# Betti numbers as characteristic numbers?

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.

• Characteristic numbers are cobordism invariant, but Betti numbers are not. – Gregory Arone Dec 25 '16 at 18:38
• 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. – Oscar Randal-Williams Dec 25 '16 at 18:41

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.