Tag Archives: topology

## 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?

## Proof of Scott’s Criterion for separability (hard math) (UPDATED)

13 Sep

UPDATED: Thanks to my dear friend Teddy (who hasn’t updated his website and is at Cornell now, not UCSB), I’ve made the converse direction of the proof more correct.  There’s definitely still a glaring defect, but that’s entirely my fault.

This is out of character for this blog- it’s not accessible for most people.  If you have taken a course in algebraic topology, you can read this post and I’ll explain everything.  Otherwise, I’m not offering enough background to understand it.  Sorry!  Blame pregnancy!

I’ve been spending the past few months slowly slogging through a big paper that my advisor recently cowrote, on an alternative proof of Wise’s Malnormal Special Quotient Theorem.  In the paper they spend a few paragraphs explaining Scott’s Criterion for separability, from Scott’s 1978 paper (need access for this).  I did not understand it when reading, but after meeting with my advisor and drawing some pictures it made a lot more sense!  So I’m going to draw some pictures for anyone trying to understand this- probably other graduate students.

Here’s the theorem as it appears in the MSQT paper.

Theorem (Scott, 1978) Suppose X is a connected complex and  $H \leq \pi_1(X)$.  Then H is separable in $\pi_1(X)$ if and only if for every finite subcomplex $A \subset X^H$, there exists an intermediate finite degree cover $\hat{X}$ such that A embeds in $\hat{X}$.

Okay let’s unpack the theorem.  First we need to say what it means for a subgroup to be separableis separable in if, whenever you pick an element which is not in H, there exists some finite index subgroup of so that H is contained in and isn’t contained in K.  Intuitively, you can “separate” from via some finite index subgroup.  There are other equivalent definitions, but this is the one we’re going with.

Recall that a finite degree cover is a covering space where each point in has finitely many preimages.

Left side is an infinite cover, the real numbers covering the circle. Middle is a happy finite cover, three circles triple covering the circle. Right is a happy finite cover, boundary of the Mobius strip double covering the circle.

Notation wise, $X^H$ just means the cover of corresponding to H, so that $\pi_1(X^H)=H$.  Also, recall that an embedding is an injective homeomorphism onto the image of the map.  So, for instance, a circle definitely embeds into the middle cover above, but not into the infinite one.  You can map a circle injectively to a subset of the real line, say to [0,1), but it’s not a homeomorphism where the two ends meet.  And by connected complex let’s say CW-complex.

Okay it’s proof time!  For notation we’re going to let $G = \pi_1(X)$.

Suppose that the condition is met, and we want to show that is separable.  Take an element not in H.  Since is the fundamental group of Xcorresponds to a (class of homotopy equivalent) loop(s) in X.  Since isn’t in H, it isn’t a loop in $latex X^H$- let’s say it’s a line segment.  By the condition, since this line segment is a compact subset of $X^H$, there exists some intermediate finite degree cover $\hat{X}$ so that the line segment embeds into it.  This finite degree cover corresponds to a finite degree subgroup K.  Since $\hat{X}$ is intermediate, is contained in K.  So is separable.

Here’s a schematic:

I feel like this picture is self-explanatory (this is a joke)

Okay let’s do from the other side now.  Suppose is a separable subgroup of G.  Pick a finite subcomplex of (the actual criterion just says compact, but we’re sticking with finite).  Look at all the elements $g_i$ of which have preimages in A- since is finite, we only have finitely many of these.  For any given $latex g_i$, since is separable we have a finite index subgroup $K_i$ which doesn’t include $g_i$ and which contains H.

I guess we still need to show embeds- do you believe me that it does?  I’m not sure I believe me.

Pick a compact subcomplex of H.  Since it’s compact, there are finitely many open sets that we need to consider, which cover A.  And since it’s a subcomplex of a CW-complex, this means we’re only looking at finitely many open cells in H.  These open cells project down to X, say in a set D.  Look at all the preimages of up in $X^H$– there’s infinitely many, since we assumed $X^H$ is an infinite cover.  And is one of the preimages of by construction.  (Also let’s make D, small enough so we have a “stack of pancakes” instead of batter all over the place).  Again, schematic:

I know it looks like three, but there are actually Infinitely many preimages of the image of A

Now if we want an intermediate cover into which embeds, it can’t include any elements of that send to itself- that is, if $g.D\cap D \neq \emptyset$we don’t want g in $\pi_1(\hat{X})$.  Because then wouldn’t embed.  How many bad are there?  Well, since deck transformations act properly discontinuously, for any point in $X^H$ there’s an open neighborhood that never gets sent to itself (besides when is the identity, of course).  And we’re in CW-complexes, so we mean an open cell.  Look at the other cells of this particular component of D.  Again by proper discontinuity, there’s only finitely many g that’ll send this to some other copy of D.

Since is separable, for any one of these we have a finite index subgroup that doesn’t include and which does contain H.  Take the intersection of all these subgroups– since they all contain H, this intersection (call it K) does too.  And K doesn’t include any of the bad g.  Back in topology-land, K corresponds to a finite degree cover of X, since the intersection of finitely many finite-index subgroups has finite index.  And this cover is intermediate by construction, and embeds in it since there aren’t any bad g.

