# Can one glue De Rham cohomology classes on a differential manifolds?

Let $$M$$ be a differential manifold and $$\mathcal H^k$$ the presheaf of real vector spaces associating to the open subset $$U\subset M$$ the $$k$$-th de Rham cohomology vector space: $$\mathcal H^k(U)=H^k_{DR}(U)$$. Is this presheaf a sheaf?
Of course not! Indeed, given any non-zero cohomology class $$0\neq[\omega]\in \mathcal H^k(U)$$ represented by the closed $$k$$-form $$\omega\in \Omega^k_M(U)$$ there exists (by Poincaré's Lemma) a covering $$(U_i)_{i\in I}$$ of $$U$$ by open subsets $$U_i\subset U$$ such that $$[\omega]\vert U_i=[\omega\vert U_i]=0\in \mathcal H^k(U_i)$$, and thus the first axiom for a presheaf to be a sheaf is violated.
But what about the second axiom?
My question:
Suppose we are given a differential manifold M, a covering $$(U_\lambda)_{\lambda \in \Lambda}$$of $$M$$ by open subsets $$U_\lambda \subset M$$, closed differential $$k-$$forms $$\omega_\lambda \in \Omega^k_M(U_\lambda)$$ satisfying $$[\omega_\lambda]\vert U_\lambda \cap U_\mu=[\omega_\mu]\vert U_\lambda \cap U_\mu\in \mathcal H^k(U_\lambda\cap U\mu)$$ for all $$\lambda,\mu \in \Lambda$$.
Does there then exist a closed differential form $$\omega\in \Omega^k(M)$$ such that we have for the restrictions in cohomology: $$[\omega]\vert U_\lambda=[\omega _\lambda]\in \mathcal H^k(U_\lambda)$$ for all $$\lambda\in \Lambda$$ ?

Remarks

1. This is an extremely naïve question which, to my embarrassment, I cannot solve.
I have extensively browsed the literature and consulted some of my friends, all brilliant geometers (albeit not differential topologists), but they didn't know the answer offhand. For what it's worth, I would guess (but not conjecture!) that such glueing is impossible.
2. If the covering of $$X$$ has only two opens then we can glue.
This follows immediately from Mayer-Vietoris's long exact sequence $$\cdots \to \mathcal H^k(M) \to \mathcal H^k(U_1) \oplus \mathcal H^k(U_2) \to \mathcal H^k(U_1\cap U_2)\to \cdots$$

Update
My brilliant friends didn't answer offhand but a few hours later, unsurprisingly, they came back to me with splendid counterexamples! See below.

• Just a comment. Your first observation sounds like a good point to be included in introductions to derived categories. Incidentally, this also makes me wonder about a description/interpretation of the cohomology sheaves of the De Rham complex in the smooth category (i.e. the quotient sheaves of the respective kernel and image sheaves under $d$). I've actually never thought about it!
– M.G.
May 12 at 12:08
• @M.G. The Poincare lemma (ncatlab.org/nlab/show/Poincar%C3%A9+lemma) says that, on a smooth manifold, the de Rham complex of sheaves is exact, so the image sheaf of $d : \Omega^{q-1} \to \Omega^q$ is the same as the kernel of $d : \Omega^{q} \to \Omega^{q+1}$ and the cohomology sheaves are zero. Concretely, this image/kernel is the sheaf of closed $q$-forms: $Z^q(U)$ is the vector space of closed $q$-forms on $U$. It is easy to see this using the description as the kernel of $d : \Omega^{q} \to \Omega^{q+1}$, since kernel is the same for sheaves and presheaves. May 12 at 14:24
• Dear @M.G., amusingly one of the brilliant geometers I allude to in my question made a very similar comment. Great minds think alike! Unfortunatately I had to tell him, as I am telling you, that I have only a very rudimentary knowledge of derived categories... May 12 at 14:24
• Dear @DavidESpeyer, thanks for the very clear explanation! For whatever reasons I had never given much thought to the De Rham complex in terms of sheaves. I guess there is always a first!
– M.G.
May 12 at 14:48

No.

Make $$M$$ by gluing three strips to two discs to form a thrice-punctured sphere. Take three open sets $$U_\lambda$$, each made by both discs and two of the strips. Then each $$U_\lambda$$ is homeomorphic to annulus and thus has $$1$$-dimensional $$H^1$$.

The pairwise intersections, made from one strip connecting two discs, are contractible and so their $$H^1$$ vanishes. Thus, for $$k=1$$, the agreement condition on the pairwise intersections is vacuous.

If your claim held, then we could choose a $$1$$-form on $$M$$ that restricts to an arbitrary cohomology class on each of the three $$U_\lambda$$, making the first de Rham cohomology of $$M$$ at least three-dimensional. But in fact it is only two-dimensional. Instead, there is a relation where the integrals around three clockwise loops around the three punctures sum to $$0$$, because these loops form the boundary of a particular subset of $$M$$.

It's true if the intersections $$U_\lambda \cap U_\kappa \cap U_\mu$$ are empty for all distinct $$\lambda,\kappa,\mu$$, by iteratively applying the Mayer-Vietoris sequence or applying a single exact sequence in sheaf cohomology.

• I think it should also be true for good covers, by applying what Bott & Tu call the "Mayer-Vietoris principle" in Section 8 of their book Differential Forms in Algebraic Topology. May 12 at 14:58
• @MarkGrant The statement is trivially true if all open sets are contractible, as then the cohomology groups vanish so any closed global differential form does the trick, which is a bit weaker than being a good cover. May 12 at 15:11
• Thank you, dear Will: this is, as always with you, an excellent answer. May 13 at 16:51
• @Mark Grant: Indeed the Mayer-Vietoris theorem says exactly that for a covering with two open pieces, glueing is always possible. So Will's nice counter-example is the most economical possible. May 13 at 16:55

This answer provides a positive answer to a refinement of the original question.

