## December 3, 2010

### The Three-Fold Way (Part 1)

#### Posted by John Baez It’s a wonderful fact that nature is described using complex Hilbert spaces. We can take a beam of electrons and split it. If we do it right, each electron goes both ways! Then we can insert a tightly wound coil of wire between the two beams. By running some current through this wire, we can make a magnetic field that’s mostly trapped inside the coil. By this method, we can multiply the part of the electron taking one route by $i$, as compared to the part that takes the other route. And we can check that this is true by studying the interference patterns that appear as the beams recombine! Indeed we can do this for any complex number on the unit circle, say $exp(i \theta)$.

But what’s so great about the complex numbers? You can set up a theory of Hilbert spaces based on any normed division algebra. And as you’re undoubtedly sick of hearing, there are three choices: the real numbers $\mathbb{R}$, the complex numbers $\mathbb{C}$, and the quaternions $\mathbb{H}$. So mathematically, at least, there are three possible kinds of quantum mechanics!

Only three? There could be more, but Solèr’s theorem picks out these three from among a vast set of alternatives, based on some simple axioms about how infinite-dimensional Hilbert spaces should work.

What about the finite-dimensional case? The Jordan–von Neumann–Wigner theorem classifies the possibilities in an approach based on algebras of observables. The Koecher–Vinberg theorem starts from seemingly different assumptions, but leads to the exact same conclusions. Both these theorems leave room for some exotic possibilities involving octonions and spin factors — but the overall message of all these results seems to be: real, complex and quaternionic quantum mechanics are equally good.

However, for some reason — or perhaps no good reason — nature is best described by complex quantum mechanics. We can take an electron and multiply it by $i$, so real quantum mechanics is out. But we can’t multiply it by $j$ or $k$ — or at least that’s what everyone says. So quaternionic quantum mechanics is out too, apparently.

This has led people to look for mathematical ways in which complex quantum mechanics is ‘better’ than the real or quaternionic theories. Lucien Hardy proved a nice result along these lines:

But I want to tell you about two others.

One of the greatest discoveries in physics is Noether’s theorem, which sets up a one-to-one correspondence between observables and one-parameter groups of symmetries. In its original form, this theorem holds in a particular approach to classical physics. But by now it’s become a general idea that applies to quantum physics as well. Energy corresponds to time translation; momentum corresponds to translation in space; angular momentum corresponds to rotation — these have become a basic part of our understanding of the world, transcending the details of any particular theory we have of the world. If the world has some symmetry, we expect there to be a corresponding observable.

In quantum mechanics, observables are self-adjoint operators. These correspond in a one-to-one way to continuous one-parameter groups of unitary operators, thanks to Stone’s theorem. So, Stone’s theorem is the quantum version of Noether’s theorem.

One problem with real and quaternionic quantum theory is that this idea breaks down, or at least becomes more subtle. Let’s see how.

Suppose $\mathbb{K}$ is any associative normed division algebra: in other words, $\mathbb{R},\mathbb{C}$ or $\mathbb{H}$. I’ve told you how to define $\mathbb{K}$-Hilbert spaces, and — you can take my word for this — a lot of what you know about complex Hilbert spaces works for real and quaternionic ones, too.

For example, let $H$ and $H'$ be $\mathbb{K}$-Hilbert spaces. We say a linear operator $T: H \to H'$ is bounded if there exists a constant $K \to 0$ such that $\|T v\| \le K \, \|v\|$ for all $v \in H$. We define the adjoint of a bounded operator $T: H \to H'$ in the usual way: $\langle u, T^\dagger v \rangle = \langle T u, v \rangle$ for all $u \in H$, $v \in H'$. It is easy to check that $T^\dagger$, defined this way, really is an operator from $H'$ back to $H$.

