In his 9 May 2005 Week 215, John Baez says:

"... The science fiction writer Greg Egan has been helping me understand Klein's quartic curve ...1) Greg Egan, Klein's quartic curve ...

... topologically speaking, it's a 3-holed torus. It's tiled by triangles, with 7 meeting at each vertex.

... Dual to ... a tiling of the hyperbolic plane by regular heptagons with 3 heptagons meeting at each vertex ... is a tiling of the hyperbolic plane by equilateral triangles with 7 triangles meeting at each vertex.

We can take a quotient space of this by a certain symmetry group and get a 3-holed torus tiled by 56 triangles with 7 meeting at each vertex. This is what Egan drew!

... First, you can take any vertex and twist it, causing the 7 triangles that meet at this vertex to cycle around. ... This gives a 7-element symmetry group.

Second, the whole thing looks like a tetrahedron, so it inherits the rotational symmetries of a tetrahedron. This gives a ... 12-element symmetry group.

... there's also a 2-fold symmetry that corresponds to turning the tetrahedron inside out! And Egan made a wonderful

movieof this. If a picture is worth a thousand words, this is worth about a million ...... So, we get a total of 7 x 24 = 168 symmetries.

... the tetrahedron in Egan's movie gets

reflectedas it turns inside out. More precisely, if you follow the four corners of the tetrahedron, you'll see that two come back to where they were, while the other two get switched. So, this symmetry acts as a reflection, or odd permutation, of the 4 corners. The rotations act as even permutations of the corners.This means that the Klein quartic has 24 symmetries forming a group isomorphic to the rotation/reflection symmetry group of a tetrahedron. ...

[ In his book Complex Regular Polytopes (2nd edition, Cambridge 1991), Coxeter says the "... fundamental region is ... a spherical triangle ( p q r ) having angles pi/p, pi/q, pi/r ... every finite group of rotations is ...[ 1:2 ]... homomorphic to a finite group of unit quaternions including the 'central' quaternion -1 ... ... every finite reflection group has a subgroup of index 2 which is a rotation group, generated by products of pairs of reflections. ... A convenient symbol for this rotation group of order 2s is (p,q,r) ... the tetrahedral group (3,3,2) of order 12, the octahedral group (4,3,2) of order 24, the icosahedral group (5,3,2) of order 60 ... are subgroups of index 2 in ... [3,3] [4,3] [5,3] respectively ... ... We shall ... use the symbol GL(2,3) as an abbreviation for ( C4 / C2 ; <4,3,2> / <3,3,2> ) ... where ... <3,3,2> ... is a normal subgroup of ... <4,3,2> ... whose quotient group is ...[ C4/2 ]... the order of ... ( C4 / C2 ; <4,3,2> / <3,3,2> ) ... is ...[ 2 ]... times the order of ... <3,3,2> ... ... Apart from the little complication caused by the common element -1 of ... C4 ... and ... <4,3,2> ... we have here an instance of a 'subdirect product' ... Corresponding to each group [p,q] of order 4s, generated by reflections in three planes of Euclidean 3-space, there is a group ... of order 8s, generated by reflections in three lines of the unitary plane. In this manner ... [3,3] yields ( C4 / C2 ; <4,3,2> / <3,3,2> ) = GL(2,3) ...". (3,3,2) = A4 is order 12 tetrahedral rotations (without reflections) [3,3] is order 24 tetrahedral rotations and reflections <3,3,2> = SL(2,3) order 24 binary tetrahedral (quaternionic) group = vertices of 24-cell in quaternionic space 336 elements of SL(2,7)include <4,3,2> = order 48 binary octahedral (quaternionic) group = vertices of 24-cell and its dual 24-cell <3,3,2> = SL(2,3) order 24 binary tetrahedral (quaternionic) group = vertices of 24-cell [3,3] = order 24 Euclidean tetrahedral reflection group and cover 2:1 all 168 elements of PSL(2,7) = SL(3,2) = Klein Quartic including 24-element octahedral (4,3,2) = S4 and 12-element tetrahedral (3,3,2) = A4 This brings to mind some further questions: Since 1 copy of binary tetrahedral = 1 x 24 = 24-element SL(2,3) and 7 copies of binary octahedral = 7 x 48 = 336-element SL(2,7) then: Can you say that the 4 faces of the tetrahedron are like the 4 dimensions of quaternions with the 3 of SL(2,3) being the 3 quaternion imaginaries ? Is the 1 copy due to the 1 associative triangle that can be constructed from the 3 imaginary quaternions ? Can you say that the 8 faces of the octahedron are like the 8 dimensions of octonions with the 7 of SL(2,7) being the 7 octonion imaginaries ? Are the 7 copies due to the 7 associative triangles that can be constructed from the 7 imaginary octonions ? ]

