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## Counting like Gauss, Part II

24 May

Here are some pictures from last week:

The picture on the left describes technique 1, which we did last time to count how many points are in an octahedron with vertices at (0,0,n), (0,0,-n), (0,n,0),(0,-n,0), (n,0,0), and (-n,0,0).  The edges of the octahedron are lines between vertices that live in the same octant, and the faces live in each octant (so there are eight faces).  Now we’ll do Technique 2: counting the points in each octahedral shell, which is also called the spherical number.

Let’s look at just the yz plane, that is, where x=0.  So we want the integer points in the yz plane that lie on the n-shell of the octahedron.  Let’s stop and think about what we said in the last paragraph: there are eight faces because there are eight identical octants.  What if we count up all the points in one octant, then multiply by 8, and subtract any points we double counted?  Sounds great to me!  Let’s now just focus on when x, y, and are all greater than or equal to 0.

So now we have these points on a line such that x=0, and the line runs between (0,n,0) and (0,0,n).  Well thanks to point-slope formula, or slope-intercept form, or however you want to write your equation, we know that these points satisfy the equation of the line: y+z=n.  How many ways can I cut up n into two numbers and such that both are at least 1?  (I’m going to say they both have to be at least one because the vertices (0,n,0) and (0,0,n) will be special in the next paragraph).  Well if I choose y, that automatically chooses z.  I have n-1 choices for y (they are {1,…,n-1}), so that means I have exactly n-1 ways to choose and z.  Then there are n-1 points on the line.  I can either count and see I have 12 edges in my octahedron, so insides of edges contribute 12n-12 to the total count, or I can see I have three edges in my positive octant, multiply by 8 (because we have eight octants), and then divide by two since each edge lies in two octants.  Either way, we contribute 12n-12 to the final count.

By the same logic, I have exactly three vertices in my positive octant, each of which appears in 4 neighboring quadrants.  These contribute 6 to the final count (3*8/4=6 if we do the double counting method, or again we can just look at how many vertices are in an octahedron).

Now I’ve counted the edges and vertices in my octahedral shell, so I only need to figure out how many points lie inside the triangular face.  Again I can write an equation for the face: x+y+z=n.  Then I need to figure out how many ways I can partition n balls into three boxes, each with at least ball inside.  This is a partition problem.

We’ve counted the outside, now we need to find the number of integer points inside the triangle.

Let’s try to tackle this the way we did the last one.  But maybe let’s use an example, like the number 6.  How many ways can I cut up 6?  Well, I can choose 3 for x, which leaves 3 leftover to split into two numbers y and z.  From above I know there are 3-1 ways to choose, which I know since I can pick 2,1 or 1,2 for and z.  I can also choose 4 for x, which leaves 2 leftover, or 2 for x, which leaves 4 leftover, or 1 for x.  Let’s organize this into a table.

I’ve considered investing in a mouse, but haven’t quite bitten the bullet yet

Notice then that the number of ways I have to partition 6 into three bins, each of which has value at least 1, is 1+2+3+4.  This logic generalizes: I can pick any number from 1 to n-2 for x, and each of those options gives me n-x leftover to split into and z, and from above that’s n-x-1 such options.  So I have $\sum_{x=1}^{n-2} n-x-1=\sum_{x=1}^{n-2} x$.  Thanks to nine-year-old Gauss via fable, I know how to sum up the numbers from 1 to k: it’s a formula, $1+\ldots+k=\frac{(k+1)(k)}{2}$.  There’s a bunch of ways to explain this formula, here’s an easy to read collection.  So if I apply this formula to my sum, I have $\frac{(n-1)(n-2)}{2}$.  This is all the points inside the face that lie completely in the octant, so if I multiply this by 8 I have contributed 4(n-1)(n-2) points.

Then in the octahedral shell, I can add up all the contributions and get 6+12n-12+4(n-1)(n-2), which simplifies to $4n^2+2$.

We actually want to count all the lattice points inside an octahedron.  The octahedron is a bit special, since any lattice point inside it will lie on one of these octahedral shells, like a bunch of nesting Russian dolls.  So now I need to do $1+ \sum_{k=1}^n (4k^2+2)$, where I added the 1 at the beginning to represent the single point (0,0,0).  Luckily I also have a formula for the sum of squares, though without fun folklore to go with it.  The formula is $\sum_1^n k^2 = \frac{n(n+1)(2n+1)}{6}$.  Plugging that into my formula, I have the number of points in the octahedron is $1+4(\frac{n(n+1)(2n+1)}{6})+2n = \frac{4}{3}n^3+2n^2+\frac{8}{3}n+1$, which is EXACTLY WHAT WE GOT LAST WEEK!!!!

