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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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    @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.2012-01-17

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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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    @t.b. There is at least one merit of controversy: People will have deeper understandings of the subject.2012-07-11
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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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    you're right Theo. I was with the compact case in mind. I'll think a little about the non-compact case.2011-09-18