Sheaf cohomology with compact supports (and Verdier duality?)

Consider a manifold and a complex where cochains are sections of vector bundles and coboundary maps are differential operators, which are locally exact except in lowest degree (think de Rham complex). I'd like to know the relationship between the cohomology of this complex and the cohomology of the formal adjoint complex with compact supports (for the de Rham complex, this is again the de Rham complex, but with compact supports, and the relationship is given by Poincaré duality).

Update: Just added a bounty to raise the question's profile. The biggest obstacle, as came out of the discussion on an unsuccessful previous answer, to a straightforward application of Verdier duality is that it's hard to see how to connect the dual sheaf $\mathcal{V}^\vee$ with the sections of the dual density vector bundles $\Gamma(\tilde{V}^{\bullet*})$. The basic construction of $\mathcal{V}^\vee$ requires, for an open $U\subset M$, the assignment $U\mapsto \mathrm{Hom}_\mathbb{Z}(\Gamma(U,V^\bullet),\mathbb{Z})$, where $\mathrm{Hom}_\mathbb{Z}$ is taken in the category of abelian groups, which is MUCH bigger than $\Gamma(U,V^{\bullet*})$ itself.

Let me be more explicit, which unfortunately requires some notation. Let $M$ be the manifold, $V^i\to M$ be the vector bundles (non-zero for only finitely many $i$) and $d^i \colon \Gamma(V^i) \to \Gamma(V^{i+1})$ be the coboundary maps. Then $H^i(\Gamma(V^\bullet),d^\bullet) = \ker d^i/\operatorname{im} d_{i-1}$. By local exactness I mean that for every point $x\in M$ there exists an open neighborhood $U_x$ such that $H^i(\Gamma(V^\bullet|_{U_x}), d^\bullet) = 0$ for all except the smallest non-trivial $i$. Now, for each vector bundle $V\to M$, I can define a densitized dual bundle $\tilde{V}^* = V^*\otimes_M \Lambda^{\dim M} T^*M$, which is just the dual bundle $V^*$ tensored with the bundle of volume forms (aka densities). For any differential operator $d\colon \Gamma(V) \to \Gamma(W)$ between vector bundles $V$ and $W$ over $M$, I can define its formal adjoint $d^*\colon \Gamma(\tilde{W}^*) \to \Gamma(\tilde{V}^*)$, locally, by using integration by parts in local coordinates or, globally, by requiring that there exist a bidifferential operator $g$ such that $w\cdot d[v] - d^*[w]\cdot v = \mathrm{d} g[w,v]$. Thus, the formal adjoint complex is defined by the coboundary maps $d^{i*}\colon \Gamma(\tilde{V}^{(i+1)*}) \to \Gamma(\tilde{V}^{i*})$.

There is a natural, non-degenerate, bilinear pairing $\langle u, v \rangle = \int_M u\cdot v$ for $v\in \Gamma(V)$ and $u\in \Gamma_c(\tilde{V}^*)$, where subcript $c$ refers to compactly supported sections. Because $\langle u^{i+1}, d^i v^i \rangle = \langle d^{i*} u^{i+1}, v^i \rangle$ this paring descends to a bilinear pairing in cohomology $$\langle-,-\rangle\colon H^i(\Gamma_c(\tilde{V}^{\bullet*}),d^{(\bullet-1)*}) \times H^i(\Gamma(V^\bullet),d^i) \to \mathbb{R} .$$

Finally, my question can be boiled down to the following: is this pairing non-degenerate (and if not what is its rank)?

As I mentioned in my first paragraph, the case $V^i = \Lambda^i T^*M$ with $d^i$ the de Rham differential is well known. Its formal adjoint complex is isomorphic to the de Rham complex itself. Essentially, Poincaré duality states that the natural pairing in cohomology is non-degenerate. I am hoping that a more general result can be deduced from Verdier duality applied to the sheaf $\mathcal{V}$ resolved by the complex $(\Gamma(V^\bullet),d^\bullet)$. I know that the sheaf cohomology $H^i(M,\mathcal{V})$ can be identified with $H^i(\Gamma(V^\bullet),d^\bullet)$. I also know that the abstract form of the duality states that the algebraic dual $H^i(M,\mathcal{V})^*$ is given by the sheaf cohomology with compact supports $H^i_c(M,\mathcal{V}^\vee)$ with coefficients in the "dualizing sheaf" $\mathcal{V}^\vee$. Unfortunately, I'm having trouble extracting the relationship between $\mathcal{V}^\vee$ and my formal adjoint complex $(\Gamma_c(\tilde{V}^{\bullet*}), d^{(\bullet-1)*})$ from standard references (e.g., the books of Iversen or Kashiwara and Schapira).

