Suppose that $S$ is a compact orientable surface. In this case, the top de Rham cohomology space $H^2(S)\cong \mathbb{R}$, with the isomorphism given by integration on $2$-forms along $S$.

Now, one can define integral cohomology classes as those cohomology classes $a$ so that $\int_S a \in \mathbb{Z}$. On the other hand, one can also define integral cohomology classes as those classes corresponding to integral Cech cohomology classes $\check{H}^2(S,\underline{\mathbb{Z}})$ in the following way:

If $\omega$ is a closed $2$-form, we can find an open cover $\mathfrak{U}$ of $S$ and $1$-forms $\alpha_U$ on each $U$ so that $\omega|_U = d\alpha_U$. Now, choose functions $f_{UV}$ so that $df_{UV} = \alpha_U - \alpha_V$. The cocycle

$$ f_{UVW} = f_{UV} + f_{VW} - f_{UW} $$

satisfies that $df_{UVW}=0$, so $f_{UVW} \in \check{H}^2(S,\underline{\mathbb{R}})$.

With this in mind, I claim that $[\omega]$ is integral if and only if one can choose $(\alpha_U)$ and $(f_{UV})$ such that $(f_{UVW}) \in \check{H}^2(S,\underline{\mathbb{Z}})$.

I want to see why these two definitions coincide. More precisely, I would like to see an explicit proof on why, given that $\int_S \omega \in \mathbb{Z}$, I can choose the $f_{UV}$ so that $f_{UVW} \in \mathbb{Z}$. Please, I prefer an explicit proof of this fact, rather than invoking Poincaré Duality/de Rham theorem.

  • $\begingroup$ The statement looks wrong: maybe you want $[\omega]$ integral if and only if the data above can be chosen such that $f_{UVW}$ is integral. Otherwise, adding some constant to one of the $f_{UV}$ easily destroys integrality. More important: you seem to ask for a proof of a special case of de Rham's theorem here, but you want to avoid the theorem itself. I would suggest to have a look into the book by Bott and Tu, where de Rham's theorem is explained, if you have a copy at hand. $\endgroup$ Commented Aug 10, 2020 at 15:18
  • $\begingroup$ Exactly, in the last paragraph I explain that what I want to know is why if $\omega$ is integral the $f_{UV}$ can be chosen so that $f_{UVW}$ is an integer. I think this already gives a proof of de Rham’s, since you can recover $\omega$ from the $f_{UVW}$ using a partition of unity. The problem is how to deal with integrality. I already looked at Boot-Tu and I do not think they do that there (though maybe I should look again). For context, this question came to me while studying prequantization in Woodhouse’s book on Geometric Quantization. $\endgroup$
    – G. Gallego
    Commented Aug 10, 2020 at 15:37
  • $\begingroup$ This fact is also claimed in Kobayashi’s paper on “Principal fibre bundles with the 1 dimensional toroidal group” (page 35) and in this post on Math StackExchange math.stackexchange.com/questions/6099/… (The comment by David Bar Moshe. My question is essentially the same as the one by Juan Rojo). $\endgroup$
    – G. Gallego
    Commented Aug 10, 2020 at 15:41

2 Answers 2


$\def\RR{\mathbb{R}}\def\ZZ{\mathbb{Z}}$I've considered assigning this when I've taught sheaf cohomology but it always seemed a little too hard. Let's see if I can do it. I'll be a little more general while I am at it and do the case of a smooth compact oriented $n$-fold. Choose a triangulation $S$ of the $n$-fold; let $F_j$ be the set of $j$-dimensional faces.

For each vertex $u \in F_0$, let $U(u)$ be the star shaped open neighborhood of $u$ as in the OP's answer. The $U(u)$ give an open cover of $X$. For any $u_0$, $u_1$, ..., $u_j$ in $F_0$, the intersection $U(u_0) \cap \cdots \cap U(u_j)$ is empty if $(u_0, \dots, u_j)$ are not the vertices of a face, and this intersection is a contractible open set which I'll call $U(\sigma)$ if $(u_0, \dots, u_j)$ are the vertices of a face $\sigma$ in $F_j$. Thus, the Cech complex of $\underline{\RR}$ is identified with the simplicial cohomology complex $$\RR^{F_0} \to \RR^{F_1} \to \cdots \RR^{F_{n-1}} \overset{d_{n-1}}{\longrightarrow} \RR^{F_n}.$$

For any $n-1$ dimensional face $\tau$, there are two $n$-faces $\sigma_1$ and $\sigma_2$ containing $\tau$. Letting $e_{\tau}$ be the basis function corresponding to $\tau$, we have $d_{n-1}(e_{\tau}) = e_{\sigma_1} - e_{\sigma_2}$. (I am being sloppy about signs, but the fact that we are on an oriented manifold will make it all work out in the end.) So (using that our manifold is connected) the cokernel of $d_{n-1}$ is clearly $\RR$, and an explicit map from $\RR^{F_n}$ to the cokernel sends a function $f \in \RR^{F_n}$ to $\sum_{\sigma \in F_n} f(\sigma)$.

