show/hide this revision's text 2 Added hypothesis to deal with dim V > 1 case correctly.

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. We This can prevent this be prevented by demanding that any every point in $\partial M$ has a an "elementary" neighbourhood where $U \cong \mathbb{H}^{n}$ such that

  • the division subdivision of $\partial U$ into faces looks like is diffeomorphic to a small neighbourhood in the boundary complete fan (a subdivision of an $n$-simplex (in other words\mathbb{R}^{n-1}$ into simplicial cones),
  • $V$ has a basis $\{e_i\}$ such that for each face $F$ meeting $U$,$V_F$ is spanned by a subset,
  • for each $e_i$, the product interior in $U$ of the boundary union of an open $k$-simplex and the faces $\mathbb{R}^{n-k}$ for some F$ such that $k$); e.ge_i \not\in V_F$ is connected.only 3 faces are allowed to meet at an edge

    • Essentially, 1. says that the subdivision of codimension 2 in $\partial M$ (while is sensible, 3. prevents situations like in the example 4 faces meet a vertex)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 every point in $\partial M$ has possesses such a "simplex boundary" neighbourhood for the subdivisionelementary" neighbourhoods, thenwhere 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 $\mathbb{R}^n$ or and for the half-space $\mathbb{H}^n$."elementary" neighbourhoods.

    For $U \cong \mathbb{R}^n$ this is just usual Poincaré duality tensored with $V$. For $U \cong \mathbb{H}^n$, we can restrict to the case where $\partial U$ is a "simplex boundaryan "elementary" neighbourhood for the subdivision of $\partial M$. If we let $V_{\partial U}$ denote the intersection of $V_F$ over all faces $F$ meeting $\partial U$, then U$,$H^0_{V, $H^k_{V, \{V_F\}}(U) = \cong V_{\partial U}$ while integration gives a well-defined map $H^n_{c, V^*bigoplus_i H^{k}_{V_i, \{ \text{ann} V_F\}}(U)^* \to V^*/\text{ann}V_{\partial U} {V_F \cong V_{\partial U}^*$, which is an isomorphism. Checking that the other cohomology groups (with boundary conditionscap V_i\}}(U) of $U$ vanish is easy when $\dim V = 1$ (for $H^{n-1}_{c, $H^k_{c, V^*, \{ {\text{ann} V_F\}}(U) = \text{ann} V_F\}}(U)$ we use that the interior of the union of the faces bigoplus_i H^{k}_{c, V_i^*, \{\text{ann} (V_F \cap V_i) \}}(U), $F$ in $\partial U$ with where $V_F = 0$ is connected, which V_i$ is where the "simplex boundary" condition comes in), and span of the clearest way to prove it general seems to be by induction on element $\dim V$. For the induction step, use e_i$ of the exact sequence basis from condition 2.The terms on the short exact sequence$right hand side all vanish, except that if $0 e_i \to\Omega^*_{W, in V_F$ for all $F$ meeting $\partial U$ then $H^0_{V_i, \{ V_F {V_F \cap W\}}(U) \to \Omega^*_{V, \{ V_F\}V_i\}} \to \Omega^*_{V/W, cong V_i$ and $H^{n}_{c, V_i^*, \{ V_F/W {\text{ann} (V_F \}}(U) cap V_i)\}}(U) \cong V_i^*$(3. is used to 0$$where show that $W H^{n-1}_{c, V_i^*, \subset V$ is any subspace {\text{ann} (and V_F \cap V_i)\}}(U) = 0$). So the analogous sequence duality holds for complexes with compact support)the "elementary" neighbourhoods.
    show/hide this revision's text 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. We can prevent this by demanding that any point in $\partial M$ has a neighbourhood where the division into faces looks like a small neighbourhood in the boundary of an $n$-simplex (in other words, the product of the boundary of an open $k$-simplex and $\mathbb{R}^{n-k}$ for some $k$); e.g. only 3 faces are allowed to meet at an edge of codimension 2 in $\partial M$ (while in the example 4 faces meet a vertex).

    I think that if $M^n$ is oriented with boundary and every point in $\partial M$ has such a "simplex boundary" neighbourhood for the subdivision, 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 $\mathbb{R}^n$ or the half-space $\mathbb{H}^n$.

    For $U \cong \mathbb{R}^n$ this is just usual Poincaré duality tensored with $V$. For $U \cong \mathbb{H}^n$, we can restrict to the case where $\partial U$ is a "simplex boundary" neighbourhood for the subdivision of $\partial M$. If we let $V_{\partial U}$ denote the intersection of $V_F$ over all faces $F$ meeting $\partial U$, then $H^0_{V, \{V_F\}}(U) \cong V_{\partial U}$ while integration gives a well-defined map $H^n_{c, V^*, \{ \text{ann} V_F\}}(U)^* \to V^*/\text{ann}V_{\partial U} \cong V_{\partial U}^*$, which is an isomorphism. Checking that the other cohomology groups (with boundary conditions) of $U$ vanish is easy when $\dim V = 1$ (for $H^{n-1}_{c, V^*, \{ \text{ann} V_F\}}(U)$ we use that the interior of the union of the faces $F$ in $\partial U$ with $V_F = 0$ is connected, which is where the "simplex boundary" condition comes in), and the clearest way to prove it general seems to be by induction on $\dim V$. For the induction step, use the exact sequence from the short exact sequence $$ 0 \to\Omega^*_{W, \{ V_F \cap W\}}(U) \to \Omega^*_{V, \{ V_F\}} \to \Omega^*_{V/W, \{ V_F/W \}}(U) \to 0$$ where $W \subset V$ is any subspace (and the analogous sequence for complexes with compact support).