The simplest example I can come up with is cyclopropane (C_{3}H_{6}, a triangle of carbon molecules) with two hydrogens removed, although I have no idea how stable this would be! I think the graph would have 54 vertices and 144 edges and an action of . The 2-cells depend on the model of how hydrogens are allowed to move, but assuming they’re only allowed to skip over the bonds (rather than over the centre of the triangle) then there are are number of squares attached, perhaps 24.

With precisely 5 minutes this morning for such fun activity as answering Azimuth puzzles…

I’m sorry you’re so busy these days! Anyway, you’ve done more than your share of answering Azimuth puzzles—I think you answered *all* the Network Theory puzzles until your current state of overwork overtook you. So I would never make fun of *you* for not trying this one. It’s all the *other* people who’re in trouble.

… can’t we establish for Puzzle 1 an isomorphism between the states and transitions of the trigonal bipyramidal molecule and those of the ethyl cation?

That’s indeed the best way to do it. After all, so far I’ve basically *defined* the Desargues graph to be the graph of states and transitions for the ethyl cation. Sure, I showed you a picture of it too—but it’s hard to do much with that.

Whether an atom is in the axis of the former corresponds to whether it’s in the pair end of the latter.

Hey, that sounds promising. In case someone didn’t follow that, David seems to be hinting that the split of chlorines into 3 equatorial ones and 2 axial ones in this trigonal bipyramidal molecule:

is analogous to the split of hydrogens into a group 3 and a group of 2 in this ethyl cation:

If so, the next step would to make sure that this analogy sets up a 1-1 correspondence between the *states* in the two problems, and also the *transitions*.

Consider the transitions, for example. Does a hydrogen hopping from one carbon to another in the ethyl cation correspond to a ‘pseudorotation’ of the trigonal bipyramidal molecule? Remember, pseudorotations are a bit funky. They work like this:

]]>… I’m considering sending the post to my graph theory students.

If you do, you might tell them that the Desargues-Levi graph is a close relative of the Petersen graph, which they may already know. The Petersen graph is the universal counterexample to all dumb guesses in graph theory:

The Desargues-Levi is the ‘bipartite double cover’ of the Petersen graph, formed by replacing each Petersen graph vertex by a pair of vertices and each Petersen graph edge by a pair of crossed edges.

I think I used to know how you could get the Petersen graph directly from thinking about the ethyl cation. Maybe just treat the two carbons as indistinguishable?

As for the validity of distinguishing between different identical particles: my approach would be to say that of course you’re allowed to distinguish between them, just as long as you remember the symmetries.

Well, Distler’s point was that the Hilbert space of quantum states of phosphorus pentachloride (for example) is smaller than I made it sound.

One thing I’ve always wondered is when in a given physical situation with such a symmetry, which is the right quotient? The usual one, or the homotopy one?

People mainly discuss this issue in the treatment of ‘gauge’ symmetries in classical field theory. The traditional approach was to take the usual quotient, but it’s starting to seem that the homotopy quotient is right. A good example is whether you should use the moduli *space* of elliptic curves when studying about strings whose worldsheet has genus 1, or the moduli *stack*. The moduli space is an ordinary quotient, while the moduli stack is a homotopy quotient. As I explained back in “week125” of *This Week’s Finds*, the Picard group of the moduli stack of elliptic curves is , and this is related to the magic role of the numbers 12 and 24 in this theory—for example, how the vacuum energy is -1/24, which is part of why the theory is only consistent in 24 + 2 = 26 dimensions.

Another question I have is about the construction of these graphs. Why have we stopped at edges? Could there be 2-cells?

Heh. I’m not doing the n-category schtick anymore. But if we were studying something like a 2d random field theory, instead of a Markov process or Markov chain (where there’s just 1 dimension, namely time), that could be natural.

]]>One thing I’ve always wondered is when in a given physical situation with such a symmetry, which is the right quotient? The usual one, or the homotopy one? Let’s say the symmetry is finite, indeed it could be just a single reflection. Can one ever see the homology of in a physical system?

Another question I have is about the construction of these graphs. Why have we stopped at edges? Could there be 2-cells?

]]>You got it. If I were talking about H_{2} instead of H_{2}O, that article you linked to on spin isomers of hydrogen would be exactly right. But—for no particularly good reason—my question was about H_{2}O.

In fact, I got a bit mixed up… it turns out I understand the answer for H_{2} a lot better than the answer for H_{2}O. So let me see if I can explain it for H_{2}.

There are two kinds of H_{2} molecules: **orthohydrogen**, where the spins of the two protons are lined up, and **parahydrogen**, where they are anti-aligned:

Since the two protons are identical fermions, the wavefunction of the hydrogen molecule has to change sign when you switch them.

In orthohydrogen the protons’ spin state is symmetric: it doesn’t change when you switch them. So their position state *must* change sign when you switch them. This means that their total orbital angular momentum can’t be 0: if it were, it would be invariant under rotations, so it wouldn’t change sign when you switch the two protons! But it can be 1, or in fact any odd integer.

On the other hand, in parahydrogen the protons’ spin state is antisymmetric: it changes sign when you switch them. That’s not obvious from the picture above, which shows one with spin up and the other with spin down. That’s a defect of the picture! In reality, their spin state is

So, it’s antisymmetric. That means their position state must *not* change sign when you switch them. That means the the total orbital angular momentum of the protons *can* be 0, or in fact any even integer.

In short, parahydrogen can have total angular momentum zero, but orthohydrogen can’t—it has to be ‘rotating’. This means the ground state of parahydrogen has lower energy. And the energy difference is significant; according to the Wikipedia article divided by Boltzmann’s constant is about 175 kelvin.

If instead we had a ‘pseudohydrogen’ molecule HH’ built from a proton and a *schmoton* and two electrons, it would still come in ‘ortho’ and ‘para’ forms, but the argument I sketched would no longer apply. So, either form could have any total angular momentum it wanted: 0, 1, 2,…

So for a pseudohydrogen molecule, unlike an ordinary hydrogen molecule, the ground state energy of the para form would equal that of the ortho form.

Now, what about water? How would pseudowater differ from water?

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