# Translation-invariant metric on locally compact group

Let $G$ be a locally compact group on which there exists a Haar measure, etc..

Now I am supposed to take such a metrisable $G$, and given the existence of some metric on $G$, prove that there exists a translation-invariant metric, i.e., a metric $d$ such that $d(x,y) = d(gx,gy)$ for all $x,y,g \in G$. How to go about this?

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I know of an answer to this, but It's 5 pages long... This is Birkhoff-Kakutani, no? planetmath.org/encyclopedia/BirkhoffKakutaniTheorem.html –  kneidell Sep 18 '11 at 20:55
Birkhoff-Kakutani is something else, according to terrytao.wordpress.com/2011/05/17/the-birkhoff-kakutani-theorem .. It addresses the question of metrizability in addition .. All I am asking is, is it possible to prove the existence of invariant metric, given already some other metric? –  Lit Sep 18 '11 at 21:00
I'm pretty sure that there is quite an easy explicit way but I simply don't remember the trick at the moment. You can extract a construction from Theorem 3 on page 8 in Koszul's Lectures on Groups of Transformations but that's not something you should be able to come up with yourself, I guess. –  t.b. Sep 18 '11 at 23:13
@kneidell: Yes, that's usually called the Birkhoff-Kakutani theorem. However, it's already implicit in Weil's work on uniform structures, he just didn't spell it out. Weil proved that a countably generated uniform structure is metrizable and the metric you get from his construction applied to the left uniform structure of a first countable group is in fact translation invariant. –  t.b. Sep 18 '11 at 23:19
@t.b.you should write your observation that if the group is metizable, it is first countable, so its left uniform structure is countable generated, so metrizable, and the resulting metric is left invariant (a long observation! :) ) as an answer, so that we can vote it up. –  Mariano Suárez-Alvarez Jan 17 '12 at 7:46

Though the answer was essentially given in the comment area, I think it's nice to have the full proof here. We follow largely Bourbaki.

Notation Let $X$ be a set. Let $U$ and $V$ be subsets of $X \times X$. We define $UV$ = {$(x, y) \in X \times X;$ there exists $z \in X$ such that $(x, z) \in U$ and $(z, y) \in V$}.

Let $U, V, W$ be subsets of $X \times X$. We define $UVW = (UV)W$.

Similarly we define $U^n, n = 1, 2, ...$

We define $U^{-1} = \{(x, y) \in X; (y, x) \in U\}$.

Lemma Let $X$ be a uniform space. Let $U_n, n = 1, 2, ...$ be a sequence of entourages on $X$. Suppose $(U_{n+1})^3 \subset U_n$ for each $n$. We define a function $g:X \times X \rightarrow [0, \infty)$ as follows.

• If $(x, y) \in \bigcap U_n$, then $g(x, y) = 0$.

• If $(x, y) \in U_n - U_{n+1}$, then $g(x, y) = 1/2^n$.

• If $(x, y) ∈ X\times X - U_1$, then $g(x, y) = 1$.

Let $(x, y) \in X\times X$. Let $z_0, z_1, ..., z_p$ be a finite sequence of elements of $X$ such that $x = z_0, y = z_p$. Then $\sum_{i=0}^{p-1} g(z_i, z_{i+1})$ $\geq$ $(1/2)g(x, y)$.

Proof(Bourbaki): We use induction on $p$. If $p = 1$, the assertion is clear.

Suppose $p > 1$. Let $a = \sum_{i=0}^{p-1} g(z_i, z_{i+1})$. Since $g(x, y) \leq 1$, if $a \geq 1/2$, then $a \geq (1/2)g(x, y)$. Hence we can assume $a < 1/2$. Let $h$ = max {$q; \sum_{i=0}^{q-1} g(z_i, z_{i+1}) \leq a/2$}.

Then $\sum_{i=0}^{h} g(z_i, z_{i+1}) > a/2$.

Hence $\sum_{i=h+1}^{p-1} g(z_i, z_{i+1}) \leq a/2$.

