February 8, 2011

The Three-Fold Way (Part 5)

Posted by John Baez

You can now see the paper these blog entries are based on:

But the blog entries have more jokes! So far, I’ve explained how certain complex representations of groups can be seen as arising from real or quaternionic representations. This gives a sense in which ordinary complex quantum theory subsumes the real and quaternionic theories. But there’s also a sense in which all three theories have equal priority. This idea can be seen already at the level of Hilbert spaces, even before group representations enter the game.

For this we need to think about categories of Hilbert spaces. As usual, let $\mathbb{K}$ be either $\mathbb{R}$, $\mathbb{C}$ or $\mathbb{H}$. Now, let $Hilb_\mathbb{K}$ be the category where:

I want to show you how any one of the categories $Hilb_\mathbb{R}$, $Hilb_\mathbb{C}$ and $Hilb_\mathbb{H}$ can be embedded in any other. This means that a Hilbert space over any one of the three normed division algebras can be seen as Hilbert space over any other, equipped with some extra structure!

So if you ask which is fundamental: real, complex or quaternionic quantum theory, there’s a certain sense in which the answer is: take your pick!

How are $Hilb_\mathbb{R}$, $Hilb_\mathbb{C}$ and $Hilb_\mathbb{H}$ related?

Of course the complex HIlbert space $\mathbb{C}^n$ has an underlying real Hilbert space $\mathbb{R}^{2n}$, and the quaternionic Hilbert space $\mathbb{H}^n$ has an underlying complex Hilbert space $\mathbb{C}^{2n}$. But there’s a slightly more sophisticated way to say what’s going on.

Let’s start with the chain of inclusions $\mathbb{R} \hookrightarrow \mathbb{C} \hookrightarrow \mathbb{H} .$ Thanks to the first inclusion, any complex vector space has an underlying real vector space. In other words: if we have a complex vector space, and we deliberately forget how to do scalar multiplication of vectors by complex numbers, and only remember how to multiply them by real numbers, it becomes a real vector space! Similarly, any quaternionic vector space becomes a complex one, thanks to the second inclusion.

The same is true for Hilbert spaces. To make the underlying real vector space of a complex Hilbert space into a real Hilbert space, we take the real part of the original complex inner product, defined by $\mathrm{Re}(a + b i) = a .$ Everyone knows that; a bit less familiar is how the underlying complex vector space of a quaternionic Hilbert space becomes a complex Hilbert space. Here we need to take the complex part of the original quaternionic inner product, defined by $\mathrm{Co}(a + b i + c j + d k) = a + b i .$ One can check that these constructions give functors $Hilb_\mathbb{H} \to Hilb_\mathbb{C} \to Hilb_\mathbb{R} .$

A bit more formally, we have a commutative triangle of homomorphisms:

\begin{aligned} \mathbb{C} &\overset{\quad \beta \quad}{\to} & \mathbb{H} \\ \alpha \uparrow & \nearrow \gamma \\ \mathbb{R} \end{aligned}

There is only one choice of the homomorphisms $\alpha$ and $\gamma$. There are many choices of $\beta$, since we can map $i \in \mathbb{C}$ to any square root of $-1$ in the quaternions. However, all the various choices of $g$ are the same up to an automorphism of the quaternions. That is, given two homomorphisms $\beta,\beta' : \mathbb{C} \to \mathbb{H}$, we can always find an automorphism $\theta : \mathbb{H} \to \mathbb{H}$ such that $\beta' = \theta \circ \beta .$ So, nothing important depends on the choice of $\beta$. Let us make a choice—say the standard one, with $\beta(i) = i$ — and use that.

Our commutative triangle of homomorphisms gives a commutative triangle of functors:

\begin{aligned} Hilb_{\mathbb{C}} &\overset{\quad \beta^*}{\leftarrow} & Hilb_{\mathbb{H}} \\ \alpha^* \downarrow & \swarrow \gamma^* \\ Hilb_{\mathbb{R}} \end{aligned}

Now, recall that a functor $F: C \to D$ is faithful if given two morphisms $f,f' : c \to c'$ in $C$, $F(f) = F(f')$ implies that $f = f'$. It is easy to see that the functors $\alpha^*, \beta^*$ and $\gamma^*$ are all faithful. When $F: C \to D$ is faithful, we say that $C$ is faithfully embedded in $D$, and we can think of objects of $C$ as objects of $D$ equipped with extra structure. For our three functors above we can describe this extra structure quite explicitly. This lets us describe Hilbert spaces for a larger normed division algebra in terms of Hilbert spaces for a smaller one. None of this particularly new or difficult: the key ideas are all in Adams’ Lectures on Lie Groups.

