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Consider the fundamental group $\pi_1(\mathbb{CP}^1\backslash \{a_1, \ldots, a_n\})$. It is said that there is a representation: $\pi_1(\mathbb{CP}^1\backslash \{a_1, \ldots, a_n\}) \to GL(n, \mathbb{C})$. I am confused with the order of these two groups. Since $\pi_1(\mathbb{CP}^1\backslash \{a_1, \ldots, a_n\}) \simeq \mathbb{Z}^{n-1}$, the order of this group is infinite. What is the order of the group $GL(n, \mathbb{C})$? Is the order of the group $GL(n, \mathbb{C})$ larger than the order of $\mathbb{Z}^{n-1}$?

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That is not what the fundamental group of that space is. Also, yes, the order of $GL(n,\mathbb{C})$ is $2^{\aleph_0}$, which is strictly larger than the order of $\mathbb{Z}^{n-1}$, and also strictly larger than the order of what the fundamental group actually is. – Chris Eagle Mar 6 '12 at 19:55
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It should be pointed out that a representation doesn't have to have the same order as the group. For example, the map g $\mapsto$ Id is a rep. – Jay Kopper Mar 6 '12 at 20:14

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Well, the cardinality of $\mathrm{GL}_n(\mathbb{C})$ is the same as that of the continuum $\mathbb{R}$, which is bigger than both that of $\mathbb{Z}^{n-1}$ and the fundamental group you are interested in (the free group on $n-1$ generators); these have the cardinality of $\mathbb{Z}$.

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Didn't notice Chris' comment in time... – Steve Mar 6 '12 at 19:59
Probably important to really emphasize to the thread creator is that a disk with $n-1$ holes is homotopy equivalent to the wedge of $n-1$ circles. Hence its fundamental group is the free group on $n-1$ generators, NOT TO BE CONFUSED with the free ABELIAN group on $n-1$ generators. (Same universal property, but in wildly different categories) – you Mar 6 '12 at 21:52
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Hey you, it wasn't emphasized enough? I thought I was being polite! – Steve Mar 7 '12 at 13:16
It's just something that I didn't fully realize right away when I was seeing this stuff for the first time. So now I yell it at people :) By the way, how do you italicize? – you Mar 7 '12 at 14:29

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