Why the circle for Pontryagin duality? [duplicate]

Question

For a locally compact group $G$, we define the Pontryagin dual as $\hat G = Hom(G,\mathbb T)$ where $\mathbb T$ is the circle group and the homomorphisms are continuous group maps. This duality has a lot of nice properties and shows up all over the place so it is probably the right definition.

However, without the benefit of hindsight, why would one choose to define the dual of a group with respect to $\mathbb T$ (instead of some other locally compact group, say). Is there a reason to promote $\mathbb T$ to a special place among all locally compact groups?

A little more broadly, are there other groups $H$ that also lead to a good theory of duality if we define $\hat G_H = Hom(G,H)$?

One possible answer would be to explin the historical context/necessity which led to the definition. But there can also be other motivations and I would be open to both kinds of answers.

On a closely related note, one can also ask a similar question for the Cartier duality in algebraic geometry. Probably, the two will have similar answers.

Answer 1

The group $\mathbb T$ is special. On this site and math.stackexchange the same question has been asked before.

See the second question at When does Pontryagin duality generalize? and Why unitary characters for the dual group in Pontryagin duality if $G$ is not compact? for an explanation.

Answer 2

This is a long comment rather than an answer. Also, I will be very sloppy about the details because I don't have a strong knowledge about them.

I think the fact that Pontryagin duality works with maps $\mathrm{Hom}(\,\_\,,\mathbb{T})$ where $\mathbb{T}=U(1)$ has to do with the following facts:

• Cartier duality, which is a very natural/reasonable thing to do, works with the multiplicative group, wich is $\mathbb{C}^\times$ over the complex numbers.
• $\mathbb{T}\subset\mathbb{C}^\times$ is the compact real form of the reductive group $\mathbb{C}^\times$, and the representation theory of a complex reductive algebraic group is "the same" as that of its compact real form.

[Important edit: as the comment by Yemon Choi reminded me, in what follows the algebraic group $G$ has to be diagonalizable, or at least a version of Cartier duality that I'm aware of is stated for such group schemes in Waterhouse Introduction to affine group schemes. This is very restrictive!]

So what's Cartier duality? Working over $\mathbb{C}$, there is an involution on the category $\mathbf{CHopf}$ of commutative Hopf algebras $$\mathbf{CHopf}\to\mathbf{CHopf}$$ $$A:=(V,m,\Delta)\mapsto A^\vee:=(V^\vee,\Delta^\vee,m^\vee)$$

sending a Hopf algebra $A$ with underlying $\mathbb{C}$-vector space $V$, multiplication $m$ and comultiplication $\Delta$ (and other data that I'm suppressing from the notation) to the Hopf algebra structure $A^\vee$ on the dual vector space $V^\vee=\mathrm{Hom}(V,\mathbb{C})$ with multiplication given by $\Delta^\vee$ and comultiplication by $m^\vee$. The category $\mathbf{aaGrp}$ of affine complex algebraic groups is opposite to the category $\mathbf{CHopf}$ via $A=(V,m,\Delta)\mapsto G:=\mathrm{Spec}(V,m)$ with group operation given by the morphism $\mathrm{Spec}(\Delta):G\times G\to G$. Vice versa, an algebraic group $(G,\mu)$ goes to $V=\mathcal{O}(G)$, the algebra of global functions on $G$, with its natural algebra structure, and comultiplication given by $\mu^{\sharp}:\mathcal{O}(G)\to\mathcal{O}(G\times G)=\mathcal{O}(G)\otimes\mathcal{O}(G)$.

Under this antiequivalence $\mathbf{CHopf}\to\mathbf{aaGrp}$, the involution $A\to A^\vee$ corresponds to $G\mapsto \mathbf{aaGrp}(G,\mathbb{C}^\times)$.

This is to say that "Pontryagin-dualizing" with respect to $\mathbb{T}$ somehow comes from the usual duality of $\mathbb{C}$-vector spaces.