## January 15, 2023

### The Tenfold Way (Part 6)

#### Posted by John Baez

I’ve been studying Bott periodicity on and off since 1979, when I did a term paper on Clifford algebras in an undergrad course on group representation theory and physics taught by Valentine Bargmann. He was 71 at the time. Some of the students mocked him for being a bit slow — but if we’d known he’d been Einstein’s assistant from 1937 to 1946, we might have shown him a bit more respect, and asked him what working with Einstein was like!

I still have that term paper somewhere. Now I’m getting a bit slow, and I still don’t understand Bott periodicity quite as well as I want. So I have some questions. But in this part I’ll mainly just explain a bunch of stuff.

Right now I’m more interested in the algebraic and geometrical aspects than their homotopy-theoretic consequences. So let’s think about Clifford algebras.

Let $Cliff_n$ be the free algebra over $\mathbb{R}$ generated by $n$ anticommuting square roots of $-1$. In its simplest algebraic form, Bott periodicity says that $Cliff_{n+8}$ is isomorphic to the algebra of $16 \times 16$ matrices with entries in $Cliff_n$:

$Cliff_{n+8} \cong M_{16}(Cliff_n)$

The only way I know to show this involves figuring out all the Clifford algebras. Luckily the first 8 are really interesting — I’ll talk about them later.

I’m interested in representations of Clifford algebras, so let $Rep(Cliff_n)$ be the category of real representations of $Cliff_n$. We’ll see what these are like in a minute: they play a fundamental role in representation theory. But there are really just 8 of them!

The reason is that when one algebra consists of all $k \times k$ matrices with entries in another algebra, their categories of representations are equivalent. Because they have equivalent categories of representation, we say these algebras are Morita equivalent.

So, $Cliff_{n+8}$ is Morita equivalent to $Cliff_n$, and we have an equivalence of categories

$Rep(Cliff_{n+8}) \simeq Rep(Cliff_n)$

For each $n$, we have an inclusion

$Cliff_n \hookrightarrow Cliff_{n+1}$

sending the generators of $Cliff_n$ to the first $n$ of the generators of $Cliff_{n+1}$. This lets us restrict any representation of $Cliff_{n+1}$ to a representation of $Cliff_n$, giving a functor

$Rep(Cliff_{n+1}) \to Rep(Cliff_n)$

I want to describe the Clifford algebras, their representation categories and these functors in detail. I would like to draw a large clock with 8 hours, one for each of the 8 Morita equivalence classes of Clifford algebras. I would like to write descriptions of the categories $Rep(Cliff_0), \dots , Rep(Cliff_8)$ on these 8 hours, and descriptions of the functors $Rep(Cliff_{n+1}) \to Rep(Cliff_n)$ labeling arrows going counterclockwise between these categories. This would be a nice version of the so-called ‘Bott clock’. But I don’t have the patience to do this right now. So I’ll adopt a more linear approach:

• $Cliff_0 \cong \mathbb{R}$. $Rep(Cliff_0)$ is the category of real vector spaces.

• $Cliff_1 \cong \mathbb{C}$. $Rep(Cliff_1)$ is the category of complex vector spaces. The functor $Rep(Cliff_1) \to Rep(Cliff_0)$ sends each complex vector space to its underlying real vector space.

• $Cliff_2 \cong \mathbb{H}$. $Rep(Cliff_2)$ is the category of ‘quaternionic vector spaces’ — that is, left $\mathbb{H}$-modules. The functor $Rep(Cliff_1) \to Rep(Cliff_0)$ sends each quaternionic vector space to its underlying complex vector space.

So far this is a nice simple progression, like the green shoots of grass growing in the spring. You might naively expect it to keep on going forever with octonions and hexadecanions and so on, just like grass keeps growing forever taller into the sky… but no, it doesn’t. Things change at this point:

• $Cliff_3 \cong \mathbb{H} \oplus \mathbb{H}$. $Rep(Cliff_3)$ is the category of $\mathbb{Z}_2$-graded quaternionic vector spaces — that is, quaternionic vector spaces $V$ that are split as a direct sum of two summands $V_0 \oplus V_1$. The inclusion $Cliff_2 \hookrightarrow Cliff_3$ maps any quaternion $q$ to the pair $(q,q)$. Thus, the functor $Rep(Cliff_3) \to Rep(Cliff_2)$ takes any $\mathbb{Z}_2$-graded quaternionic vector space and forgets the grading.

