Let $X$ be a compact differentiable manifold of dimension $m$ or, if you prefer, a smooth complex projective manifold of complex dimension $n=m/2$.

The Euler characteristic $\chi(X):=\Sigma_{i=0}^{m}(-1)^ib_i(X)$ -the alternating sum of Betti numbers of $X$- is related to the Euler class $e(T_X)$ of the tangent bundle $T_X$ by $$\chi(X)=\int_Xe(T_X)$$ where $\int_X$ denotes contraction against the fundamental class $[X]$.

Is it possible to write each Betti number $b_i(X)$ as a "characteristic number", i.e. as an integral $$b_i(X)\stackrel{\text{?}}{=}\int_X\gamma_i(T_X)$$ of some "characteristic class" $\gamma_i$ of vector bundles? If not, why not?

This question is possibly related.

  • 10
    $\begingroup$ Characteristic numbers are cobordism invariant, but Betti numbers are not. $\endgroup$ Dec 25, 2016 at 18:38
  • 7
    $\begingroup$ By embedding a finite CW complex in Euclidean space, thickening it up to a (framed) manifold, and taking its double, you can get any betti numbers but have trivial tangent bundle and hence trivial characteristic classes. $\endgroup$ Dec 25, 2016 at 18:41

1 Answer 1


No. The Stiefel-Whitney and Pontryagin numbers of a closed oriented manifold are cobordism invariants, but the Betti numbers are not.

More explicitly, all closed oriented $3$-manifolds are frameable (and null cobordant), so all of their characteristic numbers vanish, but e.g. the $3$-torus and $3$-sphere have different Betti numbers.


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