If $M$ is a compact oriented manifold with boundary then by Poincaré duality the cohomology of $\Omega(M)$ (de Rham cohomology of $M$) is dual to the cohomology of $\Omega_0(M)$, where $\Omega_0(M)$ denotes differential forms vanishing on $\partial M$. This question is about a generalization of this fact to more complicated boundary conditions.

Suppose $M$ is a compact oriented $C^\infty$ manifold with corners. Its boundary is decomposed (by the corners) to faces (of dimension $\dim M -1$).

Let $V$ be a finite-dimensional vector space, and let us choose for every face $F\subset\partial M$ a subspace $V_F\subset V$. Let us consider the complex (of $V$-valued differential forms with boundary conditions given by $V_F$'s) $$\Omega(M)_{V,\{V_F\}}= \{\alpha\in\Omega(M)\otimes V;\quad \alpha|_F\in\Omega(F)\otimes V_F\text{ for all faces }F\}\subset\Omega(M)\otimes V.$$

The "naive dual" of $\Omega(M)_{V,\{V_F\}}$ is $\Omega(M)_{V^*,\{\text{ann} V_F\}}$ with the pairing given by integration over $M$ ($\text{ann} V_F\subset V^*$ denotes the annihilator of $V_F$).

Is there a condition under which the pairing between the cohomologies of $\Omega(M)_{V,\{V_F\}}$ and of $\Omega(M)_{V^*,\{\text{ann} V_F\}}$ is perfect? What is the reason for the fact that the pairing is not always perfect?

Some remarks:

  • The "dual complex" $\Omega(M)_{V^*,\{\text{ann} V_F\}}$ can be described as $$\{\alpha\in\Omega(M)\otimes V^*;\quad \int_M\langle\alpha\wedge d\beta\rangle= (-1)^{\deg\alpha +1}\int_M\langle d\alpha\wedge \beta\rangle\quad \forall\beta\in \Omega(M)_{V,\{V_F\}} \}.$$

  • If $V_F=0$ for all $F$'s then we get the standard Poincaré duality for manifolds with boundary.

  • It is possible that manifolds with corners is not the right picture; the question might be about any "reasonable" division of $\partial M$ into "faces" (of dimension $\dim\partial M$).

  • $\begingroup$ You write: "If $V_F = 0$ for all $F's$ we get the standard Poincare duality for manifolds with boundary." But what is the role of $V$ in that statement? $\endgroup$
    – John Klein
    Feb 14, 2011 at 21:16
  • $\begingroup$ @John Klein: Then $V$ is not important. One cohomology is de Rham's tensored with $V^*$ and the other one is (isomorphic to) de Rham's with compact support in the interior of $M$ tensored with $V$. We can safely put $V=\mathbb R$ in this case. More-dimensional $V$ is only needed for those more complex boundary conditions. For $V=\mathbb{R}$ we can only have $V_F=0$ or $V_F=\mathbb{R}$. In that case, if say $\partial M$ is divided by a hypersurface to two faces, one with $V_F=0$ and the other with $V_F=\mathbb{R}$, then we do get perfect pairing (at least I hope :) $\endgroup$
    – Pavol S.
    Feb 14, 2011 at 21:58
  • $\begingroup$ I guess I don't really have a feel for your question. What is the relation between $V$ and the geometry of the stratification of the manifold? What if $V_F = V$ for all faces $F$? How did this question arise? $\endgroup$
    – John Klein
    Feb 18, 2011 at 3:44
  • $\begingroup$ If $V_F=V$ for all $F$'s we still get de Rham cohomology tensored with $V$, and de Rham with compact support in the interior tensored with $V^*$, so the duality holds. The motivation for the question was from symplectic form on the moduli space of flat connections on a surface, with various boundary conditions, and Lagrangian subspaces coming from cobordisms. $\endgroup$
    – Pavol S.
    Feb 22, 2011 at 14:12

1 Answer 1


An example where the duality fails is when $M^n$ is the closed unit ball $B^3 \subset \mathbb{R}^3$, and its boundary $S^2$ is divided into four quarters by 2 great circles. If $V = \mathbb{R}$, $V_F = V$ for 2 opposite quarters $F$ and $V_F = 0$ for the other two, then $H^1_{V, \{ V_F \}}(M) = 0$ while $H^2_{V^*, \{\text{ann} V_F\}} \cong \mathbb{R}$ (essentially, they are $H^1_c$ and $H^2_c$, respectively, of the product of an open 2-disc and a closed interval).

