# Quaternions and Tensor Categories

As you can tell from the title of this post, I am trying to drag John Baez over to our blog.

Let $Q$ be the ring of quaternions, i.e., $\mathbb{R} \langle i,j,k \rangle$ with the standard relations. Let $Q$-mod be the category of left $Q$-modules. This has an obvious tensor structure (including duals), inherited from the category of $\mathbb{R}$ vector spaces. Actually, that structure doesn’t quite work; I’ll leave to you good folks to work out what I should have said.

Let $q=a+bi+cj+dk$ be a quaternion. Anyone who works with quaternions knows that there are two notions of trace. The naive trace, $4a$, is the trace of multiplication by $a$ on any irreducible $Q$-module, using the obvious tensor structure. But there is a better notion, the reduced trace, which is equal to $2a$. Similarly, there is a naive norm, $(a^2+b^2+c^2+d^2)^2$, and there is a reduced norm $a^2+b^2+c^2+d^2$.

This all makes me think that there is a subtle tensor category structure on $Q$-mod, other than the obvious one, for which these are the trace and norm in the categorical sense. Can someone spell out the details for me?

By the way, a note about why I am asking. I am reading Milne’s excellent notes on motives, and I therefore want to understand the notion of a non-neutral Tannakian category (page 10). As I understand it, this notion allows us to evade some of the standard problems in defining characteristic $p$ cohomology; one of which is the issue above about traces in quaternion algebras.

## 29 thoughts on “Quaternions and Tensor Categories”

1. PB says:

Hello,
Could you tell/teach (if not too long or difficult) what are the trace and norm in the categorical sense ?
Thank you :)

2. Hi David,

I’m confused about what your talking about. There is no tensor structure on Q-mod. The Quaternions are a simple algebra and so any (let’s say finite dimensional) Q-module is a direct sum of N copies of Q (acting on itself from, say, the left).

In particular there is no non-zero object with abelian automorphism group. But in any Monoidal category the monoidal unit has to have abelian automorphism group by the Eckmann-Hilton Argument.

Thus the only monoidal structures on Q-mod need the zero object as the monoidal unit (direct sum is such a monoidal structure). However having the monoidal unit be the zero object is incompatible with it being a tensor category structure, since a tensor structure must distribute over direct sums.

3. Allen Knutson says:

I’m not going to attempt the details, but I remember hearing (surely from John Baez) that the quaternions are a commutative algebra in the tensor category of super-duper vector spaces… the trick being, of course, that the commutor for tensor products hides the “actual” noncommutativity.

I would hope that within that framework one could define a tensor structure on Q-mod, and that Chris’ objection would be bypassed because the automorphism group preserving the super-duper decomposition of Q^1 would indeed be abelian.

4. Allen,

Could you actually define super-duper vector spaces?

After discussing with Nick, my theory is that super-duper vector spaces are the category of $\mathbb{Z}/2\times \mathbb{Z}/2$ representations, with the R-matrix given by -1 when commuting different non-trivial representations past each other and 1 otherwise. Is this right?

This makes $\mathbb{Z}/2\times \mathbb{Z}/2$ into a ribbon Hopf algebra in an funny way (actually computing the R-matrix and ribbon element I leave as an exercise to the reader). In this case, presumably the “good” trace is some sort of quantum trace.

5. Allen Knutson says:

Is this right?

Yeah, that’s it.

I hear the octonions are also commutative associative, in super-duper-duper vector spaces.

6. Hmm, I probably didn’t ask quite the right question. (Although I’d be glad to hear more about super-duper vector spaces and Ben’s ribbon element.)

Here is the sort of behavior that, to my limited understanding, happens in the category of motives. Motives are a symmetric monoidal category. Usually the Hom spaces are taken to be $\mathbb{Q}$ vector spaces but, to simplify exposition, I’ll tensor everything up to $\mathbb{R}$.

There is an object in the theory of motives, which I’ll call $E$, whose endomorphism ring is the quaternion algebra, and for which the trace of $a + bi + cj + dk$ is $2a$. (In particular, the identity has trace $2$.)

Using the commutator, we can define $\bigwedge^2 E$. This has endomorphism ring $\mathbb{R}$ and $\bigwedge^2 (a+bi+cj+dk) = a^2 + b^2 + c^2 + d^2$.