• $Cliff_4 \cong M_2(\mathbb{H})$. $Rep(Cliff_4)$ is equivalent to the category of quaternionic vector spaces again, since $M_2(\mathbb{H})$ is Morita equivalent to $\mathbb{H}$. The inclusion $Cliff_2 \hookrightarrow Cliff_3$ maps any pair of quaternions $(q,q')$ to the diagonal matrix $\left( \begin{array}{cc} q & 0 \\ 0 & q' \end{array} \right)$ so the functor $Rep(Cliff_4) \to Rep(Cliff_3)$ sends any quaternionic vector space $V$ to the $\mathbb{Z}/2$-graded quaternionic vector space $V \oplus V$. I will call this functor ‘doubling’.

The last two functors are adjoints of each other! I’ll say more about these adjoints later, but we no longer feel like we’re going ‘forward’ in an unambiguous sense: it feels like we’ve turned back. Indeed:

• $Cliff_5 \cong M_4(\mathbb{C})$. $Rep(Cliff_5)$ is equivalent to the category of complex vector spaces, again by Morita equivalence. What about the inclusion $Cliff_4 \hookrightarrow Cliff_5$? You can think of quaternions as special $2 \times 2$ matrices of complex numbers. You can do it in various way, but the theory of Clifford algebras picks out a specific one, $\mathbb{H} \hookrightarrow M_2(\mathbb{C})$. This in turn gives the inclusion we want, $M_2(\mathbb{H}) \hookrightarrow M_4(\mathbb{C})$. What about the functor $Rep(Cliff_5) \to Rep(Cliff_4)$? It’s really a functor from complex vector spaces to quaternionic vector spaces. It sends any complex vector space $V$ to $\mathbb{C}^2 \otimes_{\mathbb{C}} V$, which becomes a quaternionic vector space using the inclusion $\mathbb{H} \hookrightarrow M_2(\mathbb{C})$.

• $Cliff_6 \cong M_8(\mathbb{R})$. $Rep(Cliff_6)$ is equivalent to the category of real vector spaces, again by Morita equivalence. What about the inclusion $Cliff_5 \hookrightarrow Cliff_6$? You can think of complex numbers as special $2 \times 2$ matrices of real numbers. You can do it in various way, but the theory of Clifford algebras picks out a specific one, $\mathbb{C} \hookrightarrow M_2(\mathbb{R})$. This in turn gives the inclusion we want, $M_4(\mathbb{C}) \hookrightarrow M_8(\mathbb{R})$. What about the functor $Rep(Cliff_6) \to Rep(Cliff_5)$? It’s really a functor from real vector spaces to complex vector spaces. It sends any real vector space $V$ to $\mathbb{R}^2 \otimes_{\mathbb{R}} V$, which becomes a complex vector space using the inclusion $\mathbb{C} \hookrightarrow M_2(\mathbb{R})$.

As you can see, these two steps have a very similar flavor! The second functor is called ‘complexification’ so the first should be called ‘quaternionification’. In fact, these functors are adjoint to the very first two on our list. So we are now going backwards.

But despite having arrived back at the category of real vector spaces, we are not quite done!

• $Cliff_7 \cong M_8(\mathbb{R}) \oplus M_8(\mathbb{R})$. $Rep(Cliff_7)$ is equivalent to the category of $\mathbb{Z}/2$ graded real vector spaces, because $M_8(\mathbb{R}) \oplus M_8(\mathbb{R})$ is Morita equivalent to $\mathbb{R} \oplus \mathbb{R}$. The inclusion $Cliff_6 \hookrightarrow Cliff_7$ maps any $8 \times 8$ real matrix $T$ to the pair $(T,T)$. Thus, the functor $Rep(Cliff_8) \to Rep(Cliff_7)$ takes a $\mathbb{Z}/2$-graded vector space and forgets the grading.

