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## Dodecahedral construction of the Poincaré homology sphere

19 Apr

Update: Thanks as usual to Anschel for catching my typos!

This semester some grad students put together a learning seminar on the Poincaré homology sphere, where each week a different person would present another of the 8 faces from this classic (1979) Kirby-Scharlemann paper.  It was a fantastic seminar that I recommend to any grad students interested in algebraic geometry, topology, geometric group theory, that sort of thing. I did the last description, which is actually description number 5 in the paper.  You can read this post as a definition of the Poincaré homology sphere, without me telling you why mathematicians would care (but it has properties that makes mathematicians care, I promise).

First, start with a dodecahedron: this is one of the five Platonic solids, which are three-dimensional objects that can be created by gluing together regular (all sides are the same, all angles are the same) polygons so that the same number of polygons meet at any corner.  The fast example of a Platonic solid is a cube (three squares meet at each corner), and a non-Platonic solid is a square pyramid (4 polygons meet at the top, but only three at each corner).  If you glue two square pyramids together, you do get a Platonic solid, the octahedron.

Glue two pyramids together along their squares sides, and now four triangular faces meet at each vertex and you have a Platonic solid: the octahedron.

So after all that build up, here’s a dodecahedron: 12 pentagons glued together the only way you can: start with one pentagon, glue five to it (one on each edge), glue those together into a little pentagonal cap with a toothy bottom.  If you make two of these caps, you can glue them together; the teeth fit into each other just right.  This is the first step in this AWESOME VIDEO below (seconds 30-45 or so):

To make a the Poincare dodecahedral space, let’s first review the torus.  A long time ago, we learned about how to make a torus: take a square, identify opposite edges while preserving orientation.

First we glue the green arrow edges up together and get a cylinder, then the blue edge arrows together…

I’m a torus!

If you only identify one pair of edges and flip the orientation, you get a Mobius strip.  If you do that to both pairs of edges, you get a Klein bottle, which you can’t actually make in three dimensions.

Mobius strip picture from wikipedia

This torus/Mobius/Klein side note is just to review that we know how to glue edges together.  So look at the dodecahedron.  Each pentagonal face has a pentagonal face exactly opposite it, but twisted by 1/10 of a turn (2pi/10).  So if you identify each face with the opposite one, doing just the minimal turn possible, you get the Poincare homology sphere.  We started with 12 faces in our dodecahedron, so this glued-up space will have 6 faces.  It also has 5 vertices and 10 edges (vs. 20 vertices and 30 edges pre-gluing).  I can’t draw it for you because it’s a 3-manifold.  But here is a funny video of walking through it!

If you draw out all the identifications and you know some group theory, you can find the fundamental group of the thing, and you can prove to yourself that it is a 3-manifold and nothing funky happens at edges or vertices.

The dual to the dodecahedron is the icosohedron.  “Dual” means you put a vertex into the middle of each face of the dodecahedron, and connect edges of the dual if the corresponding faces share an edge in the dodecahedron.

Image from plus.maths.org

So you can see that the dual to the cube is the octohedron , and the tetrahedron is its own dual.  That’s all five Platonic solids!

Top row: tetrahedron, cube, octahedron.  Bottom row: dodecahedron, icosohedron.

There’s more to the story than this!  Let’s think about spheres.  The 1-sphere is a circle in the plane, aka 2-space.  Equivalently, the 1-sphere is all points that are equidistant from 0 in 2-space.  Similarly, the 2-sphere is all points equidistant from 0 in 3-space.   This gives you a notion of the 3-sphere.  How can we picture the 3-sphere?  We can use stereographic projection.

Here are the examples of stereographic projection of the circle and the 2-sphere onto the line and 2-space, respectively.  You cut out a single point from the north pole of the sphere, and attach the space to the south pole as a tangent.  Given some point on the sphere, run a line from the north pole through that point: it hits the space at exactly one point, and that’s the stereographic projection of the sphere-point.  Notice that the closer you get to the north pole, the farther out your projection goes.  If we pretend there’s one extra point (infinity) added to the plane, we can identify the n-sphere with n-space plus a point at infinity.  Look at this link and buy things from it if you want!

Projecting from the sphere to the plane: the bottom hemisphere of the red sphere maps to the pink circle in the plane, the top half maps to the rest of the plane.

What do circles that go through the north pole look like?  Just like when we projected the circle to the line, they look like infinite lines.

So we can see the three sphere as 3-space, plus a point at infinity.   Similarly here, circles that go through the north pole look like infinite lines.

Our math claim is that $\mathbb S^3/I^* \cong \partial P^4$, or that if I act on the 3-sphere by the binary icosohedral group, I get this exact dodecahedral space as the quotient.  Binary icosohedral goup is just some extension of the group of symmetries of the icosohedron, which is the same as the group of symmetries of the dodecahedron.  So we want to see a way to see this action.  The awesome video up top shows us how to start.  I’ll describe the contents of the video; you should read the next paragraph and re-watch the video after/while reading it:

Start with one dodecahedron.  Stack another on top of it, lining up the pentagons so you can glue one to another (that means the one on top is a 2pi/10 turn off from the bottom one).  Now make a tower of ten dodecahedra, all glued on top of each other.  Make a second tower of ten dodecahedra, and glue it to the first one (so it’ll twist around a bit).  Glue the top and bottom of the first tower together (they’ll line up because we did a 2pi total turn); this’ll automatically glue the top and bottom of the second tower together.  Nestle six towers like this together, so the toruses created from the towers all nestle together.  Now you have a torus of 60 dodecahedra.  Make a second torus of 60 dodecahedra.  Put the second torus through the hole of the first, so you get a solid ball.  (Here’s the weird 4-dimensional part!)  That is a 3-ball!  (The first torus also goes through the hole of the second one).  So now we have tesselated the 3-ball with dodecahedra; this is called the 120-cell.