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The de Rham complex, critically, is not locally exact in degree 0 - consider the constant functions! Do you wish your complexes to fail to be locally exact in some degree? Otherwise, the complex is $0$ in the derived category of sheaves, and the cohomology vanishes. I think you mean to consider complexes that fail to be locally exact. The failure of local exactness is measured by $\operatorname{ker} d^i/\operatorname{im} d^{i+1}$ in the category of sheaves. I think your best hope is to use Verdier duality for that sheaf, not the whole complex. – Will Sawin Sep 20 '13 at 3:55
Will, yes, that was a bit sloppy. Like in the de Rham case, my complexes are expected to fail to be locally exact in the lowest non-trivial degree. That is, they provide a fine resolution of some sheaf (like locally constant functions in the de Rham case). I'm curious about your remark. Could you expand on how to apply Verdier duality to the resolved sheaf, and then resolve the Verdier dual sheaf itself using vector bundles? – Igor Khavkine Sep 20 '13 at 8:01
If I had a complete solution, I would have posted it as an answer. But I'll think about it and see if I can get it to work. – Will Sawin Sep 20 '13 at 15:11

You will run into some issues with differential equations with singularities.

Consider the differential operator $x\frac{d}{dx} -t$ from the trivial rank $1$ vector bundle on $\mathbb R$ to itself, for some constant $t$. The adjoint map is $-xd/dx -1 - t$, which is another operator in the same class.

If $t$ is not a nonnegative integer, then this complex is locally exact. We have to check that the differential equation $xdy/dx - ty =f(x)$ has solutions for smooth $f$. We will check this for $t<0$, but I think it is also true when $t$ is not a nonnegative integer. A solution is:

$$y = \frac{ f(0) }{t} + x^t \int_{0}^x \left(f(z)-f(0)\right) z^{-1-t} dz$$

This gives a global solution also, so it is globally exact. Moreover, since any solution to the differential equation is a multiple of $x^{t}$, if $t$ is not a nonnegative integer, there are no nonzero solutions, so the regular cohomology is trivial.

If $t$ is a negative integer, then the dual complex will have nontrivial $H^0$ due to the nonzero solutions, otherwise, say for $-1<t<0$, we can easily show that the dual complex, with $t$ also in that range, will have nontrivial compactly supported $H^1$. Indeed, since there are no solutions to the homogeneous version of the differential equation, our solution of the inhomogeneous version is unique, and we can easily find a compactly supported $f(x)$ where the unique solution $y$ is not compactly supported.

On the other hand, suppose we have a smoothness condition - specifically, that the kernel of the first map is a locally constant sheaf. In other words, a local solution to the differential equations that define the first map can be extended uniquely along any path, with possible monodromy.

Given a differential equation, a common trick is to add enough extra functions to make the equation first order. We can just as easily do this with a complex of vector bundles with differential equation operators - add variables to each map in the reverse order. This process is a homotopy equivalence of complexes, as is its dual.

Take the locally constant kernel sheaf, view it as a vector bundle iwth flat connection, and tensor it with the de Rham complex. We will build a map from this complex to the original one. This is plausible because they are both injective resolutions of the same thing, but we need to check it can be done with vector bundle maps. This is trivial in degree $0$. If we have built maps for the first $n$ degrees, we compose the $n$th map with $d$ and get a first-order function on $\Omega^{n-1}$ that vanishes, locally, on the image of anything from $\Omega^{n-2}$. Such a function, by the linear algebra of differential forms at a single point, depends only on $d$ of the form on $\Omega_{n}$.

This bundle map is a quasi-isomorphism of sheaf complexes. If we can check that its dual is also a quasi-isomorphism, we win - duality in an arbitrary locally free complex can be reduced to duality for the de Rham complex. By using mapping cones, it is sufficient to check that if a first-order complex is locally exact, its dual is also locally exact.

Let $V_0 \to V_1 \to \dots V_n$ be a locally exact first-order complex. We will actually be able to find a homotopy to $0$. $d: V_0 \to V_1$ is a first-order differential equation with no local solutions. If it has no solutions,it must have a formal reason. Specifically, if $f_1,..f_k$ are local coordinates for $V_0$, then by taking linear combinations of the differential equations, their derivatives, and the commutation relations,we must be able to obtain $f_1,\dots,f_k$. Otherwise we could solve it along curves and extend consistently to the whole space.

But this linear combination just gives an operator $k: V_1 \to V_0$ such that $kd$ is the identity. Now we have a differential equation $k\oplus d$ on $V_1$, still linear, that has no solutions. Repeating the process, we eventually get a homotopy between the bundle and $0$. Applying this homotopy to the dual, it will be locally exact as well.

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Are you saying that such examples prevent the non-degenerate pairing between sections of the vector bundles of the original complex and its dual from descending, in general, to a non-degenerate pairing on the cohomologies? I'd still like to know how to prove that the pairing is non-degenerate on the cohomologies at least in a few simple cases where it is true. – Igor Khavkine Sep 21 '13 at 1:57
Yes - because it can prevent the cohomology groups from having the same rank. I have an idea for one case that I will try to write up soon. – Will Sawin Sep 21 '13 at 2:02