Let’s define a $\mathbb{K}$-linear operator $U: H \to H'$ to be unitary if $U U^\dagger = U^\dagger U = 1$. I should warn you that when $\mathbb{K} = \mathbb{R}$, people say usually say ‘orthogonal’ instead of ‘unitary’ — and when $\mathbb{K} = \mathbb{H}$, people sometimes use the term ‘symplectic’. But let’s use the same word, ‘unitary’, for all three cases.

We define a one-parameter unitary group to be a family of unitary operators $U(t) : H \to H$, one for each $t \in \mathbb{R}$, such that $U(t + t') = U(t) U(t')$ for all $t, t' \in \mathbb{R}$. Let’s say this group is continuous if for each vector $v \in H$, $U(t)v$ depends continuously on the parameter $t$.

To avoid distractions, let me assume now that $H$ is finite-dimensional. In this case every operator is automatically bounded. And in this case, every continuous one-parameter unitary group can be written as $U(t) = \exp(t S)$ for a unique operator $S$. It is easy to check that $S$ is skew-adjoint: it satisfies $S^\dagger = -S$. Conversely, any skew-adjoint $S$ gives a continuous one-parameter unitary group by the above formula. This is a version of Stone’s theorem that applies to real and quaternionic Hilbert spaces as well as complex ones.

However, in quantum theory we usually want our observables to be self-adjoint operators, obeying $A^\dagger = A$.

Now, when $\mathbb{K} = \mathbb{C}$, we can write any skew-adjoint operator $S$ in terms of a self-adjoint operator $A$. This gives the usual correspondence between one-parameter unitary groups and self-adjoint operators, which we know and love from ordinary complex quantum mechanics.

How do we do it? Easy: we set $S(v) = A(v) i .$ Huh? Remember, we’re taking $\mathbb{K}$-Hilbert spaces to be right $\mathbb{K}$-modules: that’s why I’m writing the $i$ on the right here. It’s just an arbitrary convention, and right now I’m beginning to regret this convention: this equation looks dorky. But when I use the other convention, I also regret that. It’s no big deal, just a small annoying itch.

But never mind that — here’s the main point:

When $\mathbb{K} = \mathbb{R}$ we have no number $i$, so we cannot express our skew-adjoint $S$ in terms of a self-adjoint $A$. Nor can we do it when $\mathbb{K} = \mathbb{H}$. This time the problem is not a shortage of square roots of $-1$. Instead, there are too many—and more importantly, they do not commute! We can try to set $A(v) = S(v)i$, but this operator $A$ will rarely be linear. The reason is that because $S$ commutes with multiplication by $j$, $A$ anticommutes with multiplication by $j$, so $A$ is only linear in the trivial case $S = 0$.

Let’s do the calculation in detail, in case my verbal argument wasn’t clear. We have $A(v) = - S(v) i$ so $A(v j) = - S(v j) i = - S(v) j i = S(v) i j = - A(v) j$ In short, when we pull a $j$ out, we get an unwanted minus sign. So, $A$ is linear only when $S = 0$.

That’s one problem. A second problem, special to the quaternionic case, concerns tensor products of Hilbert spaces.

In ordinary complex quantum theory, when we have two systems, one with Hilbert space $H$ and one with Hilbert space $H'$, the system made by combining these two systems has Hilbert space $H \otimes H'$. Here the tensor product of Hilbert spaces relies on a more primitive concept: the tensor product of vector spaces.

Remember, our $\mathbb{K}$-vector spaces are one-sided $\mathbb{K}$-modules, which I arbitrarily took to be right $\mathbb{K}$-modules in some vain attempt to avoid as many silly-looking equations as possible. In an algebra class, you learn that you can tensor two one-sided modules of an algebra and get another such module if your algebra is commutative. But the quaternions are not!

• Stephen Adler, Quaternionic Quantum Mechanics and Quantum Fields, Oxford U. Press, Oxford, 1995.

The tensor product of bimodules of a noncommutative algebra is another bimodule over that algebra. So, when tensoring two quaternionic Hilbert spaces, Adler essentially chooses a way to make one of them into a bimodule… without being very explicit about this.