I might make a more technical second post on this topic explaining in detail the action, but suffice it to say that we have an action by a group that has 120 elements, so that if we quotient out this 120-cell by the action, we end up with just one dodecahedron with the faces identified the way we want them to (opposite faces identified by a twist).  What is this group of 120 elements?  It’s derived from the symmetries of the icosahedron, which has the same symmetries as the dodecahedron!

Final interesting notes on this: we identified opposite sides by just one turn.  If you do two turns (so a 4pi/10 turn), you get the Seifert-Weber dodecahedral space.  If you do three turns, you get real projective space.

Jeff Weeks article on shape of space, a.k.a. is the universe a Poincare homology sphere?

Thurston book on geometry and topology

Fun website: Jeff Week’s geometry games

## Combinatorics fun with complexes

5 Apr

Last weekend I spoke at the Graduate Student Combinatorics Conference in Clemson, in one of those parallel sessions, so there were two other speakers slotted for the same 25 minute slot.  But those two didn’t show up, so I ended up being a “plenary speaker”!  Everyone came to my talk, which is funny because I’m in geometric group theory, not combinatorics.  But I got a lot of compliments afterward even though I only got through 2/3 of the prepared material (oops that’s what happens when you don’t practice/finish the talk 2 hours before showtime).  Related: I don’t think this is humblebragging, I believe in real bragging- you’re awesome, shouldn’t you tell people about it?

Anyways, I enjoyed the first graduate talk I saw, by a grad student at KU named Bennet Goeckner [I still am not sure about etiquette for blogging but I think I’ll start using peoples’ names instead of just linking to their websites.  If they get mad at me for google hits then I’ll go back to just links.]  It was based on a paper he coauthored with three professors, so I thought I’d tell you a bit about it.

I didn’t know anything about combinatorics before going to this conference.  Like, I didn’t even know it was a field of study, despite posting about an open problem in it almost exactly two years ago.  So I was very happy that in this first talk he defined an abstract simplicial complex, which is a basic object of study in combinatorics (at least, it came up in a ton of later talks).  This is a subset D of the set of subsets of {1, 2, 3,…, n} that follows a rule: if a is in D, and b is a subset of a, then b is also in D.  Example: if the set {1,2,3} is in D, then that means all of these sets are in D too: {1},{2},{3},{1,2},{2,3},{1,3}.  Here’s why we call it simplicial: you can see all of this information if you draw a triangle (a.k.a. a 2-simplex)!

Big picture of triangle contains all the info on the right (+smile!)

Might as well be thorough: the set of subsets is called the power set, so the power set of {1,2,3} is what we listed above, plus the entire set and the empty set.  Note that in our example, we have 8 sets in the power set of a 3-element set.  This is not a coincidence!  In general, a power set will have $2^n$ elements, if the original set had elements.

Also, a n-simplex is the convex hull of (n+1) points in space: so 3 points in 2 space makes a triangle.  4 points in 3 space makes a tetrahedron, a.k.a. a 3-simplex.  5 points in 4 space makes a 4-simplex.

0,1,2, and 3-simplices

So whenever you have abstract objects that satisfy the simplicial condition, you can build an abstract simplicial complex out of them.

Here’s an example from the talk: X=<1 2 3, 2 3 4>.  Convince yourself that this is two triangles glued together along an edge labeled by 2 and 3.  We can build a lattice that encodes the subset information in a different way then the triangles picture.  I also love this example because the lattice looks like a heart, and I ❤ lattices!

Red lines indicate that the bottom set is contained as a subset in the top set

We say a simplicial complex is partitionable if you can cut it up into Boolean intervals that end in the top layer (but start at some layer).  The picture shows you the partitioning, and you can kind of tell by looking what a Boolean interval is (it describes the skeleton of a n-cube for some n).

This simplicial complex was partitionable… but my heart isn’t (it belongs to GGT)

It’s a little hard to show that things aren’t partitionable.  Here’s an example that probably showed up in the talk but I didn’t write it down: Y= <1 2 3, 3 4 5>.

Simplex and lattice, plus a happy person wearing a bowtie!

If we make one of the partitions that contains the bottom empty set and one of the top sets, we can’t make the rest into partitions that start at the top.

No way to partition remaining 5 sets

Their paper answers a conjecture from 1979, which asked if all Cohen-Macaulay simplicial complexes are partitionable (Cohen-Macaulay has something to do with homology, which we haven’t done here but my friend Jeremy has a mathy post about it).  They said haha no!  They took a counterexample in something else, called Ziegler’s Ball, chopped it up a little bit, glued a bunch of copies of it to itself, and built something surprisingly nice (with 16 vertices) that is not partitionable.  This has applications in commutative algebra, besides being a fun combinatorial thing.  The paper is relatively short and approachable if you’re a grad student looking for something fun to read, and they ask three questions for further research at the end!

## What is a manifold? What is not a manifold?

29 Mar

I just went to a talk and there was one tiny example the speaker gave to explain when something is not a manifold.  I liked it so much I thought I’d dedicate an entire blog post to what was half a line on a board.

I defined manifolds a long time ago, but here’s a refresher: an n-manifold is a space that locally looks like $\mathbb{R}^n$.  By locally I mean if you stand at any point in the manifold and draw a little bubble around yourself, you can look in the bubble and think you’re just in Euclidean space.  Here are examples and nonexamples of 1-manifolds:

Red and orange are manifolds: locally everything looks like a line.  But yellow and green are not.

At any point on the orange circle or red line, if we look locally we just see a line.  But the yellow and green both have bad points: at the yellow bad point there are 2 lines crossing, which doesn’t happen in $\mathbb{R}$, and in the green bad point there’s a corner.

I messed up a little on the orange one but imagine that that is a smooth little curve, not a kink.

