30 Jul

This talk happened in March and I still remember it (and I was super sleep deprived at the time too).  Immediately after the talk, another grad student and I were chatting in the hallway and marveling at how good it was.  He said something like “I feel like a better person for having gone to that talk.”

A few days later, I ran into the speaker and told her that I had loved her talk, and she said “I’m super unintimidating so feel free to email me or ask me if you have any questions.”

During the talk, at one point she said (again, up to sleep-deprived memory coarseness)

“It’s more important that you learn something than that I get through my talk.  There’s no point in rushing through the material if you don’t take something away from this.”

All of these quotes are to say that this was probably the best talk I’ve seen (and I’ve seen lots of talks).  Particularly because of that last quote above.  Speaker put audience before ego, and that is a rare and beautiful thing (the other contender for best talk I’ve ever seen was by someone who recently won a big award for giving incredible talks).

The good news is that this is something anyone can do – mathematics at this level is a matter of practice and good habits, and not “talent” or “genius”.

OK, done fangirling!  On to the math!

We’ll be talking about a property of groups, so brush up from a previous blog post or wikipedia.  First we need to define a  (total) ordering on a group: a binary relation ≤ that satisfies three properties (which you’d expect them to satisfy):

1. Transitivity: if a≤b and b≤c, then a≤c
2. Totality: for any a, b in the group, a≤b and/or b≤a
3. Antisymmetry: if a≤b and b≤a, then a=b.

A few examples and nonexamples:

• The usual ≤ (less than or equal to) on the real numbers is an ordering.  For the rest of this post, I’ll freely switch between using ≤ to denote being in the group, and being in the real numbers (it should be clear when we’re talking about real numbers).
• Comparing heights of people is not an ordering: it’s not antisymmetric (see picture)
• Ordering words in the dictionary is an ordering: if you’re both before/at the same place and after/at the same place as me, then we must be the same word.
• Consider the group $\mathbb{Z}_2\times\mathbb{Z}_2$, which you can think of as a collection of ordered pairs $\latex \{ (0,0), (0,1), (1,0), (1,1)\}$.  If we define an ordering by (x,y)≤(a,b) if x+a≤y+b, then we’d break antisymmetry.  If we defined it by (x,y)≤(a,b) if x<a and y≤b, then we’d break totality (couldn’t compare (0,1) and (1,0)).

Top: reals are good to go. Middle: just because we’re the same height doesn’t mean we’re the same person! Bottom: (0,1) and (1,0) don’t know what to do

• Can you come up with a relation that breaks transitivity but follows totality and antisymmetry?

Notational bit: we say that a<b if a≤b and a does not equal b.

Now we say a group is left orderable if it has a total order which is invariant under left multiplication: this means that a<b implies ga<gb for every g in the group.

Let’s go back to the reals.  If you use multiplication (like 3*2=6) as the group operation, then the usual ordering is not a left(-invariant) order: 2<3, but if you multiply both sides by -2 you get -4<-6, which isn’t true.  However, if you use addition (like 3+2=5) as the group operation, then you see that the reals are left orderable: 2<3 implies 2+x<3+x for every real x.  This is a good example of the fact that a group is a set and a binary operation.  It doesn’t make sense to say the real numbers are left orderable; you need to include what the group operation is.

Here’s an interesting example of a left orderable group: the group of (orientation-preserving) homeomorphisms on the real numbers. (Orientation preserving means that a<b means that f(a)<f(b), all in the reals sense).  If you don’t feel like clicking the link to prev. post, just think of functions.  To prove that the group is left orderable, we just have to come up with a left-invariant order.  Suppose you have two homeomorphisms g and defined on the reals.  If g(0)<f(0), then say g<f.  If f(0)<g(0), then say f<g.  If g(0)=f(0), then don’t define your order yet.  If g(1)<f(1), then say g<f.  And so on, using 2, 3, 4…  Looks like a good left order, right?  WRONG!

Pink and blue agree on all the integer points, but not in between

If g and f agree on all the integers, they could still be different functions.  So we haven’t defined our order.  We need a better left order.  What can we do?  I know, let’s use a fact!

FACT: the rationals (numbers that can be written as fractions) are countable and dense (roughly, wherever you look in the reals, you’re either looking at a rational or there are a bunch in your peripheral vision).

So now we do the same thing, but using the rationals.  Enumerate them (remember, they’re countable) so use $q_1,q_2,q_3\ldots$ in place of 1,2,3… above.  It’s another fact that if g and f agree on all rationals, then they’re equal to each other.  Let’s make sure we have an ordering:

1. Transitivity: If f≤g and g≤h, then that means there’s some numbers (call them 2 and 3) so that f(2)<g(2) and g(3)<h(3).  But since we had to go to 3 to compare g and h, that means g(2)=h(2).  So f(2)<h(2), so f≤h.
2. Totality: if I have two different homeomorphisms, then there has to be a rational somewhere where they don’t agree, by the second fact.
3. Antisymmetry: We sidestepped this by defining < instead of ≤.  But it works.

Here’s a “classical” THEOREM: If G is a countable group, then it is left orderable if and only if it has an injective homomorphism to $\text{Homeo}_+(\mathbb{R})$.

Remember, injective means that each output matches to exactly one input.  Since we showed that there’s a left order on the group of orientation preserving homeomorphisms on the reals, we’ve already proven one direction: if you have an injection, then take your order on G from the order of the homeomorphisms that you inject onto.  So if h is your injection and g, k are your group elements, say that g<k if h(g)<h(k) in $\text{Homeo}_+(\mathbb{R})$.

One thing Mann does in her paper is come up with an example of an uncountable group that doesn’t do what the theorem says (she also does other stuff).  Pretty cool, huh?  Remarkably, the paper seems pretty self-contained.  If you can read this blog, you could probably do good work getting through that paper (with lots of time), which is more than I can say for most papers (which require lots of background knowledge).

That brings me to the “also”: I’ve been quite tickled to be asked about applying to grad school/what grad school entails a handful of times, some of those times by people who found me via this blog.  So please email me if you’re interested in whatever I have to say on the subject!  I’ve applied to grad school twice and have friends in many different departments and areas.  I hear I can be helpful.

### 7 Responses to “Best talk I’ve seen: Left orderable groups. Also, ask me about grad school!”

1. a reader July 30, 2015 at 10:02 pm #

Who gave the talk?

• Steve Clarke July 30, 2015 at 11:31 pm #

I believe it was Kathryn Mann

https://math.berkeley.edu/~kpmann/#fun

• yenergy July 31, 2015 at 12:35 am #

Thanks, it was her! I linked to her paper later in the post but didn’t make it explicit that she was also the speaker

2. Steve Clarke July 31, 2015 at 12:02 am #

Great math discussion, and comments about teaching and learning. I really enjoyed this post!

• yenergy July 31, 2015 at 12:36 am #

Thanks for stopping by and reading!

3. j2kun July 31, 2015 at 3:47 am #

The other side of the coin on bad talks: I have been told by many that if you’re giving a job talk, and at the end of the talk there’s anybody in the audience who is not lost, then you won’t get the job. I guess it’s that kind of culture that breeds bad talks even among those who can give excellent talks.

• yenergy July 31, 2015 at 9:50 am #

Rough! But also good to know. I remember an approachable talk at UIC once where the (tenured) speaker jokingly said that though we would all follow the talk, we could rest assured he is in fact great. So if even tenured folks add a disclaimer…

This has come up before in this blog-how do we change a culture? I for one love good talks…