Let $\Omega^p$ be the sheaf of smooth $p$-forms, and let $Z^p$ be the subsheaf of closed $p$-forms. Note that $Z^0 = \underline{\RR}$, so we have just computed that $H^n(X, Z^0) \cong \RR$. The Poincare lemma gives short exact sequences $Z^p \to \Omega^p \to Z^{p+1}$ for $0 \leq p \leq n$, so we get boundary maps $$H^0(X, Z^n) \to H^1(X, Z^{n-1}) \to \cdots \to H^n(X, Z^0) \cong \RR.\quad (\ast)$$ In the case of a surface, the OP has given explicit descriptions of these maps in his answer.

By the usual argument with partitions of unity, $H^q(X, \Omega^p)$ vanishes for $q>0$, so all these maps are isomorphisms except the first one. The first map, in turn, is surjective with kernel $d H^0(X, \Omega^{n-1})$. So the image of the first map is $H^n_{DR}(X)$, and all the other $H^q(X, Z^{n-q})$ are isomorphic to $H^n_{DR}(X)$. Our goal, given an $n$-form $\omega$, is to show that the composition of all these maps gives $\int_X \omega$.

Note that a class in $H^q(X, Z^{n-q})$ is given by a Cech representative $( \eta_{\sigma} )_{\sigma \in F_q}$, where $\eta_{\sigma}$ is a closed $(n-q)$-form on $U(\sigma)$.

Choose a regular CW subdivision $S^{\perp}$ of $X$ dual to the triangulation. That means the poset of faces of $S^{\perp}$ is dual to that of $S$ and each $j$-face $\sigma$ in $S$ crosses the dual $n-j$ face $\sigma^{\perp}$ transversely in one point. An explicit way to do this is to take the barycentric subdivision of $S$ and draw the "obvious" dual faces. If we choose an ordering of $F_0$, that gives an orientation to every face $\sigma$ of $F_q$, and then we can use the global orientation of $X$ to orient $\sigma^{\perp}$.

I claim that the composite isomorphism $(\ast)$ from $H^{q}(X, Z^{n-q})$ sends $(\eta_{\sigma})_{\sigma \in F_q}$ to $$\sum_{\sigma \in F_q} \int_{\sigma^{\perp}} \eta_{\sigma}.$$

Let's see what this means for $q=n$. Each $\eta_{\sigma}$ is a closed $0$-form on $U(\sigma)$. A closed $0$-form is locally constant function and $U(\sigma)$ is connected, so we just have a real number for each $\sigma$ in $F_n$ and we can thus think of $\eta$ as a vector in $\RR^{F_n}$. Each $\sigma^{\perp}$ is just a point in the interior of $\sigma$. So we are just summing up the values of $\eta$ on the $n$-faces, and this is the map $\RR^{F_n} \to \RR$ that we described before.

Let's next see what this means for $q=0$. Each $\eta_{\sigma}$ is an $n$-form on $\sigma$, and the condition that $(\eta_{\sigma})$ is a Cech co-cycle says that $\eta_{\sigma}$ is the restriction of a global $n$-form $\omega$ on $X$. The $n$-faces $\sigma^{\perp}$, for $\sigma \in F_0$, partition $X$. So $$\sum_{\sigma \in F_q} \int_{\sigma^{\perp}} \eta_{\sigma} = \sum_{\sigma \in F_q} \int_{\sigma^{\perp}} \omega|_{\sigma^{\perp}} = \int_X \omega.$$

Thus, we just need to show that, if $(\eta_{\sigma})$ represents a class in $H^q(X, Z^{n-q})$ and $\delta_q$ is the boundary map $H^q(X, Z^{n-q}) \to H^{q+1}(X, Z^{n-q-1})$, then $$\sum_{\sigma \in F_q} \int_{\sigma^{\perp}} \eta_{\sigma} = \sum_{\tau \in F_{q+1}} \int_{\tau^{\perp}} \delta_q(\eta)_{\tau}. \quad (\dagger)$$

Whew! Okay, let's remember how the boundary map in sheaf cohomology works. Let $(\eta_{\sigma})$ be a cocycle for $H^q(X, Z^p)$. Since each $\sigma$ is contractible, we can lift each $\eta_{\sigma}$ to a $p-1$ form $\theta_{\sigma}$ with $d(\theta_{\sigma}) = \eta_{\sigma}$. Let $\tau$ be a $q+1$ face of our triangulation. Then $$\delta_q(\eta)_{\tau} = \sum_{\sigma \subset \tau} \pm \theta_{\sigma},$$ where the sign involves the relative orientation of $\sigma$ and $\tau$.

