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As can be seen here, the fundamental group of $\text{GL}_n(\mathbb{R})$ is $\mathbb{Z}/2\mathbb{Z}$ (for $n \ge 3$). (For $n=2$ it is $\mathbb{Z}$).

Is there a way to find an explicit representing path for the non-trivial element of $\pi(\text{GL}_n(\mathbb{R}))$? i.e describing a non-contractible (closed) path?

(I guess for $n=2$ it's just like in $SO(2)\cong \mathbb{S}^1$?)

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Asaf Shachar
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  • There is a deformation retract of the connected component $GL_n(\Bbb R)_+$ of the identity onto $SO(n)$, so it suffices to find such such a path in the latter. – Travis Willse Jan 24 '16 at 10:13

1 Answers1

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One can show the inclusion $O(n) \to GL(n)$ is a homotopy equivalence, and the inclusion $SO(2) \to SO(n)$ induces a surjective map on $\pi_1$ (and that $SO(n) \to SO(n+1)$ actually induces isomorphisms for $n>2$). The first statement is proven essentially by row reduction, and the second is an easy calculation using the homotopy exact sequence for the fibration $SO(n) \to SO(n+1) \to S^n$. The fibration is just the evaluation map on a fixed unit vector.

Given all this, it should be straightforward to construct an explicit non nullhomotopic loop in $GL(n)$. Just take the generator for $\pi_1(SO(2))$ and push it forward via the most obvious inclusion $SO(2) \to GL(n)$.

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