First we consider the extra structure possessed by the underlying real Hilbert space of a complex Hilbert space:

Theorem: The functor $\alpha^* : Hilb_\mathbb{C} \to Hilb_\mathbb{R}$ is faithful, and $Hilb_\mathbb{C}$ is equivalent to the category where:

• an object is a real Hilbert space $H$ equipped with a unitary operator $J: H \to H$ with $J^2 = -1$.
• a morphism $T : H \to H'$ is a bounded real-linear operator preserving this exta structure: $T J = J' T$.
This extra structure $J$ is often called a complex structure.

Next we consider the extra structure possessed by the underlying complex Hilbert space of a quaternionic Hilbert space. For this we need to generalize the concept of an antiunitary operator. First, given $\mathbb{K}$-vector spaces $V$ and $V'$, we define an antilinear operator $T : V \to V'$ to be a map with $T(v x + w y) = T(v)x^* + T(w)y^*$ for all $v,w \in V$ and $x,y \in \mathbb{K}$. Then, given $\mathbb{K}$-Hilbert spaces $H$ and $H'$, we define an antiunitary operator $T: H \to H'$ to be an invertible antilinear operator with $\langle T v, T w \rangle = \langle w, v \rangle$ for all $v,w \in H$.

Theorem: The functor $\beta^* : Hilb_\mathbb{H} \to Hilb_\mathbb{C}$ is faithful, and $Hilb_\mathbb{H}$ is equivalent to the category where:

• an object is a complex Hilbert space $H$ equipped with an antiunitary operator $J : H \to H$ with $J^2 = -1$;
• a morphism $T : H \to H'$ is a bounded complex-linear operator preserving this extra structure: $T J = J' T$.

This extra structure $J$ is often called a quaternionic structure. We have seen it already in our study of the Three-Fold Way.

Finally, we consider the extra structure possessed by the underlying real Hilbert space of a quaternionic Hilbert space. This can be understood by composing the previous two theorems:

Theorem: The functor $\gamma^* : Hilb_\mathbb{H} \to Hilb_\mathbb{R}$ is faithful, and $Hilb_\mathbb{H}$ is equivalent to the category where:

• an object is a real Hilbert space $H$ equipped with two unitary operators $J,K : H \to H$ with $J^2 = K^2 = -1$ and $J K = -K J$;
• a morphism $T: H \to H'$ is a bounded real-linear operator preserving this extra structure: $T J = J' T$ and $T K = K' T$.

This extra structure could also be called a quaternionic structure, as long as we remember that a quaternionic structure on a real Hilbert space is different than one on a complex Hilbert space! Of course if we define $I = JK$, then $I, J,$ and $K$ obey the usual quaternion relations.

The functors discussed so far all have adjoints, which are in fact both left and right adjoints: \begin{aligned} Hilb_{\mathbb{C}} &\overset{\quad \beta_*}{\to} & Hilb_{\mathbb{H}} \\ \alpha_* \uparrow & \nearrow \gamma_* \\ Hilb_{\mathbb{R}} \end{aligned} These adjoints can easily be defined using the theory of bimodules. As vector spaces, we have:

$\begin{array}{ccl} \alpha_* (V) &=& V \otimes {{}_\mathbb{R}\mathbb{C}_{\mathbb{C}}} \\ \beta_*(V) &=& V \otimes {{}_\mathbb{C}\mathbb{H}_{\mathbb{H}}} \\ \gamma_*(V) &=& V \otimes {{}_\mathbb{R}\mathbb{H}_{\mathbb{H}}} \\ \end{array}$

In the first line here, $V$ is a real vector space, or in other words, a right $\mathbb{R}$-module, while ${{}_\mathbb{R}\mathbb{C}_{\mathbb{C}}}$ denotes $\mathbb{C}$ regarded as a $\mathbb{R}$-$\mathbb{C}$-bimodule. Tensoring these, we obtain a right $\mathbb{C}$-module, which is the desired complex vector space. The other lines work analogously. It is then easy to make all these vector spaces into Hilbert spaces. I can’t resist mentioning that our previous functors can be described in a similar way, just by turning the bimodules around:

$\begin{array}{ccl} \alpha^*(V) &=& V \otimes {{}_\mathbb{C}\mathbb{C}_{\mathbb{R}}} \\ \beta^*(V) &=& V \otimes {{}_\mathbb{H}\mathbb{H}_{\mathbb{C}}} \\ \gamma^*(V) &=& V \otimes {{}_\mathbb{H}\mathbb{H}_{\mathbb{R}}} \end{array}$ But instead of digressing into this subject (called Morita theory), all I want to do now is mention that the functors $\alpha_*, \beta_*$ and $\gamma_*$ are also faithful. This lets us describe Hilbert spaces for a smaller normed division algebra in terms of Hilbert spaces for a bigger one!