We want to show $(\dagger)$. Plugging in the above description of the Cech co-boundary, the right hand side is $$\sum_{\tau \in F_{q+1}} \int_{\tau^{\perp}} \sum_{\sigma \subset \tau} \pm \theta_{\sigma}.$$ Pulling the sum out of the integral and switching order of summation, we have $$\sum_{\sigma \in F_q} \sum_{\tau \supset \sigma} \int_{\tau^{\perp}} \pm \theta_{\sigma}.\quad (\heartsuit)$$

Now, the subdivisions $S$ and $S^{\perp}$ are dual, so $\tau \supset \sigma$ if and only if $\sigma^{\perp} \subset \tau^{\perp}$ or, in other words, $\tau^{\perp} \subset \partial(\sigma^{\perp})$. All the signs work out perfectly, so that $(\heartsuit)$ is $$ \sum_{\sigma \in F_q} \int_{\partial(\sigma^{\perp})} \theta_{\sigma}.$$ By Stokes' theorem, $$\int_{\partial(\sigma^{\perp})} \theta_{\sigma} = \int_{\sigma^{\perp}} d(\theta_{\sigma}) = \int_{\sigma^{\perp}} \eta_\sigma.$$ We have now recovered the left hand side of $(\dagger)$.

The OP only asked for top cohomology, but I think other cohomological degrees are similar. Once again, we have maps $$H^0(X, Z^k) \to H^1(X, Z^{k-1}) \to \cdots \to H^k(X, Z^0)$$ giving isomorphisms $$H^k_{DR}(X) \cong H^1(X, Z^{k-1}) \cong \cdots \cong H^k(X, Z^0) \cong H^k(X, \RR). \quad (\diamondsuit)$$ We'd like to know that a class $\omega$ in $H^k_{DR}(X)$ is represented by a class in $H^k(X, \ZZ)$ if and only if $\omega$ pairs to an integer against every integer chain in $H_k(X, \ZZ)$; it is enough to test against chains coming from the triangulation $S$. Let $c = \sum_{\rho \in F_k} c_{\rho} \rho$ be a $k$-chain. We want to how to pair all the spaces in $(\diamondsuit)$ against $c$. Let $\eta$ be a $q$-cocycle for $Z^{k-q}$. I believe the same argument as before shows that $\langle c, \eta \rangle$ is $$\sum_{\sigma \in F_q} c_{\rho} \sum_{\rho \in F_k} \int_{\sigma^{\perp} \cap \rho} \eta_{\sigma}. $$ In particular, if $q=k$, then $\sigma^{\perp} \cap \rho$ is a single point when $\rho = \sigma$ and otherwise $0$. So, viewing the Cech cohomology $H^k(X,\mathbb{R})$ as the cohomology of $$\RR^{F_0} \to \RR^{F_1} \to \cdots \RR^{F_{n-1}} \overset{d_{n-1}}{\longrightarrow} \RR^{F_n}$$ and the simplicial cohomology $H_k(X, \ZZ)$ as the homology of $$\ZZ^{F_0} \leftarrow \ZZ^{F_1} \leftarrow \cdots \leftarrow \ZZ^{F_n},$$ the pairing between $H^k(X,\mathbb{R})$ and $H_k(X, \ZZ)$ is induced by the obvious pairing between $\RR^{F_k}$ and $\ZZ^{F_k}$.

We then want to show that, if a cocycle in $\RR^{F_k}$ pairs integrally against all cycles in $\ZZ^{F_k}$, then that cocyle is cohomologous to one in $\ZZ^{F_k}$. That sounds like some easy linear algebra, although I don't see a one line proof.

  • $\begingroup$ Awesome answer! Just one little question. If we call $\psi$ the map sending an element in $\mathbb{R}^{F_n}$ to the sum of all the faces, it is obvious that $\mathbb{im} d_{n-1} \subset \ker \psi$, is the other inclusion also true? I guess the two are isomorphic, but are they equal? I haven’t been able to construct a preimage for $d_{n-1}$ given that the sum is 0. It is not obvious to me. $\endgroup$
    – G. Gallego
    Commented Aug 12, 2020 at 8:50
  • $\begingroup$ Ok, I guess page 41 in Munkres answers my question. $\endgroup$
    – G. Gallego
    Commented Aug 12, 2020 at 11:13
  • 1
    $\begingroup$ Glad you found an answer! Let me point out that you asked to avoid Poincare duality, but it was almost impossible to involve proving it: The same complex computes $H_k$ with respect to $S$ and $H^{n-k}$ with respect to $S^{\perp}$. So the Munkres reference talks about $H_0$, but the exact same computation comes up in your setting where you wanted to compute $H^n$. $\endgroup$ Commented Aug 13, 2020 at 14:47

I think I have an answer, although there is one step that still bugs me.

