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Using this fact:

Every $X \in \operatorname{SL}(n,\Bbb R)$ can be decomposed as $X=\exp(A)\exp(S)$, with $A, B \in M_n(\Bbb R), \operatorname{tr}(A)=\operatorname{tr}(B)=0$ (i.e $A,B \in \mathfrak{sl}(n,\mathbb R)$) $A$ skew-symmetric, $S$ symmetric

How can I prove that $\operatorname{SL}(n,\Bbb R)$ is connected?

some_math_guy
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1 Answers1

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Hint: function $t \mapsto \exp((1-t)S)\exp((1-t)A)\quad\quad$ gives a path from $\exp(S)\exp(A)\quad$ to the identity matrix $I$. Use this to show that $\mathrm{SL}(n, \Bbb{R})$ is path-connected and hence connected.

Rob Arthan
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  • for $ t=0$ I get $\exp S \exp A=X$. For t=1, the identity. – some_math_guy Sep 25 '23 at 20:48
  • Don't I need two generic points to prove path-connectness? I mean $\alpha(0)=C, \alpha(1)=D$, matrices in $SL(n,\Bbb R)$ Instead of the particular X and I? – some_math_guy Sep 25 '23 at 20:50
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    $S$ and $A$ are arbitrary and hence so is $X = \exp(S)\exp(A)$. So given arbitrary $X$ and $Y$, there is a path $\eta$ from $X$ to $I$ and a path $\zeta$ from $Y$ to $I$. If you compose $\eta$ with the reversal of $\zeta$ you get a path from $X$ to $Y$. – Rob Arthan Sep 25 '23 at 22:24
  • Using this method, does itreally matter that A and S are skew-symmetric and symmetric? It doesn not seem to be used anywhere – some_math_guy Sep 26 '23 at 11:30
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    The fact that $A$ and $S$ are skew-symmetric and symmetric is used to see that $(1-t)A$ and $(1-t)S$ are also skew-symmetric and symmetric so that the path stays inside $\mathrm{SL}()n, \Bbb{R})$. (This uses the (true) converse of the given fact.) – Rob Arthan Sep 26 '23 at 19:55