Alas, there is no canonical way to make a quaternionic Hilbert space $H$ into a bimodule. Indeed, given one way to do it, you can get lots of new ones by introducing a new way to do left multiplication by scalars while keeping the right multiplication unchanged: $(x v)_{new} = \alpha(x) v , \qquad (v x)_{new} = v x ,$ where $v \in H$, $x \in \mathbb{H}$ and $\alpha$ is an automorphism of the quaternions. Every automorphism is of the form $\alpha(x) = g x g^{-1}$ for some unit quaternion $g$, so the automorphism group of the quaternions is $SU(2)/\{\pm 1\} \cong SO(3)$. Thus, we can ‘twist’ a bimodule structure on $H$ by any element of $SO(3)$, obtaining a new bimodule with the same underlying quaternionic Hilbert space. If I remember correctly, this idea is lurking in Adler’s work. I just wish the underlying math had been explained in a way that I could understand better.

When Toby Bartels did some work on quaternionic functional analysis, he used bimodules explicitly, right from the start:

More recently, Chi-Keung has investigated these questions in more depth. His paper provides a nice overview of many different approaches I hadn’t known about:

But he points out an interesting fact: if we take a quaternionic Hilbert space to be a bimodule of the quaternions (equipped with an $\mathbb{H}$-valued inner product satisfying some sensible conditions), then the category of quaternionic Hilbert spaces is equivalent to the category of real Hilbert spaces! So, at least from a category-theoretic perspective, it seems like we’re back to real quantum mechanics!

The reason is not hard to see, if you know a little algebra. It’s easier to see for vector spaces, rather than Hilbert spaces. A bimodule of the quaternions is the same as a left $\mathbb{H} \otimes \mathbb{H}^{op}$ module, where $\mathbb{H}^{op}$ is the quaternions with the multiplication turned around. But because $\mathbb{H}$ is a $\ast$-algebra, $\mathbb{H}^{op} \cong \mathbb{H}$ Moreover, $\mathbb{H} \otimes \mathbb{H}$ is isomorphic to $M_4(\mathbb{R})$, the algebra of $4 \times 4$ real matrices. So, the category of $\mathbb{H}$-bimodules is equivalent to the category of $M_4(\mathbb{R})$-modules.

But it’s well-known that $M_4(\mathbb{R})$ is Morita equivalent to $\mathbb{R}$, meaning that their categories of modules are equivalent!

Thus, the category of $\mathbb{H}$-bimodules is equivalent to the category of real vector spaces.

In short: so far it’s not looking too good for real and quaternionic quantum mechanics. In both of these, we lack the correspondence between observables and symmetries that we’re used to. And in the quaternionic case, we can’t describe systems built from two parts… unless we pull a trick that basically takes us back to real quantum mechanics!

But in the posts to come, I’ll propose a way to tackle these problems. The trick is to treat real, complex and quaternionic quantum theory as part of a single unified structure: the ‘three-fold way’. This nice thing is that the necessary math is already known — it’s been sitting there for many years, patiently waiting for us to appreciate it.

Posted at December 3, 2010 12:39 AM UTC

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### Re: The Three-Fold Way (Part 1)

You need to find a way of algebraically playing with some “I” (squared to -1, maybe +1 as well) that doesn’t appear ad hoc and carries some sensible interpretation.

That would be Geometric/Clifford algebra, then?

Posted by: Maya Incaand on December 3, 2010 12:25 PM | Permalink | Reply to this

### Re: The Three-Fold Way (Part 1)

I don’t have much to say except that I’m really interested in these questions, and look forward to the next installment!

Posted by: Jamie Vicary on December 3, 2010 7:13 PM | Permalink | Reply to this

### Re: The Three-Fold Way (Part 1)

Likewise!

I’d be interested in John’s (or anyone else’s) take on two older, and perhaps in some sense standard, answers to the question of real vs. complex QM:

(1) Only for complex QM are bipartite states uniquely determined by the probabilities they assign to “local” observables. (This goes back at least to a paper by Araki of around 1980, and has been observed by many people since then.)

