## Why is the dual of a torus the same as its fundamental group?

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The set of continuous homomorphisms from a torus ${\mathbb T}^n = ({\mathbb R}/{\mathbb Z})^n \to {\mathbb R}/{\mathbb Z}$ can be identified with ${\mathbb Z}^n$ if we assign to each $k = (k_1, \ldots k_n) \in {\mathbb Z}^n$ the character $x \mapsto k \cdot x$.

The fundamental group of homotopy classes of loops ${\mathbb R}/{\mathbb Z} \to {\mathbb T}$ can also be identified with ${\mathbb Z}^n$ because each equivalence class (with base point at the origin) can be represented by $x \mapsto (k_1 x, k_2 x, \ldots, k_n x)$ for some $k \in {\mathbb Z}^n$.

My question is basically how much of a coincidence this isomorphism is. For one thing, it can't be too natural because the fundamental group pushes forward under a map, whereas the character group pulls back. So the natural question should be whether there is a natural relationship at the level of the first cohomology group instead of the fundamental group.

Of course, totally disconnected groups have interesting dual groups even though their cohomology is uninteresting as far as I know. And ${\mathbb R}^n$, being isomorphic to its dual but contractible, does not seem to exhibit a similar relationship.

But for a Lie group, say, I would like to know if there's a natural relationship between its representation theory (e.g. irreducible representations) on the one hand and its topology (e.g. cohomology) on the other hand. It might be no deeper than "well, the cohomology groups are representations".

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I guess you may be referring to Pontryagin duality. My impression is that results on the representations of Lie groups by Harish-Chandra and others are an attempt to generalize Pontryagin duality to the non-commutative category. – Agol Feb 16 2011 at 21:22

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Re: "cohomology is uninteresting" see en.wikipedia.org/wiki/Group_cohomology – Sam Nead Feb 16 2011 at 21:55

## 3 Answers

The two are naturally dual lattices. The fundamental group of a torus $T$ can be canonically identified with the group (known as the cocharacter lattice) of $\it homomorphisms$ from the circle group to $T$, or equivalently the kernel of the (universal cover=exponential map) homomorphism from the Lie algebra $t$ to $T$. The characters of the torus on the other hand can be identified naturally with a subsest of $t^*$, i.e. characters of the Lie algebra which are integral on the kernel of $t\to T$, i.e. the dual to the cocharacters.

As for the general question about Lie groups it seems way too general (note that a connected group acts trivially on its cohomology groups, so the relation is not that..) One direction to read about is the relation between the cohomology of the group, that of its classifying space, group cohomology and invariant polynomials (or "Casimirs") on the Lie algebra.

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Thanks!  Very helpful. – Phil Isett Feb 16 2011 at 4:42
@David: Do you know a good place to read about those things and their relations? – ndkrempel Feb 18 2011 at 1:08

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I am not sure if this is something you're looking for:

There is a natural bijection $[X,K(G,n)] \rightarrow H^n(X;G)$ for any commutative group $G$ and any CW-complex $X$.

In particular, for $G = \mathbb{Z}$ and $X=\mathbb{T}^n$ we have the bijection $[\mathbb{T}^n,\mathbb{R}/\mathbb{Z}]\rightarrow H^1(\mathbb{T}^n)\cong H_1(\mathbb{T}^n)\cong\pi_1(\mathbb{T}^n$).

[] To establish the bijection with $Hom(\mathbb{T}^n,S^1)$, we can use Pontryagin duality to establish $Hom(\mathbb{T}^n,S^1)=Hom(\widehat{S}^1, \widehat{\mathbb{T}}^n)=Hom(\mathbb{Z},\mathbb{Z}^n)=Hom(\mathbb{Z}^n,\mathbb{Z})$. But $[\mathbb{T}^n,S^1]=H^1(\mathbb{T}^n)=Hom(\pi_1\mathbb{T}^n,\mathbb{Z})=Hom(\mathbb{Z}^n,\mathbb{Z})$, giving us the desired bijection.

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I don't see why the set of homotopy classes $[\mathbb{T}^n,\mathbb{R}/\mathbb{Z}]$ should be equal to the set of group homomorphisms $\mathbb{T}^n \to \mathbb{R}/\mathbb{Z}$. (Also, some argument is needed to show that $\pi_1$ is abelian, e.g. the Eckman-Hilton lemma.) – Dan Petersen Feb 16 2011 at 8:35

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The fact that $\pi_1$ is abelian is well-known and trivial to show by the product-formula... But yes there should be more to say on why the set of homotopy classes is equal to the set of continuous group homomorphisms... for this, a rather twisted way to go about it I will place in an edit. – Chris Gerig Feb 16 2011 at 20:42
Some of your equalities should be isomorphisms. – S. Carnahan♦ Feb 17 2011 at 3:07

This is just a minor elaboration on David Ben-Zvi's answer. You can see the duality between the fundamental group of $T$ and the character lattice by composing based loops $\mathbb{R}/\mathbb{Z} \to T$ with characters $T \to \mathbb{R}/\mathbb{Z}$. You end up with based loops in $\mathbb{R}/\mathbb{Z}$, whose homotopy classes are completely determined by an integer invariant, namely the degree. You can view the degree algebraically as the induced homomorphism on $\pi_1(\mathbb{R}/\mathbb{Z},0) \cong H_1(\mathbb{R}/\mathbb{Z}, \mathbb{Z})$, or geometrically as the winding number if you choose an isomorphism with an oriented circle in the plane. This yields a perfect pairing between $\pi_1(T,0)$ and $X^*(T)$.

In the nonabelian situation, one can see that the fundamental group of a connected topological group doesn't determine that much about the representation theory. There are plenty of simply connected groups with fairly complicated representation theory. However, the K-theory of the classifying space can say a lot about representations.

Regarding the title of the question, one should be careful about calling two groups "the same", even if they are isomorphic.

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