We begin with the functor $\alpha_*$, which is called complexification:

Theorem: The functor $\alpha_* : Hilb_\mathbb{R} \to Hilb_\mathbb{C}$ is faithful, and $Hilb_\mathbb{R}$ is equivalent to the category where:

• an object is a complex Hilbert space $H$ equipped with a antiunitary operator $J : H \to H$ with $J^2 = 1$;
• a morphism $T: H \to H'$ is a bounded complex-linear operator preserving this exta structure: $T J = J' T$.

The extra structure $J$ here is often called a real structure. We have seen it already in our study of the Three-Fold Way.

Next let’s look at the functor from complex to quaternionic Hilbert spaces. It has no name, as far as I know:

Theorem: The functor $\beta_* : Hilb_\mathbb{C} \to Hilb_\mathbb{H}$ is faithful, and $Hilb_\mathbb{C}$ is equivalent to the category where:

• an object is a quaternionic Hilbert space $H$ equipped with a unitary operator $J$ with $J^2 = -1$;
• a morphism $T: H \to H'$ is a bounded quaternion-linear operator preserving this extra structure: $T J = J' T$.

This result is less well-known than the previous ones, so let me sketch a proof:

Proof: Suppose $H$ is a quaternionic Hilbert space equipped with a unitary operator $J$ with $J^2 = -1$. Then $J$ makes $H$ into a right module over the complex numbers, and this action of $\mathbb{C}$ commutes with the action of $\mathbb{H}$, so $H$ becomes a right module over the tensor product of $\mathbb{C}$ and $\mathbb{H}$, considered as algebras over $\mathbb{R}$. But this is isomorphic to the algebra of $2 \times 2$ complex matrices. The matrix \left(\begin{aligned} 1 & 0 \\ 0 & 0 \end{aligned} \right) projects $H$ down to a complex Hilbert subspace $H_\mathbb{C}$ whose complex dimension matches the quaternionic dimension of $H$. By applying arbitrary $2 \times 2$ complex matrices to guys in this subspace we get back everything in $H$, so $\beta_* H_\mathbb{C}$ is naturally isomorphic to $H$.   █

Composing $\alpha_*$ and $\beta_*$, we obtain the functor from real to quaternionic Hilbert spaces. I’ve seen this called quaternification — or occasionally ‘quaternization’, but that means something else in chemistry!

Theorem: The functor $\gamma_* : Hilb_\mathbb{R} \to Hilb_\mathbb{H}$ is faithful, and $Hilb_\mathbb{R}$ is equivalent to the category where:

• an object is a quaternionic Hilbert space $H$ equipped with two unitary operators $J, K$ with $J^2 = K^2 = -1$ and $J K = -K J$.
• a morphism $T: H \to H'$ is a bounded quaternion-linear operator preserving this extra structure: $T J = J' T$ and $T K = K' T$.

Again, let me sketch a proof:

Proof: The operators $J, K$ and $I = J K$ give make $H$ into a left $\mathbb{H}$-module. Since this action of $\mathbb{H}$ commutes with the existing right $\mathbb{H}$-module structure, $H$ becomes a module over the tensor product of $\mathbb{H}$ and $\mathbb{H}^{op} \cong \mathbb{H}$, considered as algebras over $\mathbb{R}$. But this tensor product is isomorphic to the algebra of $4 \times 4$ real matrices! So, the matrix \left(\begin{aligned} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{aligned}\right) projects $H$ down to a real Hilbert subspace $H_\mathbb{R}$ whose real dimension matches the quaternionic dimension of $H$. By applying arbitrary $4 \times 4$ real matrices to guys in this subspace we get back everything in $H$, so $\gamma_* H_\mathbb{R}$ is naturally isomorphic to $H$.   █

Finally, it is worth noting that some of the six functors we have described have additional nice properties:

• The categories $Hilb_\mathbb{R}$ and $Hilb_\mathbb{C}$ are symmetric monoidal categories, meaning roughly that they have well-behaved tensor products. The complexification functor $\alpha^*: Hilb_\mathbb{R} \to Hilb_\mathbb{C}$ is a symmetric monoidal functor, meaning roughly that it preserves tensor products.
• The categories $Hilb_\mathbb{R}, Hilb_\mathbb{C}$ and $Hilb_\mathbb{H}$ are dagger-categories, meaning roughly that any morphism $T : H \to H'$ has a Hilbert space adjoint $T^\dagger : H' \to H$ such that $\langle T v, w \rangle = \langle v, T^\dagger w \rangle$ for all $v \in H$, $w \in H'$. All six functors preserve this dagger operation.
• For $Hilb_\mathbb{R}$ and $Hilb_\mathbb{C}$, the dagger structure interacts nicely with the tensor product, making these categories into dagger-compact categories, and the functor $\alpha^*$ is compatible with this as well.