(2) Mackey’s observed that one can pass to the complexification of a real Hilbert space, at the cost of introducing some additional observables (the real-linear self-adjoint operators). Perhaps this strategy, whatever its merits as applied to a single system, doesn’t play well with the demand for a satisfactory tensor product?

Posted by: Alex Wilce on December 5, 2010 12:52 AM | Permalink | Reply to this

### Re: The Three-Fold Way (Part 1)

Re: $\mathbb{H} \otimes \mathbb{H}^{{op}} \cong {M}_2(\mathbb{R})$.

There is a lot to be said here, and I’m very unconvinced. First of all, I don’t know how you’re taking the tensor product. Presumably, what you mean when you say “$\mathbb{H}$-$\mathbb{H}$ bimodule”, you mean that the center of $\mathbb{H}$ should act the same on both sides. Then you’re secretly performing $\otimes_{\mathbb{R}}$.

So now I’m confused. Over $\mathbb{R}$, the quaternions are four-dimensional (hence the name “quaternion”). Then $\mathbb{H} \otimes \mathbb{H}^{{op}}$ should be sixteen-real-dimensional, not four. Maybe you meant to write ${M}_4(\mathbb{R})$ rather than $2$?

Yes, this must be what you mean. In fact, I remember from some class that you can give the action of $\mathbb{H} \otimes \mathbb{H}^{{op}}$ on $\mathbb{H} = \mathbb{R}^4$ by $(x\otimes z)\cdot y = x y z$.

Anyway, I totally agree that the ($\mathbb{R}$-linear) category of $\mathbb{H}$-$\mathbb{H}$ bimodules is equivalent as an $\mathbb{R}$-linear category to the category of real vector spaces. Why does it follow that the tensor products are the same?

Notably, $\mathbb{R}$-vect is symmetrically braided, and in general bimodule categories are not. Indeed, by a result of Schauenburg (“The monoidal center construction and bimodules”, 1999), the center of the category of $\mathbb{H}$-$\mathbb{H}$ bimodules is precisely the center of $\mathbb{R}$-vect, which is just $\mathbb{R}$-vect again, and the equivalence is given by $V \mapsto \mathbb{H} \otimes V$.

Is this the functor that implements the Morita equivalence? Actually, yes. So these are equivalent as braided categories. But how surprising is this: every bimodule is a direct sum of $\mathbb{H}$s. And yet I thought you said that there were bimodules not of this form. What you actually said was that if I only know one side of the action, that’s not enough information to determine the bimodule. (“we can ‘twist’ a bimodule structure on $H$ by any element of ${SO}(3)$” is a bit of a lie: the twisted thing is isomorphic to the original doodad by multiplication by something, and the choice of isomorphisms is an $\mathbb{R}^\times$.)

Anyway, maybe another proof is to directly argue that Vect only admits one monoidal structure. This is also true, and maybe more trivial than using the Schauenburg result. But it also takes some thinking. It is equivalent to this answer, which notably fails over many rings (e.g. group rings of abelian groups).

The reason I emphasize this is because for many purposes, the tensor product is very important. It’s what distinguishes two points from half a point, for example — the categories of sheaves on these two stacks are the same, but the monoidal structures are different. As always, it’s “multiplication” (in this case, tensor) from which geometry arises. Monoidal structures also control lots of other questions — “bilinearity”, “dimension” — and so it’s important to ask which one you have.

Posted by: Theo on December 4, 2010 5:22 AM | Permalink | Reply to this

### Re: The Three-Fold Way (Part 1)

Theo wrote:

There is a lot to be said here, and I’m very unconvinced. First of all, I don’t know how you’re taking the tensor product.

Throughout these posts on state-observable duality and the three-fold way, ‘vector space’ without further qualification is supposed to mean ‘real vector space’, ‘algebra’ is supposed to mean ‘not necessarily associative algebra over $\mathbb{R}$’, and the tensor products are taken over $\mathbb{R}$.

