2
$\begingroup$

Let's first start with the example from where this question arose. I consider the metric space $\mathbb{R}$ and the Lebesgue-measure $\lambda$ on $\mathbb{R}$ as well as the transitive group action $+\colon (\mathbb{R},+)\times \mathbb{R}\to\mathbb{R}$, for which the measure and the metric are both invariant. When integrating over a set $M$ with respect to $\lambda$ I can fix a point $x\in\mathbb{R}$ (the metric space rather than the group) and, since the action is transitive, integrate over a set $M'$ with respect to some measure $\mu$ on the additive group $(\mathbb{R},+)$, such that $M'+x=M$ and $\mu(M')=\lambda(M)$. This is possible as well as completely independent of the choice of $x$ by choosing $\mu$ to be the Lebesgue-measure on $(\mathbb{R},+)$.

Now I want to generalise this statement to arbitrary metric spaces and groups. I have a metric space $X$ with a radon-measure $\mu$ as well as a transitive group action from a topological group $\cdot\colon G\times X\to X$ which leaves the metric and the measure invariant. Analogous to $\mathbb{R}$ I chose another condition that for any $\varepsilon>0$ there exists a neighbourhood $U$ of $1\in G$ such that the metric is left small when translating with elements from $U$, i.e. $\forall x\in X,g\in U\colon d(gx,x)<\varepsilon$. This so far were reasonable assumptions since they give me the required continuity of the group action. Trying to find a suitable measure on $G$ just like above I came to 3 distinct conditions, that could give me what I require.

  1. Assume that $G$ is commutative. This will make sure that $\mu(Ux)=\mu(Uy)$ for every $x,y\in X,U\subseteq G$ since $y=gx$ for some $g$ since I can simply commute the $g$ to the front and disregard it due to invariance.
  2. Assume that the action is free, i.e. $G$ and $X$ are basically the same since for any $x\in X$ the function $f\colon G\to X,g\mapsto gx$ would be a bijection, giving me a very natural choice for the measure on $G$.
  3. Assume that not only above condition, $\forall\varepsilon>0 \exists U \text{ negihbourhood of }1\in G\forall x\in X,g\in U\colon d(gx,x)<\varepsilon$ but even more: $\forall\varepsilon>0 \exists U \text{ negihbourhood of }1\in G\forall x\in X\colon g\in U \iff d(gx,x)<\varepsilon$, which would also solve some problems.

The questions now is: Which of these 3 conditions is most natural/weakest? Which would you go for in the most general case? Do I even need any of them or would this work with the above statements, without 1,2 or 3? Thanks for any help!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.