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I need some reference with a simple proof for the fact that:

SL(2,$\mathbb{R}$) is unimodular.

Thank you so much.

Skid Row
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  • http://www.math.jhu.edu/~fspinu/423/9.pdf – Travis Willse Jun 14 '17 at 11:37
  • Thank you! The proof is very short but I don't understand. I need a simpler proof using only measure theory in undegraduate school (even if it is long). – Skid Row Jun 14 '17 at 12:03
  • @SkidRow This definitely is not U.S. stuff, but even more interesting: what is your definition of "unimodular group"? As far as I know, you must use either Haar measure (as in the link), or modular functions, so: what is it? – DonAntonio Jun 14 '17 at 12:06
  • I mean the same definition as in the link: a group is unimodular if the Haar measure (by definition left-invariant) is right-invariant as well. – Skid Row Jun 14 '17 at 12:07
  • @SkidRow But then what is it you don't understand there? They show there a derivation on the lie algebra $;\mathfrak{sl},(2,\Bbb R);$ and use the fact that this groups equals its commutator subgroup... – DonAntonio Jun 14 '17 at 12:09
  • I don't know about sl(2,$\mathbb{R}$) and [X,Y], I think I need to learn Lie algebra to understand these notions. – Skid Row Jun 14 '17 at 12:12

2 Answers2

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The best reference is probably the book

Lang, Serge. $SL_2(R)$. Addison-Wesley Publishing Co., Reading, Mass.-London-Amsterdam, 1975. xvi+428 pp.

Mikhail Katz
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    Determinant equal to one is not the definition, as far as I know, of unimodular group. Lang itself, in page two, defines unimodular group by means of Haar measure. – DonAntonio Jun 14 '17 at 12:07
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The link in the comments is dead and the accepted answer is not particularly satisfying for me, so I might as well post another answer for future reference.

Fact 1: $\mathfrak{sl}(2, \mathbb{R})$ is semisimple.

Fact 2: Semisimple Lie algebra is perfect (i.e. $\mathfrak{g} = [\mathfrak{g}, \mathfrak{g}]$). So if the Lie group with perfect Lie algebra is nonabelian, any homomorphism from its lie algebra to an abelian Lie algebra is trivial.

Fact 3: If G is connected, any Lie group homomorphism ϕ : G → H is determined by the induced Lie algebra homomorphism dϕ : g → h

By 2, a homomorphism from $\mathfrak{sl}(2, \mathbb{R})$ to the lie algebra of $\mathbb{R^*}$ is trivial, so a homomorphism from $SL(2, \mathbb{R})$ to $\mathbb{R^*}$ is trivial. Hence the modular function is trivial.

It's easy to find the reference for the 3 facts. I guess this is probably the solution in the dead link, based on the discussion there.

Nancium
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