For precise definitions of the terms here, click on the links. I’ve spent a lot of time trying explain how these concepts unify physics with topology and other subjects:

and this wonderful book:

• Bob Coecke, editor, New Stuctures for Physics, Lecture Notes in Physics 813, Springer, Berlin, 2000, pp. 95–174.

The three-fold way is best appreciated with the help of these category-theoretic ideas. A more $n$-categorical treatment of symplectic and orthogonal structures can be found in my old paper on 2-Hilbert spaces:

If I’m feeling exceptionally energetic I may say more about the $n$-categorical aspects, Morita theory, and so on. More likely, I’ll wrap up the story next time by saying how the Three-Fold Way solves some of the problems of real and quaternionic quantum theory.

Posted at February 8, 2011 3:32 AM UTC

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Re: The Three-Fold Way (Part 5)

It seems to me that there is a typo on top of page 28 of the arXiv paper:
Instead of Hilb_R ⊗ Hilb_R → Hilb_C
it should say Hilb_R ⊗ Hilb_C → Hilb_C.

Posted by: Dmitri Pavlov on February 8, 2011 6:25 AM | Permalink | Reply to this

Re: The Three-Fold Way (Part 5)

Thanks, Dmitri — you’re right!

You can find an updated version on my website; corrections will take a while to reach the arXiv.

Posted by: John Baez on February 8, 2011 9:09 AM | Permalink | Reply to this

Re: The Three-Fold Way (Part 5)

Here’s a little puzzle that confuses me occasionally. We can think of a real Hilbert space as a quaternionic Hilbert space in two ways: directly, or indirectly via complex Hilbert spaces.

Directly: A real Hilbert space can be obtained from a quaternionic Hilbert space with two unitary operators $J, K$ with $J^2 = K^2 = -1$ and $J K = - K J$.

Indirectly: A real Hilbert space can be obtained from a complex Hilbert space with an antiunitary operator $K$ such that $K^2 = -1$, and a complex Hilbert space from a quaternionic Hilbert space with a unitary operator $J$ such that $J^2 = -1$.

What’s going on?

Posted by: John Baez on February 8, 2011 9:17 AM | Permalink | Reply to this

Re: The Three-Fold Way (Part 5)

This stuff is amazingly cool, but I’m still very confused.

Did you mean $K^2 = 1$?

Posted by: Tom Ellis on February 8, 2011 11:03 PM | Permalink | Reply to this

Re: The Three-Fold Way (Part 5)

John wrote:

A real Hilbert space can be obtained from a complex Hilbert space with an antiunitary operator $K$ such that $K^2 = −1$.

Tom wrote:

Did you mean $K^2=1$?

Yes, I did in the quoted sentence, but not in the later sentence about getting real Hilbert spaces from quaternionic ones. Thanks for catching that typo! I’ll use my superpowers to go back and correct it, so future puzzle-readers don’t get confused.

Posted by: John Baez on February 9, 2011 1:22 AM | Permalink | Reply to this

Re: The Three-Fold Way (Part 5)

Also, did you mean Indirectly followed by Directly?

Posted by: Mark Meckes on February 9, 2011 2:11 PM | Permalink | Reply to this

Re: The Three-Fold Way (Part 5)

Aargh. Fixed.

Posted by: John Baez on February 10, 2011 12:52 AM | Permalink | Reply to this

Re: The Three-Fold Way (Part 5)

When you’re identifying objects of Hilb_H with real Hilbert spaces you say a morphism is a “complex-linear operator”, and likewise you identify objects of Hilb_C with quaternionic Hilbert spaces you say a morphism is a “complex-linear operator”.

Are these typos (for “real-linear” and “quaternionic-linear” respectively) or am I missing an important part of the picture here? I don’t understand how they could be anything *but* typos, so if I’m wrong then I’m in dire need of having this pointed out!

Posted by: Tom Ellis on February 8, 2011 10:39 PM | Permalink | Reply to this

Re: The Three-Fold Way (Part 5)

Tom wrote:

When you’re identifying objects of $Hilb_{\mathbb{H}}$ with real Hilbert spaces you say a morphism is a “complex-linear operator”, and likewise you identify objects of $Hilb_{\mathbb{C}}$ with quaternionic Hilbert spaces you say a morphism is a “complex-linear operator”.

