Tag Archives: group theory

The (2,3,7) Triangle Group

8 Nov

Hi!  Today we’re going to use some stuff we learned about a long time ago (non-Euclidean geometry, manifolds, and groups) and put it together to explore a particular group.  This is based on a talk given by my dear friend and co-organizer Michelle Chu.  “Co-organizer?  Of what?” you ask.  Well thanks for asking!  Last weekend we did held the first Texas Women in Mathematics Symposium – we had over 60 people come, lots of talks, lots of networking, and lots of food.  By the end of the day I got to add “annual” to that description, and it seems like a lot of schools were interested in hosting it in future years.  Maybe some time I’ll write a post about how to found an annual conference (this is my second!).

Anyways, let’s first talk about tilings by triangles.  We first choose some integers p, q, r and set the three angles of a base triangle equal to $\frac{\pi}{p}, \frac{\pi}{q}, \frac{\pi}{r}$.  Now reflect over each of the three sides to start tiling your space.  Turns out this tiling will lead to a group.  Here’s an example with p=q=4 and r=2 (so we have a right isosceles triangle):

Start with the pink triangle, and reflect it over each of the three sides to make the colored triangles as shown.

Now do the reflections again.  I kept the pink base triangle and grayed out the first images.  Note that I colored the bottom left image yellow, for reflecting over the vertical side of the bottom orange triangle, but I also could color it orange, for reflecting over the horizontal side of the left yellow triangle.  This means that yellow+orange = orange+yellow in the group.

A third iteration of the same process; there are more relations here (that I didn’t color)

I picked a particularly good example, so that my triangles could tile the Euclidean plain.  We learned some time ago about non-Euclidean geometries: the space is spherical if the sum of triangle angles is more than $\pi$, and hyperbolic if triangles are thin and their sum of angles is less than $\pi$.  So based on how I choose my p, q, and r, I’ll find myself tiling different spaces.  Here’s an example of one iteration on a sphere for p=q=2 and r=5:

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It’s pretty easy to find the integer solutions for p, q, r to tile each space.  The only triangles that tile the flat plane are when (p,q,r) = (2,3,6), (2,4,4), and (3,3,3).  We already saw (2,4,4), and I’ll just tell you that (3,3,3) is when you use equilateral triangles (so there are 6 around each vertex), and (2,3,6) are those 30-60-90 triangles we learned about back in trigonometry class: here’s the picture from wikipedia:

Similarly there are only a few (relatively) that tile the sphere: (2,3,3), (2,3,4), (2,3,5), and (2,2, n), where is some number bigger than 1.  Of course this forms an infinite family of tilings, since you can choose n.  In the example above I chose n=5, and if is bigger the base pink triangle just gets skinnier.

But I say there’s only a few that tile the sphere because everything else tiles the hyperbolic plane.  That’s a lot of choices!  It might also make you think, “hm, maybe the hyperbolic plane is interesting.”

Let’s bring us back to groups.  How does a tiling of a space lead to a group?  Well, let the reflections over the (extended) sides of the base triangle be the generators of our group.  If I name these a, b, and c, I immediately get the relators $a^2=b^2=c^2=1$.  Next we have to figure out the rest of the relators.  I hinted at them in the pictures above.  They are $(ab)^p=(bc)^r=(ca)^q$.  Now we have a group presentation [link for definition] $R\triangle(p,q,r)\cong \langle a, b, c \ | a^2=b^2=c^2=(ab)^p=(bc)^r=(ca)^q=1\rangle$.

Also, fun coincidence: if you create the dual tree to the tiling by putting a vertex inside each triangle and connecting two vertices by a line if one triangle is the image of another under one of the reflections, you get something that looks a lot like the Cayley graph of the reflection triangle group.  The only difference is that each edge needs to be two edges (like a little loop) to reflect that each generator has order 2.

