth-submanifold-of-mn-mathbb – Kelvin Lois Nov 25 '17 at 11:14