We call 2-manifolds surfaces, and we’ve played with them a bunch (curves on surfaces, curve complex, etc. etc.).  Other manifolds don’t have fun names.  In general, low-dimensional topology is interested in 4 or less; once you get to 5-manifolds somehow everything gets boring/collapses.  It’s sort of like how if you have a circle in the plane, there’s something interesting there (fundamental group), but if you put that circle into 3-space you can shrink it down to a point by climbing a half-sphere to the North Pole.

Empty pink circle in the plane can change size, but not topology (will always have a hole).  In 3-space, it can contract to a point.

The other thing we’ll want to think about are group actions.  Remember, a group acts on a set X if there’s a homomorphism that sends a group element g to a map $\phi_g:X\to X$ such that the identity group element maps to the identity map, and group multiplication leads to composition of functions: $gh \mapsto \phi_g \circ \phi_h$.  That is, each group element makes something happen on the set.  We defined group actions in this old post.  Here’s an example of the integers acting on the circle:

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

So far we’ve seen groups and manifolds as two different things: groups are these abstract structures with certain rules, and manifolds are these concrete spaces with certain conditions.  There’s an entire class of things that can be seen as both: Lie Groups.  A Lie group is defined as a group that is also a differentiable manifold.  Yes, I didn’t define differentiable here, and no, I’m not going to.  We’re building intuitions on this blog; we might go into more details on differentiability in a later post.  You can think of it as something smooth-ish without kinks (the actual word mathematicians use is smooth).

Top is smooth and differentiable. Bottom isn’t; there are weird kinks in its frown

So what are examples of Lie groups?  Well, think about the real numbers without zero, and multiplication as the group operation.  This is a manifold-at any point you can make a little interval around yourself, which looks like $\mathbb{R}$.  How is it a group?  Well, we have an identity element 1, every element has an inverse 1/x, multiplication is associative, and the reals are closed under multiplication.

Here’s another one: think about the unit circle lying in the complex plane.  I don’t think we’ve actually talked about complex numbers (numbers in the form x + iy, where is the imaginary square root of -1) on this blog, so I’ll do another post on them some time.  If you don’t know about them, take it on faith that the unit circle in the complex plane is a Lie group under multiplication.  Multiplying by any number on the unit circle gives you a rotation, which preserves the circle, again 1 is the identity, elements have inverses, and multiplication is associative.  Circles, as we said before, are 1-manifolds.

Examples of Lie Groups: the real line minus a point, and the unit circle in the complex plane

If you have a group action on a set, you can form a quotient of the set by identifying two points in the set if any group element identifies them: that is, and become one point if there’s a group element so that g.x=y.  For instance, in the action of the integers on the circle above, every point gets identified with three other points (so 12, 3, 6, and 9 o’clock on a clock all get identified under the quotient).  Your quotient ends up being a circle as well.  We denote a quotient of a group G acting on a set X by X/G.

So here’s a question: when is a quotient a manifold?  If you have a Lie group acting on a manifold, is the resulting quotient always a manifold?  Answer: No!  Here’s the counterexample from the talk:

Consider the real numbers minus zero using multiplication as the group operation (this is the Lie group $\mathbb{R}^{\times}$) acting on the real line $\mathbb{R}$ (this is a manifold).  What’s the quotient?  For any two non-zero numbers a, b on the real line, multiplication by a/b sends to a, so we identify them in the quotient.  So every non-zero number gets identified to a point in the quotient.  What about zero?  Any number times zero is zero, so zero isn’t identified with anything else.  Then the quotient $\mathbb{R}/\mathbb{R}^{\times}$ is two points: one for zero, and one for all other numbers.

If the two points are “far apart” from each other, this could still be a 0-manifold (locally, everything looks like a point).  But any open set that contains the all-other-numbers point must contain the 0-point, since we can find real numbers that are arbitrarily close to 0.  That is, 0 is in the closure of the non-zero point.  So we have two points such that one is contained in the closure of the other, and we don’t have a manifold.  In fact our space isn’t Hausdorff, a word I mentioned a while back so I should probably define in case we run into it again.  Hausdorff is a serious way of explaining “far apart.”  A space is Hausdorff (adjective) if for any two points in the space, there exist disjoint neighborhoods of the two spaces.  So the real line is Hausdorff, because even if you take two points that look super close, like 2.000000000001 and  2.000000000002, you can find infinitely many numbers between them, like 2.0000000000015.

Any two points on the real line, if you zoom in enough, have space between them.  So the real line is Hausdorff.

If you’re curious as to when the quotient of a smooth Manifold by a Lie Group is a manifold, you should go take a class to fully appreciate the answer (the Quotient Manifold theorem). The phrasing of the Quotient Manifold Theorem below is from a book by John Lee, called Introduction to Smooth Manifolds (the version from wikipedia gets rid of one of the conditions but also gets rid of much of the conclusion).  Briefly: a smooth action means the function on M is smooth (see the picture above; we didn’t do an in-depth definition of smooth), a free action means there aren’t any fixed points, and a proper action has to do with preimages of certain types of sets.

Theorem 21.10. Suppose G is a Lie group acting smoothly, freely, and properly on a smooth manifold M. Then the orbit space M/G is a topological manifold of dimension equal to dimMdimG, and has a unique smooth structure with the property that the quotient map π:MM/G is a smooth submersion.

## Connecting hyperbolic and half-translation surfaces, part II (math)

1 Mar

Last week we saw the standard definition for a hyperbolic surface.  You can tweak this standard definition to define all sorts of surfaces, and we tweaked it for a definition of half-translation surfaces.  Here are the two definitions:

•  A hyperbolic surface is a topological space such that every point has a neighborhood chart from the hyperbolic plane and such that the transition maps are isometries.
• A half-translation surface is a topological space such that all but finitely many points* have a neighborhood chart from the Euclidean plane and such that the transition maps are combinations of translations and flips.  These finitely many points are called singularities.