By the induction assumption, $(1/2)g(x, z_h) \leq \sum_{i=0}^{h-1} g(z_i, z_{i+1}) \leq a/2$. Hence

(1) $g(x, z_h) \leq a$

Similarly, $(1/2)g(z_{h + 1}, y) \leq \sum_{i=h+1}^{p-1} g(z_i, z_{i+1}) \leq a/2$. Hence

(2) $g(z_{h + 1}, y) \leq a$.

Clearly,

(3) $g(z_h, z_{h + 1}) \leq a$

Let $k$ = min {$k ∈ \mathbb{Z}; k > 0, 1/2^k \leq a$}. Since $a < 1/2$, $k \geq 2$.

By (1), (3), (2), we get:

(a) $(x, z_h) \in U_k$.

(b) $(z_h, z_{h + 1}) \in U_k$.

(c) $(z_{h + 1}, y) \in U_k$.

Hence, $(x, y) \in (U_k)^3 \subset U_{k-1}$

Hence $g(x, y) \leq 1/2^{k-1} ≦ 2a$. QED

Theorem 1 Let $X$ be a uniform space. Suppose $X$ has a fundamental system of countable entourages. Then there exists a pseudometric $d$ on $X$ such that $d$ is compatible with the uniform structure of $X$.

Proof: Let $V_n, n = 1, 2, ...$ be a fundamental system of countable entourages. By induction and the axiom of dependent choice, we can define a sequence of entourages $U_n, n = 1, 2, ...$ which satisfies the following conditions.

(1) Each $U_n$ is symmetric, i.e. $U_n = (U_n)^{-1}$.

(2) $U_1 \subset V_1$

(3) $(U_{n+1})^3 \subset U_n \cap V_{n+1}$ for each $n \geq 1$.

Let $f(x, y)$ = inf $\sum_{i=0}^{p-1} g(z_i, z_{i+1})$ for each $(x, y) \in X\times X$, where $g:X\times X \rightarrow [0, \infty)$ is the function defined in the Lemma and the inf is taken over every finite sequence of elements $z_0, z_1, ..., z_p$ of $X$ such that $x = z_0, y = z_p$.

Clearly $f$ is symmetric and satisfies the triangle inequality. Since $f(x, y) \leq g(x, y)$, $f(x, x) = 0$. Hence $f$ is a pseudometric.

By the Lemma, $(1/2)g(x, y) \leq f(x, y) \leq g(x, y)$.

Let $W_a$ = {$(x, y) \in X\times X ; f(x, y) < a$} for any $a > 0$.

Let $a > 0$ be a real number. Let $k$ be an integer such that $k > 0$ and $1/2^k < a$. Let $(x, y) \in U_k$. $f(x, y) ≦ g(x, y) ≦ 1/2^k ＜ a$. Hence $U_k ⊂ W_a$.

Conversely, let $k > 0$ be an integer. Suppose $f(x, y) \leq 1/2^{k+1}$. Since $f(x, y) \geq (1/2)g(x, y)$, $g(x, y) \leq 1/2^k$. Hence $W_{1/2^{k+1}} \subset U_k$. QED

Theorem 2 Let $G$ be a topological group. Suppose $G$ has a fundamental system of countable neighborhoods of the identity $e$. Then there exists a pseudometric $d$ on $G$ such that $d(x, y) = d(zx, zy)$ for any $x, y, z \in G$. Moreover $d$ is compatible with the left unform structure of $G$.

Proof: Let $V_n, n = 1, 2, ...$ be a system of neighborhood of $e$. We can assume that $V_n = (V_n)^{-1}$, $(V_n)^3 \subset V_n$ for each $n$. Let $U_n$ = {$(x, y) \in G\times G ; x^{-1}y \in V_n$} for each $n$. For each $n$, $U_n$ is symmetric and $(U_n)^3 \subset U_n$. $U_n, n = 1, 2, ...$ is fundamental system of entourages of the left uniform structure of $G$.