• $Cliff_8 \cong M_{16}(\mathbb{R})$. $Rep(Cliff_8)$ is equivalent to the category of real vector spaces, by Morita equivalence, and now we are really back where we started. The inclusion $Cliff_7 \hookrightarrow Cliff_8$ maps any pair of $8 \times 8$ real matrices $(T,T')$ to the $16 \times 16$ matrix $\left( \begin{array}{cc} T & 0 \\ 0 & T' \end{array} \right)$ so the functor $Rep(Cliff_8) \to Rep(Cliff_7)$ sends any real vector space $V$ to the $\mathbb{Z}/2$-graded real vector space $V \oplus V$. This is again a form of ‘doubling’.

You’ll notice that this last pair of functors is suspiciously similar to the second pair we saw.

Now, all 8 functors we’ve seen have adjoints, which are other functors on our list.

“Left or right adjoints?” the category theorists wearily inquire, tired of people failing to say which. Both — in fact these functors all have ambidextrous adjoints, which are both left and right adjoints. I think this has something to do with the fact that Clifford algebras are semisimple. I could probably figure it out — this is not one of questions I meant to ask — but since you’re probably waiting for the questions to come along, I might as well ask:

Question 1. Suppose $A$ and $B$ are semisimple algebras over some field $k$ and $f: A \to B$ is a homomorphism. Does the ‘restriction of scalars’ functor $Rep(B) \to Rep(A)$ always have an ambidextrous adjoint?

So, I could summarize the story so far by drawing a clock with 8 hours, functors going clockwise from each hour to the next, and their ambidextrous adjoints going counterclockwise. But I will lazily draw this picture linearly, with hour 8 = hour 0 appearing both on top and on bottom. The functors I’ve already listed will point upward, and their adjoints will point down.

$Rep(Cliff_0) \simeq [\text{real vector spaces}]$

$complexification \downarrow \uparrow \text{forgetting complex structure}$

$Rep(Cliff_1) \simeq [\text{complex vector spaces}]$

$quaternionification \downarrow \uparrow \text{forgetting quaternionic structure}$

$Rep(Cliff_2) \simeq [\text{quaternionic vector spaces}]$

$doubling \downarrow \uparrow \text{forgetting grading}$

$Rep(Cliff_3) \simeq [\mathbb{Z}/2\text{-graded quaternionic vector spaces}]$

$\text{forgetting grading} \downarrow \uparrow \text{doubling}$

$Rep(Cliff_4) \simeq [\text{quaternionic vector spaces}]$

$\text{forgetting quaternionic structure} \downarrow \uparrow \text{quaternionification}$

$Rep(Cliff_5) \simeq [\text{complex vector spaces}]$

$\text{forgetting complex structure} \downarrow \uparrow \text{complexification}$

$Rep(Cliff_6) \simeq [\text{real vector spaces}]$

$\text{doubling} \downarrow \uparrow \text{forgetting grading}$

$Rep(Cliff_7) \simeq [\mathbb{Z}/2\text{-graded real vector spaces}]$

$\text{forgetting grading} \downarrow \uparrow \text{doubling}$

$Rep(Cliff_8) \simeq [\text{real vector spaces}]$

I’ll end with two small remarks.

If you’re paying close attention you may have noticed something funny: some of the categories and functors show up twice in this chart! The reason — or at least one reason — is that we’re treating the Clifford algebras as ordinary algebras, when in fact they are naturally $\mathbb{Z}/2$-graded. We should think of them as $\mathbb{Z}/2$-graded algebras generated by odd anticommuting square roots of $-1$. Two Clifford algebras that are Morita equivalent as ordinary algebras can be Morita inequivalent as $\mathbb{Z}/2$-graded algebras. This eliminates the redundancy we’re seeing here. For more on this, see:

• The Tenfold Way (Part 4): super division algebras and the Brauer–Wall groups of $\mathbb{R}$ and $\mathbb{C}$.

