Here is the link to the paper: arxiv.org/abs/0901.0678

It has 5 sections, so it seems smartest if we tackle one section at a time. It's about 10 pages a section.

What kind of timeline should we establish?

I'm thinking we should have a section every week? That's a little more than a month to go over ~50 pages. It seems pretty reasonable to me.

What do you think?

Seems good!

Cool, and so we can always ask questions here of each other as we go through it, and probably aim to have the section read by Sunday of each week. Do you agree? What else should we do?

That sounds decent. I will probably tex up some notes while I read it.

That's smart, I might do the same, even though I'm very slow at TeX.

We could do a ShareLaTeX document if you want, just to prevent duplicated effort, but I've never actually used ShareLaTeX for sharing, so I don't know how that goes, and it might be more instructive to TeX individually.

Probably tex individually, esp. if you know more measure theory. I will probably inject what I learn in respect to that throughout mine as I learn it.

But I can definitely share my notes every week if you wanted.

Yeah, okay, I think TeXing and then sharing is probably the smartest.

Alright, so I think having the first section read by next Sunday (May 7?) should work?

Sounds good. Gives people some time to join without being behind as well.

I guess we are starting with chapter 2, right?

well section 2

Because section 1 just the introduction page. :P

Lol, yeah, of course :P

Huh I wonder why the pdf is duplicated twice.

Yeah, I noticed the same thing... I was shocked with the PDF was 120 pages long, but then the TOC said it was 55 pages... I don't know why it duplicated.

I have never heard of this commensurable thing before.

Also: is there a difference between saying infinite countable group and countably infinite group?

On the second point: I don't think there really is, but it feels more natural to say "countable infinite group" since "countable" is a type of "infinite group".

On the first one: me neither, but reading it I think that's kind of where the richness of this theory comes from: two groups being commensurable is a good way of seeing that the groups are related, but obviously not 'equal' in the same way group theory would usually consider (i.e. they're not isomorphic).

So the basic idea of Definition 2.1 is that if $\Lambda$ and $\Gamma$ are measure equivalent you have some space $\Omega$ with measure $m$ and the two groups both act in such a way that they preserve the measure on $m$. Moreover, we require that if we take a fundamental domain (i.e. one element of each orbit) under each action, then each fundamental domain is of finite measure. (The fact that $m = \bigsqcup\limits_{\gamma \in \Gamma} \gamma X$ follows from the fact $X$ is a fundamental domain).

The idea of index in this context comes from comparing the relative "number" of orbits, or (kind of equivalently) the relative "size" of their orbits.

Put concretely, it is $[\Gamma, \Lambda]_\Omega = \frac{m(X)}{m(Y)}$

So the equivalence is entirely in terms of their actions behaving the same way on some infinite measure space.

The remarks about "ergodic couplings" are kind of opaque to me, but this made it slightly more clear: math.stackexchange.com/questions/421142/…

One last thing before I go to bed: the stuff about lattices has some basic things about Haar measure and topological groups as prerequisites, and I found these which cover those prerequisites really quickly (if you don't already know about that sort of thing).

I've encountered them via topology and Lie groups.

I assume the group action preserves the measure.

I am getting actually mixed signals about this when I google it. The action should either have in addition to the usual group action axioms that the induced maps from \Omega to \Omega by some group element g are measureable, or alternatively that they are measurable AND preserve the measure in the sense that the preimage of a measurable set A by g has the same value under the measure m as A does.

i.e. m(g^{-1}(A)) = m(A)

It seems to me that we want the action to preserve measure, meaning that we want the measure to actually be equal in the image and preimage.