So what’s special about the (2,3,7) triangle group?  We know from above that it tiles the hyperbolic plane.  Check out this great picture from wikipedia of the tiling:

Maybe we’ll take a second here to point out that you can see the p, q, r values in the tilings, both in this picture and the other wikipedia picture above for (2,3,6): pick your favorite triangle, and look at its three vertices.  Each vertex is adjacent to other triangles, and since there are $2\pi$ angle around any vertex, we can figure out that p,q,r are just $\frac{n}{2}$.

Also, of all the integers you can pick for p, q, r, it turns out that 2, 3, and 7 maximize the sum $\frac{\pi}{p}+\frac{\pi}{q}+\frac{\pi}{r}$ while staying less than $\pi$.  [It ends up giving you $\frac{41\pi}{42}$ for those of you keeping track at home.]

So we maximize something with the numbers 2, 3, 7.  Well it turns out we do more than that- we also minimize the volume of the resulting quotient-we touched on and defined those in this post about manifolds.  And this is unique (up to conjugation/fiddling), and makes the smallest possible quotient.  Huzzah!

On a personal note, I’ve had a demonic cold keeping me in bed for the past two weeks, so forgive me if I messed up (pretty sure I did on the reflections; I’ll try to fix those soon).  Also, hope you voted today!  I voted a week and a half ago.

What is a covering space?

14 Nov

We’ve briefly covered fundamental groups before, and also I’ve talked about what geometric group theory is (using spaces to explore groups and vice versa). One way to connect a group to a space is to look at a covering space associated to that group. So in this post, we’ll come up with some covering spaces and talk about their properties. This is in preparation for talking about separability (we already have an advanced post about that).

Aside: you might catch me slipping into the royal we during my math posts.  This is standard practice in math papers and posts, even if a paper is written by a single author.  Instead of saying “I will show” and proving stuff to you the reader, we say “we will show” and we go on a journey together.  I’m sure that’s not why mathematicians do this, but I like to think of it that way.

Also, sometimes I say “group” when I’m obviously referring to a space, and then I mean the Cayley graph of that group (which changes depending on generating set, but if it’s a finite generating set then all Cayley graphs are quasi-isometric).

Let’s start with an example, and then we’ll go on to the definition.  Here’s an old picture to get us in the mood:

This blue curve goes around the circle three times.

This picture was from the short fundamental groups post: you’re supposed to see that the blue spiral up above represents a curve going three times around the circle below.  Now consider this next picture:

Blue line covering the happy circle below

Here the blue spiral goes on forever in both directions.  If you unwound it, you’d get a line stretching on forever in both directions, which we’ll call the real line (the same number line you’re used to, with real numbers along it).  This picture sums up the intuition that the real line covers the circle: for any point on the circle, there are a bunch of points on the real line directly above it that project down to that point.  In fact, it does more than that:

Pink parts of blue line cover the pink part of the circle

For any point on the circle, there’s a neighborhood (the pink part) so that up in the real line, there are a bunch of neighborhoods that map down to that pink part.  And those neighborhoods aren’t next to each other nor all up in each other’s business: they’re disjoint.  So here’s the definition:

A covering space X of a space Y is a space with a map p: X->Y such that any point in Y has a neighborhood N whose preimage in $p^{-1}(N)\subset$ X is a collection of disjoint sets which are homeomorphic to N.

So why is this helpful?  Well, in our example we can say that the real line covers the circle, from the pink pictures.  We could also say that the circle wound around itself three times covers the circle, from the first picture in this post:

The three highlighted parts up above are homeomorphic to the the pink part on the bottom circle’s chin

The picture I just drew might not convince you, because every point on the bottom space needs to have a neighborhood that lifts up to the top space, and what about the left most point of the circle?  Well, up above that neighborhood just winds around between the top and bottom copies:

Still a cover: each of those pink things up above are homeomorphic to the bottom cheek

The fundamental group of the circle is the integers, so maybe using geometric group theory (or algebraic topology, really) we can come up with conclusions about the integers using facts about the circle or the line, and vice versa.  In fact, there’s a correspondence between group structures and covering spaces!  With some conditions, covering spaces correspond to subgroups of fundamental group.