... Egan was able to spot a hidden cube lurking in his picture ... each corner of his tetrahedral gadget is made of a little triangular prism with one triangle facing out and one facing in: for example, the pink triangle staring you right in the face, or the light blue one on top. Since 4 x 2 = 8, there are 8 of these triangles. Abstractly, we can think of these as the 8 corners of a cube ... The way these 8 triangles come in pairs corresponds to how the vertices of a cube come in diagonally opposite pairs ... you can even draw apictureof a cube on the Klein quartic by drawing suitable curves that connect the centers of these 8 triangles ... If we gave the Klein quartic the metric it inherits from the hyperbolic plane, the edges of the cube would be geodesics. ...... The Klein quartic is tiled by 56 triangles. 8 of them give the cube we've just been discussing. In Egan's picture these triangles look special, since they lie at the corners of his tetrahedral gadget. But this is just an illusion caused by embedding the Klein quartic in 3d space. In reality, the Klein quartic is perfectly symmetrical: every triangle is just like every other. So in fact there are lots of these cubes, and every triangle lies in some cube. ... In fact, each triangle lies in just

onecube. So, there's precisely one way to take the 56 triangles and divide them into 7 bunches of 8 so that each bunch forms a cube. So: the symmetry group of the Klein quartic acts on the set of cubes, which has 7 elements.... this symmetry group also acts on the Fano plane, which has 7 points. This suggests that cubes in the Klein quartic naturally correspond to points of the Fano plane. And Egan showed this is true! ... The Fano plane also has 7 lines. What 7 things in the Klein quartic do these lines correspond to?

Anticubes!You see, the cubes in the Klein quartic have an inherent handedness to them. You can go between the 8 triangles of a given cube by following certain driving directions, but these driving directions involve some left and right turns. If you follow the mirror-image driving directions with "left" and "right" switched, you'll get ananticube. Apart from having the opposite handedness, anticubes are just like cubes. In particular, there's precisely one way to take the 56 triangles and divide them into 7 bunches of 8 so that each bunch forms an anticube. Here's a picture:... Each triangle has a colored circle and a colored square on it. There are 7 colors. The colored circle says which of the 7

cubesthe triangle belongs to. The colored square says which of the 7anticubesit belongs to. If you stare at this picture for a few hours, you'll see that each cube is completely disjoint from precisely 3 anticubes. Similarly, each anticube is completely disjoint from precisely 3 cubes. This is just like the Fano plane, where each point lies on 3 lines, and each line contains 3 points! So, we get a vivid way of seeing how every figure in the Fano plane corresponds to some figure in the Klein quartic curve. This is why they have the same symmetry group.... Here we are beginning to see how two superficially different geometries are secretly the same:

FANO PLANE KLEIN'S QUARTIC CURVE 7 points 7 cubes 7 lines 7 anticubes incidence of points and lines disjointness of cubes and anticubesHowever, we're only half done! We've seen how to translate simple figures and indicence relations in the Fano plane to complicated ones in Klein's quartic curve. But, we haven't figured out translate back!

KLEIN'S QUARTIC CURVE FANO PLANE 24 vertices ??? 84 edges ??? 56 triangular faces ??? incidence of vertices and edges ??? incidence of edges and faces ???Here I'm talking about the tiling of Klein's quartic curve by 56 equilateral triangles. We could equally well talk about its tiling by 24 regular heptagons, which is the Poincare dual. Either way, the puzzle is to fill in the question marks. I don't know the answer!

To conclude - at least for now - I want to give the driving directions that define a "cube" or an "anticube" in Klein's quartic curve. Say you're on some triangle and you want to get to a nearby triangle that belongs to the same cube. Here's what you do:

- hop across any edge,
- turn right,
- hop across the edge in front of you,
- turn left,
- then hop across the edge in front of you.
Or, suppose you're on some triangle and you want to get to another that's in the same anticube. Here's what you do:

- hop across any edge,
- turn left,
- hop across the edge in front of you,
- turn right,
- then hop across the edge in front of you.
(If you don't understand this stuff, look at the picture above and see how to get from any circle or square to any other circle or square of the same color.)

You'll notice that these instructions are mirror-image versions of each other. They're also both 1/4 of the "driving directions from hell" that I described last time. In other words, if you go LRLRLRLR or RLRLRLRL, you wind up at the same triangle you started from. You'll have circled around one face of a cube or anticube! In fact, your path will be a closed geodesic on the Klein quartic curve... like the long dashed line in Klein and Fricke's original picture ...

...".

In his 20 April 2005 Week 214, John Baez says:

"... each heptagon [ in the picture immediately above from Week 215 ] ... has been "barycentrically subdivided" into 14 right triangles. But don't worry about that yet; concentrate on the heptagons. There's a blue heptagon in the middle, 7 red ones touching that, 7 yellow ones touching those, then 7 green ones falling off the edge of the picture, and 2 blue ones broken into bits all around the corners of the picture. That's a total of 24 heptagons.We wrap this thing up into a 3-holed torus using the numbers on the edges of the picture:

- connect edges 1 and 6
- connect edges 3 and 8
- connect edges 5 and 10
- connect edges 7 and 12
- connect edges 9 and 14
- connect edges 11 and 2
- connect edges 13 and 4
In other words, connect edges n and n+5 mod 14. To connect them the right way, make sure that triangles of the same color never touch each other. ... Ignore the little triangles; just pay attention to the heptagons! Then:

- Start on any edge of any heptagon and march along in either direction.
- Then, when you get to the end, turn left.
- Then, when you get to the end, turn right.
- Then, when you get to the end, turn left.
- Then, when you get to the end, turn right.
- Then, when you get to the end, turn left.
- Then, when you get to the end, turn right.
- Then, when you get to the end, turn left.
- Then, when you get to the end, turn right.
- You should now be back where you started!!!
These are like the driving directions the devil gives people who ask the way out of hell. LRLRLRLR and you're right back where you started. But the resulting Platonic surface is heavenly. It has lots of symmetries. Each of the 24 heptagons has 7-fold rotational symmetry - and amazingly, all these rotations extend to a symmetry of the Platonic surface!

