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I'm looking for a quick way to compute the Pontryagin dual group of the n-dimensional torus $\mathbb{T}^n$ (with $\mathbb{T} := \mathbb{R} / \mathbb{Z}$). The only way I know is from "Dikran Dikranjan - Introduction to Topological Groups" and is very tedious.

I thought so: $\mathbb{T}^n$ is compact then $\widehat{\mathbb{T}^n}$ has the discrete topology. The set $E$ of all the functions

$\mathbb{T}^n \to \mathbb{C} : x \mapsto e^{2 \pi a \cdot x}$

with $a \in \mathbb{Z}^d$ is a subset of $\widehat{\mathbb{T}^d}$ which separates the points of $\mathbb{T}^d$ (I mean: for any $x,y \in \mathbb{T}^d$ with $x \neq y$ exists some $\chi \in E$ such that $\chi(x) \neq \chi(y)$. Now I was hoping that Stone-Weierstass theorem could help me to prove that $E$ is dense in $\widehat{\mathbb{T}^d}$ and so that $\widehat{\mathbb{T}^d} = E$, because $\widehat{\mathbb{T}^d}$ is discrete. However I don't know how to apply Stone-Weierstass theorem in this case.

I thank you for any ideas!

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up vote 7 down vote accepted

The key case is the 1-dimensional torus. We want to show every continuous homomorphism $\chi \colon {\mathbf R}/{\mathbf Z} \rightarrow {\mathbf T}$ has the form $x \bmod {\mathbf Z} \mapsto e^{2\pi inx}$ for some integer $n$.

Any character of ${\mathbf R}/{\mathbf Z}$ can be pulled back to a character of ${\mathbf R}$ by composing with the canonical map ${\mathbf R} \rightarrow {\mathbf R}/{\mathbf Z}$. Let $\chi' \colon {\mathbf R} \rightarrow {\mathbf T}$ be the resulting character of ${\mathbf R}$, i.e., $\chi'(x) = \chi(x \bmod {\mathbf Z})$. All continuous homomorphisms ${\mathbf R} \rightarrow {\mathbf T}$ have the form $x \mapsto e^{2\pi ixy}$ for some real number $y$. Accepting this for a moment, we'd have $\chi'(x) = e^{2\pi ixy}$ for some $y$. Thus $\chi(x \bmod {\mathbf Z}) = e^{2\pi ixy}$. Taking $x = 1$, we get $1 = e^{2\pi iy}$. Therefore $y$ is an integer. Writing $y$ as $n$, for psychological purposes, $\chi(x \bmod {\mathbf Z}) = e^{2\pi inx}$ for an integer $n$, which is what you wanted.

Our task now is to compute the characters of ${\mathbf R}$. For this I like the method from Conway's "A Course in Functional Analysis". For any continuous homomorphism $\gamma \colon {\mathbf R} \rightarrow {\mathbf T}$, we want to show $\gamma(x) = e^{2\pi ixy}$ for some real number $y$. We will use differential equations. Because $\gamma(0) = 1$, by continuity $\int_0^a \gamma(t)dt \approx a$ for small positive $a$, so $\int_0^a\gamma(t)dt \not= 0$ for an appropriate $a$. Fixing such $a$, for all real $x$ we have $$\int_x^{x+a} \gamma(t)dt = \int_0^a \gamma(t+x)dt = \int_0^a \gamma(t)\gamma(x)dt = \gamma(x)\int_0^a \gamma(t)dt.$$ Therefore $$ \gamma(x) = \frac{\int_x^{x+a} \gamma(t)dt}{\int_0^a\gamma(t)dt}. $$ The right side, by the fundamental theorem of calculus, is a differentiable function of $x$, and therefore $$ \gamma'(x) = \frac{\gamma(x+a) - \gamma(x)}{\int_0^a\gamma(t)dt} = \frac{\gamma(a)-1}{\int_0^a\gamma(t)dt}\gamma(x), $$ so $\gamma'(x) = s\gamma(x)$ for a complex number $s$. By the theory of ODE, $\gamma(x) = Ce^{sx}$. Since $\gamma(0) = 1$, we have $C = 1$, so $\gamma(x) = e^{sx}$. Since $|\gamma(x)| = 1$, we have $e^{{\rm Re}(s)x} = 1$ for all real $x$, so ${\rm Re}(s) = 0$. Write $s = 2\pi iy$ for a real number $y$ and we're done.