I hope this was as exciting for you as it was for me.  My friend Jeremy Kun and an anonymous number theorist commented last week about the Gauss Circle Problem and the error terms in it.  I just wanted to add that Gauss offered up the problem and the error guess, and it took 100 years until Sierpinski came up with a proof for something not that close to the error guess, and now it’s been like 200 years and we don’t quite have a proof for what we think the answer is yet.  Rumor has it that several years ago someone posted a proof to the arXiv, then quietly withdrew it.

Partner has pointed out that there’s a wikipedia entry for this problem but it doesn’t go into as much depth or have as cute pictures as my two blog posts.

## Counting points is fun! Gauss did it, why can’t we?

17 May

I’m writing to you from the Institute for Advanced Study (!!!! Starstruck!!!!  Einstein!  1930!  More importantly, Noether!)  I’m in the middle of a two week program  that’s been going on since 1994 for women and mathematics, and I wanted to share with you an easy-to-state, hard to solve combinatorics problem.

First, some two dimensional background and set up.  The Gauss Circle Problem has been around for a looong time (200+ years) and is not actually solved.  Let’s live in the xy plane, and mark points where and are integers in the plane.  We get a lattice of points:

Why didn’t I fill them all out?  Lattice find out!

The Gauss circle problem asks: if I make a circle with radius centered at the origin, how many lattice points are inside that circle?  We can count the first few:

There’s four points on the pink circle and one point inside, so G(1)=5.  The orange circle has four points on it, and it also adds the four points between the pink and orange circles.  So G(2)=4+4+5=13.  And the beige circle has eight points on it, and also adds more inside points, so G(3)=29.  Gauss thought that $G(n)=\pi n^2+ O(n)$.  We haven’t talked about this big O notation (it’s literally called big O notation; we’re very creative nomenclators), and I’m not going to do so here: take it as an “error term up to linearity”.  That means that while G(273) is around 74529π, it’s off by a factor that’s not “too far” from 273, a.k.a. probably closer to 273 than to 74529.

How do you count G(3)?  One way you can do it is count G(2), then count all the added points, like we did above.  You can also find a smaller region to look at, then multiply to fill in the circle and subtract double counted points.  In the example above, let’s just look at one quadrant.

I picked the northeast quadrant because that’s where I am!

It’s pretty easy to see how many points are in there: I count 11 integer lattice points.  Now we multiply by 4 to get 44 lattice points.  But we double counted!  Imagine multiplication by two as picking up that pie slice, and rotating it, and putting it back down in the tin in the northwest corner.  I’ve overlapped the first slice by one edge, so if I want to count the number of lattice points I’ll have to subtract the four points on that edge from my total.  Similarly, multiplying by four double counts all of the edge points on the green lines, and counts the origin four times.  So I need to subtract 3 for the quadruple counted origin, and 12 for the double counted edges: that leaves 44-15=29, which is good because it’s what I had above!

For the next problem, we’re going to use a black box called Pick’s Theorem.  It’s not an extremely dark black box since there’s a proof of it on that wikipedia page, but I’m not going to explain the proof here, just say the statement: If you have a polygon with integer coordinates in the plane, then A=i+b/2-1, where A is the area of the polygon, is the number of lattice points inside the polygon, and is the number of lattice points on the boundary of the polygon.  Notice this works for polygons, not circles (circles is the Gauss circle problem, still unsolved).

If I let A, i, and represent what they do for my initial polygon, I can write a formula for the number of lattice points after I scale the first polygon by n$G_P(n)=i_n+b_n=A_n+\frac{b_n}{2}+1$ by Pick’s theorem.  Since A stands for area and I’m scaling linearly by a factor of n, $A_n=n^2a$.  Similarly, b stands for boundary which is a linear factor, so $b_n=nb$.  Plugging these both in to the original equation gives $G_P(n)=n^2A+\frac{nb}{2}+1$, which is great!  You only need to find and for the first polygon to find the number of lattice points in any scaling of it.