Precision note: according to Wikipedia, we need to add the adjective “Hausdorff” to our topological space.  We won’t worry about this or give a precise definition of it; you can just know that Hausdorff has something to do with separating points in our space.

Half-translation spaces come with something nifty that occurs in Euclidean space.  You know how when you look at a piece of notebook paper, it has all these nice parallel lines on it for writing?  Or if you look at a big stack of paper, each sheet makes a line which is parallel to the hundreds of others?

This is from clip art panda.  How useful!

Mathematicians call that a foliation: each sheet of paper is called a leaf.  This is an intuitive definition; we’re not going to go into a technical definition for foliation.  Just know that Euclidean space comes with a foliation of all horizontal lines y=r, where is some real number.  Then since transition maps of half-translation spaces come from either straight translations or flips, the foliation carries over to the half-translation space (though orientation might have flipped, we don’t care about those in this application).

All the slides are generously shared by Aaron Fenyes

Notice in the picture in the lower left that there are a few points where the horizontal foliation doesn’t quite work.  Those are the singularities that show up in the definition of a half-translation surface (we need them if we want our surface to be anything besides an annulus).

At those singular points, we glue together patches of Euclidean space.  The orange color in this picture shows the path of the critical leaf as it winds all the way around the surface some number of times.

Last week we had those nify gifs to show us how to think about curvature as positive, negative, or zero.  Here’s the example of zero curvature, because the last arrow is the same as the first arrow:

We can actually get precise numbers instead of just signs for curvature.

Here the triangle encompasses -π/3 curvature.  Notice that it embeds straight down into the hyperbolic surface, so we see an actual triangle down in the lower left.  If we made this triangle bigger and bigger, eventually it’d wrap around the surface and we wouldn’t see a triangle, just a bunch of lines hinting at a triangle up in the hyperbolic plane.  That’s the next picture.

Curvature can range from -π to π.  Here’s an example of an extremely negatively curved triangle which has -π curvature:

Such a triangle in hyperbolic space has all three corners on the boundary/at infinity.  This is called an ideal triangle.  So all ideal triangles encompass -π curvature.  You can see also how in the surface, we have a collection of lines whose preimage is the ideal triangle.

We can also use the same process to find curvature in other places.  For instance, if we make a little hexagon around a singularity of a half-translation surface, we can go around it with the same parallel transport process to figure out how much curvature the singularity contains.  We’ll make use of that horizontal foliation we saw earlier.

This looks very similar to our ideal triangle: the arrow starts off pointing up, and ends up pointing exactly the opposite direction.  So this singularity has -π curvature too, just like the ideal triangles.

Now for the math part!  Here’s the question: given a hyperbolic surface, how can we construct an associated half-translation surface?

Answer: we’ll use those foliations that we had before, as well as something called a geodesic lamination: this is when you take a closed subset of your surface, and give it a foliation.  So it’s like a foliation, only there’ll be holes in your surface where you didn’t define how the pages stack.  The first example of a geodesic lamination is a plain ol’ geodesic curve in your surface: the curve itself is a closed subset, and the foliation has exactly one leaf, the curve itself.  After this example they get real funky.

You don’t even have to take me to funkytown; geodesic laminations are already there!

Given a book, we might want to know how many pages we’ve read once we stick our finger in somewhere.  Luckily there are page numbers, so we can subtract the page number we started at from the page number we’re standing at.  Similarly, given a foliation, we might want to have a measure on it, transverse to the leaves.  If we have one, it’s called a measured foliation.  These exist.

So let’s start with our hyperbolic surface, and choose a maximal measured geodesic lamination on it.  Maximal means that the holes are the smallest they could possibly be.  Turns out this means they’re the images of ideal triangles under the atlas.

Told you they were funky.

Also, there are only finitely many of these triangle-shaped holes down in the surface (we’re sweeping some math under the rug here).  Now we need to get from this surface to a half-translation surface.  We’ll keep that foliation given by the lamination, and we need to get rid of those complementary triangles somehow.  So the lamination’s foliation will become the horizontal foliation of the half-translation surface, and the ideal triangles will correspond to singular points.  We can’t just collapse the ideal triangles to singular points, because as we saw earlier, images of ideal triangles are really funky and wrap around the surface.  We need to find smaller triangles to turn into singular points.  Here’s the picture:

Upstairs, we made a new purple foliation (transverse to the lamination) of the complementary ideal triangles, by using arcs of circles perpendicular to the boundary circle (these circles are called horocycles).  So now we have teensier triangles in the middle of the ideal triangles, called orthic triangles.  To make a half-translation surface, we’ll quotient out the horocycles, which means that in each ideal triangle, we identify an entire purple arc with one point.

Quotienting out horocycles a.k.a. identifying the pink lines all as individual pink points.  That means each side of the orthic triangle is identified with a point, so the orthic triangle disappears.

In this way we get tripods from triangles.  The middles of these tripods are singular points of the half-translation surface.  The measure from the measured lamination gives a measure on the foliation of the half-translation surface.

But Euclidean space actually comes with horizontal and vertical distances defined (remember, half-translation surfaces locally look like Euclidean space).  So far we have a way to get one direction.  How do we get the transverse distance?  We use the fact that we chose a geodesic lamination of our hyperbolic surface.  Geodesics are curves of shortest length; in particular they have length.  So if I’m in my translation surface and moving along a leaf of the foliation, I can look back at where I was in the lamination of the hyperbolic surface and use that distance.  [There’s some rug math here too.]  So we’ve made neighborhoods in the half-translation surface look like Euclidean space.

