# Euler characteristic, Gauss-Bonnet, and a product formula

I know very little about the Pfaffian or how it works, and I'm new at Riemannian geometry in general. But I was wondering if there is some way to make this "intuitive" argument for the fact that a fiber bundle satisfies $\chi(\text{bundle}) = \chi(\text{fiber})\chi(\text{base})$ rigorous.

1. Say you've got a surjective smooth submersion $\widetilde{M} \rightarrow M$ that's also a (smooth?) fiber bundle. (Assume all the manifolds are compact... though I wish there were some weaker condition than that for the Gauss-Bonnet theorem to apply. It'd be nice if one could make sense of it whenever the manifold had finitely generated cohomology).
2. EDIT: See update below. Slap Riemannian metrics on $\widetilde{M}$ and $M$ so that the submersion is a Riemannian submersion. (I just feel like one would need this in step 4. The other thing that would make sense is if we could somehow put a Riemannian metric on $\widetilde{M}$ that would be the 'twisted product' of Riemannian metrics on $F$ and $M$- whatever that means).
3. By the Generalized Gauss-Bonnet Theorem, $\int_{\widetilde{M}}Pf(\Omega_{\widetilde{M}}) = (2\pi)^n \chi(\widetilde{M})$.
4. And here's the thing that seems the hardest to make rigorous (though who knows, I may have already been wrong at step 2): I want to say that $\int_{\widetilde{M}}Pf(\Omega_{\widetilde{M}}) = \int_M\int_F Pf(\Omega_F)Pf(\Omega_M) = (2\pi)^k(2\pi^{n-k})\chi(F)\chi(M)$.

In order for this to even start making sense there would have to be some sense in which an analog of Fubini's theorem works for twisted products. I'd also need the Pfaffian to satisfy the identity that I want it to satisfy in this case (which I think follows from the fact that if we have a direct sum of matrices, $A \oplus B$, then $Pf(A \oplus B) = Pf(A)Pf(B)$.) So: is there any way to actually make this argument stand? (I think that it would work for a trivial fibration, but that's no fun.)

So far I only know how to prove this product formula using a spectral sequence, so it'd be cool if this were a legitimate way of visualizing it in this special case.

Updates: First of all, it turns out that there is a Fubini's theorem for local products (see Generlized Curvature by Jean-Marie Morvan). Instead of Step 2 I think what we want is to put a metric on $M$, and then put metrics $h_x$ on $F_x$ (the fiber over $x$) that somehow 'varies smoothly with $x$' (hopefully that's possible) and then define a metric on $\widetilde{M}$ by breaking every vector into its horizontal and vertical pieces and applying the metric on $M$ and $F$ separately. Pretty much by definition this should make $\Omega_{\widetilde{M}}$ be the matrix $\Omega_M \oplus \Omega_F$ (I think?) and then the result should follow.

So I still have details to work out- and I'll keep thinking about it, but in the meantime if any of you have seen this before, then a reference would be really helpful! I'm afraid I've done something silly, or I'm assuming something I can't...

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I vote for the theorem to be called "Chern-Gauss-Bonnet" instead of "Generalized Gauss-Bonnet" in the text of the question! –  Q.Q.J. Mar 4 '11 at 10:54
Not related to this approach, but there's a very nice little subcase/generalization of this fact called the Riemann-Hurwitz formula: en.wikipedia.org/wiki/Riemann-Hurwitz_formula –  Aaron Mazel-Gee Mar 4 '11 at 18:33
@Aaron: Neat! @QQJ: You're probably right, but I feel bad bumping up this question for the minor edit. –  Dylan Wilson Mar 6 '11 at 9:27

I understand you're trying to prove using the Pfaffian but what's wrong with the triangulation proof of what you want? Fibre bundles $F\to E\to B$ are locally trivial and triangulate your base and the fibres then it should be clear that $\chi(E)=\chi(F)\chi(B)$.

