Fundamental group of GL(n,C) is isomorphic to Z. How to learn to prove facts like this?

Solution 1:

The first thing you have to do is to note that the inclusion $U(n)\to\mathrm{GL}(n,\mathbb C)$ induces an isomorphism on the fundamental groups. This can be done by noting that a loop in $\mathrm{GL}(n,\mathbb C)$ can be deformed to one in $U(n)$ by performing the Gram-Schmidt procedure at each point of the loop, and checking that this can be done continuously and so on.

Next, considering the beginning of the long exact sequence for the homotopy groups of the spaces appearing in the fibration $$U(n-1)\to U(n)\to S^{2n-1}$$ which arises from the transitive linear action of $U(n)$ on $S^{2n-1}\subseteq\mathbb C^{n}$ you can prove, by induction, that the inclusion $U(1)\to U(n)$ induces an isomorphism on fundamental groups. Then you can explicitly describe $U(1)$ as a space homeomorphic to $S^1$.

Solution 2:

The method Mariano discusses in his answer is absolutely the way that mathematicians compute fundamental groups (and also higher homotopy groups) of Lie groups. Here I just want to mention how his first step applies in a more general context.

1) Concerning $\operatorname{GL}_n(\mathbb{C})$: the unitary group $U(n)$ is a maximal compact subgroup of $\operatorname{GL}_n(\mathbb{C})$, and moreover any maximal compact subgroup is conjugate to $U(n)$. (This can be seen by considering Hermitian forms, c.f. e.g. Section 1 of http://math.uga.edu/~pete/8410Chapter9.pdf.) Moreover, the Gram-Schmidt process gives a deformation retraction from $\operatorname{GL}_n(\mathbb{C})$ to $U(n)$, hence these two spaces are homotopy equivalent. And in fact even more is true: there exists a finite-dimensional Euclidean space $E$ such that $\operatorname{GL}_n(\mathbb{C})$ is homeomorphic to $U(n) \times E$: this is the QR decomposition. Moreover:

2) Everything in 1) goes over verbatim for $\operatorname{GL}_n(\mathbb{R})$ with the unitary group $U(n)$ replaced by the orthogonal group $O(n)$.

3) For any reductive group $G$ over $\mathbb{R}$ or $\mathbb{C}$, there exists a maximal compact subgroup $K$, any two such are conjugate in $G$, and $G$ is homeomorphic to the product of $K$ with a finite-dimensional Euclidean space. This last fact is a consequence of the Iwasawa decomposition, a far-reaching generalization of the QR-decomposition.

Solution 3:

For 2. and 3. I can recommend Allen Hatcher's Algebraic Topology book which can be accessed free from the author's webpage: http://www.math.cornell.edu/~hatcher/AT/ATpage.html