So that’s that!  You can also go backwards from a half-translation surface to a hyperbolic surface by blowing up the singular points into ideal triangles.  [More math, especially when the singularities of the half-translation surface are messy or share critical leaves].  Aaron claims this is folklore, but a quick google search led me to this paper (in French) and this one by the same author who connects flat laminations (on half-translation surfaces) to the geodesic ones we see in hyperbolic surfaces in section 5.

*I lied about finitely many points.  You can have infinitely many singularities in a half-translation surface; they just have to be discrete (so you should be able to make a ball around each other disjoint from the others, even if the balls are different sizes).  Examples of discrete sets: integers, $2^x, x>0$. Examples of not-discrete sets: rational numbers, $2^x, x<0$.

## Connecting hyperbolic and half-translation surfaces, part I (definitions)

24 Feb

I love talks that start with “I haven’t seen this written down explicitly anywhere, but…” because that means someone is about to explain some math folklore!  There are some statements floating around in mathland that specialists in those fields believe, so the rest of us believe them because the specialists said so, but no one knows a citation or a written proof for them.  That’s folklore.  Two weeks ago I gave a talk and someone asked a question (are RAAGS uniquely determined by their defining graphs?) and I said “probably, but I have no references for you.”  I found a reference a day later and emailed it to her, but the reference was way hard and had way more machinery than I was expecting.  The power of folklore!

Anyways, this series of posts will be based on a talk by a grad student at UT, Aaron Fenyes.  This was a great, great talk with lots of pretty slides, which Aaron has generously allowed me to put up here.  We’ll review curvature and surfaces, and then talk about how to go back and forth between two kinds of surfaces.

We’ve chatted about hyperbolic space v. Euclidean and spherical space before in terms of Euclid’s postulates, but let’s chat a bit about curvature. We say the Euclidean plane/real space has curvature 0, that hyperbolic space is negatively curved, and spherical space is positively curved.  There’s a nice way to see curvature: draw a triangle in your space (that old link also has some triangle conditions in it), and imagine standing at a point on that triangle and looking toward one corner of the triangle.  By “looking out” I mean your gaze should lie tangent to the triangle.  Remember:

Left: tangent; line hits circle at exactly one point.
Center: not tangent, line hits circle at two points
Right: not tangent, line doesn’t hit circle

Now walk toward the corner you’re facing, and then walk down the second side of the triangle still facing that direction (so you’re sidestepping), and walk around the next corner (so you’re now walking backwards) and keep going until you end up where you started.  This is called parallel transport.  If your triangle was in Euclidean space, then you’re facing the same way you were when you started.

Gif!  Note the ending blue vector is identical to the beginning purple vector.  So we have curvature 0.

If your triangle was slim, then you might find yourself facing the opposite way that you started!  Or if your triangle isn’t that curved, you’ll find yourself facing a direction counterclockwise from your original one.

Aaron Fenyes made the pictures; I gif’d it!  My first gif!

This is the summary of the gif: you can see how the ending light blue vector is pointed away from the original purple vector

From the original to the new vector: it’s a pi/3 counter-clockwise turn.

Similarly, if your triangle was fat, you’ll end up facing a direction clockwise from your original.

Here’s the picture of just the first and last arrows:

The green arrow is now clockwise from the red arrow, which means this has positive curvature.

So curvature is a way to measure how far clockwise you’ve turned after doing this parallel transport.

I love that description of curvature vs. the way I did it before, but they’re all good ways of seeing the same thing.  Next we need to review surfaces.  When we first met hyperbolic surfaces, we built them by gluing pairs of pants together, which themselves were stitched together from right angled hexagons which lived in hyperbolic space.  Redux:

Now if I take a little patch from my hyperbolic surface, I can trace back through one or two pairs of pants to find the original hexagon(s) in hyperbolic space where my patch came from.  So I have a map from hyperbolic space to my patch of hyperbolic surface, describing the metric and geometry around that patch.  This map is called a chart, and every point on a hyperbolic surface will have a chart associated with it, sending some part of hyperbolic space to a neighborhood of that point.

Here Aaron picked a blue patch and an orange patch in the surface, and the picture shows their charts from the hyperbolic plane to the patches.

This picture might make you leery: what happens when images of charts overlap, like they do here?  The preimages in the hyperbolic plane are disjoint, but they map to the same yellow area in the surface.  We want to say there’s some reasonable relationship between the yellow preimage patches in the hyperbolic plane.  That relationship is the only one we know, isometry:

There’s an isometry of the hyperbolic plane sending the orange patch to the new orange patch, so that the yellow parts overlap exactly.

If we look only at the yellow patch, we can find another way to describe the map in the picture: first, do the blue chart sending the blue patch to the surface.  Then, do the inverse of the orange chart, which sends the orange surface patch to its preimage.  Restricted to the yellow overlap patch, this is the definition of a transition map.

So here’s another way to think of hyperbolic surfaces, instead of gluing hexagons together like before.  A hyperbolic surface is a topological space such that every point has a neighborhood chart from the hyperbolic plane and such that the transition maps are isometries.

If you change where the chart is coming from, we can change the adjective before surface.  For instance, a flat surface is when the charts come from the Euclidean plane.  Now we’re going to define half translation surfaces, where the charts come from the Euclidean plane, but we have some more conditions on the transition maps.  The isometries of the Euclidean plane all come from a combination of translations and rotations.  Instead of allowing all isometries, we’ll only allow some of them:

In this picture you can see the orange and blue patches on the surface which come from the Euclidean plane.  Now we’re allowing translations and pi (180 degree) rotations only for our transition maps.  That’s why they’re called half-translation surfaces: charts from the Euclidean plane, and transition maps are translations plus half-rotations (flips).  As an aside, a translation surface is when we allow translations only, and no flips.

In the next post in this series, I’ll go through Aaron’s explanation of how we can go from hyperbolic surfaces to half-translation surfaces and back, and we’ll get to revisit our old friend the curve complex.  It’ll be fun!

