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Let $\mathbb{T}^n:=\mathbb{R}^n/\mathbb{Z}^n$ be the quotient of the group $(\mathbb{R}^n,+)$ by the subgroup $(\mathbb{Z}^n,+)$.

I'm trying to construct the Haar measure of $\mathbb{T}^n$. I constructed a measure which is finite and Radon. However I don't how to prove that the measure I constructed is translation invariant. Below I'll sketch the construction. Firstly suppose that $\lambda$ is the Lebesgue measure of $\mathbb{R}^n$ and that $\mathfrak{B}$ is the Borel $\sigma$-algebra of the unit cube $[0,1]^n$ with the subspace topology of $\mathbb{R}^n$.

  • Define $\pi :[0,1]^n\to \mathbb{T}^n$ by $\pi (x):=x+\mathbb{Z}^n$. We can define a metric $d_{\mathbb{T}^n}$ in $\mathbb{T}^n$ such that $\pi$ is continuous (see here).
  • Let $\mathfrak{B}_{\mathbb{T}^n}$ be the Borel measure of $\mathbb{T}^n$ induced by the topology induced by the metric $d_{\mathbb{T}^n}$;
  • Let $\pi _\star (\lambda |_{\mathfrak{B}}) :\pi_\star\mathfrak{B}\to\overline {\mathbb{R}} $ be the pushforward measure of the restriction $\lambda|_{\mathfrak{B}} $. It's easy to show that $\mathfrak{B}_{\mathbb{T}^n}\subseteq\pi_\star\mathfrak{B} $ since $\pi$ is continuous.

Define $\vartheta :\mathfrak{B}_{\mathbb{T}^n}\to \overline{\mathbb{R}}$ by $\vartheta (B):=\pi _\star (\lambda |_{\mathfrak{B}}) (B)$.

We can show that $\vartheta :\mathfrak{B}_{\mathbb{T}^n}\to \overline{\mathbb{R}}$ is a probability measure of $\mathbb{T}^n$ and it's also a Radon measure.

My question is: how can I show that $\vartheta $ is translation invariant?


I did almost nothing worth mentioning.

I think it's important to observe that the restriction $\pi |_{[0,1)^n}:[0,1)^n\to\mathbb{T}^n$ is bijective.

With the property of pushforward measure we can show that, given any $B\in\mathfrak{B}_{\mathbb{T}^n}$ and $a\in\mathbb{T}^n$, we have

$$\vartheta(B+a)=\int _{\mathbb{T}^n}\mathbf{1}_{B+a}d\vartheta=\int _{[0,1]^n}\mathbf{1}_{B+a}(\pi (x))d\lambda (x)\,\,\color{red}{(1)} $$

Above $\mathbf{1}_{B+a}$ is the indicator function of $B+a$.

If there's an $\alpha\in[0,1]^n$ such that $\mathbf{1}_{B+a}(\pi (x))=\mathbf{1}_{\pi^{-1}(B)+\alpha}(x)$ for all $x\in [0,1]^n$, then we can use $(1)$ to prove that $\vartheta (B+a)=\lambda (\pi ^{-1}(B)+\alpha )=\lambda (\pi^{-1}(B))=\vartheta(B)$. However I wasn't able to prove that there's such $\alpha\in[0,1]^n$.


If anyone knows an easier way to construct the Haar measure of the $n$-dimensional Torus please tell me.

Thank you for your attention!

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3 Answers 3

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One slightly indirect way to do this is to invoke the Riesz-Markov-Kakutani theorem, with corollary that, given a discrete subgroup $\Gamma$ (thinking of $\mathbb Z$) of an abelian topological group $G$ (thinking of $\mathbb R$), fixing a Haar measure $\mu$ on $G$, there is a unique measure $\nu$ on $G/\Gamma$ such that $$ \int_{G/\Gamma} \sum_{\gamma\in\Gamma} f(\gamma+\dot{g})\;d\nu(\dot{g}) \;=\; \int_G f(g)\;d\mu(g) $$ As part of this, one proves that the averaging map $C^o_c(G)\to C^o(G/\Gamma)$ is surjective.

This characterization can be used to prove essentially all of the desired properties, such as group-invariance on the quotient $G/\Gamma$, from the group-invariance on $G$.

EDIT: the proof of the corollary is short and straightforward: given $f\in C^o(G/\Gamma)$, let $F\in C^0_c(G)$ be such that $f(g) = \sum_{\gamma\in G/\Gamma} F(\gamma+g)$. For fixed $\dot{g}_o\in G/\Gamma$, let $g_o\in G$ be a representative. Then the averaged version of $F(g+g_o)$ is $f(\dot{g}+\dot{g_o})$, and $$ \int_{G/\Gamma}f(\dot{g}+\dot{g}_o)\,d\dot{g} \;=\; \int_G F(g+g_o)\;dg \;=\; \int_G F(g)\;dg \;=\; \int_{G/\Gamma} f(\dot{g})\;d\dot{g} $$ Here we use the important fact that the corresponding integral is translation-invariant if and only if the measure is translation-invariant. :)