Let’s see how this correspondence works in our example with the integers.  We know that the even integers are a subgroup of the integers, and so are $3\mathbb{Z}, 4\mathbb{Z}$, etc.  In fact, these are all of the subgroups (and the trivial subgroup, which is just the element {0}).  Above, we drew two covering spaces of the circle: the real line, where each neighborhood of the circle has infinitely many homeomorphic copies hanging out in the real line, and the circle wound around itself three times, where each neighborhood has three copies.  The number of copies is called the degree of the cover, and sometimes one says the cover is an n-fold covering.  You can wind the circle around itself times for any n, which will correspond to the $n\mathbb{Z}$ subgroup.  How does this correspondence work?  Well, looking at the degree three/3-fold picture again, if you go around the covering circle once, you’ll project down to going around the base circle three times.  So if you go around the covering circle and count, you’ll get 0, 3, 6, 9… In contrast, the real line corresponds to the trivial subgroup (and is an infinite degree cover), and it’s called the universal cover of the circle.  Every space has a unique universal cover, which is a covering space with trivial fundamental group.

Now a preview of why we’ll like this.  Sometimes spaces are tricky and not fun and it’s easier to look upstairs at a cover, and then go back downstairs.  Let’s let the downstairs space be two circles pinched together at a point.

First, you should get convinced that the picture above is a cover; I colored the homeomorphic copies in order to highlight what’s happening.  Also, pretend the branching part goes on forever, a la the Cayley graph of the free group on 2 generators:

from wikipedia

So here’s an example: let’s say we have a path downstairs that goes around the green circle several times.  And maybe we don’t want this path to hit itself over and over again, so we look at a cover upstairs so it turns into a line instead.  So instead of just being immersed (locally injective), the path is embedded (injective) in the cover upstairs.

The orange scribble downstairs goes around the green loop over and over again, hitting itself. Upstairs, it’s a line and doesn’t hit itself

Next time I write about current math research, I’ll be using covering spaces a lot!  In fact, one of the main questions is this: if you have a path downstairs that hits itself (is immersed), what’s the minimum degree cover you need to ensure that the path is embedded in the cover?  This question isn’t explicitly answered yet for loops on surfaces, but the research I’ll blog about gives some bounds on the degree.

The fundamental theorem of geometric group theory, Part II: proof

1 Oct

A refresher from last week: If a group G acts on a proper geodesic metric space X properly discontinuously and cocompactly by isometries, then G is quasi-isometric to X.  Moreover, G is finitely generated.

Yes, I put “proper geodesic metric space” in italics because I forgot it in the statement of the theorem last week.  [Aside: actually, you only need “length space” there, and proper geodesic will follow by Hopf-Rinow.  But let’s skip that and just go with proper geodesic.]  I also added the second sentence (which isn’t really a moreover, it comes for free during the proof).

At the end of last week I defined proper: closed balls are compact. A space is geodesic if there is a shortest path between any two points which realizes the distance between those points.  For instance, the plane is geodesic: you can just draw a line between any two points.  But if you take away the origin from the plane, it’s not geodesic anymore.  The distance between (1,1) and (-1,-1) is $2\sqrt{2}$, but the line should go through the origin.  There is no geodesic between those two points in this space.

Now we have all our words, so let’s prove the theorem!  I’ll italicize everywhere we use one of the conditions of the theorem.

Since our action is cocompact, we have a compact set K so that translates of it tile X.  Pick a point inside K, call it $x_0$, and a radius R so that K is entirely contained inside a ball of radius R/3 centered at $x_0$.  For notation, this ball will be labelled $B(x_0,R/3)$.

Schematic: K is the red square, special point is the yellow dot, yellow circle is radius R/3, lighter circle is radius R. Cartoon on right illustrates this relationship

We’ll pick a subset of the group G: Let $A =\{ g\in G: g.B(x_0,R)\cap B(x_0,R) \neq \emptyset\}$.  X is proper, so closed balls are compact.  Since the action is properly discontinuous, this means that is finite.  [Reminder: properly discontinuous means that only finitely many group elements translate compact sets to intersect themselves].