And that’s a proof of Scott’s criterion!  My next blog post will either be baking/cooking or a reasonable math post.

## Homeomorphisms of the torus, part IV (topology of the identity component)

30 Dec

See Part I for a definition of homeomorphism and torus and Part II for a bit more linear algebra.

I still owe a Part III for the explanation of the linear algebraic classification of homeomorphisms.  But let’s take a step away from linear algebra and look at shapes (my favorite!)

We know what homeomorphisms are (continuous functions with continuous inverses), with the famous example in the picture below: a coffee cup turns into a donut and vice-versa.

New meaning to cup of (j)Oe. From wikipedia

In fact, this pictures doesn’t just show us the homeomorphism (which says where each point of the coffee cup gets sent to in the torus and vice-versa).  It also shows us a homotopy (remember the definition from this post)- essentially, because we can see it traveling through time and back, it’s a homotopy.  And in fact, this homotopy is an isotopy– a type of homotopy where at each frozen point in time, the image is homeomorphic to what we started with.  An example of a homotopy which is not an isotopy is the map that ends up sending $x\mapsto -x$, where x is a real number.  Homotopies take place over time, so I would actually write this map as $\mathbb{R}\times [0,1] \to \mathbb{R}; (x,t)\mapsto (1-t)x+t(-x)$.  So when t=0, we have x mapping to x, and when t=1, x maps to -x.  One reason this isn’t an isotopy is because when t=1/2, all of the real line gets mapped to the point 0.  And mapping everything to 0 isn’t a homeomorphism (what would the continuous inverse be?)

A big part of geometric group theory is using shapes to come up with algebraic theorems, and using algebra to come up with shapes.  One thing you can do (which we’re doing RIGHT NOW!) is take a shape, do some algebra, and then make a new shape.  To be specific, our first shape is the torus.  Our algebra was figuring out the group of homeomorphisms of the torus, also written as Homeo(T)- T for torus.  Sometimes you’ll see $T^2$, to specify that we’re talking about the 2-torus rather than a higher dimension (more on higher dimensional tori later.  Isn’t it cool that the plural of torus is tori?  Pronounced tor-eye.)  Now we’re going to make a new shape from this group of homeomorphisms.

We’ll only consider homeomorphisms isotopic to the identity, written as $Homeo_0(T^2)$.  Starting in 1962 and finishing in 1965, badass Mary-Elizabeth Hamstrom proved in a series of papers that $Homeo_0(X)$ is contractible (homotopic to a point) if X is a two-manifold with a short list of exceptions [torus, sphere, plane, disk, annulus, disk with a hole in it, plane with a hole in it.]

Abstractly, I realize that there are many 10-year olds out there who could make a better picture than this. But I’m still so proud of myself. Exceptions to the theorem that the space of homeomorphisms isotopic to the identity is contractible.

Let’s look at our current favorite from this list, $Homeo_0(T^2)$.  If we start from the identity, which homeomorphisms can we isotope to?  Well, I can rotate my torus around its hole-axis, and that ending homeomorphism is definitely isotopic to the identity (the rotation through time is the isotopy; where the points end up is the ending homeomorphism).

Orange dot moves to red dot.

Since I can rotate by all the degrees up to 360, which brings me back to the identity, this means that $Homeo_0(T^2)$ contains a circle- each point on the circle represents rotating the torus by that many degrees.

What else can I do that’ll be isotopic to the identity?  I can rotate the torus around its center circle (running through the middle of the donut), like if I was wringing out a towel.

Again, orange dot to red dot.

Again, I can do this for 360 degrees before coming back to the identity.  So there’s another circle, different from the first one, in $Homeo_0(T)$.

I can also do any combination of these two: I can rotate 27 degrees around the hole-axis, and then 78 around the center circle.  This is true for any numbers between 0 and 360, but then 0 and 360 are the same for both.  So far, we have the picture on the left.  I colored it to indicate that 0=360 on both axes.  Look familiar?

Notice that all four corners are the same homeomorphism: the identity can be had by rotating by 360 degrees in either direction, or in both directions, or by doing nothing.  So we’ve shown that $Homeo_0(T^2)$ actually contains a torus!  This is cool because all those other $Homeo_0(X)$ were contractible.  In fact, $Homeo_0(T^2)$ is a torus- we’ve actually described all the homeomorphisms of the torus which are isotopic to the identity- combinations of rotations around the center axis and around the center circle.

Personally I find this much more exciting than classifying the homeomorphisms by trace (yeup that’s happening in Part III, nope Part IV is coming out before Part III and you’re going to like it), probably because it involves shapes rather than numbers.

Update on health: I’m taking antibiotics to help with whooping cough.  So that explains why I’ve been sick for a month.  My boyfriend walked by and asked what I was doing an hour ago, and I told him I was very busy feeling sorry for myself.  Then I wrote this blog post to be less lump-ish.