Are these typos (for “real-linear” and “quaternionic-linear” respectively) or am I missing an important part of the picture here?

Yes, they are typos for “real-linear” and “quaternion-linear”, respectively. I also caught another mistake like this, and fixed it in the blog entry and the paper.

In every case, if we are thinking of $\mathbb{K}'$-Hilbert spaces as $\mathbb{K}$-Hilbert spaces, the description should say:

• an object is a $\mathbb{K}$-Hilbert space equipped with extra structure;
• a morphism is a $\mathbb{K}$-linear map preserving this structure.

I’m guilty of cutting-and-pasting without carefully changing everything that needed to be changed.

By the way, if you still find anything confusing after all these typos have been fixed, just ask. I owe you!

Posted by: John Baez on February 9, 2011 1:39 AM | Permalink | Reply to this

Re: The Three-Fold Way (Part 5)

I know that tradition demands that a complex structure on a real vector space be denoted by $J$, but I would find it rather less confusing, especially in a context where quaternionic structures are also hanging around, if you could call it $I$ instead, and then use $J$ (or $K$, I suppose) for the additional operator that supplies a quaternionic structure. (-:

I also feel like it might also be clearer to use different letters than $J$ and $K$ for the operators that characterize complexifications and quaternifications (although again I suspect that tradition demands otherwise), since they are not a memory of “multiplication by $j$ (or $i$)” but rather a sort of “conjugation”. At least, that’s what I think they are… did you ever actually tell us what the operator $J$ is on the complexification of a real Hilbert space, and its friends on quaternified real and complex Hilbert spaces? You told us how to recover the thing we complexified (or quaternified) from the operators, but where did the operators come from to start with?

I can guess that on $V\otimes {}_{\mathbb{R}}\mathbb{C}_{\mathbb{C}}$ the operator $J$ acts by $J(x\otimes z) = x \otimes \overline{z}$. But I don’t really see what the quaternionic versions are. The conjugation operator on a complexification comes from the single (involutive) automorphism of $\mathbb{C}$ over $\mathbb{R}$; but in the other thread we were just talking about how the automorphism group of $\mathbb{H}$ is $SO(3)$, which reduces to $S^1$ once we fix an embedding of $\mathbb{C}$. So why doesn’t the quaternification of a complex Hilbert space come with a whole action by $S^1$, and the quaternification of a real Hilbert space with an action of $SO(3)$?

I don’t think I have a hope of tackling your puzzle until I understand that!

Posted by: Mike Shulman on February 8, 2011 11:39 PM | Permalink | PGP Sig | Reply to this

Re: The Three-Fold Way (Part 5)

Mike wrote:

I know that tradition demands that a complex structure on a real vector space be denoted by $J$, but I would find it rather less confusing, especially in a context where quaternionic structures are also hanging around, if you could call it $I$ instead,…

Good point. As you can tell, I gave up long ago on choosing names of letters that have any sort of mnemonic value here… it’s hard to do this systematically throughout the whole paper without various clashes. But maybe it’ll help even if it’s unsystematic.

…did you ever actually tell us what the operator $J$ is on the complexification of a real Hilbert space…?

No, I didn’t! That’s a silly omission. Of course it’s complex conjugation, or ‘Konjugation’ in German — so it deserves to be called $K$. Earlier in the paper I needed to call it $J$, so I could talk about the two cases where $J^2 = \pm 1$ in a single breath. But I suppose now I could let down my hair a bit and call it $K$.

I don’t really see what the quaternionic versions are.

I’ll do one of them:

If you quaternify a real vector space $V$ you get $V \otimes_{\mathbb{R}} \mathbb{H}$. This has a right action of $\mathbb{H}$, which is what makes it a quaternionic vector space (in my conventions)… but it also has a commuting left action of $\mathbb{H}$! So, it comes equipped with quaternion-linear operators $I, J, K$ satisfying $I^2 = J^2 = K^2 = I J K = -1$.

I should explain this in the paper! It’s a useful clue… though the paper was not intended to be a puzzle.

Posted by: John Baez on February 9, 2011 1:55 AM | Permalink | Reply to this

Re: The Three-Fold Way (Part 5)

I guess you mean that a quaternion $x$ acts on a tensor product element $v \otimes y \in V\otimes_{\mathbb{R}} \mathbb{H}$ by $x\cdot (v\otimes y) = v \otimes (x\cdot y)$?