But yes, there was a mistake: as algebras over $\mathbb{R}$, $\mathbb{H} \otimes \mathbb{H}^{op} \cong M_4(\mathbb{R})$. I’ll fix that now!

It’s easy to get a homomorphism

$\alpha : \mathbb{H} \otimes \mathbb{H}^{op} \to L(\mathbb{H})$

as you described, where

$L(\mathbb{H}) \cong M_4(\mathbb{R})$

is the algebra of real-linear transformations of $\mathbb{H}$. Just use the left and right actions of $\mathbb{H}$ on itself:

$\alpha(x \otimes z): y \to x y z$

And one can check that $\alpha$ is an isomorphism. For example, the dimensions are both 16, so it’s enough to check either that it’s one-to-one or that it’s onto. I used to know a really slick way, but right now I can’t remember it for the life of me. So, here’s a somewhat goofy way:

We’ll use the well-known fact that $SO(4)$ is in the image of $\alpha$: we get all rotations of 4d space as transformations of the form

$y \mapsto x y z$

where $|x| = |z| = 1$. So, differentiating, all of $so(4)$ is in the image of $\alpha$.

Thinking of these as matrices in the $1,i,j,k$ basis, these are all the skew-adjoint matrices — so now we just need to get all the self-adjoint matrices!

We get a basis of diagonal matrices from these transformations:

$y \mapsto 1 y 1 , \quad y \mapsto i y i , \quad y \mapsto j y j , \quad y \mapsto k y k$

Conjugating these diagonal matrices by elements of $SO(4)$, we see that all self-adjoint matrices are in the image of $\alpha$. So we’re done.

Anyway, I totally agree that the ($\mathbb{R}$-linear) category of $\mathbb{R}$-bimodules is equivalent as an $\mathbb{R}$-linear category to the category of real vector spaces. Why does it follow that the tensor products are the same?

I never claimed they were the same as monoidal categories — but I agree wholeheartedly that I should claim this, if I want to argue that ‘quaternionic bimodule quantum mechanics’ is the same as ‘real quantum mechanics’. So, thanks for raising this question — and thanks even more for answering it!

But how surprising is this: every bimodule is a direct sum of $\mathbb{H}$s. And yet I thought you said that there were bimodules not of this form. What you actually said was that if I only know one side of the action, that’s not enough information to determine the bimodule.

The reason for my somewhat confusing exposition was that

1) I’d been unhappy with Adler’s approach to quaternionic quantum mechanics for a long time, since he seems to choose an $\mathbb{H}$-bimodule structure for his left $\mathbb{H}$-modules when he wants to tensor them, without noting that there are lots of choices,

but

2) I only just recently read Ng’s remark that the category of $\mathbb{H}$-bimodules is equivalent to the category of real vector spaces,

so

3) I hadn’t come to grips with all its implications, e.g. the fact that all these $\mathbb{H}$-bimodules coming from a given left $\mathbb{H}$-module are isomorphic. This fact doesn’t absolve Adler, because even if different choices of extra structure give you isomorphic objects, the choice must still be noted. I.e., the forgetful functor from $\mathbb{H}$-bimodules to $\mathbb{H}$-modules is still not full. But, it’s an important fact to note!

So, thanks for pushing the subject forward a notch or two! By the end of these posts, I hope it’ll be clear that bimodules are really important in this game.

Monoidal structures also control lots of other questions — “bilinearity”, “dimension” — and so it’s important to ask which one you have.

Right. In the axiomatic approach to quantum-like theories that’s been emphasized by Samson Abramsky, Bob Coecke and their hordes of collaborators, it’s the ‘$\dagger$-compact category’ that matters — and the symmetric monoidal structure is crucial here. So, it’s quite interesting that the monoidal category of $\mathbb{H}$-bimodules is monoidally equivalent to that of real vector spaces, allowing us to make it symmetric monoidal, etc.