Now we’ll show that G is finitely generated, and it’s generated by A.  Choose some group element g in G.  Draw a geodesic in between your special point $x_0$ and its g-translate $g.x_0$.  Now we’ll mark some points on that geodesic: mark a point every R/3 away from $x_0$, all the way to the end of the geodesic.  You’ll have [(length of the segment)/(R/3) rounded up] many points marked.  Let’s call that number n.

There are n blue points, and they’re all R/3 away from each other. Notice the last blue point might be closer to g.x_0, but it’s definitely less than or equal to R/3 away.

Here’s the clever part.  Remember that K tiles X by G-translates (cocompactness), so each of those blue points lives inside a G-translate of K.  Since $x_0$ lives inside K, that means there’s a nearby translate of $x_0$ to each blue point.  And since K fits inside a R/3 ball, each translate is less than or equal to R/3 away from its blue point.

The green points are translates of x_0: I also colored x_0 and g.x_0. The yellow circle indicates the the green point is within R/3 of its blue point.

We can bound how far apart the consecutive green points are from each other: each one is within R/3 of its blue point, which are all R/3 apart from their neighbors.  So the green points are at most R/3+R/3+R/3= R from each other.

Middle portion is exactly R/3 long. So by the triangle inequality, the green points are less than or equal to R from each other.

Remember that the green points represent G-translates of $x_0$.  In the picture above I numbered them $g_0.x_0=x_0,g_1.x_0,g_2.x_0,\ldots g_nx_0=g.x_0$.  We just said that $d(g_1.x_0,g_2.x_0)\leq R$.  Since G acts by isometries, this means that $d(g_2^{-1}g_1.x_0,x_0)\leq R$.  So $g_2^{-1}g_1$ lives inside our set A that we defined above- it moves $x_0$ within of itself.

Here’s a bit of cleverness: we can write $g=g_n=g_0^{-1}g_1\cdot g_1^{-1}g_2 \cdots g_{n-1}^{-1}g_n$, because all of the middle terms would cancel out and we’d be left with $g=g_0\cdot g_n = 1\cdot g = g$.  But each of those two-letter terms lives in A, so we just wrote as a product of elements in A.  That means that A generates G.  We said above that A is finite, so G is finitely generated.

That was the “moreover” part of the theorem.  The main thing is to show that G is quasi-isometric to X.  Let’s try the function $g\mapsto g.x_0$.

Above, we wrote as a product of elements of A, so that means that the length of is at most n.  In other words, $d_G(1,g)\leq n$.  Now we’d like to bound it by $d_X(x_0,g.x_0)$.  We found by dividing the geodesic into pieces, so we have $n\leq \frac{d_X(x_0,g.x_0)}{R/3}+1$, where we added a 1 for the rounding.  So we have one side of the quasi-isometry: $d_G(g_1,g_2)\leq \frac{3}{R}d_X(g_1.x_0,g_2.x_0)+1$ (using the action by isometries).

Now we need to bound the other side, which will be like stepping back through our clever argument.  Let M be the maximum distance that an element of translates $x_0$.  In symbols, $M=max_{a\in A} d_X(x_0,a.x_0)$.  Choose some in G, with length n.  That means we can write as a product of elements in A: $g=a_1\cdots a_n$.  Each $a_i$ moves $x_0$ at most M.  If we step between each translate, we have $d(a_i.x_0,a_{i+1}.x_0)=d(a_{i+1}^{-1}a_i.x_0,x_0)\leq M$.  There are steps from $x_0$ to $g.x_0$, and each step contributes at most M to the distance.  So $d_X(x_0,g.x_0)\leq M d_G(1,g)$.

With bounds on both sides, we can just pick the larger number to get our actual quasi-isometry.  We also need the function to be quasi-onto, but it is because the action is cocompact so there are translates of $x_0$ all over the place.

Huzzah!