Let $f(x, y)$ = inf $\sum_{i=0}^{p-1} g(z_i, z_{i+1})$ for each $(x, y) \in X\times X$, where $g:G\times G \rightarrow [0, \infty)$ is the function defined in the Lemma and the inf is taken over every finite sequence of elements $z_0, z_1, ..., z_p$ of $X$ such that $x = z_0, y = z_p$.

By Theorem 1, $f$ is a pseudometric compatible with the left uniform structure of $G$.

Let $(x, y) \in U_n$. For any $z \in G$, $(zx)^{-1}zy = x^{-1}y \in V_n$. Hence $(zx, zy) \in U_n$.

Conversely, if $(zx, zy) \in U_n$, then $(x, y) \in U_n$. Hence $g(zx, zy) = g(x, y)$. Hence $f(zx, zy) = f(x, y)$. QED

Corollary Let $G$ be a metrizable topological group. Then there exists a metric $d$ on $G$ such that $d(x, y) = d(zx, zy)$ for any $x, y, z \in G$. Moreover $d$ is compatible with the left unform structure of $G$.

Proof: Since $G$ is metrizable, it has a fundamental system of countable neighborhoods of the identity $e$. Hence, by Theorem 2, there exists a pseudometric $d$ on $G$ such that $d(x, y) = d(zx, zy)$ for any $x, y, z \in G$. Since $d$ is compatible with the left uniform structure of $G$ and $G$ is Hausdorff, $d$ is a metric.QED

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I removed that unnecessary part at the beginning of your answer (also in the AB5-thread). This is simply off-topic and has nothing to do with the present thread. Also, if you're insulted, you should maybe reconsider how you addressed others in the comment threads in quite a few of your questions... –  t.b. Jul 1 '12 at 17:33
Well, there's a reason your questions tend to generate endless comment threads. It's not only about being right it's also about how you communicate it (diplomacy). While you may be right that whatever you said is justified in the abstract, this may not be how it is heard or read. You can phrase something as "you're wrong because ..." or you can phrase it as "I'm sorry, but I don't understand: isn't ...?". While the mathematical content of the messages is exactly the same the latter is much more likely to generate a fruitful discussion. –  t.b. Jul 3 '12 at 4:58
We wouldn't have this exchange if I thought you weren't trying to be polite and if I didn't think you made some very good contributions both questions and answers. I agree with your last comment, but there's a flip-side to that coin: given that people tend to overreact if you tell them they made a mistake you need to tell them very gently. Don't tell them directly that they're wrong, don't tell them they misunderstand something. Tell them you're confused and ask a question, so they can say: "oops you're right" and do not feel the need to defend themselves. –  t.b. Jul 4 '12 at 11:03
Some people seem not to like controvertial questions, but I rather like them. I think a soft question with an answer agreed by most of the members is boring. –  Makoto Kato Jul 5 '12 at 23:54
@t.b. There is at least one merit of controversy: People will have deeper understandings of the subject. –  Makoto Kato Jul 11 '12 at 0:16

What if you take the original metric $d_0$ and define the new metric by
$d(x,y) = \int_G d_0(gx,gy) d\mu$

where $d\mu$ is the Haar measure?

The invariance of the measure implies the invariance of the integral, and hence of $d$.

EDIT: As Theo pointed out, this works only when $G$ is compact.

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Thanks .. This would do, if only the triangle inequality were checked... –  Lit Sep 18 '11 at 22:45
It follows from the triangle inequality for $d_0$ and the fact that if $0\leq \varphi \leq \psi$ then $\int_G \varphi d \mu \leq \int_G \psi d \mu$ –  Lucas Kaufmann Sep 18 '11 at 22:55
Lucas: the problem is that this integral will be infinite unless $G$ is compact (for compact $G$ you indeed get a compatible translation invariant metric this way). Just take $G = \mathbb{R}$, and start with the absolute value, for example. –  t.b. Sep 18 '11 at 22:58
you're right Theo. I was with the compact case in mind. I'll think a little about the non-compact case. –  Lucas Kaufmann Sep 18 '11 at 23:41