Posted by: John Baez on December 4, 2010 8:14 AM | Permalink | Reply to this

### Re: The Three-Fold Way (Part 1)

Theo writes:

Anyway, maybe another proof is to directly argue that Vect only admits one monoidal structure. This is also true, and maybe more trivial than using the Schauenburg result. But it also takes some thinking. It is equivalent to this answer, which notably fails over many rings (e.g. group rings of abelian groups).

I’d be most interested to see the details of this. (The argument at the end of the given link didn’t enlighten me much.) Presumably, some element of monoidality fails for, e.g., A, B –> B(A’,B’) where B( , ) is the space of bilinear forms.

Posted by: Alex Wilce on December 5, 2010 1:16 AM | Permalink | Reply to this

### Re: The Three-Fold Way (Part 1)

Theo wrote:

Anyway, maybe another proof is to directly argue that $Vect$ only admits one monoidal structure.

One where the tensor product is bilinear on homsets, I guess you mean. Of course $Vect$ has at least one other famous monoidal structure besides the usual $\otimes$, namely $\oplus$. I don’t know if there are others!

Alex wrote:

I’d be most interested to see the details of this.

Me too. Of course, for showing that tensoring with $\mathbb{H}$ gives a monoidal equivalence between the monoidal category of real vector spaces and the monoidal category of $\mathbb{H}$-bimodules, it’s a lot easier to simply check that this is true! But the other approach is interesting too.

(For some reason I feel the need to repeat that by an $\mathbb{H}$-bimodule I mean a real vector space equipped with compatible left and right actions of the real algebra $\mathbb{H}$, not just an abelian group equipped with compatible left and right actions of the ring $\mathbb{H}$. I’m not sure how different these are, but I mean the former: everything is ‘over $\mathbb{R}$’ here.)

Posted by: John Baez on December 5, 2010 2:56 AM | Permalink | Reply to this

### Re: The Three-Fold Way (Part 1)

What about Octonions? I’m sorry if you consider that an off-topic question, but Lisi is currently discussing Triality in E8 at Not Even Wrong.

Posted by: Greg Sivco on December 4, 2010 12:05 PM | Permalink | Reply to this

### Re: The Three-Fold Way (Part 1)

A far shot, even for my own lazy standards, is to wonder if the failure to accommodate octonions in some theories (here, or the C×H×H×M3 of Connes) is related to the failure to accommodate supersymmetry.

Posted by: Alejandro Rivero on December 4, 2010 3:43 PM | Permalink | Reply to this

### Re: The Three-Fold Way (Part 1)

You can see some remarks about octonions and quantum theory in my discussion of the Jordan–von Neumann–Wigner theorem, including links to the papers John Huerta and I wrote. In a nutshell, it’s clear to me how the octonions are important in superstring theory, but it’s not clear how they fit into the more general story about ‘foundations of quantum theory’ that I’m trying to tell now. So, nature still holds plenty of secrets.

You can see my thoughts about Lisi’s work over on Not Even Wrong.

Posted by: John Baez on December 5, 2010 12:52 AM | Permalink | Reply to this

### Re: The Three-Fold Way (Part 1)

Anyone understand what Atiyah is saying from the slides of this talk – From Algebraic Geometry to Physics - a personal perspective?

On slide 17 it says

Use all 4 Division Algebras

1, 2, 3 $\to$ Standard Model E.M, weak, strong

4 (octonions) $\to$ Gravity.

It gets better

Riemann Hypothesis $\to$ Quantum gravity

Zeros of Riemann Zeta functions are eigenvalues of Gravity Hamiltonian

and that’s just a ‘special case’.