Applying Pick’s theorem to right triangles aka the triangle numbers

So here’s the next problem: Find G(n) for an octahedron in three dimensions (we’ve run into the octahedron in the poincare homology sphere posts).  For a recap, the octahedron is what happens when you draw vertices at (0,0,n), (0,n,0), (n, 0,0), (-n,0,0), (0,-n,0), (0,0,-n) and draw the edges between them and then fill it in.

So now it’s a little harder and weirder to count, but we can still do it, using a mixture of the techniques above.  I’ll show two different ways to count this.

Because it’s messy I only drew the dots on the axes

Technique 1: use level sets!  My friend and co-author Andrew Sanchez came up with this clever idea, which uses the theorem we have above.  Imagine that we’re at the top of the octohedron, at point (0,0,n), and we start taking an elevator down the z-axis.  At the very top we see exactly one point.  At the next level, we see four points: (1,0,n-1), (0,1,n-1), and the same points with -1s instead of 1s.  At the next level we see a diamond: fixing z at n-2, we have a diamond between points (0,2),(2,0),(0,-2),and (-2,0).

Imagine the thin black line is the z axis, and we’re moving down it through the centers of these sets.

This is exactly $G_D(k)$ where is the diamond!  So from the formula above, we can calculate $G_D(k)=2k^2+2k+1$, by find the initial area (2) and number of boundary points (4) of the initial diamond.

Now we’ll need to add all of these points from all of the level sets.  The elevator starts with one point, which corresponds to z=n and k=0, and in the fattest part of the octahedron we have z=0 and k=n.  Then it gets skinny again.  So we can count all the points from z=0 to z=n, multiply by two, and subtract what we double counted (the fat base).  This shows up as follows: $2[\sum_{k=0}^n (2k^2+2k+1)] - (2n^2+2n+1)$.  There are a bunch of proofs that $\sum_{k=0}^n k^2= \frac{n(n+1)(2n+1)}{6}$ and that the sum of the first numbers is $\frac{n(n+1)}{2}$.  So if I go through the arithmetic (you can do it I believe in you!) I get $\frac{4}{3}n^3+2n^2+\frac{8}{3}n+1$.

Do you believe it?  Let’s do a second technique too!  Actually, hate to leave you on a cliffhanger, but a second technique will take a lot more words and this post is already long enough.  So you’ll have to wait til next week to see the technique I used with my friend Priyam Patel, whose current research I have previously blogged.  Preview: we use partitions!

## Dodecahedral construction of the Poincaré homology sphere, part II

26 Apr

Addendum: I forgot to mention that this post was inspired by this fun New Yorker article, which describes a 120-sided die.  It’s not the 120-cell; as far as I can tell it’s an icosahedron whose faces are subdivided into 6 triangles each.  The video is pretty fun.  Related to last week, Henry Segerman also has a 30-cell puzzle inspired by how the dodecahedra chain together.  In general, his Shapeways site has lots of fun videos and visual things that I recommend.

Side note: when I told my spouse that there are exactly 5 Platonic solids he reacted with astonishment.  “Only 5?!”  I’d taken this fact for granted for a long time, but it is pretty amazing, right?!

Last week we learned about how to make the Poincaré homology sphere by identifying opposite sides of a dodecahedron through a minimal twist.  I thought I’d go a little further into the proof that $S^3/I^*\cong \partial P^4$, where the latter is the way that Kirby and Scharlemann denote the Poincaré homology sphere in their classic paper.  This post is a guided meditation through pages 12-16 of that paper, and requires some knowledge of algebraic topology and group actions and complex numbers.

Honestly, I don’t know too much about $I^*$, but I do know that it’s a double cover of I, which is the group of symmetries of the icosahedron.  For instance, if you pick a vertex, you’ll find five rotations around it, which gives you a group element of order 5 in I.  Every symmetry will be a product of rotations and reflections.

Icosahedron from wikipedia, created from Stella, software at this website.

Last time we watched this awesome video to see how you can tessellate the three sphere by 120 dodecahedra, and explained that we can think of the three sphere as {Euclidean three space plus a point at infinity} using stereographic projection.

Hey that’s great!  Because it turns out that there are 60 elements in I, which means that $I^*$ has 120 elements in it.  Let’s try to unpack how acts on the three sphere.

