# Multiplicativity of Euler characteristic for non-orientable fibrations

Let $E\to B$ be a fibration with fiber F, and assume for simplicity that B is connected. Suppose moreover that B and F have Euler characteristics (perhaps they are manifolds). Then often, one can conclude that E has an Euler characteristic as well, and that

$$\chi(E) = \chi(B)\cdot \chi(F).$$

The only proof of this that I have been able to find uses a spectral sequence argument, and requires that $\pi_1(B)$ act trivially on the homology of F, so that the homology in the spectral sequence can be taken with constant coefficients. This condition is sometimes referred to as orientability of the fibration (with respect to the homology theory, normally rational homology).

Is the result known to be true any more generally than this? Is there any other known proof? Are there any examples where it is known to be false?

Assume for simplicity that $B,F$ are finite CW-complexes and let $p:E\to B$ be the bundle projection.

Suppose $B$ is obtained from a CW-complex $B'$ by attaching an $n$-cell. Suppose $\chi(B')\chi(F)=\chi (E')$ with $E'=p^{-1}(B')$. Then $H^*(E,E')\cong \tilde H(E/E')\cong H^*(D^n\times F,S^{n-1}\times F)$ [upd: some more details: $E/E'$ is the one point compactification of $p^{-1}(B\setminus B')$; now $B\setminus B'$ is an $n$-disk and so $p^{-1}(B\setminus B')\cong (D^n\setminus S^{n-1})\times F$, so the one point compactification of $p^{-1}(B\setminus B')$ is $(D^n\times F)/(S^{n-1}\times F)$. Now using excision and homotopy we see that $\tilde H^*(E/E')\cong H^*(D^n\times F,S^{n-1}\times F)$.]

So $\chi(E,E')=(-1)^n\chi(F)$. So by induction on the number of cells we get $\chi(E)=\chi(B)\chi(F)$. No assumptions on the action of $\pi_1(B)$ are necessary.

• Nice argument. If only you could do the same for signature :( – Igor Rivin Nov 7 '11 at 20:07
• Thanks! Can this be generalized to Lefschetz numbers? – Mike Shulman Nov 7 '11 at 20:55
• Mike -- good question. When the action of $\pi_1(B)$ on the homology of the fiber is trivial everything is fine: one can use the spectral sequences. Otherwise I don't see a straightforward way to compute the Lefschetz number. Take this example: set $E=B=S^1$ and let $p$ be a double cover. Set $f:B\to B$ to be the symmetry wrt say $x$-axis. We can cover $f$ by a map $g:E\to E$ that will be the identity over one of the fixed points of $f$ and that will swap the elements of the fiber over the other fixed point; the Lefschetz numbers of the maps of different fibers are different. – algori Nov 7 '11 at 21:30
• It looks like algori is assuming you have a locally trivial bundle, rather than a fibration (although the argument works also). A slight variant is to use $Z/2$ coefficients:, then when the image of $\pi_1(B)\to Aut(H_*(F;Z/2))$ is finite (eg if $F$ a finite complex), pull back the fibration to the corresponding finite cover of $B$, where the action is now trivial so e.g. the spectral sequence argument applies. Then a simple version of algori's argument shows that $\chi$ is multiplicative under finite coverings, and so divide $\chi(B)$ and $\chi(E)$ by this degree. – Paul Nov 8 '11 at 17:00
• Paul -- that's what algori was assuming, yes. Nice trick, by the way. – algori Nov 8 '11 at 18:13

Since this question seems to be attracting some renewed interest, I may as well point out that a few years after I asked it, Kate Ponto and I proved a generalization of this formula by purely homotopical/categorical methods, which applies to any fibration and yields a result about the Lefschetz number and even the Reidemeister trace. The paper is here.

Let $B$ be a finite simplicial complex. Let us consider its dual cell-decomposition. The preimage of each cell has Euler characteristic equal to that of $F.$ Now use the usual combinatorial formula for the Euler characteristic of the union of finitly many sets, $\chi(A_1 \cup \dots \cup A_k)$, taking for $A_i$, the preimage of the dual cell of the $i$-th vertex of $B$. We obtain the required equality.