Posted by: David Corfield on December 5, 2010 4:33 PM | Permalink | Reply to this

### Re: The Three-Fold Way (Part 1)

Michael Rios has argued nicely that the bi-octonions are a good place to start. Then Atiyah’s 27 becomes a 54 and the 28 a 56, and we can fit the 56 into a 57 dimensional space associated to 3x3 matrices for spatial partitions (3,8,8) or (1,9,9). The first case is qubits with ternary triality, and the second case is qutrits with binary triality. These are like 3x3 analogues of the 2x2 Freudenthal matrices, which Atiyah mentions. Here, binary triality refers to ‘points’, ‘lines’ and ‘faces’ in the usual way, and ternary triality is something new. ‘String’ theorists like Gunaydin like to study the 57 dim space in terms of a non linear action of E8.

Posted by: Kea on December 5, 2010 8:46 PM | Permalink | Reply to this

### Re: The Three-Fold Way (Part 1)

David Corfield wrote:

Anyone understand what Atiyah is saying from the slides of this talk — From Algebraic Geometry to Physics - a personal perspective?

No, but it’s interesting that a mathematician of such exquisite Apollonian restraint decided to go all Dionysian on us and publicize some wildly ambitious dreams.

Peter Woit went to this talk. Atiyah gave it at the Simons Center in Stony Brook on November 3rd, 2010. And over on Not Even Wrong, Peter wrote:

On Wednesday, the Simons Center hosted a day-long inaugural conference (videos and slides of talks should appear on their web-site at some point). Appropriately, the first talk was an inspirational one from Michael Atiyah, going over a wide variety of different mathematical ideas. One theme was the quaternions, with Atiyah pointing out that Hamilton had written down a square-root of the Laplacian many decades before this trick was re-discovered by Dirac in writing down the Dirac equation. After recalling the relation between the division algebras (real and complex numbers, quaternions, octonions) and the Hopf invariant one problem, Atiyah suggested that the Freudenthal magic square has a similar relationship to the Kervaire invariant one problem, with the recent complicated proof by Hopkins et al. an analog of the original Adams proof in the Hopf case, with the analog of Atiyah’s “postcard proof” still to be discovered. He ended with some comments about ideas of Alain Connes about non-commutative geometry and the Riemann hypothesis, and suggested that the conjectured self-adjoint operator that could explain the Riemann hypothesis might be the Hamiltonian of quantum gravity. I noticed that Atiyah was supposed to be giving a talk at the IAS today with the impressive title of “Quantum Gravity and the Riemann Hypothesis”, but it appears to have been canceled.

Posted by: John Baez on December 6, 2010 2:47 AM | Permalink | Reply to this

### Re: The Three-Fold Way (Part 1)

NOT Noether’s Theorem but Noether’s First variational Theorem

Even in the variational context, there’s a second one,
which has become increasingly important in the quantum context.

For a really good account of both,
see

Les Th'eor`emes de Noether by Yvette Kosmann-Schwarzbach

Posted by: jim stasheff on December 4, 2010 5:41 PM | Permalink | Reply to this

### Re: The Three-Fold Way (Part 1)

The equivalence between R and H is known as T duality by mathematicians like Mulase and Waldron et al, who have a beautiful series of papers on ribbon diagrams for Riemann surfaces and moduli spaces. In particular, according to their ribbon matrix model there is an explicit map between the symplectic and orthogonal ensembles. Now the modern association of three qubit triality to the dualities S, T and U is immediately categorical when ribbon diagrams are used. This is much more interesting than playing around with strictly linear structures.

Posted by: Kea on December 4, 2010 9:09 PM | Permalink | Reply to this

### Re: The Three-Fold Way (Part 1)

Those papers by Motohico Mulase and Andrew Waldron look nice, for example Duality of orthogonal and symplectic matrix integrals and quaternionic Feynman graphs, which is about the large-$N$ limit of $SO(N)$ and $Sp(2N)$ matrix models.

For mathematicians who aren’t into ‘matrix models’, a quick introduction to the jargon might make this subject a lot less scary.