I don't think you're going to find a 2-line proof of the computation of the character group of ${\mathbf R}/{\mathbf Z}$ or ${\mathbf R}$ because you need to get exponential functions from somewhere. Above we get them from our knowledge of solutions to the most basic first-order ODE, and to bring in an ODE we needed a clever idea to see that the a priori continuous function $\gamma$ is in fact differentiable.

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Excuse me if I am being too naive: Is not this saying the representation of $S_{1}$ to $S_{1}$ is isomorphic to the integers? – Kerry Feb 25 '12 at 21:42
@Changwei: you are not being too naive. The continuous homomorphisms from the unit circle to itself are the power maps $z \mapsto z^n$ for integers $n$, which compose with each other the same way that integers add (in the exponent), so the dual group of the unit circle is the integers. – KConrad Feb 25 '12 at 23:10
If you allow Stone-Weierstrass, all you need is that the "finite trigonometric polynomials" (= Laurent polynomials = finite linear combinations of $e^{2\pi i n x}$) constitute a $\bf C$-algebra of continuous functions on the circle that separates points and is closed under complex conjugations. "Separates points" is easy because $e^{2\pi i x}$ already separates points, and "closed under conjugation" is immediate. You can handle the $d$-torus the same way, using the fact that the $e^{2\pi i x_j}$ together already separate points. – Noam D. Elkies Feb 25 '12 at 23:34
@KConrad: Thanks, this is very clear. – Kerry Feb 26 '12 at 5:55
@michael-grade83: Use orthogonality of characters. Any character $\chi$ is uniformly approximable by a linear combination of the $e^{2\pi i k}$. If $\chi$ were not itself one of the $e^{2\pi i k}$ then $\chi$ would be orthogonal to functions arbitrarily close to itself, which is a contradiction. Likewise on the $d$-torus, with "$k$" interpreted as an element of ${\bf Z}^d$ rather than ${\bf Z}$. – Noam D. Elkies Feb 27 '12 at 3:52

You can read off a simple proof from Section VII.2 of Bourbaki's General Topology Part 2. Roughly, the proof goes as follows. Let $\mathbb{T}:=\mathbb{R}/\mathbb{Z}$, then any continuous homomorphism $\mathbb{T}^m\to\mathbb{T}^n$ lifts to a continuous homomorphism $\mathbb{R}^m\to\mathbb R^n$ which then is linear and maps $\mathbb{Z}^m$ into $\mathbb{Z}^n$.

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In my setting the dual group of $G$ is the group of continuous homomorphisms $G \to \mathbb{T}^\times$, where $\mathbb{T}^\times = \{z \in \mathbb{C} : |z| = 1 \}$ is the circle group. So if I understand you say that if $\chi \in \widehat{\mathbb{T}^d}$ then I can extend $\chi$ to a continuous homomorphism $\chi^\prime : \mathbb{R}^d \to \mathbb{R}$. But I don't know how to do this, however I know how extend $\chi$ to a continuous homomorphism $\chi^\prime : \mathbb{R}^d \to \mathbb{T}^times$ by periodicity, I mean $\chi^\prime (x) = \chi(\mbox{fractional part of } x)$. – user21706 Feb 25 '12 at 19:50
$\mathbb{R}/\mathbb{Z}$ is clearly continuously isomorphic to the circle group (the isomorphism being $t\mapsto\exp(2\pi i t)$), so we talk about the same thing (specialize my post to $n=1$). For the details of the proof you should look up the given source, it takes only few pages and proves much more. Fancy tools like Stone--Weierstrass are not used in the proof. – GH from MO Feb 25 '12 at 19:58
OK, thanks, anyway if someone find a more contained proof I'll be happy. – user21706 Feb 25 '12 at 20:07
What do you mean by "a more contained proof"? Bourbaki's General Topology Part 2 finds, with a detailed proof, all the continuous homomorphisms $\mathbb{R}^m\to\mathbb{T}^n$ (Proposition 3 on Page 81). From this statement (whose proof is straightforward) you can find all continuous homomorphisms $\mathbb{T}^m\to\mathbb{T}^n$ by composing them with the natural map $\mathbb{R}^m\to\mathbb{T}^m$. Finally, the special case $n=1$ gives the Pontryagin dual of $\mathbb{T}^m$ as given in your original post. It does not get simpler than that! – GH from MO Feb 25 '12 at 20:18
Yes, Bourbaki's General Topology Part 2 do too much things for me. I'm writing a paper and I only need to find the characters of $\mathbb{T}^d$ quickly. I can use Stone-Weierstrass theorem because I must use it in another part of the paper. – user21706 Feb 25 '12 at 20:24