## Playtime with the hyperbolic plane

2 Feb

Update: Thanks to Anschel for noting that I messed up the statement of the last exercise.  It’s fixed now.  Thanks to Justin for noting that I messed up a square root.  Pythagorean theorem is hard, yo.

About a year and a half ago I explained what hyperbolic space is, specifically by contrasting it with Euclidean space and spherical space.  We’ve also run into hyperbolic groups a few times, which are groups whose Cayley graphs are somehow like hyperbolic space.  More precisely, a group is hyperbolic if, whenever you have a Cayley graph of that group, triangles are $\delta-$thin, which means the third side of any triangle is contained in a $\delta$ neighborhood of the other two sides.  It’s important that the same $\delta$ works for every triangle in the space.

Here the bottom side is contained in a neighborhood of the other two sides, and the triangle looks like it belongs in Star Trek

Here each side is contained in a small neighborhood of the other two sides, and it seems like the triangle is curving inward

Note that triangles in Euclidean space are way totally far from being $\delta-$hyperbolic.  For any big number n, you can make a triangle so that the third side is not contained in an n-neighborhood of the other two sides: just take a 2n horizontal segment and a 2n vertical segment to make an isoceles right triangle.  If is bigger than 2, then the midpoint of the hypotenuse is farther than away from the other two sides.  As usual, this long paragraph could be better done in a picture.

Soooo not hyperbolic: you can make arbitrarily fat triangles in Euclidean space.  Also, the purple line should have $\sqrt{2}}n as its length, not the square root of n.[/caption] I thought today we could just play around with hyperbolicity. I'm running a small reading group on geometric group theory with some grad students, and today we got sidetracked a few times by just basic thoughts about geodesics in the hyperbolic plane. We all thought they were interesting, so here I am trying to share it with you! There are lots of other definitions of hyperbolicity, but I like$latex \delta-thin triangles. Oh I forgot to mention that a nneighborhood of a point/line/shape consists of all the points within n of that point/line/shape. So, for instance, a 3-neighborhood of a point in Euclidean space is a circle. But with a taxicab metric, that 3-neighborhood is a squarey circle. [caption id="attachment_3081" align="alignnone" width="181"] Purple points are all distance three or less from red point Anyways, I just put in that definition because it’s the first thing you’ll hear or see in a colloquium talk that involves the word “hyperbolic.” Let’s play with the upper half plane model of hyperbolic space! Here’s a repeat picture from that October 2014 post (wow that’s when baby was born! He’s walking around and getting into trouble now, btw.). Straight lines are ones that go straight up to infinity, and segments of half-circles whose diameters lie on the bottom line The graph paper lines in this picture are misleading; they contrast hyperbolic geodesics with Euclidean ones. So the gray lines are Euclidean geodesics, and the colored ones are hyperbolic. All geodesics in this model are either straight lines perpendicular to the horizontal axis, or semicircles perpendicular to the horizontal axis. All of the horizontal axis and everything that the straight up and down geodesics end at (sort of like a horizontal axis infinitely far away) represent infinity. I’ll write down the metric in case you were wondering, but we won’t need it for what we’ll be doing: $ds^2=\frac{dx^2+dy^2}{y^2}$ [I took this formulation from wikipedia]. What this says is that the hyperbolic metric is a lot like the Euclidean one, except that the higher up you go on the y-axis, the less distance is covered (because of that 1/y factor). More precisely, if you’re just looking at the straight line geodesics, the distance between two points at heights a<b is $ln(\frac{b}{a})$. All the lines have the same length ln(2). Blue: ln (8/4), green: ln (16/8) The other fact we might want to know is that things that look like Euclidean dilations (stretching something like your pupil dilates from looking in a bright light to a dark room) are isometries in this model. You can see that in the picture above: the lines look like they’re stretching longer and longer in the Euclidean metric, but they’re actually all the same length. Speaking of isometries, if you have any two geodesics (like a vertical line and a big old semi-circle somewhere else), you can find an isometry that sends one to the other. First question: what do circles look like? Whenever you have a metric space, it’s nice to know what neighborhoods look like, and the first thing you might want to consider are neighborhoods of points. Turns out circles in this model look like circles in Euclidean space, but the centers aren’t where you think they are. For instance, here’s a picture of circles with radius ln(2), which we saw in the straight lines above. The center of each circle is at the top of its surprised mouth. The next highest line segment shows that each vertical diameter is actually a diameter (twice the radius). Notice that the centers of these circles hang a lot lower in the circle than they do in the Euclidean metric! Isn’t playtime fun?! Generally when I play with math I throw out a lot of garbage ideas, and then eventually one of them is somewhat right. Other people apparently think for awhile before they put out an idea. Anyways, here are some sketches of what I thought a 2-neighborhood of a vertical line might look like: This is the most subtle joke I have ever put in this blog Maybe you looked at these and were like “Yen that is nonsense what were you thinking?!” Maybe you are my advisor or a practiced mathematician. Let’s go through the nonsense-ness of each of these pictures: The rightmost picture is a 2-neighborhood of the vertical line in Euclidean space. We know hyperbolic space is pretty drastically different from Euclidean space, so we wouldn’t expect the neighborhoods to be so similar. The middle and left pictures have similar shapes but different curviness, and yes we’d expect a hyperbolic neighborhood to look different so those are guesses based in some more intution. However, let’s try to figure out the actual size of a neighborhood of a vertical line. We can use our previous pictures, and switch to a ln(2) neighborhood. Changed my mind this is the most subtle joke I’ve put in this blog please someone get it and appreciate it please please Here I moved all our ln(2) circles so that their centers laid on the same line. A neighborhood of a line is just the union of the neighborhoods of all of the points on that line, so if we just keep making ln(2) circles along the line we’ll end up with a neighborhood of the whole line. So you can see that our actual neighborhood ended up being upside down from my middle picture above. If this explanation didn’t make sense, here’s [half] a 2-neighborhood of a Euclidean line: Note how the denser the circles, the closer their boundaries on the left get to becoming that straight line we see on the right. Actually using Euclidean intuitions and then mixing them up a bit is a great way to play with the hyperbolic plane. This next exercise was an actual exercise in the book but it is just so crazy I have to share it with you. It’s just dramatically different from Euclidean space, just like the triangles were. If you have a circle in the hyperbolic plane and project it to a geodesic segment that it doesn’t intersect (which means for any point on the circle, you find the closest point to it on the geodesic and draw a dot on the geodesic there), the projection is shorter than $ln(\frac{\sqrt{2}+1}{\sqrt{2}-1})$. Here’s the picture in Euclidean space where this makes no sense: Third place likes getting on the podium. I meant, the vertical lines show the projections from the faces to the horizontal line, and you can see they can be as big as you want if you just make bigger and bigger circles. And here’s a picture in hyperbolic space that might make you think this could possibly just maybe be true. Any circle will eventually fit inside a big huge circle that looks like the blue one in the picture, so its projection would be shorter than the projection of the blue one. That means you only have to worry about big huge circles in that particular position. And by “big huge,” I mean “of (Euclidean) radius n“. Remember, if we’re just looking at vertical lines, we know how to measure distance: it’s $ln(\frac{a}{b})$. So if you can show that the small orange circle hits the vertical line at $\sqrt{2}n-n$ and the big orange circle hits it at $\sqrt{2}n+n$, you’ll have proved the contraction property. Try using Euclidean geometry, and think about how we did the triangles case. That was fun for me I hope it was fun for you! ## Open problem in number theory 3 Dec A number theorist was trying to convince me that while you can do things easily for finite sets of prime numbers, it’s really hard to make a leap to infinitely many primes. She gave me a sketch of an example, and I thought I’d share it/some other interesting number theory things. We say that an integer is squarefree if for every prime number p, $p^2$ does not divide a. Remember, the primes are the numbers that are only divisible by themselves and 1, like 2, 3, 5, 7, etc. Every integer can be written as a product of prime numbers, so squarefree means what we think it does: no perfect square factors. Here are the first squarefree numbers which aren’t primes: 6, 10, 14, 22, 26, 30,… So you can ask a few questions as soon as you get this definition. How many squarefree numbers are there? A: infinitely many, because any product of two primes is squarefree, and there are infinitely many primes. But how big is that infinity? Remember, not all infinities are equal. Maybe a better question is, what’s the ratio of squarefree numbers vs. not-squarefree numbers? This seems a little nuts because we’ll have infinity divided by infinity, but it’s not nuts to say that half the numbers are even. So we’re not nuts! Huzzah! This is from wolfram alpha and shows 10000 numbers: the white pixels are squarefree. 1-100 are the bottom horizontal line of pixels, you can see 1, 4, 8,9,12,16,18,20 all blacked out in the lower left. A: about $6/\pi^2$ much of the numbers are squarefree, and this has been known since at least 1951. Just from the first 20, we have 60% squarefree, and $6/\pi^2$ means that as you look at bigger and bigger numbers, the proportion of squarefree numbers approaches 60.97….% But this is crazy, right? Where did that $\pi$ come from? Well, you can go prime by prime. For instance, about 1/9 of numbers are divisible by 9, so 8/9 of numbers are 9-free. And 3/4 are 4-free. So 3/4*8/9=2/3 are 4-and-9-free. You could do this for any finite set of primes, and say that the chance that islatex p^2\$-free for all of your finite set is $\prod_{p\in S} (1- \frac{1}{p^2})$.  Then we’d conjecture that you could make this an infinite product $\prod_{primes} (1-\frac{1}{p^2})$.

