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My motivation for this question stems from needing to show that the fundamental groups of the projectivised tangent bundles of $\mathbb{S}^2$ and $\mathbb{RP}^2$ are finite. I am aware of a proof that shows their respective fundamental groups have order 4 and 8 respectively, which is done through explicitly calculating $\mathbb{P}T\mathbb{S}^2$ and $\mathbb{P}T\mathbb{RP}^2$.

In this paper, the exact values 4 and 8 are used later on. For my purposes, I don't need the exact values - just the fact they are finite. Are there general methods for computing when the fundamental group of a space is finite, aside from explicitly calculating it? Is it perhaps obvious in either of the cases above?

jl2
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1 Answers1

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Here's a result that can sometimes be used to show that the fundamental group of a space is finite:

If $X$ is compact and has a universal cover, say $\tilde{X}$, then $\pi_1(X)$ is finite if and only if $\tilde{X}$ is compact.

Michael Albanese
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  • Ah yes I had used that result just a few minutes ago! I'm not sure how easy it would be to show the universal covers of $\mathbb{P}T\mathbb{S}^2$ and $\mathbb{P}T\mathbb{RP}^2$ are compact in my case though. Thanks – jl2 Aug 17 '16 at 17:11
  • Actually would this just follow from the fact that $\mathbb{P}T\mathbb{S}^2$ and $\mathbb{P}T\mathbb{RP}^2$ are compact and Hausdorff? – jl2 Aug 17 '16 at 17:13
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    No, consider $\mathbb S^1$. @jl2 –  Aug 17 '16 at 17:14
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    We'd need the additional requirement that $p^{-1}(x)$ is finite for all $x$ (where $p$ is the universal cover), right? – jl2 Aug 17 '16 at 17:17
  • @jl2 That would do it (http://math.stackexchange.com/questions/323711/fiber-bundle-is-compact-if-base-and-fiber-are), but those fibers are in bijection with the fundamental group -- so you would have to know beforehand that the fundamental groups are finite. – Elle Najt Aug 17 '16 at 17:37
  • Do you know a textbook that proves this? – Danu Aug 23 '16 at 09:57