You can "almost" prove that rather abstractly. The dual group (this is a group because the torus is abelian) is discrete. Moreover, because the torus is connected, the dual group is torsion-free. Hence the dual group is isomorphic to a product of $\mathbb{Z}$.

The only problem here is to limit the rank of this group to be 1. Obviously, this has something to do with the geometry and the topology of $T$ (such as the fact the group is monothetic, or moving into infinitesimal equivalence as in Keith's answer, which can be thought as some Lie theoretic framework), otherwise, the exact argument will work for say $\mathbb{T}^{n}$ for any $n$.

Another possibility here (which one might call cheating) is to use Peter-Weyl theorem (because $\mathbb{T}$ is compact, you don't need the full proof of the Pontryagin duality here). Then out of the blue, recalling Poincaré's inequality and get that the eig. functions of the Laplacian will be a basis to $L^{2}(T)$, and then find them and see that "magically" they are happen to be characters. This gives you an easy differential eq. with boundary conditions which you can solve explicitly.

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On the pure abstract-group level, even just knowing that the dual group is abstractly isomorphic to $\mathbb Z$ isn't enough, right? We've produced a rank-1 free subgroup of $\widehat{\mathbb T}$, but $\mathbb Z$ has proper such subgroups in abundance. (That is, we need the additional topological fact that the identity character $\mathbb T \to \mathbb C^\times$ given by $z \mapsto z$ has no proper root (that is a continuous homomorphism).) – L Spice Oct 16 '15 at 3:44
@LSpice, on a rather abstract level, any homomorphisms between two Baire groups are automatically continuous. Moreover, even in the topological level, any rank $1$ subgroup of $\mathbb{Z}$ is isomorphic to $\mathbb{Z}$, I haven't claimed that by the so-called proof I've sketched you actually get the Pontryaign pairing (if you know that pairing beforehand, you can prove the isomorphism directly). – Asaf Oct 18 '15 at 19:13
@LSpice, anyhow, the existence of roots you've mentioned is intimately related to the theory of divisible groups, whose structure theorem can be also achieved during the proof of the structure theorem of LCA groups. – Asaf Oct 18 '15 at 19:15
I'm not sure if I understand your response (particularly the second comment). I was suggesting that proving that $\widehat{\mathbb T}$ is a free Abelian group of rank 1 is not what @user21706 meant; rather, the goal is to show that the specific map $\mathbb Z \to \widehat{\mathbb T}$ given by $n \mapsto (z \mapsto z^n)$ is an isomorphism. – L Spice Oct 20 '15 at 14:18

If your purpose of calculating the dual of the torus is to have the Fourier inversion formula, I just want to add the remark that you don't need to go through all the machinery of locally compact topological groups in order to prove the basic results of Fourier analysis.

For the circle, one needs only note that the distribution $u(x) = \sum_{n \in {\mathbb Z}} e^{2 \pi i n x}$ on the torus ${\mathbb R}/{\mathbb Z}$ is invariant under multiplication by $e^{2 \pi i x}$, and is therefore a multiple of the delta-function $\delta(x)$ -- integrating against the smooth function $1$, you see that, in fact, $u(x) = \delta(x)$. (This move is simply how you usually calculate the "sum of a geometric series".) Substituting into the general formula $f(x) = \int_{{\mathbb R}/{\mathbb Z}} f(y) \delta(x - y) dy$ gives the Fourier inversion formula.