The fundamental theorem of geometric group theory (part I), topology

24 Sep

I love the phrase “THE fundamental theorem of…” It’s so over the top and hyperbolic, which is unlike most mathematical writing you’ll run into.  So you know that it’s important if you run into the fundamental theorem of anything.  By now we all have some background on geometric group theory: you’ll want to know what a group action is and what a quasi-isometry is.  (Refresher: a group G acts on a space X if each group element g gives a homomorphism of the space X to itself.  A quasi-isometry between two spaces X and Y is a function f so that distances between points get stretched by a controlled scaling amount + an additive error term).  We say a group G is quasi-isometric to a space X if its Cayley graph is quasi-isometric to X.  Remember, a Cayley graph is a picture you can draw from a group if you know its generators.

Still from Wikipedia: a Cayley graph of the symmetries of a square

There are several more terms we’ll want to know to understand the theorem, but I’ll just do one more before we start.  We say a group G acts on a space X by isometries if it acts on X, and each homomorphism is actually an isometry (it preserves distance).  So for instance, the integers acting on the real line by multiplication isn’t by isometries, because each homomorphism spreads the line out (so the homomorphism of the reals to themselves given by 3 is $x \mapsto 3x$, which stretches out distances).  But if the action is defined by addition, then you’re okay: $x\mapsto x+3$ preserves distances.

Under the red function, f(2)-f(1)=6-3=3, but 2-1=1, so this isn’t an isometry.
Under the green function, f(2)-f(1)=5-4=1, which is equal to 2-1. This is always true, so this is an isometry.

So here’s the fundamental theorem:

If a group G acts properly discontinuously, cocompactly, and by isometries on a proper metric space X, then G is quasi-isometric to X.

You can parse this so far as “If a group G acts by isometries on a space X with condition condition condition, then G is quasi-isometric to X.”  Putting aside the conditions for now, how would we prove such a theorem?  Well, to show something is quasi-isometric, you need to come up with a function so that the quasi-isometry condition holds: for all x,y in X, we need $\frac{1}{K} d_G(f(x),f(y))-C\leq d_X(x,y) \leq K d_G(f(x),f(y))+C$.

So let’s deal with those conditions!  An action is cocompact if there’s some compact subset S of X so that G-translates of S cover all of X.  Remember, each element g in G gives an isometry of X, so it’ll send S to some isometric copy of itself somewhere else in X.  In our example above, the integer 3 will send the compact subset [5,7] to the isometric copy [8,10].  In fact, our example action is cocompact: you can keep using [5,7] as your compact set, and notice that any point on the real line will eventually be covered by a translate of [5,7].  For instance, -434.32 is covered by [-435,-433], which is the image of [5,7] under the isometry given by -440.

This action is also cocompact. Here I have the plane, conveniently cut up with an integer lattice. Can you figure out what the action is? Hint: the red square is a unit square, and the pink squares are suppose to be various translates of it.

G acts on X properly discontinuously if for any two points x,y in X, they each have a neighborhood $U_x, U_y$ so that only finitely many g make $g.U_x\cap U_y\neq\emptyset$.  Let’s look at our example action again.  If I take the points 4365.234 and 564.54 in the real line, I’d like to find neighborhoods around them.  Let’s choose the intervals [4365,4366] and [564,565].  The only integers that make these hit each other are -3801 and -3800.  In particular, 2 is finite, so this indicates proper discontinuity.  If we actually wanted to prove the action is properly discontinuous, we’d want to show this is possible for all numbers, not just these two specific ones I chose.

Schematic of proper discontinuity: only finitely many g will send the yellow oval to hit the blue blob

Finally, a metric space X is proper if all closed balls are compact.  Balls are intuitively defined: they’re all the points that are at a fixed distance or less from your center.  In the plane, balls are circles, centered around points.  And compact-well, aw shucks I haven’t defined compact and we’ve been using it!  Time for some topology.  We’ll prove this theorem next time around, this post is just definitions and background.  (Sorry for the cliffhanger, but it should be clear what we’re going to do next time: make a function, show it’s a quasi-isometry).