By this crazy formula from 1737 (oh that Euler!), $\sum_1^{\infty} n^{-s} = \prod_{primes} \frac{1}{1-p^{-s}}$.  And then that sum on the left is the definition of the Riemann zeta function, and $\zeta(2) = \frac{\pi^2}{6}$, which was proven by Euler 91 years after Basel posed the question.

But that step where we conjectured that you could jump to an infinite product?  It works in this case, but through a lot of hard work!  Which I do not understand nor will explain.

Here’s a related question.  Suppose you have a polynomial (a function in the form $f(x) = a_nx^n + a_{n-1}x^{n-1}+\ldots+ a_1x + a_0$), and you put numbers into it.  How many of those resulting function values are squarefree?  If you have something silly like $f(x) = 9x+9$, all values will be divisible by a square.  So let’s cut out polynomials like that.  Do squarefree values happen infinitely often?  And if so, how often [in the infinite sense that half of numbers are even]?

Open question: Does $f(x) = x^4+2$ take infinitely many square-free values?  If so, what is the probability that f(x) is squarefree, if you randomly choose an x?

You can ask a question with just putting in primes instead of putting in all numbers for too.  A polynomial is irreducible if you can’t write it as a product of other polynomials.  For instance, $x^3-1$ is not irreducible, since $x^3-1 = (x-1)(x^2+x+1)$, but $x^2+x+1$ is irreducible.  The degree of a polynomial is the biggest number that appears as an exponent of x, so the degree of $x^3-1$ is 3.

Open question (with conjecture): If f(x) is an irreducible polynomial of degree 3 or more, how many squarefree values does it take?

Dude, number theory is full of unsolved problems that are easy to state!  When reading up for this post, I ran into this magic squares problem.  Remember a magic square is one where the sum of all the numbers in each column, in each row, and along the diagonals is all the same number.

Sum of rows, columns, and diagonals is all 15.  I’m sorry I didn’t bother to make you a magic square; these values are from wikipedia.

Open question: Build a magic 3×3 square of square numbers.