The above is really just for a conceptual point of view, since it's unnecessary to quote distribution theory for this purpose. To give a direct proof, suppose $f(x)$ is smooth; after translating, it is enough to show that $f(0) = \sum_{n \in {\mathbb Z}} \hat{f}(n) = \sum_n \int f(x) e^{- 2 \pi i n x} dx$ (which converges absolutely because you can integrate by parts enough times to prove decay of $\hat{f}(n)$). We can assume without loss of generality that $f(0) = 0$ by subtracting the constant $f(0)$ (this is how we know "the constant" in the inversion formula is correct and corresponds to testing against $1$ in the previous argument).

In the case $f(0) = 0$, we can write $f(x) = (e^{2 \pi i x} - 1) g(x)$ for some smooth function $g(x)$ (which proves the previous claim that if $e^{2 \pi i x} u = u$ then $u(x) = C \delta(x)$). But then,

$\sum_{n \in {\mathbb Z}} \hat{f}(n) = \sum_{n \in {\mathbb Z}} (\hat{g}(n - 1) - \hat{g}(n))$

but this is $0$ because $\hat{g}$ is integrable. (In particular, assuming $f$ to be $C^\infty$ was certainly unnecessary, and the Fourier inversion formula holds pointwise for functions in a certain class.)

If you like to think in terms of Stone Weierstrass, you can view this argument is saying that the map $Tf$ which takes a Fourier transform and then performs the inverse Fourier transform is the identity. It relies on the fact that $f(x) = (e^{2 \pi i x} - e^{2 \pi i x_0}) g(x)$ whenever $f$ vanishes at $x_0$ which is a strong sense in which trigonometric polynomials separate points. The conclusion that $T$ is a multiple of the identity then follows from $T$ being linear not only over ${\mathbb C}$ but also over the algebra of trigonometric polynomials.

Another remark about distribution theory here is that for the argument of KConrad above, the equation $\gamma'(x + s) = \gamma'(s) \gamma(x)$'' already makes sense when $\gamma$ is a distribution (in particular, if $\gamma$ is continuous). The way it makes sense is by integrating by parts against a test function, which is related to the integration performed the accepted proof. One can use a very similar argument to KConrad's and show that the only distributions solving $\gamma(x + y) = \gamma(x)\gamma(y)$ are exponentials as I'll (almost) show below.

Finally, observe that the fact that the Fourier inversion formula holds for $L^2$ functions implies in particular that all the characters have been accounted for. After all, if you had any other character, it would be orthogonal to all the exponential functions and hence have $\hat{\gamma} = 0$. However, this argument requires not only that the Fourier transform is an isometry on smooth functions in $L^2$ (a consequence of the inversion formula), but one also has to argue that $L^2$ functions can be approximated by smooth functions, which is typically done by mollification $f(x) \approx \int f(x + h) \eta(h) dh$ for a smooth function $\eta$ of integral $\int \eta(h) dh = 1$ and small support. The same kinds of mollifications or similar calculations are used in proofs of Stone Weierstrass theorem; by a similar reasoning, once you have enough characters to separate points, you know you have a complete set.

But, as KConrad showed with $\eta$ the characteristic function of a short interval, when $f = \gamma$ is a character, mollification shows $\int \gamma(x + h) \eta(h) dh = \gamma(x) \int \gamma(h) \eta(h) dh$, which implies in particular that $\gamma(x)$ is smooth because $\int_{{\mathbb R}/{\mathbb Z}} \gamma(x +h) \eta(h) dh = \int_{{\mathbb R}/{\mathbb Z}} \gamma(h) \eta(h - x) dh$ is always smooth (even when $\gamma$ is just a distribution).

But then once you have established that $\gamma$ is smooth, the same equation $\gamma(x) \gamma(y) = \gamma(x+y)$ which proves the orthogonality of characters can also be differentiated classically and evaluated at $y = 0$, and then (as KConrad noted) you just need to solve the ODE that defines the exponential functions. So I'll end here now that we have "come full circle".

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