You may have also noticed a category that’s missing from the above chart: the category of $\mathbb{Z}/2$-graded complex vector spaces. We’d get that if we considered complex rather than real Bott periodicity. If $\mathbb{C}\!\, liff_n$ is the complex algebra generated by $n$ anticommuting square roots of $-1$, then

$\mathbb{C}\!\, liff_0 = \mathbb{C}, \qquad \mathbb{C}\!\, liff_1 = \mathbb{C} \oplus \mathbb{C}$

and

$\mathbb{C}\!\, liff_{n+2} \cong M_2(\mathbb{C}\!\, liff_n)$

so complex Bott periodicity has period 2. The complex analogue of our big chart looks like this:

$Rep(\mathbb{C}\!\, liff_0) \simeq [\text{complex vector spaces}]$

$\text{doubling} \downarrow \uparrow \text{forgetting grading}$

$Rep(\mathbb{C}\!\, liff_1) \simeq [\mathbb{Z}/2\text{-graded complex vector spaces}]$

$\text{forgetting grading} \downarrow \uparrow \text{doubling}$

$Rep(\mathbb{C}\!\, liff_2) \simeq [\text{complex vector spaces}]$

Again, the category at the bottom is a repeat of the category at top, due to Bott periodicity. It’s not so necessary this time.

Posted at January 15, 2023 11:17 PM UTC

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### Karoubi

I have been told that one very thorough explanation of the relationship between Clifford algebras and Bott periodicity is contained in Karoubi’s PhD thesis “Algèbres de Clifford et K-théorie”

http://www.numdam.org/item/?id=ASENS19684121610

which has however two disadvantages: it is in French, and Karoubi uses a rather general setup of “topological K-theory of Banach categories”.

Fortunately, some aspects of his work are summarized (in English, and without Banach categories) in Section 42 of Dugger’s textbook-in-progress “A geometric introduction to K-theory”:

http://math.uoregon.edu/~ddugger/kgeom_070622.pdf

On the algebraic side, Karoubi systematically uses the real Clifford algebras $C^{p,q}$ attached to quadratic forms of signature $(p,q)$. For them, periodicity takes the form

(1)$C^{p+8,q}\simeq C^{p,q+8}\simeq M_16(C^{p,q})$

which decomposes as two relatively elementary isomorphisms

(2)$C^{p+4,q}\simeq C^{p,q+4}$

and

(3)$C^{p+1,q+1}\simeq M_2(C^{p,q}).$
Posted by: Simon Pepin Lehalleur on January 16, 2023 2:28 PM | Permalink | Reply to this

### Re: Karoubi

Thanks — I’ll have to check out Dugger’s textbook!

I’m pretty familiar with the Clifford algebras $C^{p,q}$ and the network of isomorphisms relating them… especially since in physics we need $C^{p,1}$ and $C^{1,p}$ to handle spinors on $(p+1)$-dimensional Minkowski spacetime. (Oddly, it can make a difference whether you use a metric like $dt^2 - dx^2 - dy^2 - dz^2$ or $dx^2 + dy^2 + dz^2 - dt^2$.) I’ve taught a course on this stuff and would like to write a book on it. Existing nice books which describe the isomorphisms you mention include Budinich and Trautman’s The Spinorial Chessboard and Michelson and Lawson’s Spin Geometry, so I’d have to do something new to make it really interesting. The tenfold way seems like a good new organizing principle.

What’s really bothering me is something I didn’t even manage to get to yet. There are two different ways to get symmetric spaces out of Clifford algebras, which wind in opposite directions around the Clifford clock, and I don’t see how they’re related. But I’ll have to explain this in future posts.

Dugger’s book looks like it may help….

Posted by: John Baez on January 16, 2023 4:48 PM | Permalink | Reply to this

### Re: The Tenfold Way (Part 6)

I’m hoping that you’ll connect to the graded Brauer groups of central simple Z/2-graded algebras over R,C (being Z/8 and Z/2 respectively). In particular while not every Clifford algebra is simple, they’re all graded simple, in that the only ideal isn’t Z/2-graded.

Also I’d like to point out Vakil’s work with Larson and Bryan on Bott periodicity algebro-geometrically (not yet published but you can find talks on YouTube).

