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?

Solution 1:

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

Solution 2:

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.