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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.