I’ve been puzzling over the other one and I can’t figure out what you can do for the quaternification of a complex vector space. The above formula doesn’t make sense for any quaternion other than a multiple of $i$ acting on $V\otimes_{\mathbb{C}} \mathbb{H}$, as far as I can tell, since $\mathbb{C}\hookrightarrow \mathbb{H}$ doesn’t land in the center of $\mathbb{H}$. So we can have a complex number $z$ such that $v z \otimes y = v \otimes z y$, but $v z \otimes (x y) \neq v \otimes (x z y)$.

Posted by: Mike Shulman on February 9, 2011 9:51 PM | Permalink | PGP Sig | Reply to this

Re: The Three-Fold Way (Part 5)

Oh! We don’t need anything that isn’t a multiple of $i$; we can use the “inner action” of $i$ itself. $J(v\otimes y) = v \otimes i y = v i\otimes y$.

Posted by: Mike Shulman on February 9, 2011 10:19 PM | Permalink | Reply to this

Re: The Three-Fold Way (Part 5)

Quaternization and Complexification, eh? Should there be a notion of module over the octonions? If so, should there be a free octonization of a quaternion/complex module? Could we call them all ionizations? Then the Chemists will have TWO words overloaded!

Posted by: some guy on the street on February 8, 2011 11:45 PM | Permalink | Reply to this

Re: The Three-Fold Way (Part 5)

some guy on the street wrote:

Could we call them all ionizations?

That’s a great idea! And it reminds me of a complicated joke I’ve been trying to perfect for years.

The symbols C, H and O are used in chemistry to denote the elements from which carbohydrates are formed. So, if we could think of nice ways to combine the division algebras $\mathbb{C}$, $\mathbb{H}$ and $\mathbb{O}$ which match the possibilities allowed in chemistry, we could develop a theory of ‘mathematical carbohydrates’ like

methanol: $\mathbb{C H}_3 \mathbb{C O H}$

formaldehyde: $\mathbb{C H}_2 \mathbb{O}$

and so on.

The obvious way we could try to combine $\mathbb{C}, \mathbb{H}$ and $\mathbb{O}$ to form larger ‘compounds’ is to tensor them over the real numbers. Unfortunately, hydrogen has 1 bond, oxygen has 2 and carbon has 4. This doesn’t match anything about $\mathbb{H}$, $\mathbb{C}$ and $\mathbb{O}$… at least not in that order!

But, it comes tantalizingly close.

The real dimensions of $\mathbb{H}$, $\mathbb{C}$ and $\mathbb{O}$ are 4, 2 and 8. If we think of them as complex vector spaces their dimensions halve: they have dimensions 2, 1, and 4. And those are the numbers of ‘bonds’ we want… but in the wrong order.

So, to save the day we really need the symbols for the complex numbers, quaternions and octonions to be H, O, and C, respectively. We can claim someone made a big mistake. Or we could change the notation in chemistry. But for now, let me change the math notation. So:

• The complex numbers are a module over themselves. Since we’re calling the complex numbers H, we say that H has 1 ‘bond’.
• Since a quaternion can be identified with 2 complex numbers (not canonically, but oh well), this makes them into a module over the algebra of pairs of complex numbers. Since we’re calling the quaternions O, we say that O has 2 ‘bonds’.
• Since an octonion can be identified with 4 complex numbers (not canonically, but oh well), this makes them into a module over the algebra of 4-tuples of complex numbers. Since we’re calling the octonions C, we say that C has 4 ‘bonds’.

So now if I write something like

CH4

it has a meaning. We start by taking the direct sum of four copies of H, which is a module over H4, along with one copy of C, which is also a module over H4. Then we tensor them over H4: this attaches them with 4 ‘bonds’.

There could be problems with this scheme… which I prefer to let you discover on your own. But I’d really like someone to improve it, to make something both mathematically interesting and chemically realistic!

It’s also nice that chemists use ‘R’ to stand for a radical.

Posted by: John Baez on February 9, 2011 4:59 AM | Permalink | Reply to this

Re: The Three-Fold Way (Part 5)

Just to check I understand what’s going on: while a Hilbert space over any of these three division algebras induces a corresponding one over any of the others, there’s still something special about $\mathbb{C}$. Namely, the functors $\alpha_\ast$ and $\beta^\ast$ have disjoint images inside $Hilb_{\mathbb{C}}$, so a complex Hilbert space either “is real,” or “is quaternionic,” or “is honestly complex” with no overlap. That’s the opposite of true in the other two cases, since as you pointed out, $\gamma^\ast$ factors through $\alpha^\ast$, and also $\gamma_\ast$ factors through $\beta_\ast$.

I’m a little surprised at your use of the word “embedded” for a functor that is faithful but not full. I don’t usually think of the category of groups as “embedded” in the category of sets. But I guess that’s a discussion that’s been had to death elsewhere….

