[Math] Poincaré duality with boundary conditions

at.algebraic-topologydg.differential-geometry

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 $\dim\partial M$).

Best Answer

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.

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