Just like groups are the fundamental object of study in algebra, open sets are the fundamental object of study in topology.  You’re already familiar with one type of open set, the open interval (say, (6,98), which includes all numbers between 6 and 98 but doesn’t include 6 and 98).  I just described another above: balls.  So, open circles in the plane are open sets.  Sets are closed if their complement is open: that is, the rest of the space minus that set is open.  In the real line example, [6,74] is closed because $(-\infty,6)\cup(74,\infty)$ is open (it’s the union of infinitely many open sets, say (74,76) with (75,77) with (76,78) and on and on).

Notice that I haven’t yet defined what an open set is.  That’s because it’s a choice- you can have the same space have different topologies on it if you use different definitions of open sets.  I’ll point you to these wikipedia articles for more examples on that.

A set is compact if every covering of it by open sets has a finite subcover.  That means that any time you write your set S as a union of open sets, you can choose finitely many of those and still be able to hit all the points of S.  From above, the set $(74,\infty)$ is not compact, because you can’t get rid of any of the sets in that infinite covering and still cover the set.  On the real line, a set is compact if it’s closed and bounded (this is the Heine-Borel theorem, a staple of real analysis).

So that’s enough for today.  More next time (like a proof!)  Also, I’m using my husband’s surface to blog this, which means I did all the pictures using my fingers.  It’s like finger painting.  What d’you think?  Better than usual pictures, or worse?

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.

Universal acylindrical actions

25 Jun

I’m at a fantastic summer graduate school at MSRI (the Mathematical Sciences Research Institute, a.k.a. “math heaven”) right now and re-met a friend I’d seen at a few earlier conferences.  I saw that she’d posted a preprint up on arXiv recently, so I thought I’d try to blog about it!

Remember that a group is a collection of elements paired with some kind of operation between them (the integers with addition, rational numbers with multiplication, symmetries of a square with composition).  For that operation, you put in two group elements and get another group element out.  You can imagine different functions with different inputs and outputs.  Like you might have a function where you put in Yen and late night, and it outputs pumpkin.  Or you could put one group element in, and a location, and get a different location [like if you put in the group element -2 to the location (3,3), maybe you get (1,1)].  More precisely, a group action on a space is a homomorphism* which takes in a group element and a point in the space and outputs a (possibly different) point on that space.  For instance, you can give an action of the integers on the circle by saying that rotates the circle by $n/2\pi$.

Each integer rotates the circle by pi/2 times the integer. Looks like circle is getting a little sick of the action…

In the picture above, if you input the integer 2 and the original purple dot, you get the new location of the dot (180 degrees from its original location, aka pi away).  If you say the original purple dot is location and the new location is y, the notation is that 2.x=y.  A homomorphism is a function that respects this: f(xy)=f(x)f(y).

We say a space is hyperbolic if it locally “looks like” hyperbolic space (there’s a particularly nice function between it and hyperbolic space).  The title of Carolyn’s paper is “Not all acylindrically hyperbolic groups have universal acylindrical actions,” so we need to learn what “acylindrical” means (look, we’ve already learned what groups and actions are, and we know the words “not”,”all”,and “have”!  We’re doing great!)

Here’s the precise definition, and then I’ll break it down:

An action of a group on a hyperbolic space is called acylindrical if, for any $\epsilon >0,$ there exist numbers M,N>0 such that for every pair of points x,y with distance d(x,y)>M, the number of group elements that translate both x,y by less than epsilon is bounded by N: $|\{g: d(x,g.x)\leq \epsilon, d(y,g.y)\leq \epsilon\}| \leq N$.

Here’s the non math-y intuition for this: if you have a pool noodle and you spin one end around, the other one generally will fly away from where it used to be.

Here’s the math-y intuition for this: choose two points that are M-far apart.  Make a little $\epsilon$-circle around each, then connect the two with a cylinder.  The condition says that only a few group elements preserve the cylinder (that means that when acts on all the points in the cylinder, it maps them back into other points in the cylinder).  So if you have a bunch (perhaps infinitely many) elements that preserve one circle, most of them send the other circle/rest of the cylinder away.

A group is called acylindrically hyperbolic if you can find a hyperbolic space on which the group acts acylindrically.  In practice, such groups actually act on a whole bunch of different spaces acylindrically.