Posted by: Allen Knutson on January 17, 2023 4:34 AM | Permalink | Reply to this

### Re: The Tenfold Way (Part 6)

I talked earlier about the graded Brauer groups (aka Brauer–Wall groups) of $\mathbb{R}$ and $\mathbb{C}$, and how to unify them in the ‘Brauer–Wall monoid’:

• The Tenfold Way (Part 1): a broad overview.

• The Tenfold Way (Part 2): the threefold and tenfold way, and the Brauer and Brauer–Wall groups of $\mathbb{R}$ and $\mathbb{C}$.

• The Tenfold Way (Part 3): the Brauer monoid and Brauer–Wall monoid of a field.

• The Tenfold Way (Part 4): super division algebras and the Brauer–Wall groups of $\mathbb{R}$ and $\mathbb{C}$.

• The Tenfold Way (Part 5): A short paper summarizing a few aspects of the tenfold way.

I didn’t tackle that $\mathbb{Z}$/2-graded aspect here, but I want to. If we include it, the ‘redundancy’ I mentioned will go away, and we’ll get 10 truly different categories.

Also I’d like to point out Vakil’s work with Larson and Bryan on Bott periodicity algebro-geometrically (not yet published but you can find talks on YouTube).

Posted by: John Baez on January 17, 2023 9:18 PM | Permalink | Reply to this

### Re: The Tenfold Way (Part 6)

By the way, one thing I’m confused about is this:

Here I’m including each Clifford in the next

$i: Cliff_n \hookrightarrow Cliff_{n+1}$

by mapping the generating square roots of $-1$ in $Cliff_n$ to the first $n$ of the generating square roots of $-1$ in $Cliff_{n+1}$.

In the next part I’m alluding to another inclusion

$j: Cliff_n \hookrightarrow Cliff_{n+1}$

Namely, I’m using the fact that the even part of $Cliff_{n+1}$ is isomorphic to $Cliff_n$. This is not a specific inclusion, just an inclusion up to automorphisms of $Cliff_n$. But no matter how we choose $j$, it must differ from $i$!

Why? Here’s why:

$i$ is a homomorphism of superalgebras and $j$ cannot be, because $i$ maps the odd generators of $Cliff_n$ to odd generators of $Cliff_{n+1}$, while $j$ maps all of $Cliff_n$ to the even part of $Cliff_{n+1}$.

However, in the next part I use this ‘bad’ property of $j$ to get symmetric spaces (see the last paragraph there), so it’s actually good in some ways.

Note that $i$ is a $\ast$-algebra homomorphism, because the $\ast$-structure on a Clifford algebras is defined by making the generating square roots of $-1$ be skew-adjoint.

Can $j$ be chosen to be a $\ast$-algebra homomorphism? If not, I’m in trouble, because in the next part I’m acting like it sends unitary elements of $Cliff_n$ to unitary elements of $Cliff_{n+1}$.

Let’s look at a standard formula for $j$, for example from Theorem 3.7 of Michelson and Lawson’s Spin Geometry. If $e_1, \dots, e_n$ are the generating square roots for $Cliff_n$ and $e_1, \dots, e_{n+1}$ are those for $Cliff_{n+1}$, they take

$j(e_i) = e_{n+1} e_i$

Note that

$j(e_i^\ast) = - j(e_i) = - e_{n+1} e_i = e_i e_{n+1} = e_i^\ast e_{n+1}^\ast = (e_{n+1} e_i)^\ast$

so $j$ is indeed a $\ast$-algebra homomorphism! Whew!

But I need to think more about the difference between $i$ and $j$ as I try to develop a unified story here.

Posted by: John Baez on January 17, 2023 10:58 PM | Permalink | Reply to this

### Bott periodicity, the operator algebra way

At the moment, my favourite proof of complex Bott periodicity in topological K-theory is the operator algebra proof of Cuntz. It is nicely summarized in this short note by Tobias Fritz and uses a different kind of algebraic gadget than the complex Clifford algebra, namely the Toeplitz $C^*$-algebra.

