I want to prove that $SL(n,\mathbb{R})$ is not simply connected. For this, How can I prove that the fundamental group of $SL(n,\mathbb{R})$ is $\mathbb{Z}$ for $n=2$ and $\mathbb{Z_2}$ for $n>2$.
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Using Gram-Schmidt, $\mathrm{SL}_n(\mathbb{R})$ is a deformation retract of $\textrm{SO}(n)$, therefore: $$\pi_1(\textrm{SL}_n(\mathbb{R}),\cdot)\cong\pi_1(\textrm{SO}(n),\cdot).$$ Now, recall that for $n\geqslant 3$, $\textrm{SO}(n)$ is homeomorphic to $\mathbb{R}\textrm{P}^{n-1}$ and that $\mathbb{S}^{n-1}\twoheadrightarrow\mathbb{R}P^{n-1}$ is the universal covering space of $\mathbb{R}P^{n-1}$ and it has degree $2$, whence: $$\pi_1(\textrm{SO}(n),\cdot)\cong\mathbb{Z}/(2).$$
C. Falcon
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4$SO(n)$ is not homeomorphic to $RP^{n-1}$. – Moishe Kohan May 27 '17 at 16:21
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@MoisheCohen Thank you for pointing that out! – C. Falcon May 27 '17 at 16:21
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$SL_n(\mathbb{R})$ has the same fundamental group as $GL_n^+(\mathbb{R})$, which has been computed in this duplicate, using Gram-Schmidt.
Dietrich Burde
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