# How does Pontryagin duality fit into the general cohomology theory framework?

Pontryagin duality implies the isomorphic relation of the function space $C(G)$ on a locally compact group $G$ to the function space on it's dual group $\hat G \overset{\sim}{=}\text{Hom}(G,T)$, where $T$ is the circle group. (This isomorphism is the generalization of the Fourier transform in the case $G=\mathbb R$ with translations $t\mapsto t+\Delta t$, and $\hat G \overset{\sim}{=}\{\omega|t\mapsto\text{e}^{2\pi i\ \omega\cdot t}\} \overset{\sim}{=}\mathbb{R}$.)

I read here, that this result is of relevant in the history of cohomology theory.

But I don't really know how to fit it into the picture. For the duality result, you seem to only need the definiton of the group, the rest follows from natural constructions, while the cohomology thoeries I'm aware of seem to be more general. They have a $\text d$-operator, which I miss in the topological group theorem case, and there also is a base space in the latter case.

Is maybe $G$ to be taken as base space? Or is $G$ just to be seen as the fibre object without a complicated base? The second idea arises becuase $\text{Hom}(G,T)$ maps groups into the small circle group. In a way this is anologous to the cotangent space which eats vectors and maps to the reals or the complex numbers. Is there a relation only in as far as people in the 30's dicovered the concept of a character is a relevant one. The cohomology results don't seem to care directly about the function space $C$ on this object.

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## 1 Answer

I think all that is meant is that Poincare duality on a compact, smooth, oriented $n$-manifold $M$ can be phrased as saying that the groups $H^i(M,\mathbb Z)$ and $H^{n-i}(M, \mathbb R/\mathbb Z)$ are naturally Pontrjagin dual to one another. This is a way of phrasing Pontrjagin duality for integer valued comohology that doesn't require mentioning Tor or Ext functors, which would be required for the version of Pontrjagin duality where you compare $H^i(M,\mathbb Z)$ and $H^{n-i}(M,\mathbb Z)$.

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