In a sense, the reason that the duality fails is that near the intersection of the two great circles, the set of boundary points where the forms are allowed to be non-zero is disconnected, and that no matter how small a neighbourhood we choose in $B^3$ for the intersection point, its cohomology will therefore not be entirely elementary. This can be prevented by demanding that every point in $\partial M$ has an "elementary" neighbourhood $U \cong \mathbb{H}^{n}$ such that

  1. the subdivision of $\partial U$ into faces is diffeomorphic to a complete fan (a subdivision of $\mathbb{R}^{n-1}$ into simplicial cones),
  2. $V$ has a basis $\{e_i\}$ such that for each face $F$ meeting $U$, $V_F$ is spanned by a subset,
  3. for each $e_i$, the interior in $U$ of the union of the faces $F$ such that $e_i \not\in V_F$ is connected.

Essentially, 1. says that the subdivision of $\partial M$ is sensible, 3. prevents situations like in the example above, and 2. makes sure we can state 3. sensibly when $\dim V > 1$ (see example in Trial's comment below). I think that if $M^n$ is oriented with boundary and possesses such "elementary" neighbourhoods, then $$H^k_{V, \{V_F\}}(M) \cong H^{n-k}_{c, V^*, \{ \text{ann} V_F\}}(M)^*$$ where the subscript $c$ indicates the cohomology of a complex with compact supports. It should be possible to prove this using induction on a good cover (and the duality between the Mayer-Vietoris sequences for normal and compactly supported de Rham cohomology) like for standard Poincaré duality, provided that the statement is true for open subsets $U \subset M$ diffeomorphic to $\mathbb{R}^n$ and for the "elementary" neighbourhoods.

For $U \cong \mathbb{R}^n$ this is just usual Poincaré duality tensored with $V$. For an "elementary" neighbourhood $U$, $$H^k_{V, \{V_F\}}(U) = \bigoplus_i H^{k}_{V_i, \{V_F \cap V_i\}}(U) $$ $$H^k_{c, V^*, \{\text{ann} V_F\}}(U) = \bigoplus_i H^{k}_{c, V_i^*, \{\text{ann} (V_F \cap V_i) \}}(U), $$ where $V_i$ is the span of the element $e_i$ of the basis from condition 2. The terms on the right hand side all vanish, except that if $e_i \in V_F$ for all $F$ meeting $\partial U$ then $H^0_{V_i, \{V_F \cap V_i\}} \cong V_i$ and $H^{n}_{c, V_i^*, \{\text{ann} (V_F \cap V_i)\}}(U) \cong V_i^*$ (3. is used to show that $H^{n-1}_{c, V_i^*, \{\text{ann} (V_F \cap V_i)\}}(U) = 0$). So the duality holds for the "elementary" neighbourhoods.

  • $\begingroup$ I think this is a counterexample even though your condition is satisfied (if I understood it correctly): $M$ is tetrahedron, $\dim V =2$, we choose 3 different 1-dim subspaces of $V$ as $V_F$'s for 3 of the faces, and for the 4th $F$ we put (e.g.) $V_F=V$. $\endgroup$
    – Pavol S.
    Feb 22, 2011 at 14:05
  • $\begingroup$ I think I see the error in my argument. The short sequence in the induction on $\dim V$ need not be exact, because elements of the image of $\Omega^*_{V, \{ V_F\}} \to \Omega^*_{V/W, \{ V_F/W \}}(U)$ may be forced to vanish along an edge between faces. This possibility could be excluded by requiring that for any edge $E$ (of any codimension), $\{ V_F : F \text{ meets } E \}$ must look like a set of coordinate planes (i.e. $V$ has a basis such that each $V_F$ is spanned by a subset) and taking $W$ to be a coordinate axis. (The hypothesis is trivial for codimension 1 edges). $\endgroup$ Feb 23, 2011 at 22:35

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