Posted by: Mike Shulman on February 9, 2011 4:16 AM | Permalink | PGP Sig | Reply to this

Re: The Three-Fold Way (Part 5)

Mike wrote:

Just to check I understand what’s going on: while a Hilbert space over any of these three division algebras induces a corresponding one over any of the others, there’s still something special about $\mathbb{C}$.

Good point!

Namely, the functors $\alpha_*$ and $\beta^*$ have disjoint images inside $Hilb_{\mathbb{C}}$.

Not really, I think. An irreducible unitary complex group representation can’t arise from both a real one and a quaternionic one — that’s the big ‘disjointness’ result.

But all complex Hilbert spaces arise from real ones (by complexification, $\alpha_*$), and all even-dimensional complex Hilbert spaces arise from quaternionic ones (by forgetting structure, $\beta^*$). So there’s a lot of overlap here!

In my 2-Hilbert spaces paper I explained what’s really going on. There’s a kind of symmetric monoidal abelian category called a ‘symmetric 2-H*-algebra’ that’s equipped with enough structure that you can carry out all the arguments I did in these blog posts for $\mathrm{Rep}(G)$, but without ever mentioning the compact group $G$.

In any symmetric 2-H*-algebra, every simple object that’s self-dual is either real, or quaternionic, but not both.

When $G$ is a compact group, $Rep(G)$ is a symmetric 2-H*-algebra, and the simple object are just the irreducible representations. $Hilb_\mathbb{C}$ is also a symmetric 2-H*-algebra, but the only simple object is $\mathbb{C}$. This object is real but not quaternionic — so the theorem is still true… but it’s a lot less exciting!

I’m a little surprised at your use of the word “embedded” for a functor that is faithful but not full.

Whoops! I was never really a category theorist, so remembering standard terminology was never my strong suit — so this usage was not the result of some deep conviction on my part, simply forgetfulness.

Okay, I’ll probably get rid of the word ‘embedded’.

By the way, thanks for asking all these questions and probing into all the nooks and crannies of this theory. It’s great fun for me having someone really pay attention to what I’m saying and understand it all. This $\mathbb{R}/\mathbb{C}/\mathbb{H}$ world is like a delightful little playground, especially with the help of some category theory, but it’s more fun with someone else around!

Posted by: John Baez on February 9, 2011 6:17 AM | Permalink | Reply to this

Re: The Three-Fold Way (Part 5)

So there’s a lot of overlap here!

Ah, yes, of course; clearly I was not awake when I wrote that. In part 5 we’re only talking about vector spaces, whereas in Part 4 we were talking about irreducible representations. Thanks for the tip about symmetric 2-H*-algebras; maybe I’ll go look it up.

It’s great fun for me having someone really pay attention to what I’m saying and understand it all.

It’s nice to have something fun to talk about that I can actually understand (mostly)! Like others, I’ve missed having you around here recently.

Hypercomplex numbers were one of the first “neat things” I learned about in high school that started to get me excited about math. It’s wonderful how there seems to be yet more to discover in there even now.

Posted by: Mike Shulman on February 9, 2011 7:50 PM | Permalink | Reply to this

Re: The Three-Fold Way (Part 5)

Thanks, Mike, I think that’s exactly the kind of insight I was hoping for here.

Posted by: Mark Meckes on February 9, 2011 2:17 PM | Permalink | Reply to this

Re: The Three-Fold Way (Part 5)

It often seems to happen in category theory that if you’re looking for a construction of something, if you can express formally what it is you want, then the construction will be canonically determined for you. I spent far too much time today trying to figure out a way to canonically determine all these structures, but I think I’ve got something resembling an answer.

First notice that since each of the six functors defined above has both a left and a right adjoint, it preserves limits and colimits, and moreover each of them is conservative (isomorphism-reflecting). Therefore, by the monadicity theorem, they are all monadic (and also comonadic). This means there is an abstract-nonsense answer to the question of what “structure” we need to equip the objects of their codomain with to make them equivalent to an object of their domain: namely, an algebra structure for the induced monad (or a coalgebra structure for the induced comonad).

In the case of the forgetful functors $\alpha^\ast$, $\beta^\ast$, and $\gamma^\ast$, the induced monad is pretty obvious and gives what you would expect. For instance, the monad on $Hilb_{\mathbb{R}}$ generated by $\gamma^\ast$ is just tensoring over $\mathbb{R}$ with $\mathbb{H}$. Since $\mathbb{H}$ is generated as an $\mathbb{R}$-algebra by $J$ and $K$ such that $J^2=K^2 = -1$ and $J K=-K J$, that gives us the definition of quaternionic structure on a real vector space.

