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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
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