I was hoping to find a version of Cuntz’s proof for real Bott periodicity, but I couldn’t locate a full treatment. As in the periodicity of real Clifford algebras, the situation is naturally bigraded, the periodicity should decompose into two statements, and one of them is proven in Proposition 1.20 of this paper using the real Toeplitz C^*-algebra.

This doesn’t seem completely unrelated to the tenfold way, because there seems to be a line of research (a recent paper , a book , a survey) on topological insulators looking at the “bulk-boundary correspondence” using index theory for the topological K-theory of (real and complex) C^*-algebras and Toeplitz extensions.

Posted by: Simon Pepin Lehalleur on January 18, 2023 11:52 AM | Permalink | Reply to this

### Re: The Tenfold Way (Part 6)

Concerning the ambidexterity of the adjunctions, this must be related to the fact both categories involved are dagger categories. This should imply that if a dagger functor has an adjoint on one side, then this is also an adjoint on the other side (in a canonical way constructed from the dagger).

I think that this situation is typical for categories of complex *-representations of *-algebras, and more specifically for categories of unitary representations of groups.

Posted by: Tobias Fritz on January 19, 2023 2:52 AM | Permalink | Reply to this

### Re: The Tenfold Way (Part 6)

Thanks! I hadn’t thought about using a dagger structure. Semisimple algebras over $\mathbb{R}$ are all isomorphic to direct sums of matrix algebras over $\mathbb{R}, \mathbb{C}$ and $\mathbb{H}$, so they can all be given the structure of real $\ast$-algebras. Does that provide enough structure to make their categories of representations into dagger categories? Hmm, it seems not instantly, it seems we need to check stuff like this:

1) Define a $\ast$-representation of a real $\ast$-algebra on a real Hilbert spaces to be one with $\rho(a^\ast) = \rho(a)^\dagger$. Show the category of $\ast$-representations of a real $\ast$-algebra on finite-dimensional real Hilbert spaces is a $\dagger$-category. (I haven’t actually checked this; it just sounds plausible.)

2) Every representation of an $n \times n$ matrix algebra over $\mathbb{R}, \mathbb{C}$ or $\mathbb{H}$ on a finite-dimensional real vector space is isomorphic, as a representation, to a $\ast$-representation of this algebra on a real Hilbert space. (To do this, note that any irreducible representation of an $n \times n$ matrix algebra over $\mathbb{R}, \mathbb{C}$ resp. $\mathbb{H}$ is isomorphic to the irreducible representation on the space $\mathbb{R}^n, \mathbb{C}^n$ resp. $\mathbb{H}^n$, which becomes a $\ast$-representation when we equip this space with its standard real Hilbert space structure.)

3) Thus, the category of finite-dimensional real representations of a matrix algebra over $\mathbb{R}, \mathbb{C}$ or $\mathbb{H}$ is equivalent to its category of finite-dimensional real $\ast$-representations.

4) Generalize from these matrix algebras to direct sums of them: any semisimple algebra over $\mathbb{R}$ can be given a $\ast$-structure by doing it one direct summand at a time, and use this to show its category of finite-dimensional real representations is equivalent to its category of finite-dimensional real $\ast$-representations.

The latter is a dagger category, so we’re done…. whoops, wait — we can’t always transfer dagger structures across equivalences!

So maybe we should just decide the category of $\ast$-representations is what we want to think about. Or structure the argument a bit differently, so we keep the same category of representations but equip each object with the structure of a $\ast$-representation.

For the case of Clifford algebras we can at least avoid the fact that a matrix algebra has many different $\ast$-algebra structures, since Clifford algebras naturally become $\ast$-algebras when we decree that the generating square roots of $-1$ are skew-adjoint.

By the way, all this reminds me of Qiaochu Yuan’s remark that for a functor to have an ambidextrous adjoint is not a mere property but an extra structure: there are sometimes nonisomorphic ways to extend it to an ambidextrous adjunction!

He mentions the example of the functor $f^\ast: Rep(H) \to Rep(G)$ between categories of finite-dimensional complex representations of finite groups determined by a homomorphism $f: G \to H$. In this case we can choose invariant inner products on the representations and make them into unitary representations, getting a $\dagger$-category of representations.

