# Why does the Gauss-Bonnet theorem apply only to even number of dimensons?

One can use the Gauss-Bonnet theorem in 2D or 4D to deduce topological characteristics of a manifold by doing integrals over the curvature at each point.

First, why isn't there an equivalent theorem in 3D? Why can not the theorem be proved for odd number of dimensions (i.e. what part of the proof prohibits such generalization)?

Second and related, if there was such a theorem, what interesting and difficult problem would become easy/inconsistent? (the second question is intentionally vague, no need to answer if it is not clear)

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Roughly speaking, because $\chi(\mbox{odd dimensional manifold}) = 0$. –  Neal Jun 26 '12 at 12:56
Look at page 3, math.berkeley.edu/~alanw/240papers00/zhu.pdf –  Ehsan M. Kermani Jun 26 '12 at 13:14
This reminded me of Kleiner's isoperimetric comparison theorem: the Euclidean isoperimetric inequality holds in nonpositively curved 3-manifolds. This interesting and difficult problem is still open in dimensions above 4. Kleiner comments on p.39 that the missing ingredient is an estimate that one obtains from Gauss-Bonnet in 3D but not in higher dimensions. –  user31373 Jun 26 '12 at 17:01

I think the reason is that the original Chern-Gauss-Bonnet theorem can be treated topologically as $$\int_{M} e(TM)=\chi_{M}$$ and for odd dimensional manifolds, the Euler class is "almost zero" as $e(TM)+e(TM)=0$. So it vanishes in de rham cohomology. On the other hand, $\chi_{M}=0$ if $\dim(M)$ is odd. So the theorem now trivially holds for odd dimension cases.
Another perspective is through Atiyah-Singer index theorem. The Gauss-Bonnet theorem can be viewed as a special case involving the index of the de rham Dirac operator: $$D=d+d^{*}$$ But on odd dimensional bundles, the index of $D$ is zero. Therefore both the left and right hand side of Gauss-Bonnet are zero.