Singular moduli spaces and the Rubik’s cube

I’m currently in Maryland at a conference to celebrate the 62nd birthday of John Millson (why 62? Beats me. I guess people like him so much they couldn’t wait for 65).

Ravi Vakil is talking about “Murphy’s Law” in algebraic geometry, which was ably summed up by Harris and Morrison (not our Morrison) as “There is no geometric configuration so horrible that it does not appear in a Hilbert scheme.” A Hilbert scheme is a moduli space of subvarieties of $\mathbb{P}^n$, so it’s a very natural object to look at in algebraic geometry, and thus it’s very upsetting that it’s as singular as you can possibly imagine. This means all kinds of horrifying objects must exist, like surfaces in characteristic 31 which deform to characteristic 0 to 43rd order, but no further. Yuck!

Now, there’s lots more to say about this from a high level perspective (like a long list of moduli spaces satisfying Murphy’s law), but I wanted to talk about an example of a singular moduli space you already know.

So. Grab the nearest Rubik’s cube (or vivid mental image thereof). Now look at it in it while all the sides are aligned (when it’s actually cube shaped). What is the space of “deformations” at the point (the Zariski tangent space)? There’s 9 of them, since in each of the planes of rotation, there are 3 freely rotating pieces. But if you rotation a little bit in one of the planes, suddenly, the other directions of rotation lock up, and you have a lot less symmetry, 3 dimensions from the direction you rotated, and one each from rotating the whole cube, which gives us 5. As you may have noticed, 5 is not 9, so by the usual definition of a singular point, the aligned configuration must be a singular point.

Congratulations! You’ve shown that the moduli space of configurations of the Rubik’s cube is singular. As you can see, no moduli space is safe.

In fact, this just the tip of the iceberg. It’s a well-known principle that if an object has a deformation with less symmetry than the original object, it will be a singular point of the moduli space, which is basically what we are applying here. What Ravi’s theorem says is there are lots other, much less obvious singularities out there in the world.

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8 thoughts on “Singular moduli spaces and the Rubik’s cube”

1. 64, I suppose, would make sense. Kirillov, for example, had a thing of some sort in honor of his $2^6$ birthday. (I’m calling it a “thing of some sort” because I don’t remember the details, but his office is down the hall from mine so I pass the piece of paper that advertised it quite often.)

2. “26” should be two to the sixth in the previous comment, of course; I didn’t realize the comments here didn’t accept the HTML sup tag.

3. “26″ should be two to the sixth in the previous comment, of course; I didn’t realize the comments here didn’t accept the HTML sup tag.

That’s good. I was starting feel insecure about having turned 26 without enough fanfare. Though I do still have 8 months or so to plan for $3^3$.

4. Were you at Chris Woodward’s talk? Can you give us a report on it?

5. Scott Carnahan says:

The cube is a great example. The obstructed deformations are really physically obstructed.

I guess you’re considering the pair (cube, embedding in R^3), since otherwise you should mod out by a global rotational symmetry. From a practical standpoint of having the cube in your hands, the embedding is good to keep in mind, but I think people often look at symmetries that keep the centers of the faces fixed, making the tangent spaces 6 and 2 dimensional, respectively.

6. Scott, the Rubik’s cube is *always* embedded in R^3.

7. Ben, Scott’s point is that the “Rubik’s cube group” is usually taken to exclude any moves that permute the center cubies. Thus anyone used to talking about math and the cube (or even anyone who knows the standard names of the moves) would initially think of Scott’s space, not yours.

8. anon says:

the deformation space of a rubik’s cube is simply reducible. the irreducible components are nonsingular themselves.

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