He also mentions the so-called ‘Nakayama isomorphism’ between the left and right adjoints of $f^\ast$, and if you click on his link it leads to a question about my student Jeffrey Morton’s paper 2-vector spaces and groupoids, asked by some poor suffering person who doesn’t know what adjoint functors are.

Posted by: John Baez on January 19, 2023 2:55 AM | Permalink | Reply to this

### Re: The Tenfold Way (Part 6)

Whoops, I had not noticed the difference between the category of representations and the category of $\ast$-representations! I guess I’m too used to thinking about $\ast$-representations only.

Yes, 1) is indeed correct. The general statement is that if $t : F \Rightarrow G$ is a natural transformation between dagger functors between dagger categories, then also $t^\dagger : G \Rightarrow F$, defined as the componentwise dagger, is a natural transformation. This is because the naturality condition of $t^\dagger$ at $f : X \to Y$ is $t^\dagger_Y \circ G(f) = F(f) \circ t^\dagger_X,$ and this is precisely the dagger of the naturality condition of $t$ at $f^\dagger : Y \to X$, namely $G(f^\dagger) \circ t_Y = t_X \circ F(f^\dagger).$

Posted by: Tobias Fritz on January 19, 2023 2:58 AM | Permalink | Reply to this

### Re: The Tenfold Way (Part 6)

Thanks for spelling out the proof of 1), Tobias!

I’m starting to think that the tenfold way is deeply $\dagger$-categorical. Not only did I want ambidextrous adjoints here in Part 6, I needed to think of Clifford algebras as $\ast$-algebras in Part 7, and I defined symmetric spaces by considering all the ways of extending a $\ast$-representation of $Cliff_k$ to a $\ast$-representation of $Cliff_{k+1}$ in Part 8. Of course the importance of $\mathrm{O}(n)$, $\mathrm{U}(n)$ and $\mathrm{Sp}(n)$ in Bott periodicity should have been a big clue in the first place, since these are groups of transformations of real, complex and quaternionic Hilbert spaces obeying $g g^\ast = g^\ast g = 1$.

Posted by: John Baez on January 19, 2023 3:10 AM | Permalink | Reply to this

### Re: The Tenfold Way (Part 6)

Nice! Perhaps you’ll want to go even one step further and even consider those categories as W$^\ast$-categories, which are the many-object generalizations of von Neumann algebras. W$^\ast$-category theory is actually simpler than ordinary category theory! And I think that what you’re seeing is a glimpse of this fact. Not only is every W$^\ast$-adjunction ambidextrous, but also every W$^\ast$-functor is continuous and cocontinuous. There also seems to be an adjoint functor theorem which should imply that every W$^\ast$-functor between categories of $\ast\!$-representations is an ambidextrous adjoint. I haven’t worked out the details of this yet, and I’m also not sure if it will still hold when you only consider finite-dimensional $\ast\!$-representations. (I have an unfinished draft on all this which will hopefully see the light of day next month.)

On the other hand, perhaps W$^\ast$-categories are overkill if you only work with finite-dimensional $\ast\!$-representations. As with von Neumann algebras, W$^\ast$-categories really start to shine when you have infinite-dimensional things flying around. So they may be less relevant when your groups are compact.

Posted by: Tobias Fritz on January 19, 2023 8:12 AM | Permalink | Reply to this

### Re: The Tenfold Way (Part 6)

Thanks! I used to love W${}^\ast$-algebras, so I should love W${}^\ast$-categories too. But in this project I want to avoid analysis and focus on algebra, because I think some very simple things are going on here that people haven’t fully understood yet. So I only want to think about categories enriched over finite-dimensional vector spaces, and hope that groups like

$\mathrm{U} = \lim_{n \to \infty} \mathrm{U}(n)$

arise here mainly as a way to summarize what’s going on in the finite-dimensional case. Sort of how you can get the infinite loop space $BGL$ from the category of finite-dimensional vector spaces.

But still, if this project works it should have spinoffs for infinite dimensions.

Posted by: John Baez on January 19, 2023 4:17 PM | Permalink | Reply to this

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