First one has to pick a triangulation of $S$. Let us call $V$, $E$ and $F$ the sets of vertices, edges and faces. By choosing now the open covering $\mathfrak{U}$ associated to this triangulation (i.e. the one given by the "stars" of the vertices, see p. 42 at Griffiths-Harris), we get that, while the open sets are in bijection with the vertices, the double intersections correspond to the edges and the triple to the faces.

Thus, in this case the Cech complex looks like this

$$ \mathbb{R}^{|V|} \overset{d_1}{\longrightarrow} \mathbb{R}^{|E|} \overset{d_2}{\longrightarrow} \mathbb{R}^{|F|}, $$

where $d_1( (k_v)_{v \in V} ) = (k_{v} - k_{w})_{vw\in E}$ and $d_2( (k_{vw})_{vw\in E} ) = (k_{vw} + k_{wu} - k_{vu})_{vwu \in F} .$

By carefully writing the matrix of $d_2$ one gets that $\mathrm{coker} d_2 \cong \mathbb{R}$. On the other hand, if one defines $\psi: \mathbb{R}^{|F|} \rightarrow \mathbb{R}$ in such a way that

$$ \psi( (k_{vwu})_{vwu \in F} ) = \sum_{vwu \in F} k_{vwu}, $$

one checks that $\mathrm{im} d_2 \subset \ker \psi$ and $\ker \psi \neq \mathbb{R}^{|F|}$, so $\ker \psi = \mathrm{im} d_2$. What this implies is that $\psi$ gives an isomorphism

$$ \psi: H^2(\mathfrak{U},\underline{\mathbb{R}}) \longrightarrow \mathbb{R}. $$

The key now is to find an explicit isomorphism

$$ I: H^2(S,\mathbb{R}) \longrightarrow H^2(\mathfrak{U}, \underline{\mathbb{R}}) $$

so that $\psi(I(a)) = \int_S a$.

This would show that the $a$ with $\int_S a \in \mathbb{Z}$ are precisely those that $I(a) \in H^2(\mathfrak{U}, \underline{\mathbb{Z}})$.

I think that now the way to construct explicitly the isomorphism $I$ is like in Woodhouse's "Geometric Quantization" (A6) and also in Weil's paper "Sur les theoremes de de Rham". On one hand, the way one gets a cocycle from a $2$-form is like in the statement of my question. On the other hand, the way to recover $\omega$ from the $f_{UVW}$ is by defining

$$ \omega_f = \sum_{U,V,W} f_{UVW} h_{W} dh_U \wedge dh_V, $$

where the $h_U$ are a partition on unity on the covering $\mathfrak{U}$. In Woodhouse it is shown that $[\omega_f]=[\omega]$.

However, I still do not understand why (or ever if it is true that) $\int_S \omega_f = \sum_{U,V,W} f_{UVW}$.

(BTW, also check out this question: https://mathoverflow.net/questions/329836/integral-of-top-forms-in-terms-of-Čech-representative ).

  • $\begingroup$ I'd think using Stokes' patchwise repeatedly should work, no? There is also the question of normalisation which will be fixed by the same calculation. $\endgroup$ Commented Aug 11, 2020 at 11:21
  • $\begingroup$ @AlexArvanitakis Well, I think that should be the idea, but to be honest I don’t know how to do it exactly... $\endgroup$
    – G. Gallego
    Commented Aug 11, 2020 at 11:45
  • $\begingroup$ I think it's a bit fiddly in general (in cases your cover is misbehaving). Try doing a sphere first with the obvious cover (with exactly three (EDIT: two + the overlap) open sets) $\endgroup$ Commented Aug 11, 2020 at 12:10
  • $\begingroup$ Well, in my case is for a triangulation, which is fairly simple. I am trying it for a tetrahedron with no success. The cover you suggest has no triple intersections and thus no 2-cocycles. $\endgroup$
    – G. Gallego
    Commented Aug 11, 2020 at 13:49
  • $\begingroup$ indeed actually that's a bad one--noncontractible, sorry (so in fact $\omega$ would not be exact on each open set). You can get a contractible one with, I count, 4 open sets $\endgroup$ Commented Aug 11, 2020 at 14:28

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