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  • $\begingroup$ Could you please give me a reference that contains a proof of this corollary? $\endgroup$
    – rfloc
    Dec 28, 2021 at 20:50
  • $\begingroup$ Could you please clarify one point about the theorem you mentioned? I'm thinking about it and I don't know how to verify, for example, the identity $\int _{\mathbb{T}^n}\varepsilon_k (\tilde x)d\vartheta (\tilde x)=\int _{[0,1]^n}e^{2\pi i\langle x,k\rangle}d\lambda (x) $ with $\varepsilon_k:\mathbb{T}^n\to\mathbb{C}$ given by $\varepsilon _k(x+\mathbb{Z}^n):=e^{2\pi i\langle x,k\rangle }$. Is there an easy way to obtain such identity in a trivial way using you theorem? $\endgroup$
    – rfloc
    Jan 7, 2022 at 1:18
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    $\begingroup$ In brief, take $L^1$ limits of $C^0_o$ functions... Everything is defined in terms of $C^0_o$ functions, and we know how integrals behave under $L^1$ limits... :) $\endgroup$ Jan 7, 2022 at 1:23
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Below I present a direct approach, which is a completion of what you started. The main addition is, roughly speaking, the following observation: translations at the torus level does not change the Borel $\sigma$-algebra, but they do change the Borel $\sigma$-algebra of the unit cube (as well as the unit cube). Thus in order to derive invariance of the pushforward measure from the invariance of Lebesgue measure one needs to keep track of the changes in the Borel $\sigma$-algebras and the unit cubes.


Let us denote by $\pi:\mathbb{R}^n\to\mathbb{T}^n$ the natural projection. Then $\Sigma=[0,1[^n\subseteq \mathbb{R}^n$ is a strict fundamental domain for $\pi$ (i.e., $\pi|_\Sigma:\Sigma\to \mathbb{T}^n$ is a bimeasurable bijection as in my answer to your other question here: The domain of the Haar measure of the $n$-dimensional Torus ). Note that if $t\in \mathbb{R}^n$, then $t+\Sigma =\{t+\sigma\,|\, \sigma\in \Sigma\}\subseteq \mathbb{R}^n$ is also a strict fundamental domain (To see this note that any $t\in\mathbb{R}^n$ can be written uniquely as $z_t+\sigma_t$, where $z_t\in\mathbb{Z}^n$ and $\sigma_t\in \Sigma$, and that $\pi$ is $\mathbb{Z}^n$-periodic; thus one can rearrange the pieces of $t+\Sigma$ according to the partition $\mathbb{R}^n=\biguplus_{z\in\mathbb{Z}^n}z+\mathbb{Z}^n$). Let us define a Borel probability measure on $\mathbb{T}^n$ like you did, so that we put

$$\vartheta=\left(\pi|_\Sigma\right)_\ast(\lambda_\Sigma),$$

where $\lambda_\Sigma=\operatorname{leb}_{\mathbb{R}^n}(\,\cdot\, \cap \Sigma)=\operatorname{leb}_{\mathbb{R}^n}(\,\cdot\, \,|\, \Sigma)$ is the Lebesgue/Haar measure on $\mathbb{R}^n$ restricted/conditioned on the fundamental domain $\Sigma$.

Next observe that one has analogously a pushforward measure using the strict fundamental domain $t+\Sigma$ for any $t\in\mathbb{R}^n$, and each one of these pushforward measures is equal to $\vartheta$. Indeed, for $t\in \mathbb{R}^n$ and $B\subseteq \mathbb{T}^n$ Borel measurable,

\begin{align*} \left(\pi_{t+\Sigma}\right)_\ast(\lambda_{t+\Sigma})(B) &= \lambda_{t+\Sigma}(\left(\pi_{t+\Sigma}\right)^{-1}(B))\\ &= \operatorname{leb}_{\mathbb{R}^n}((t+\Sigma)\cap\pi^{-1}(B))\\ &\stackrel{(\ast)}{=} \operatorname{leb}_{\mathbb{R}^n}(\Sigma\cap\pi^{-1}(B))\\ &=\lambda_\Sigma(\left(\pi|_\Sigma\right)^{-1}(B))\\ &=\vartheta(B). \end{align*}

Here in the equality marked with $(\ast)$ we used the fact that $\pi^{-1}(B)+\mathbb{Z}^n=\pi^{-1}(B)$, that is, $\mathbf{1}_{\pi^{-1}(B)}$ is $\mathbb{Z}^n$-periodic (See Integral of periodic function over the length of the period is the same everywhere or An integrable and periodic function $f(x)$ satisfies $\int_{0}^{T}f(x)dx=\int_{a}^{a+T}f(x)dx$. for the $n=1$ case; the general case follows by iterated integrals).

