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I am currently working on this algebraic topology problem and got stuck:

Suppose $X$ is a finite CW complex with $\pi_1(X)$ a nontrivial finite group. Show that its universal cover $\widetilde{X}$ is not contractible. (Hint: Use Lefschetz fixed theorem)

I guess I am supposed to apply Lefschetz fixed point theorem to the deck transformations but from there I don't know how to proceed.

Najib Idrissi
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No_way
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2 Answers2

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There is a way to prove without Lefschetz fixed point theorem. If $X$ is a finite CW-complex, so is $\widetilde{X}$. If $|\pi_1(X)|=d$, then $d\cdot\chi(X)=\chi(\widetilde{X})$. If $\widetilde{X}$ is contractible, then $\chi(\widetilde{X})=1$. Then $d=1$, contradicting the nontriviality of $\pi_1(X)$.

Edit: One can construct a triangulation for $X$ such that it can be lifted to one for $\widetilde{X}$. Using definition of covering spaces, the number of $i$-dimensional simplices in the triangulation of $\widetilde{X}$ is $d$ times that of $i$-dimensional simplices in the triangulation of $X$. So we have $d\cdot\chi(X)=\chi(\widetilde{X})$.

Alex Fok
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    Wow, that's neat! I wouldn't have come up with $d\cdot\chi(X)=\chi(\widetilde{X})$ myself, but now it seems very natural. – No_way Jun 06 '15 at 11:12
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If $X$ is a finite CW complex and $E \to X$ is a covering space with finite fibers, then $E$ is a finite CW complex too. This is because a covering space of a CW-complex has a CW-complex structure too, and a covering space of a compact space is compact (a CW-complex is finite iff it's compact).

So suppose $X \sim K(G,1)$ with nontrivial $G$ is a finite CW-complex. It follows that $\widetilde{X}$ is a finite (contractible) CW-complex, so you can apply the Lefschetz fixed point theorem. Take a nontrivial deck transformation $f$ of $\widetilde{X}$ (which exists because $G$ is nontrivial); it has no fixpoint. But the Lefschetz number of $f$ is $$\Lambda_f = \sum_{k \ge 0} (-1)^k \operatorname{tr}(f_*|_{H_*(\widetilde{X}; \mathbb{Q})}),$$ and since $\widetilde{X}$ would be contractible in this situation, it's equal to the trace of $f$ on $H_0(\widetilde{X}; \mathbb{Q}) \cong \mathbb{Q}$, and $f_* = \operatorname{id}_{\mathbb{Q}}$ has trace $1$. So the Lefschetz number is nonzero and $f$ must have a fixed point, contradiction (cf. Any map of a contractible complex to itself has a fixed point).

Community
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Najib Idrissi
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  • I just realized the title doesn't exactly match the question. But if $X$ isn't a $K(G,1)$ then $\widetilde{X}$ wouldn't be contractible anyway (the higher homotopy groups are the space for a space and its covering spaces). – Najib Idrissi Jun 06 '15 at 10:38
  • Thanks for your answer! I should have thought about that the higher homology of a contractible space is 0... – No_way Jun 06 '15 at 11:10
  • Well the title is equivalent to the question. – No_way Jun 06 '15 at 11:10