Let $G$ be a connected, semisimple subgroup of $GL(n,\mathbb{R})$. Here, semisimple means that $G$ has no normal, connected abelian subgroup other than the trivial group.

In Mostows book Strong rigidity of locally symmetric spaces, on p. 11, he writes

Let $G$ be an analytic subgroup of $GL(n,\mathbf{R})$. [...] Assume moreover that $G$ is semi-simple [...]. Since $G$ is its own commutator subgroup, each element of $G$ has determinant one.

I'm trying to understand this: Does the following proof work?

  • $G/[G,G]$ is semisimple as a quotient of the semisimple group $G$ (is this correct?)
  • $G/[G,G]$ is connected since $G$ is connected (quotients of connected spaces are connected)
  • Since $G/[G,G]$ is abelian, it must be trivial, i.e. $[G,G]=G$.

Its also not clear to me why $\det(g) = 1$ for each $g \in G$. Why does this follow?


Note that the determinant of a commutator is always $1$. The result follows. $$\det [g,h]=\det (g^{-1}h^{-1}gh)=1$$

  • $\begingroup$ So the three bullet points above in my proof are correct? $\endgroup$
    – abenthy
    Jul 20 '17 at 9:53
  • 1
    $\begingroup$ Yes, they are correct. $\endgroup$ Jul 20 '17 at 10:02
  • $\begingroup$ Could you maybe explain to me why the quotient of a semisimple lie group is semisimple? $\endgroup$
    – abenthy
    Dec 13 '17 at 11:02

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