# Poincaré duality for currents and non-closed forms

In page 8 of Quantization of Higher Abelian Gauge Theory in Generalized Differential Cohomology by Szabo, the author claims that Poincaré duality holds for non-closed forms as long as the other form (pairing with the former one) is distributional, i.e., it's a de Rham current.

Let me be more precise. Let $X$ be a manifold of dimension $n$. Given a de Rham current of compact support $j_e \in H^m_{dR, c} (X)$ that is exact in $H^m_{dR} (X)$ (I don't think this exactness part is relevant, anyway …), the author claims that there exists a homology class $[W_e] \in H_{m - n} (X, \mathbb{R})$ such that $$\int_X a \wedge j_e = \int_{W_e} a$$ for every $a \in \Omega^{n - m} (X)$.

This fact is indeed need in the paper I mentioned since in page 6 he uses this property (without mentioning it previously) to deduce Dirac charge quantization condition from $\exp (-\int_X A\wedge j_e) = \exp (-\int_{W_e} A)$ for possibly non-closed $A$.

I would like a proof or a reference for this fact about currents (mentioned above)

## 1 Answer

The standard reference is de Rham's Differentiable Manifolds, which develops the whole theory in terms of currents. But more quickly, see the beginning of Section 7.3 of Nicolaescu's notes http://www3.nd.edu/~lnicolae/Lectures.pdf

A current is a functional, so it must eat differential forms and spew out numbers. This integral $\int_X a\wedge j_e$ is nothing but notation, to represent $\int_{W_e} a$ (here we implicitly pull-back the form $a$ to $W_e$ and use either compactly-supported forms or characteristic functions). See Chapter III of de Rham's book. Now when $j_e$ is a form itself, then the wedge product (and integral) are defined as usual.

• Thanks for your answer. Apparently you're referring to Poincaré duality of currents as stated in mathoverflow.net/questions/201580/… . The problem is that this only holds for closed $a$, i.e., for a cohomology class. My question concerns the non-closed case, which appears in the paper mentioned above whenever the base space in non-contractible. – user40276 May 26 '16 at 2:46
• That's the result of reading a physics paper (which uses math incorrectly and vaguely). – Chris Gerig May 26 '16 at 5:58