First, we think of how the three sphere acts on the three sphere.  By “three sphere,” I mean all the points equidistant from the origin in four space.  The complex plane is a way to think of complex numbers, and it happens to look exactly like $\mathbb{R}^2$.  So if I take the product of two copies of the complex plane, I’ll get something that has four real dimensions.  We can think of the three sphere as all points distance 1 from the origin in this $\mathbb{C}^2$ space.  So a point on the three sphere can be thought of as a pair of points (a,b) , where and are both complex numbers.  Finally, we identify this point with a matrix $\left( \begin{array}{cc} a & b \\ -\bar{b} & \bar{a} \end{array} \right)$, and then we can see how the point acts on the sphere: by matrix multiplication!  So for instance, the point (a,b) acts on the point (c,d) by $\left( c \ d \right) \left( \begin{array}{cc} a & b \\ -\bar{b} & \bar{a} \end{array} \right)= (ac - \bar{b}d, bc + \bar{a}d)$, where I abused a bit of notation to show it in coordinate form.

What does this actually do?  It rotates the three sphere in some complicated way.  But we can actually see this rotation somewhat clearly: set equal to 0, and choose a to be a point in the unit circle of its complex plane.  Because is a complex unit, this is the same as choosing an angle of rotation θ [$a=e^{i\theta}$].

Remember how we put two toruses together to make the three-sphere, earlier in the video?  Each of those toruses has a middle circle so that the torus is just a fattening around that middle circle.  Now think about those two circles living in our stereographic projection.  One is just the usual unit circle in the xy plane of $\mathbb{R^3}$, and the other is the axis plus the point at infinity.  So how does (a, 0) act on these circles?  We can choose the basis cleverly so that it rotates the xy unit circle by θ, and ‘rotates’ the axis also by θ.  That means that it translates things up the axis, but by a LOT when they’re far on the z-axis and by only a little bit when they’re small.

We rotate the blue circle by the angle, and also rotate the red circle.  That means the green points move up the z-axis, but closer to the origin they move a little and farther away they move a lot.

Side note: this makes it seem like points are moving a lot faster the farther you look from the origin, which is sort of like how the sun seems to set super fast but moves slowly at noon (the origin if we think of the path of the sun as a line in our sky + a point at infinity aka when we can’t see the sun because it’s on the other side of the Earth).

Similarly, if we don’t have b=0, we can do some fancy change of coordinate matrix multiplication and find some set of two circles that our (a,b) rotate in some way.  In either case, once we define how the point acts on these two circles we can define how it acts on the rest of the space.  Think of the space without those two circles: it’s a collection of concentric tori (these ones are hollow) whose center circle is the blue unit circle, and whose hole centers on the red axis.  If you have a point on one of those tori, we move it along that torus in a way consistent with how the blue and red circles got rotated.

This is a schematic: the blue and green circles get rotated, so the purple point on the pink torus gets rotated the way the blue circle does, and then up the way the green circle does.

What does this have to do with I?  Fun fact: the symmetries of the icosahedron are the same as the symmetries of the dodecahedron!  (Because they’re duals).  So let’s look back at that tessellation of the 120-cell by dodecahedra, and stereographically project it again so that we have one dodecahedron centered at the origin, with a flat face at (0,0,1) and (0,0,-1), and a tower of ten dodecahedra up and down the z-axis (which is a circle, remember).

The origin dodecahedron.

Now imagine rotating around the green axis by a click (a 2pi/5 rotation).  This is definitely a symmetry of the dodecahedron.  It rotates the blue circle, and by our action as we described earlier, it also rotates the green circle, taking the bottom of our dodecahedron to the top of it (because $|e^{-\pi i/5}| = |e^{\pi i/5}| =1$).  So this identifies the bottom and top faces with that minimal rotation.  We said earlier that this rotation has order 5 in I, which means that it has some corresponding group element in $I^*$ with order 10.  10 is great, because that’s the number of dodecahedra we have in our tower before we come back to the beginning: so if we keep doing this group element rotation, we end up identifying the top and bottom of every dodecahedron in our z-axis tower of 10.

This is definitely a screenshot of that youtube video above, plus a little bit of paint so I could indicate the origin dodecahedron.

Similarly, using change of coordinate matrix basis madness, we can figure out how the rotations around the centers of each of the faces acts on the 120-cell (hint: each one will identify all the dodecahedra in a tower just like our first one did).  With 120 elements in $I^*$, each element ends up identifying one of the dodecahedra in the 120 cell with our origin dodecahedron, including that little twist we had when we defined the space.

So that’s it!  That’s how you go from the tessellation of the 120-cell to the Poincare homology sphere dodecahedral space.  Huzzah!

## 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, $2^x, x<0$.