Integral classes in de Rham cohomology

If $M$ is a differentiable manifold, De Rham's theorem gives for each positive integer $k$ an isomorphism $Rh^k : H^k_{DR}(M,\mathbb R) \to H^k_{singular}(M,\mathbb R)$. On the other hand, we have a canonical map $H^k_{sing}(M,\mathbb Z) \to H^k_{singular}(M,\mathbb R)$ . Allow me to denote (this is not standard) its image by $\tilde {H}^k_{singular }(M,\mathbb Z)$. My question is : how do you recognize if, given a closed differental $k$- form $\omega$ on $M$, its image $Rh^k([\omega]) \in H^k_{singular}(M,\mathbb R) $ is actually in $\tilde {H}^k_{singular}(M,\mathbb Z)$.

I would very much appreciate a concrete answer, ideally backed up by one or more explicit calculations. Thank you for your attention.

Solution 1:

By Poincare duality a $k$-form will be integral iff its integral over all (smooth) singular $k$-cycles is an integer.

Solution 2:

I think Robin's result is also clear if we look at the definition of the isomorphism between De Rham and (${\cal C}^\infty$) singular cohomologies. (Ok, we are both saying the same thing, but I like to say it this way.)

Remember (see Dupont, Curvature and characteristic classes, Springer LNM 640) the definition of the isomorphism at the cochain level:

$$ I : \Omega^n (M) \longrightarrow C^n (M) \ , $$

Here $\Omega^n(M)$ are the degree $n$-forms on $M$, $C^n(M)$ the ${\cal C}^\infty$ singular $n$-cochains. Then, for $\omega \in \Omega^n (M)$, the value of $I(\omega )$ on a ${\cal C}^\infty $ singular $n$-simplex $\sigma$ is defined as

$$ I (w)_{\sigma} = \int \sigma^* \omega \ , $$

where the integral is taken over the standard $n$-simplex $\Delta^n$.

Hence, for $[I(w)]$ to be an integral class it suffices that all values of $I(w)$ on cycles $\tau = \sum_i \lambda_i\sigma_i$ are integer numbers; that is, if and only if $I(w)_\tau = \sum_i \lambda_i\int \sigma_i^*\omega \in \mathbb{Z}$ for all cycles $\tau$.