Recall that two closed differential $$k$$-forms $$ω_0$$, $$ω_1$$ on a smooth manifold $$M$$ have the same de Rham cohomology class if and only if they are concordant, i.e., there is a closed differential $$k$$-form $$τ$$ on $$\def\R{{\bf R}} \R⨯M$$ such that the pullbacks of $$τ$$ to $$\{0\}⨯M$$ and $$\{1\}⨯M$$ are equal to $$ω_0$$ and $$ω_1$$ respectively.

Thus, the given data can be reformulated as a collection of closed differential forms on $$\{U_λ\}_{λ∈Λ}$$ whose restrictions to pairwise intersections $$U_λ∩U_μ$$ are concordant.

In order to get a good descent-type statement, we make two modifications that are standard in sheaf theory:

• We introduce the additional data of (a specific choice of) a concordance $$ω_{λ,μ}$$ between $$ω_λ$$ and $$ω_μ$$ on the open subset $$U_λ∩U_μ$$.

• More generally, for every $$(n+1)$$-tuple $$T$$ of indices in $$Λ$$ we introduce an $$n$$-dimensional concordance, given by a closed differential $$k$$-form $$ω_T$$ on $$Δ^n⨯(U_{T_0}∩⋯∩U_{T_n})$$, which must be compatible with forms assigned to various faces of $$T$$.

It is this type of data that can be glued together. In fact, a much more general statement is true, where the sheaf of closed differential $$k$$-forms is replaced by any simplicial presheaf on the site of smooth manifolds:

Theorem (Theorem 1.1 in arXiv:1912.10544):

Suppose $$F$$ is a presheaf of simplicial sets on the site of smooth manifolds and smooth maps of manifolds, equipped with the usual Grothendieck topology of open covers. Define the simplicial presheaf $$\def\B{{\rm B}} \B F$$ via the formula $$\def\op{{\rm op}} \def\hocolim{\mathop{\rm hocolim}} \B F(M) = \hocolim_{n∈Δ^\op} F(Δ^n⨯M).$$ If $$F$$ is an ∞-sheaf (i.e., satisfies the homotopy descent condition), then so is $$\B F$$. Furthermore, $$\B F$$ is representable by the space $$\B F(\R^0)$$: the canonical map $$\def\R{{\bf R}} \def\Map{\mathop{\rm Map}} \B F(M)→\R\Map(M,\B F(\R^0))$$ is a weak equivalence.

This implies the desired statement: the data of forms on $$U_λ$$ together with concordances on $$U_λ∩U_μ$$ etc., defines a Čech descent data for the simplicial presheaf $$\B F$$, where $$F$$ is the sheaf of closed differential $$n$$-forms. According to the above theorem, this descent property of $$\B F$$ allows us to glue this data to a single section of $$\B F$$ (and therefore of $$F$$) over $$M$$, as desired.

Taking other simplicial presheaves $$F$$ produces similar gluing statements for other geometric objects, e.g., principal $$G$$-bundles with connection, bundles $$d$$-gerbes with connection, etc.

In particular, we see that the original statement is true if all triple intersections $$U_λ∩U_μ∩U_ν$$ are empty, since in this case there are no higher concordances to choose.

Here is the solution obtained by one of my brilliant geometer friends evoked in the question:
Take $$X=S^2$$, the unit $$2$$-sphere with equation $$x_1^2+x_2 ^2+x_3^2=1$$, and cover it by the three open strips $$U_i=\{(x_1,x_2,x_3)\in S^2\vert \vert x_i \vert \leq \frac 35 \}$$.

1. The $$U_i$$'s do cover $$S^2$$: a point on the unit sphere can't have its three
coordinates $$\geq \frac 35$$ .
2. For all $$i\neq j$$ it we see, by projecting on the coordinate planes, that $$U_i\cap U_j$$ is the disjoint sum of two antipodal open spherical quadrilaterals homeomorphic to squares, so that $$\mathcal H^1(U_i\cap U_j)=0$$.
3. Each $$U_i$$ deformation retracts to its central great circle, so that each $$H_{DR}^1(U_i)=\mathbb Z$$.

It is then clear that given arbitrary nonzero De Rham cohomology classes De Rham $$0\neq [\omega_i]\in H_{DR}^1(U_i)$$ the glueing condition is vacuously satisfied because of 1.
Nevertheless these cohomolgy classes can't be glued to a cohomology class in $$S^2$$ since $$\mathcal H^1(S^2)=0$$.
Important remark
Let $$X$$ be $$\mathbb C, U_0$$ be the open complement in $$X$$ of the closed disk $$\bar D=\{z\in \mathbb C\vert \vert z\vert \leq1 \}$$ and add a few open discs $$U_1,\cdots, U_n$$ of radius $$\lt 1$$ covering $$\bar D$$ in order to obtain an open covering $$U_0,U_1,\cdots U_n$$ of $$X$$.
Now let $$[0]\neq[\omega _0]\in \mathcal H^1(U_0)\cong \mathbb Z$$ be a nonzero cohomology clas and define (no choice here!) $$0=[\omega_i]\in \mathcal H^1(U_i)=0$$.
The compatibility conditions are trivially satisfied since all intersections $$U_i\cap U_j (i\neq j)$$ are contractible, so that $$\mathcal H^1(U_i\cap U_j )=0$$.
Nevertheless we can't glue our cohomology classes $$[\omega_i]$$ to a global cohomology class $$[\omega] \in \mathcal H^1(X)$$ since the only global cohomology class on $$X$$ is $$0\in \mathcal H^1(X)=0$$, which does not restrict to $$[\omega _0]\neq 0\in \mathcal H^1(U_0)$$.