• Layra Idarani, SG-invariant polynomials, 4 January 2018.

All that is a continuation of a story whose beginning is summarized here:

• John Baez, Quantum mechanics and the dodecahedron.

So, there’s a lot of serious math under the hood. But right now I just want to marvel at the fact that we’ve found a wavefunction for the hydrogen atom that not only has a well-defined energy, but is also invariant under this 7,200-element group. This group includes the usual 60 rotational symmetries of a dodecahedron, but also other much less obvious symmetries.

]]>• Layra Idarani, SG-invariant polynomials.

Namely, he takes any Coxeter group acting on its vector space and answers this puzzle:

**Puzzle.** What’s the dimension of the space homogeneous harmonic polynomials of degree on that are invariant under the even part of

**Answer.** Any Coxeter group has a list of numbers associated to it, and the answer to the puzzle is how many ways you can write as an unordered sum of s for and at most one copy of

These lists of numbers associated to Coxeter groups are explained in week186 and week187, but Layra says what these lists are. For the symmetry group of the icosahedron the list is

2, 6, 10

so the answer to the puzzle is: the number of ways we can write as a sum of 6s, 10s and at most one 15. Which is what we’d already seen!

But the list for is

2, 8, 12, 14, 18, 20, 24, 30

so we can now do this case as well. The dimension of the space of degree- harmonic polynomials on that are invariant under the even part of the Weyl group of is the number of ways you can write as an unordered sum of 2s, 8s, 12s, 14s, 18s, 20s, 24s, 30s and at most one 120.

]]>On the other hand, 6, 10 and 15 are the first 3 nontrivial binomial coefficients. And it just so happens that any number that can’t be written as a sum of these can’t be written as a sum of nontrivial binomial coefficients at all!

This is pretty easy to see. The next binomial coefficient, 21, doesn’t help get you new sums because 21 = 6 + 15. The one after that, 28, doesn’t help get you new sums because 28 = 6 + 6 + 6 + 10. And all the binomial coefficients after that don’t help at all because they’re bigger than 30, and every number greater than or equal to 30 is a sum of 6’s, 10’s, and 15’s.

(In fact you only need to use at most one 15 in this sum.)

All this reminds me of another interesting property of the number 30. It’s the largest natural number such that all natural numbers smaller than it and relatively prime to it are actually prime—except, of course, for 1:

7, 11, 13, 17, 19, 23, 29

This list is contained in your list. I don’t see any significance except that the niceness of the icosahedron is closely connected to it having a very ’round’ number of rotational symmetries, namely 60. The Babylonians liked this number because it was divisible by 2, 3, 5 and even 12… so of course sadistic Babylonian school-teachers gave homework where students had to divide by 7. (A tablet has been found with this homework problem on it.)

]]>Am I right in saying that SL(2,R) is the group of conformal transformation on the complex plane?

Not quite. The only conformal transformations of the complex plane are translations, rotations and dilations. It’s much better to include a point at infinity and consider the Riemann sphere. SL(2,C) acts as conformal transformations of the Riemann sphere via fractional linear transformations

SL(2,R) is the subgroup that preserves the upper half-plane H. The upper half-plane has a metric making it into the ‘hyperbolic plane’, and SL(2,R) acts as isometries. If you’re a physicist, you want to think of SL(2,R) as the double cover of SO(2,1), which acts on 3d Minkowski spacetime, and the upper half-plane as the hyperboloid

which is naturally a copy of the hyperbolic plane. To get number theory into the game we need to look at SL(2,Z) and its congruence subgroups like the groups Γ(n) that I was just talking about.

]]>So to form these binary tetrahedral/octahedral/icosahedral groups, we first reduce from SL(2,R) to SL(2,Z) , then look at quotient groups?

Right. This is part of a big branch of math called the theory of modular curves. A **modular curve** is what you get when you take the upper half-plane and mod out by a discrete subgroup of SL(2,R). It’s called a curve because it’s of *complex* dimension 1. The theory of modular curves is important in number theory: for example, it’s what Wiles used to prove Fermat’s Last Theorem.

Now I really don’t have intuition for how quotients act on the Riemann sphere. Any pedagogical sources for this?

I don’t think you can do much better than the blog series by my friend Tim Silverman, who started this discussion. He wrote a bunch of posts, but the first one already gets you to the Platonic solids:

- Tim Silverman, Pictures of Modular Curves (I), 6 October 2010.

[fn1] Am I right in saying that SL(2,R) is the group of conformal transformation on the complex plane? But now we’re talking about SL(2,Z) instead—is that just to focus on integer lattices in the complex plane instead?

]]>I hadn’t thought of finite subgroups of SL(2,R) which also happen to be finite subgroups of SO(3).

*Those* groups aren’t very interesting. I said the binary tetrahedral, octahedral and icosahedral groups are finite subgroups of SU(2) that are also finite quotient groups of SL(2,Z).

I know essentially nothing about the finite subgroups of SL(2,R).

They’re all cyclic groups,

Any finite subgroup of a Lie group G is compact so it must sit inside a maximal compact subgroup K ⊂ G. The *existence* of a maximal compact subgroup is a nontrivial theorem, but it’s true, and it’s good to learn the maximal compact subgroups of all your favorite Lie groups. For SL(2,R), every maximal compact subgroup is conjugate to SO(2). So, every finite subgroup of SL(2,R) is conjugate to a subgroup of SO(2). So, every finite subgroup of SL(2,R) is a cyclic group!