Denote by $l^{\mathbb{R}^n}_\bullet$, $l_\bullet$ and $l^{\mathbb{T}^n}_\bullet$ the left translation actions of $\mathbb{R}^n$ on itself, $\mathbb{R}^n$ on $\mathbb{T}^n$, and $\mathbb{T}^n$ on itself, respectively, so that we have:

enter image description here

We claim that $\vartheta$ is $l_\bullet$-invariant. This implies $\vartheta$ is also $l^{\mathbb{T}^n}_\bullet$-invariant, that is, $\vartheta$ is a left Haar measure on $\mathbb{T}^n$ (One can of course argue analogously for right translations; the fact that $\mathbb{T}^n$ is abelian or compact would guarantee that the same measure would be produced). Fix $t\in\mathbb{R}^n$. Then we have

enter image description here

where all filled-in arrows are isomorphisms of (standard) probability spaces and the dashed arrow is an isomorphism of (standard) measurable spaces. Invariance of $\vartheta$ is equivalent to the dashed arrow being an isomorphism of (standard) probability spaces, which is clear by the commutativity of the diagram (and functoriality of pushforwards):

\begin{align*} (l_t)_\ast(\vartheta) &=\left(\pi|_{t+\Sigma}\circ l^{\mathbb{R}^n}_t\circ (\pi|_{\Sigma})^{-1}\right)_\ast(\vartheta)\\ &=\left(\pi|_{t+\Sigma}\right)_\ast\circ \left(l^{\mathbb{R}^n}_t\right)_\ast\circ \left((\pi|_{\Sigma})^{-1}\right)_\ast(\vartheta)\\ &=\left(\pi|_{t+\Sigma}\right)_\ast\circ \left(l^{\mathbb{R}^n}_t\right)_\ast (\lambda_{\Sigma})\\ &=\left(\pi|_{t+\Sigma}\right)_\ast (\lambda_{t+\Sigma})\\ &=\vartheta. \end{align*}


For a more general account of this argument see Prop.4.1.3 on p. 45 of Witte Morris' Introduction Arithmetic Groups (https://arxiv.org/abs/math/0106063). I followed this book especially in my use of the phrase "strict fundamental domain".

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  • $\begingroup$ But ... there's no necessity to use a fundamental domain. What about situations where it's difficult to certify such? Etc....? $\endgroup$ Jan 6, 2022 at 18:31
  • $\begingroup$ @paulgarrett Under fairly general conditions there is always a strict fundamental domain for a quotient, and further when strict fundamental domains exist the pushforward measures are again choice-independent (so in fact above $\vartheta$ is also independent of what $\Sigma$ is). I agree that for the construction of the Haar measure on the torus maybe this is too much. In my defense my argument is a direct continuation and completion of the OP's argument. Also I should note arguments analogous to what I present above come in handy in measure rigidity quite a bit, in my opinion. $\endgroup$
    – Alp Uzman
    Jan 6, 2022 at 20:59
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    $\begingroup$ Ah, well, yes, sure, we (collectively) know that lots of fundamental domains exist, and, yes, sometimes there is interesting information in the details, but for me it has turned out that the measure on a quotient is much more useful than finding a fundamental domain. Surely it depends on one's goals. For me, it was a relief at some point to not have to understand any details of fundamental domains of arithmetic groups on hermitian symmetric spaces, despite Siegel's and others' writing about it at length! :) $\endgroup$ Jan 6, 2022 at 21:19
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    $\begingroup$ There are some examples (of actual use to me...) where a quotient is more fluent than a fundamental domain. For example, the classic question: given a Lebesgue-measurable set $E\subset \mathbb R$ of measure $>1$, show that there are $x,y\in E$ such that $x-y\in \mathbb Z$. This is very easy in a quotient set-up, but I think less clear thinking about $[0,1]$ and so on. It was a bit of a revelation to me some years ago. Anyway, yes, tastes vary. :) $\endgroup$ Jan 6, 2022 at 21:22
  • $\begingroup$ Thank you for the remarks. $\endgroup$
    – Alp Uzman
    Jan 6, 2022 at 21:27
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A quick way of producing Haar measure on the $n$-torus is to use differential forms. This requires the machinery of integration on smooth manifolds, but this is not unexpected as the torus is a Lie group (and not just a topological group). Let $(x^1, ..., x^n)$ be the standard coordinate vectors on $\mathbb{R}^n$ and $\omega = dx^1 \wedge ... \wedge dx^n$ be the standard volume form. Then for any $x \in \mathbb{R}^n$, if $T_x$ denotes translation by $x$, $T_x^*\omega = \omega$, so that $\omega$ is translation invariant.

In particular, $\omega$ is invariant under the action of $\mathbb{Z}^n$, and so descends to a volume form $\omega_{\mathbb{T}^n}$ on $\mathbb{T}^n$. This form can be seen to be translation invariant by lifting a translation to $\mathbb{R}^n$. As $\omega_{\mathbb{T}^n} $ is translation invariant, the associated Radon measure $\mu_{\mathbb{T}}$ (specified by $\mu(U) = \int_U \omega_{\mathbb{T}^n}$ for $U$ open) is as well.

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