@littmath@mathstodon.xyz

· 38mo ago

Replying to

After a huge amount of work, by Boalch, Hitchin, Kitaev, Dubrovin-Mazzocco, and others, the classification in this case was finished by Lisovyy-Tykhyy, relying on an intensive computer calculation. Here’s part of their classification: 12/n

13

10

3

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

For the experts, the group generated by the half-twists is the braid group, and it acts on Xₙ. 8/n

9

15

1

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

Everything here is joint work with Aaron Landesman and Josh Lam -- if you're an expert, you can read the paper here: https://arxiv.org/abs/2308.01376 2/n

arXiv.org

Finite braid group orbits on $SL_2$-character varieties

Let X be a 2-sphere with n punctures. We classify all conjugacy classes of Zariski-dense representations $$ρ: π_1(X)\to SL_2(\mathbb{C})$$ with finite orbit under the mapping class group of X, such that the local monodromy at one or more punctures has infinite order. We show that all such representations are "of pullback type" or arise via middle convolution from finite complex reflection groups. In particular, we classify all rank 2 local systems of geometric origin on the projective line with

8

2

1

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

Please let me know if you have any comments or questions! I'm very happy to discuss more. 21/n, n=21.

8

15

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

Let’s define Xₙ={A₁, ⋯, Aₙ ∣ ∏ Aᵢ=id}/∼ to be the set of solutions, up to equivalence. Xₙ is an interesting geometric space—for example, if n=4 and one restricts the Aᵢ appropriately, one gets the Cayley cubic. 5/n

7

18

1

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

We’re trying to understand the finite orbits—that is, the solutions to ∏ Aᵢ=id with the most symmetry. In other words, we want to classify solutions where, no matter how we twist, we always come back to where we started (up to equivalence). 9/n

5

14

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

Where does this question come from? In some sense it goes back to the beginning of the 20th century. When n=4 (that is, we have 4 2x2 matrices), these finite orbits are the same as algebraic solutions to the Painlevé VI equation. 11/n

5

11

2

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

We show that in every other case, (A₁, …, Aₙ) is related, by an algebra-geometric operation called “middle convolution” (discovered by Katz in the 90s) to a “finite complex reflection group.” 15/n

5

6

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

As a starting point, we’re trying to understand n-tuples of 2x2 matrices A₁, …, Aₙ whose product is the identity matrix. This question goes back to the middle of the 19th century, as I'll explain later in the thread, but let's just play around for now. 3/n

4

2

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

Given one solution (A₁, …, Aₙ), there’s an easy way to make more solutions—simply change coordinates. That is, replace A₁, …, Aₙ with BA₁B⁻¹,⋯, BAₙB⁻¹ for some matrix B. This is boring, so we’ll regard A₁, …, Aₙ as *equivalent* to BA₁B⁻¹,⋯, BAₙB⁻¹. 4/n

4

2

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

Here’s an example (and the picture in the previous tweet is another). Both are taken from this paper of Yuriy Tykhyy: https://arxiv.org/pdf/2010.08477.pdf 10/n

arxiv.org

4

12

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

Our result is: a complete classification, for any number of 2x2 matrices, as long as one of them has infinite order. 13/n

4

23

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

The nature of our classification is as follows. There are some cases where a classification is already complete—if (A₁, …, Aₙ) generate a finite or dihedral group (classical), if they are “pullbacks” (Diarra), or if they share a common eigenspace (Cousin-Moussard). 14/n

4

7

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

These are, morally, symmetry groups of “complex regular polyhedra.” For example, the symmetry group of the icosahedron, pictured below, is such a reflection group. 16/n

4

5

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

For example, this solution comes (via middle convolution) from the symmetries of the icosahedron! One can make other solutions, from, for example, the symmetries of the 4-dimensional cube. 17/n

4

19

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

Then at least n-6 of the Aᵢ must be scalar matrices, i.e. scalar multiples of the identity. And this is sharp. 20/n

4

2

1

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

The spaces Xₙ have a huge amount of symmetry—in other words, we haven’t exhausted all ways to produce new solutions to our equation ∏ Aᵢ=id. Here’s one: given (A₁, …, Aₙ), consider (A₁, …, AᵢAᵢ₊₁Aᵢ⁻¹, Aᵢ, …, Aₙ). 6/n

3

17

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

In other words, we’ve swapped the i-th and (i+1)st matrices—but to make sure our equation ∏ Aᵢ=id Is still satisfied, we have to conjugate the (i+1)st by the i-th. This is called a “half-twist.” 7/n

3

16

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

And finite complex reflection groups were classified by Shephard and Todd in the ‘50s. So this really reduces any question about our tuples (A₁, …, Aₙ) to a combinatorial question about this list of finite groups. 18/n

3

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

Let me just finish with a very concrete, and I think surprising, consequence. Suppose we have our 2x2 matrices with finite orbit, (A₁, …, Aₙ), not all upper triangular, such that ∏ Aᵢ=id. To make things simple to state, let’s assume some Aᵢ is 1 1 0 1. 19/n

3

2

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

@dimpase@mathstodon.xyz $R^1$ is a derived functor; $f^*$ is pullback through a morphism $f$, and $f_*$ is pushforward! But yeah, it's some complicated operation defined in terms of sheaf cohomology.

2

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

Math content aside, one thing I’m curious about is—if you’re not an expert but tried to read this, did you get anything out of it? I tried to aim it at someone who just knows a bit of linear algebra, but not sure I succeeded.

2

3

2

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

@julesh@mathstodon.xyz Thanks!

1

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

@sophieschmieg@infosec.exchange From my POV the geometric interpretation is the reason the Q is interesting. One remark is that as you add more symmetries, there will be fewer finite orbits, which should make the question easier; I think it’s not too hard to see that with your symmetries, there won’t be any interesting finite orbits.

1

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

@sophieschmieg@infosec.exchange Good question! The main reason is that the action I describe has a geometric meaning—it corresponds to the action of the mapping class group of a punctured sphere (the braid group) on representations of pi_1(punctured sphere). More algebraically, it’s natural to only consider symmetries that preserve (or permute) the conjugacy classes of the A_i, and the ones I listed generate all of those.

1

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

PS. There are lots of interesting open questions here! We still don’t have a classification when all the A_i have finite order. And for r x r matrices, r>2, basically nothing is known. Help!

0

1

0

littmath @littmath@mathstodon.xyz

· 38mo ago

Replying to

@dimpase@mathstodon.xyz too complicated for social media, unfortunately…

0

1

0