The $(-)_\ast$ monads require a little more thought; let’s take the easiest case of $\alpha_\ast$. If $W\in Hilb_{\mathbb{R}}$, we can identify $\alpha_\ast W$ with $W\oplus W$ as a real vector space, with complex action defined by $(x,y)i = (y,-x)$. (Maybe that’s not exactly the way you would have guessed, but it’s what worked for me.) If we consider $\alpha_\ast$ as right adjoint to $\alpha^\ast$ (which we must do in order to have an induced monad $\alpha_\ast \alpha^\ast$ on $Hilb_{\mathbb{C}}$), then I think the counit $\alpha^\ast \alpha_\ast W \to W$ is $(x,y)\mapsto x$ and the unit $V \to \alpha_\ast \alpha^\ast V$ (for $V\in Hilb_{\mathbb{C}}$) is $x \mapsto (x,x i)$.

But for $V\in Hilb_{\mathbb{C}}$, we have another isomorphism (of real vector spaces) $\alpha_\ast \alpha^\ast V\cong V\oplus V$, defined by $(x,y) \mapsto \frac{1}{2}[ x - y i, x + y i]$ with inverse $[a,b] \mapsto (a+b,a i - b i)$. (I’ll use round and square brackets to distinguish which copy of $V\oplus V$ a given pair refers to.) The induced complex structure on the “square” copy of $V\oplus V$ is $[a,b]i = (a+b,a i - b i) i = (a i - b i, -a - b) = [a i, - b i].$ So as a complex vector space, this is actually $V\oplus V^\ast$, where $V^\ast$ is the complex conjugate $\mathbb{C}$-module. In these coordinates, the unit $V \to \alpha_\ast \alpha^\ast V$ is $x\mapsto (x,x i) = [x,0]$.

Thus, an algebra structure for the monad $\alpha_\ast \alpha^\ast$ on $V\in Hilb_{\mathbb{C}}$ is a map $V \oplus V^\ast \to V$, which restricts to the identity on the first component (this being the unit condition), and satisfies an associativity condition. The second component $V^\ast \to V$ gives our antilinear map $K\colon V\to V$, and the associativity condition makes it satisfy $K^2 = 1$, so we recover exactly the notion of real structure on a complex vector space.

Now again by generalities about monads and adjunctions, for any real vector space $W$, there is an induced $\alpha_\ast \alpha^\ast$-structure on $\alpha_\ast W$, which is the $\alpha_\ast$-image of the counit $\alpha^\ast \alpha_\ast W \to W$. Recall that this was $(x,y)\mapsto x$, so under $\alpha_\ast$ it becomes $((x,y),(z,w)) \mapsto (x,z)$ or $[(a,b), (c,d)] = ((a,b) + (c,d), (a,b)i - (c,d)i) = ((a+c,b+d), (b-d,c-a)) \mapsto (a+c, b-d).$ Setting $a=b=0$, we obtain $K((c,d)) = (c,-d)$, so that on a complexification $K$ does indeed act by “conjugation.”

Finally, since $\alpha_\ast$ is monadic, the monadicity theorem tells us exactly how to recover a real vector space from an $\alpha_\ast \alpha^\ast$-algebra: we take the coequalizer in $Hilb_{\mathbb{R}}$ of the two maps $\alpha^\ast \alpha_\ast \alpha^\ast V \;\rightrightarrows\; \alpha^\ast V.$ One of these is the algebra structure; the other is the counit of the adjunction. Thus, one of them sends $[a,b] \mapsto a + K(b)$, while the other sends $[a,b] = (a+b,a i - b i) \mapsto a+b$. So we end up with the quotient of $\alpha^\ast V$ by the relations $a+b = a+K(b)$, or equivalently $b = K(b)$ for all $b$. Since $K$ is a projection, this equivalently picks out its fixed set. (Presumably if we did the comonadic version, we would find an equalizer which picks out the $K$-fixed points directly.)

I’m sure you can do something similar with $\beta_\ast$ and $\gamma_\ast$, although I haven’t pushed the calculations all the way through. (There are a lot of plus and minus signs to keep track of, and I already spent a lot of time on this that I didn’t really have…). And presumably if we trace through the resulting (co)equalizers in those two cases, we’d get something equivalent to your descriptions above in terms of matrix algebras and projection matrices.

Seeing it this way makes me feel like I understand it better—although that may not be true for the intended audience of your paper! (-:

However, I still don’t really have an answer to your puzzle, although I feel like I ought to be able to extract one from all of this.

Posted by: Mike Shulman on February 10, 2011 5:35 AM | Permalink | PGP Sig | Reply to this

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