More precisely, assume that you have triangulated finely enough so that the simplices in your base are already in local charts. For simplicity, let $B$ be a $k$-dimensional manifold. If $\Delta_k\subset B$ then the number of vertices in $\pi^{-1}(\Delta_k)$ is the product of the number of vertices of $\Delta_k$ and the number of vertices $v(F)$ of $F$. The number of edges of $\pi^{-1}(\Delta_k)$ equals $e(\Delta_k)v(F)+v(\Delta_k)e(F)$. You continue in this way and an alternate sum tells you that $\chi(\pi^{-1}(\Delta_k))=\chi(\Delta_k)\chi(F)=\chi(F)$. When you glue fibres over charts (in this case faces of $\Delta_k$) then the Euler characteristic formula $\chi(X_1\cup X_2)=\chi(X_1)+\chi(_2)-\chi(X_1\cap X_2)$ tells you that $\chi(\pi^{-1}(\Delta_k\cup\Delta_k'))=\chi(\Delta_k)\chi(F)+ \chi(\Delta_k')\chi(F)-\chi(\Delta_{k-1})\chi(F)$. Then another alternating sum tells you that you should ultimately sum $1$'s (with signs) over the simplices of your base and finally multiply $\chi(F)$ which gives you the formula you want.

EDIT : In case this answer isn't not helpful, feel free to comment. I initially wanted to put it up as a comment but ended up clicking on the answer button!

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Thank you- but I really did want to see if this Pfaffian thing would work. I recently saw the triangulation proof in a paper of Mostow- it's very nice and it does give a bit more intuition than the spectral sequence proof. –  Dylan Wilson Mar 3 '11 at 23:18
This might be totally stupid, but isn't it not known whether all manifolds are triangulable? I read in an (admittedly old-ish) book that this wasn't known, but maybe it's been proven since then. –  Aaron Mazel-Gee Mar 4 '11 at 18:33
@Aaron - You're right in saying that not all manifolds are triangulable (the Kirby-Siebenmann class of $M$ has to vanish if $M$ has to have piecewise linear structure) but in this question the manifolds under consideration are all smooth and hence admit triangulations. –  Somnath Basu Mar 4 '11 at 19:48

The Gauss-Bonnet theorem has, in my opinion, two parts. The first part is how to associate to a smooth oriented vector bundle $V \to X$ on a manifold a differential form representing the Euler class. For that, one needs a metric on $V$ and a metric connection. Take the curvature und plug it into the Pfaffian - that gives a closed differential form, the Euler form, whose de Rham class only depends on $V$ and its orientation. If $V=TM$ and $M$ is Riemannian, you can pick the Levi-civita connection, but that is not really relevant here.

The second step is that $\int_M e (TM)= \chi(M)$, and this is algeraic topology, involving Poincare duality.

Let $f:E \to B$ be a smooth fiber bundle with $d$-dimensional compact fibres (=proper submersion) and assume that $B$ is closed and oriented. Let $T_v E := ker (df)$ be the vertical tangent bundle. There is a short exact sequence of vector bundles $0 \to T_v E \to TE \to f^{\ast} TB \to 0$ and a splitting is given e.g. by a Riemann metric on $E$.

The Euler class $e$ of vector bundles satisfies $e(V \oplus W) = e(V) e(W)$, which follows for examples from direct sum formula for the Pfaffians (in fact, the above identity holds even for the differential form representatives). There is an integration over the fibre

$$\int_{E/B} \omega,$$

which produces a $k-d$-form on $B$ out of a $k$-form on $E$. For a construction, see Berline, Getzler, Vergne ''Heat kernels and Dirac operators'' (it is in the background chapter 1). There is a Fubini principle for integration over the fibres, which reads:

$$\int_E e(TE) = \int_E e(T_v E) f^{\ast} e (TB) = \int_B (\int_{E/B} e(T_v E)) e (TB).$$

The form $\int_{E/B} e(T_v E)$ is a $0$-form, i.e. a function, and you can test its values by restriction to the fibres. The value is the Euler characteristic of the fibre, by Gauss-Bonnet. Thus

$$\int_B (\int_{E/B} e(T_v E)) e (TB)= \chi(F) \int_B e(TB)= \chi(F) \chi(B).$$

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I still don't understand how we pick euler classes on each fiber in a 'continuous' way... what am I missing? –  Dylan Wilson Mar 6 '11 at 9:30
Or is the problem avoided by starting with a metric on the total space, restricting it to the fibers, and then 'pushing it down' to the base? –  Dylan Wilson Mar 6 '11 at 9:31
Pick a (bundle) metric on $T_v E$. This gives you a definite form on $E$ representing $e(T_v E)$. Then pick a metric on $TB$ and pull it back to $f^{\ast} TB$. Finally, pick a splitting of $TE$, which gives a metric on $TE$. This metric is well-adapted to the problem. –  Johannes Ebert Mar 6 '11 at 10:11