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.

• Kurt: I think your question would be easier to read if you drop those "singular" subscripts. Sure enough everyone knows that $H^k(M)$ without qualifiers stands for ordinary cohomology. Also, why do you need two letters ($Rh^k$) for the isomorphism between De Rham and singular cohomologies? I think Dupont's notation (just $I$, for "integration") is pretty standard and using standard notations helps everyone to understand better what we mean. – d.t. Oct 5 '10 at 14:36
• Robin has answered your question below. As for calculations, you may want to try these yourself in the case of the circle (e.g. by determining explicitly a closed one-form on the circle who integral around the circle is 1, and proving that this one-form is unique up to adding an exact one-form). – Matt E Oct 5 '10 at 15:31

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

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 $\tau$.

• Sorry for the necropost (nice post, BTW, +1) ; please ignore if the material is not fresh any more. Is it fair to say that $\int \sigma^* \omega \ ,$ is a parametrization of the simplex $\sigma$? I'm just curious if we can see this integration as the types of integrals of forms we do in $\mathbb R^n$ , where we integrate an n-form $w$ over an n-manifold by parametrizing by , say, $g(x_1,..,x_n)$ and then evaluating $\int w(g(x_1,..,x_n)$? Is the effect of $\sigma*$ similar to that of parametrizing, so that, e.g., if $\sigma_i$ were a standard simplex, $\sigma*$ would be the identity? – user99680 May 5 '14 at 3:29
• What about open manifolds, where we do not have Poincare duality? – Danu Apr 3 '18 at 10:59

Here is an explicit procedure based on the isomorphism between the de-Rham and Cech cohomologies for smooth manifolds based on R. Bott and L.W. Tu's book: Differential forms in algebraic topology.

The description will be given for a three form but it can be generalized along the same lines to forms of any degree. We suppose that the manifold has a finite good cover.

The data needed is the transition functions between the coordinate charts (which will be denoted by: $U_\alpha$, $U_\beta$, etc.) and of course, the coordinate expression of the given form on each chart.

Given a three form F, then by the Poincare lemma, on $U_\alpha, F = dB_\alpha$, where $B_\alpha$ are two forms on $U_\alpha$.

Thus by the Poincare lemma on $U_\alpha \cap U_\beta$:

$B_\alpha-B_\beta = dA_{\alpha\beta}$ , where $A_{\alpha\beta}$ are one forms on $U_\alpha \cap U_\beta$.

Since on $U_\alpha \cap U_\beta \cap U_\gamma$:

$d(A_{\alpha\beta}+A_{\beta\gamma}+A_{\gamma\alpha})=0$

Then by the Poincare lemma

$A_{\alpha\beta}+A_{\beta\gamma}+A_{\gamma\alpha} = d\phi_{\alpha\beta\gamma}$

where: $\phi_{\alpha\beta\gamma}$ are zero forms on $U_\alpha \cap U_\beta \cap U_\gamma$.

Again, Since on: $U_\alpha \cap U_\beta \cap U_\gamma \cap U_\delta$:

$d(\phi_{\alpha\beta\gamma}-\phi_{\beta\gamma\delta}+\phi_{\gamma\delta\alpha}-\phi_{\delta\alpha\beta})=0$

Then:

$\phi_{\alpha\beta\gamma}-\phi_{\beta\gamma\delta}+\phi_{\gamma\delta\alpha}-\phi_{\delta\alpha\beta} = C_{\alpha\beta\gamma\delta}$

where: $C_{\alpha\beta\gamma\delta}$ are constants.

The differential form F is integral iff:

$C_{\alpha\beta\gamma\delta}= 2 \pi n_{\alpha\beta\gamma\delta}$

where $n_{\alpha\beta\gamma\delta}$ are integers on all quadruple intersections.

• A couple of things: (i) why bother with the $2\pi$? and (ii) there is enough freedom in choosing the $A$s, $B$s, $\phi$s etc. so that even with an integral cohomolgy class one may end up with $n$s that aren't integers, although defining a Cech cocycle homologous to an integer one. – Robin Chapman Oct 6 '10 at 16:52
• I agree with the previous comment, you have to choose your As, Bs and phi's very carefully if you want C to be integer. The correct statement would be that F is integer if and only if this careful choice can be done. Actually I am trying to prove this by explicitly finding the correct As Bs and phi's and it is giving me headache, anyone knows a reference for this? – juan rojo Nov 29 '15 at 17:58

This has nothing to do with poincare poincare duality or even de rham cohomology.

Let $X$ be any space. By the universal coefficient theorem the canonical map $$H^p(X,\mathbb{Z}) \to Hom_{\mathsf{Ab}}(H_p(X,\mathbb{Z}),\mathbb{Z})$$

Is surjective. Moreover (also by the universal coefficient theorem if you like) we have

$$H^p(X,\mathbb{R}) \cong H_p(X,\mathbb{R})^*$$

Together we get that $\alpha \in H^p(X,\mathbb{R})$ comes from an integral cohomology class iff the induced linear form $\tilde{\alpha}:H_p(X,\mathbb{R}) \to \mathbb{R}$ is a linear extension of a form $H_p(X,\mathbb{Z}) \to \mathbb{Z}$. In other words:

A real cohomology class is integral iff the corresponding linear form takes integer values when restricted to the lattice $H_p(X,\mathbb{Z})/Torsion \subset H_p(X,\mathbb{R})$