Now suppose that you’ve got an element in G and you want to see how that particular element acts.  We say is loxodromic if you can find a space and a point in it so that the map $\mathbb{Z}\to X$ that sends an integer to the orbit of the point $n\mapsto g^n.s$ is a quasi-isometry– roughly, if you draw all the points that gets mapped to if you apply over and over again, you get something that looks like a line.

The older tree is the same as the younger tree up to scaling (multiplication) and adding some constants (the leaves). This is an example of a quasi-isomeTREE.  [Also pretend both trees go on forever.]

Just for fun here’s a picture of something that’s not a quasi-isometry:

The ribbon on the right goes on forever in both directions, so it’s not quasi-isometric to the tree

You might’ve noticed above that we say an element is loxodromic if we can find space on which it acts in this particular way.  But we also said that a group can act on several different spaces.  So even if an element acts loxodromically on one space, that doesn’t necessarily mean it acts loxodromically on another space (even if the group acts on that other space).  We actually call an element generalized loxodromic if there exists some space on which it acts loxodromically.  Then if you can find an action so that all generalized loxodromic actions are, in fact, loxodromic, you’ve found a universal acylindrical action.  So this paper gives an example of an acylindrically hyperbolic group that doesn’t have such an action.

Blog notes: For the summer I’m going to blog every Thursday (day was chosen arbitrarily).  Also, I went back and tagged all the gluten-free recipes as gluten-free.  And you should know that whenever I mention a person in this blog by name or link to them, that means that I admire them/am inspired by them.

What is the fundamental group?

29 May

I’m at Yale for my fifth reunion, and it’s my birthday!  Happy birthday to me!  I’m a little overwhelmed by seeing all the old friends/catching up/showing off pictures of the baby so I’m hiding in my suite and pumping milk and writing a quick post about the fundamental group.

We’ve talked before about what a group is– a set of elements with some operation that takes two elements to another one (like addition with the group of integers takes 5+3 to 8 or multiplication takes 1*9 to 9) which satisfies some group axioms.  Given a geometric or topological object, we can associate a group with it my defining these elements and an operation, and making sure that they satisfy the axioms.

First we fix a basepoint of our space, which means that you pick a point and say that’s the one, that’s the special one I want.  Then our group elements with be isotopy classes of loops (this means you can wiggle loops to be the same, as in the red ones below) that go through the basepoint.

Red curves are homotopic to each other; blue curve is not

The group operation is concatenation– first you do one loop starting and ending at the basepoint, then you do the next loop starting and ending at the basepoint.  You can homotope away the middle connection to more clearly see the resulting loop.

Here’s the example:

Red and green make… more red. I didn’t want to make a brown curve

I don’t want to FOMO my reunion (I already ran away to take a nap) so we’ll make this super fast and just look at the fundamental group of the circle and of the torus.

How can I make a loop around a circle?  Well, there’s one obvious way- make one full circle and end up where you starting.  You can homotope to something that backtracks for a bit and then comes forward again, and you could go around two different directions (counterclockwise vs. clockwise).  So let’s call these +1 and -1.

Red goes around once counterclockwise, even though it backtracks a bit, and orange goes around once clockwise.  Imagine the colors down on the black circle.

If you put the red and orange curves together, concatenating like we did above, you’d fully backtrack over yourself, which means you could homotope to just a point.  Let’s call that 0 (can you guess where we’re going here?)

This blue curve goes around the circle three times.

You can go around any integer number of times, but no fractions because you won’t end up back at the basepoint.  So this is a rough schematic of why the fundamental group of the circle is the integers.

I want to go back to my reunion, so I’ll just tell you that the fundamental group of the torus is $\mathbb{Z}^2$, a.k.a. ordered pairs of integers, as I hinted in a previous post using the following picture.

Left: follow the numbers to see the knot. Right: look at the bottom-most green line.

Sorry for the short and delayed post.  It’s my birthday, YOLO.  (I’m not that sorry about this post being delayed but I am sorry if it is unclear/too short please comment/let me know if you need more explanation).