So: simplifying a bit, the ‘matrix models’ discussed here are ways to compute things like

$\frac{\int e^{- tr(X^2 + A X) }d X}{\int e^{- tr(X^2) } d X }$

where $X$ ranges over the space of $N \times N$ self-adjoint real, complex or quaternionic matrices, and $A$ is a fixed matrix of this sort. Physicists know beautiful diagrammatic methods for working out asymptotic expansions for these quantities as power series in $1/N$. They call these the ‘$1/N$ expansions of $SO(N)$, $SU(N)$ and $Sp(2N)$ matrix models’.

Not surprisingly, all this is closely related to what mathematicians call the Gaussian Orthogonal Ensemble, Gaussian Unitary Ensemble, and Gaussian Symplectic Ensemble. These were the three stars of Dyson’s old paper ‘The Three-Fold Way’. I’ll be talking about another aspect of that paper next time.

I don’t actually see how the ‘duality’ between the $SO(N)$ and $Sp(2N)$ matrix models is related to the idea Theo and I have been talking about, namely the equivalence between the category of $\mathbb{H}$-bimodules and the category of real vector spaces. It’s clearly related to the old idea that the representation category of $Sp(2N)$ acts like the representation category of ‘$SO(-2N)$’.

But maybe these ideas are related somehow. In fact they have to be related somehow, it’s just a matter of how.

Posted by: John Baez on December 5, 2010 1:52 AM | Permalink | Reply to this

### Re: The Three-Fold Way (Part 1)

Perhaps a clue on pages 21 and 22, where they show that the duality comes down to the Poincare duality on graphs. Would not the bimodule ‘twists’ allow us to find an ‘adjoint functor’ to the forget-Pauli-matrices map?

Of course, I am biased by the idea that things like (Pauli) MUB diagrams (and fields with one element) are more fundamental than traditional vector spaces, where linearity and ordinary analysis just gets in the way of studying Stone duality more abstractly.

Posted by: Kea on December 5, 2010 3:39 AM | Permalink | Reply to this

### Re: The Three-Fold Way (Part 1)

This looks interesting as a typed approach to QM. Avoids all the Hilbert spaces etc. Any comments from the categorists?

It’s a PhD thesis by Ross Duncan from 2006.

Posted by: solrize on December 7, 2010 7:00 AM | Permalink | Reply to this

### Re: The Three-Fold Way (Part 1)

Solrize writes:

I’ve been writing about this sort of thing here at the n-Café for years. I like it!

For more detailed comments, you might try these:

You sound a bit like a computer scientist, since you seem to like ‘types’. If so, you might like the second paper above.

I wouldn’t say Duncan’s approach is trying to ‘avoid’ Hilbert spaces; instead, it fits them in a nice general context. That’s also roughly what Mike and I were trying to do.

The three-fold way, which I’m starting to discuss here, will turn out to be deeply related to the ‘strongly compact categories’ which you’ll find in Duncan’s thesis.

Posted by: John Baez on December 7, 2010 10:32 AM | Permalink | Reply to this

### Re: The Three-Fold Way (Part 1)

I think the publication date for New Structures for Physics should be 2011, not 2000.

Posted by: Blake Stacey on December 7, 2010 8:17 PM | Permalink | Reply to this

### Re: The Three-Fold Way (Part 1)

Yes, you’re right — though I think Mike Stay already got his copy a while ago.

Posted by: John Baez on December 8, 2010 5:33 AM | Permalink | Reply to this

### Re: The Three-Fold Way (Part 1)

He did, ah that’s good. You didn’t get it yet?

Posted by: bob on December 8, 2010 2:14 PM | Permalink | Reply to this

### Re: The Three-Fold Way (Part 1)

No. Did I remember to ask that it be sent to Singapore? It may be sitting in Riverside.

Posted by: John Baez on December 8, 2010 2:53 PM | Permalink | Reply to this
Read the post The Three-Fold Way (Part 5)
Weblog: The n-Category Café
Excerpt: Learn about the functors going between the categories of real, complex and quaternionic Hilbert spaces.
Tracked: February 8, 2011 3:23 AM

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