14

It is known that if $X$ is a finite CW complex and if $Y \to X$ is a $n$-sheeted covering then $Y$ is a finite CW complex and $\chi(Y)=n \cdot \chi(X)$.

More generally, Euler characteristic can be defined as $\chi(X)= \sum_i(-1)^i \text{rank}(H_i(X))$ when all but finitely many homology groups are trivial (ie. when $X$ is of bounded finite type). In this case, does the preceding statement still hold?

If $X$ is of bounded finite type and if $Y \to X$ is a $n$-sheeted covering, is it true that $Y$ is of bounded finite type and $\chi(Y)=n \cdot \chi(X)$?

A weaker problem is: Is there a homological proof for CW complexes?

Seirios
  • 33,157

1 Answers1

11

From the Leray-Serre spectral sequence for a covering map $Y\to X$, which is a fibration with discrete fibers, we get an isomorphism $H^p(Y)\cong H^p(X,\mathcal H^0)$, where $\mathcal H^0$ denotes the local system of coefficients which at each point of $X$ has group equal to $H^0(p^{-1}(x))$.

If the covering is of $n$ sheets, then $\mathcal H^0$ is locally $\mathbb Z^n$, with the fundamental group of $X$ acting by permutation of the standard basis according to the monodromy permutation representation.

Now the local system $\mathcal H^0$ corresponds to a sheaf $\mathcal F$ on $X$, and for sensible $X$ (paracompact, say), one can compute singular cohomology with coefficients in the local system as sheaf cohomology with coefficients on the sheaf $\mathcal F$. If $X$ has a good finite cover $\mathcal U$ (in the sense of the book of Bott-Tu) then one can compute sheaf cohomoogy $H^p(X,\mathcal H^0)$ as the Cech cohomology $H^p(\mathcal U,\mathcal H^0)$. Looking at the complex which computes this by definition, we see that the Euler characteristic of $H^p(\mathcal U,\mathcal H^0)$, and therefore of $H^\bullet(X,\mathcal H^0)$, is $n$ times that of $H^\bullet(X,\mathbb Z)$. Notice that the existence of good covers implies being of bounded finite type, as you say (but I think it even implies that the space of of the homotopy type of a CW-complex, namely the nerve of the good covering... so all this might not get us much)

The fact that the Euler characteristic of a sensible space with coefficients on a local system of coefficients which locally looks like $\mathbb Z^n$ is $n$ times that of the space should be written down somewhere, but I cannot find it now. There is this answer by Matt but he does not give a reference.

  • There is an assumption in your argument: under what conditions will $X$ have a finite good cover? I don’t think that bounded finite type-ness will suffice for that. – John Klein Feb 25 '22 at 19:33
  • @JohnKlein, well, that I am assuming that is very clear since I wrote if $X$ has a good finite cover…, isn't it? :-) FWIW, I also don't think bounded type is enough to ensure there is one, and I have never found anything but rather draconian hypotheses that ensure there is one, sadly. – Mariano Suárez-Álvarez Mar 01 '22 at 22:00
  • Well, having a good finite cover is practically the same as having the homotopy type of a finite complex. The result is true for bounded type if the fibration is orientable (Spanier p. 481).

    In a related setting, we can assume the spaces of the fibration are dominated by finite complexes. In that case, I have sketched what I hope to be a correct proof in that case. See https://mathoverflow.net/a/417080/8032

    – John Klein Mar 02 '22 at 00:15