Moreover, if you do any of these rotations and then do another one centered at some other heptagon, you get another rotation centered at another heptagon. So, our Platonic surface has a symmetry group with 7 x 24 = 168 elements! These are all the symmetries except for reflections. If we include reflections, the symmetry group has size 168 x 2 = 336.

Now let's talk about those little triangles. Since our surface is made of 24 heptagons, each chopped into 14 right triangles, there are a total of 24 x 14 = 336 triangles. This is just the number of symmetries!

This is no coincidence. We can specify a symmetry by saying where it sends our favorite right triangle. Since it can go to any other triangle, there are 336 possibilities.

By the way, this trick works for ordinary Platonic solids as well. For example, if we take a dodecahedron and barycentrically subdivide all 12 pentagons, we get 10 x 12 = 120 right triangles. If we pick one of these as the "identity element", we can specify any symmetry by saying which triangle this triangle gets sent to. So, the set of triangles becomes a vivid picture of the 120-element symmetry group of the dodecahedron. It's called the "Coxeter complex". This idea generalizes in many directions, and is incredibly useful.

Anyway... there is much more to say about this stuff. For example,

if we take our hyperbolic plane tiled with heptagons and count them grouped according to how far they are from the central one, we get the sequence 7, 7, 14, 21, 35, 56, 91, .... These are 7 times the Fibonacci numbers!If we think of the hyperbolic plane as the unit disc in the complex plane, this surface becomes a "Riemann surface", meaning that it gets equipped with a complex structure. This was Felix Klein's viewpoint when he discovered all this stuff in about 1878. He showed this surface could be described by an incredibly symmetrical quartic equation in 3 complex variables: u^3 v + v^3 w + w^3 u = 0 where we count two solutions as the same if they differ by an overall factor. So, it's called "Klein's quartic curve". ...

... Klein's quartic curve turns out to have the maximum number of symmetries of any 3-holed Riemann surface.

... There's only one way to make a sphere into a Riemann surface - it's called the Riemann sphere. You can think of it as the complex numbers plus a point at infinity. This has infinitely many symmetries. They're called conformal transformations, and they all look like this:

az + b z |---> -------- cz + dThey form a group called PSL(2,C), since it's the same as the group of 2x2 complex matrices with determinant 1, mod scalars. It's also the same as the Lorentz group!

There are different ways to make a torus into a Riemann surface, some with more symmetries than others (see "week 124"). But, there are always translation symmetries in both directions, so the symmetry group is always infinite.

On the other hand, a Riemann surface with 2 or more holes can only have a finite group of conformal transformations. In fact, in 1893 Hurwitz proved that a Riemann surface with g holes has at most 84(g - 1) For g = 3, this is 168. So, Klein's quartic surface is as symmetrical as possible! (We don't count reflections here, since they don't preserve the complex structure - they're like complex conjugation.)

... Sitting inside PSL(2,C) is PSL(2,Z), where we only use integers a,b,c,d in our fractional linear transformation

az + b z |---> -------- cz + dThis subgroup acts on the upper half-plane H, which is just another way of thinking about the hyperbolic plane.

Sitting inside PSL(2,Z) is a group G(7) consisting of guys where the matrix

a b c dis congruent to the identity:

1 0 0 1modulo 7. This is an example of a "congruence subgroup"; these serve to relate complex analysis to number theory in lots of cool ways. In particular, Klein's quartic curve is just H / G(7) Since G(7) is a normal subgroup of PSL(2,Z), the quotient group PSL(2,Z)/G(7) = PSL(2,Z/7) acts as symmetries of Klein's quartic curve. And, this group has 168 elements!

In fact, this group is the second smallest nonabelian simple group. The smallest one is the rotational symmetry group of the icosahedron, which has 60 elements. This group is actually PSL(2,Z/5), and Klein had run into it in his work on the icosahedron and quintic equations (see "week 213"). So, it's actually far from sheer luck that he then moved on to PSL(2,Z/7) and ran into his wonderful quartic curve. By the way, this 168-element group is also known as PSL(3,Z/2) - the symmetry group of the "Fano plane". This is a name for the projective plane over Z/2. The Fano plane is closely related to the octonions ... in fact, our 168-element group acts on the set of octonion multiplication tables...

... Having the number 56 on my brain, I can't resist noting that if you take Klein's quartic curve tiled by heptagons, and you count the vertices, you get 24 x 7 / 3 = 56 since each vertex is shared by 3 heptagons. ...".