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 $\lambda\in \Lambda$ ?


  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 $$\cdots \to \mathcal H^k(M) \to \mathcal H^k(U_1) \oplus \mathcal H^k(U_2) \to \mathcal H^k(U_1\cap U_2)\to \cdots$$

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

  • $\begingroup$ 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! $\endgroup$
    – M.G.
    May 12 at 12:08
  • 1
    $\begingroup$ @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. $\endgroup$ May 12 at 14:24
  • 1
    $\begingroup$ 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... $\endgroup$ May 12 at 14:24
  • $\begingroup$ 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! $\endgroup$
    – M.G.
    May 12 at 14:48

4 Answers 4



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.

  • 2
    $\begingroup$ 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. $\endgroup$
    – Mark Grant
    May 12 at 14:58
  • 4
    $\begingroup$ @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. $\endgroup$
    – Will Sawin
    May 12 at 15:11
  • $\begingroup$ Thank you, dear Will: this is, as always with you, an excellent answer. $\endgroup$ May 13 at 16:51
  • 3
    $\begingroup$ @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. $\endgroup$ 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
My contribution to this answer is: zero, nada, zilch, que dalle...


Here is the great answer given by another of my brilliant friends:
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 $[\omega _0]\neq 0\in \mathcal H^1(U_0)$.
Here too the answer is entirely due to my geometer friend.


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