# 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!