I hope I’ve managed to convey some interesting number theory to you today!  Personally I’m not that into number theory, but I tried to not let that color this post.

Look at this bibimbap I made for dinner the other day!  My first time making bibimbap.

A photo posted by Yen Duong (@yenergyyyy) on Dec 1, 2015 at 7:18am PST

//platform.instagram.com/en_US/embeds.js

## Current research: lifting geodesics to embedded loops (and quantification)

19 Nov

Last week we learned about covering spaces, and I made a promise about what we’d talk about in this post.  For those who are more advanced, this all has to do with Scott’s separability criterion, so you can take a look back at that post for a schematic.  I’ll put the picture in right here so this post isn’t all words:

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.

In my friend Priyam Patel’s thesis, she has this main theorem:

Theorem (Patel): For any closed geodesic g on a compact hyperbolic surface $\Sigma$ of finite type with no cusps, there exists a cover $\tilde{\Sigma}\to\Sigma$ such that g lifts to a simple closed geodesic, and the degree of this cover is less than $C_{\rho}\ell_{\rho}(g)$, where $C_{\rho}$ is a constant depending on the hyperbolic structure $\rho$.

We know what geodesics are, and we say they’re closed if the beginning and end are the same point (so it’s some sort of loop, which might intersect itself a bunch).  But wait, Yen, I thought that geodesics were the shortest line between two points!  The shortest path from a point to itself is not leaving that point, so how could you have a closed geodesic?  Nice catch, rhetorical device!  A closed geodesic is still going to be a loop, but it won’t be the shortest path between endpoints because there are no endpoints.  Instead, just think locally: if a closed geodesic has length l, then if you look at any two points x and y less than l/2 apart from each other, the closed geodesic will describe an actual geodesic segment between x and y.  It’s locally geodesic.

What about hyperbolic surfaces of finite type with no cusps?  Well, we say a surface $\Sigma$ is of type (g, b, n) if it has genus (that’s the number of holes like a donut), boundary components, and punctures or cusps.

Pink: (4,0,0)
Orange: (3,0,2)
Green: (1,2,1)
Ignore the eyes they’re just for decoration

Boundary components are sort of like the horizontal x-axis for the half plane: you’re living your life, totally happy up in your two-dimensional looking space, and then suddenly it stops.  This is also what a boundary of a manifold is: where the manifold locally looks like a half-space instead of all of $\mathbb{R}^n$.  Surfaces are 2-manifolds.

Finally, I drew punctures or cusps suggestively- these are points where you head toward them but you never get there, no matter how long you walk.  These points are infinitely far from the rest of the surface.

I think we know all the rest of the word’s in Priyam’s theorem *(hyperbolic structure is a hyperbolic metric).  The important thing to take from it is that she bounds the degree of the cover above by a constant times the length of the curve.  This means that she finds a cover with degree smaller than her bound (you can always take covers with higher degree in which the curve still embeds, but the one she builds has this bound on it).

Just looking at this old picture again so you can have a sort of idea of what we’re thinking about

She’s looking for a minimum degree cover and finds an upper bound for it in terms of length of the curve.  Let’s write that as a function, and say $f_{\rho}(L)$ gives you the minimum degree of a cover in which curves of length embed (using the hyperbolic structure $\rho$).   What about a lower bound?

Here’s where a theorem (C in that paper) by another friend of mine, Neha Gupta, and her advisor come in:

Theorem (Gupta, Kapovitch): If $\Sigma$ is a connected surface of genus at least 2, there exists a $c=c(\rho, \Sigma)>0$ such that for every $L\geq sys(\Sigma), f_{\rho}(L)\geq c(\log(L))^{1/3}$.

So they came up with a lower bound, which uses a constant that depends on both the surface and the structure.  But it looks like it only works on curves that are long enough (longer than the systole length, which we’ve seen before in Fanoni and Parlier’s research: the length of the shortest closed geodesic on the surface).  Aha!  If you’re a closed geodesic, you’d better be longer than or equal to the shortest closed geodesic.  So there isn’t really a restriction in this theorem.  Also, that paper is almost exactly 1 year old (put up on arxiv on 11/20/2014).

Now we have $c_{\rho,\Sigma}(\log(L))^{1/3}\leq f_{\rho}(L) \leq C_{\rho}L$.

This is where it gets exciting.  We know from Scott in 1978 that this all can be done, and then Patel kickstarts the conversation in 2012 about quantification, and then two years later Gupta and Kapovich do the other bound, and boom! in January 2015, just three months after Gupta-Kapovich is uploaded to the internet, my buddy Jonah Gaster  improves their bound to get $\frac{1}{c}L\leq f_{\rho}(L)$, where his constant doesn’t even depend on $\rho$.  He does this in a very short paper, where he uses specific curves that are super hard to lift and says hey, you need at least this much space for them to not run into each other in the cover.

Here’s a schematic of the curves that are hard to lift (which another mathematician used to prove another thing [this whole post should show you that the mathematical community is tight]):

This curve in the surface goes around one part of the surface 4 times, and then heads over to a different part and circles that. This schematic is a flattened pair of pants, which we’ve seen before (so the surface keeps going, attached to this thing at three different boundary components).  I did not make this picture it is clearly from Jonah’s paper, page 4.

So that’s the story… for now!  From Liverpool (Peter Scott) to Rutgers in New Jersey (Priyam) to Urbana/Champaign in Illinois (Gupta and Kapovitch) to Boston (Jonah), with some quick nods to a ton of other places (see all of their references in their papers).  And the story keeps going.  For instance, if you have a lower bound in terms of length of a curve, you automatically get a lower bound in terms of the number of times it intersects itself ($K\sqrt{i(g,g)}\leq \ell(g)$, same mathematician who came up with the curves).  So an open question is: can you get an upper bound in terms of self-intersection number, not length?

## 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!