It’s the category that thinks the quaternions have a two dimensional real representation. How?

7. Majid and his collaborators did some work on the octonion case, e.g., 9802116. I think the term “super-duper-duper” doesn’t quite capture some of the subtlety in choosing an associator. There are also anyonic presentations of Azumaya algebras over varieties associated to Brauer group elements. I don’t know if it’s written down anywhere.

8. If you switch from R to Z_p, I think a quaternion algebra should act on the Dieudonné module for p-torsion in a supersingular elliptic curve. I don’t know how this translates to R.

9. I am slowly being dragged into things that I don’t really understand yet :). So, the basic point here is that the endomorphism algebra of a supersingular elliptic curve is a quaternion order. To give a specific example, if $p$ is a prime which is $3 \mod 4$, then the endomorphisms of $y^2 = x^3-x$ are

$\mathbb{Z} \langle j, f \rangle / \left( j^2 = -1,\ f^2 = -p,\ jf = -fj \right).$

This acts on all the Tate modules; that causes no problem because this ring becomes $\mathrm{Mat}_{2 \times 2}(\mathbb{Q}_{\ell})$ after tensoring with $\mathbb{Q}_{\ell}$. I think it also acts on the Dieudonne module, but I don’t know that theory. In all of these cases, we have to go a field larger than $\mathbb{Q}$.

The strange thing is that, in the ring of motives, there is an object with an action of this ring. That object acts like it has dimension $2$ (based on traces) and acts like the relevant field of scalars is $\mathbb{Q}$ (based on its endormophism ring being
$\mathbb{Q} \langle j, f \rangle /\left( j^2 = -1,\ f^2 = -p,\ jf = -fj \right).$

10. Ben,

Your category is actually equivalent (in a nontrivial way) to
the category of representations of ${\mathbb Z}/2\times {\mathbb Z}/2$ as a braided tensor category. It is an old fact due to Joyal
and Street that a semisimple braided tensor category where all
simple objects are invertible is determined (up to a braided equivalence) by the function on the set of simple objects whose
value on object $X$ equals the braiding on $X\otimes X$
(and this function happens to be a quadratic form on
the group of simple objects). In your definition the function is
identically 1 whence my claim.

11. I think the braiding should be -1 for $X \otimes X$ when X is not 1. That way, the category of super-duper vector spaces contains the category of super vector spaces.

12. I guess the object you’re looking for is the $\mathbb{F}_p$-motive underlying H^1 of this elliptic curve. As you pointed out, the quaternionic action can’t be seen through the etale realizations, but I think it is visible in the crystalline realization. I don’t know how the crystalline realization works, but I guess it has something to do with the Dieudonné module of the p-divisible group.

Here’s a quick overview: there is a contravariant functor that takes group schemes of p-power order over F_p to modules of finite Z_p-length over the Dieudonné ring, which in this case is Z_p[F,V]/(FV – p). This functor is an antiequivalence of categories, and respects lots of stuff like base change (which may make the Dieudonné ring noncommutative due to actions of F and V on coefficients). The p^n-torsion of an elliptic curve is (if it’s defined over F_p) taken to a rank 2 free module over Z/(p^n)Z with possibly interesting action of F and V, and it has a natural action by the opposite of the endomorphism ring of the curve. In this case, p is nilpotent, so the action doesn’t base extend to the quaternion algebra over the rationals.

If the p^n-torsion is not defined over F_p, you can use $W_n(\overline{\mathbb{F}_p})$ as the coefficient ring, with F and V commuting semilinear operators. Passing to the direct limit of p-power torsion yields the inverse limit of these modules, which is a rank two module over the maximal unramified extension of Z_p, but with the semilinear action of F and V. This should have a quaternion action that can be base-changed by rationals, but I don’t know details regarding the action. As you mentioned, it is defined over a larger field.

13. The j-invariant of a supersingular elliptic curve in char. p always lies in F_{p^2}, and can lie in F_p. But to have all the endomorphisms be defined, one must work over F_{p^2} (at least). Then the crystalline cohomology (tensored with Q, say — where here Q is the rationals, not the quaternions) will be a rank two vector space over the degree 2 unramified extension of Q_p (what number theorists often call Q_{p^2}); this appears in the crystalline theory because it is the fraction field of the ring of Witt vectors of F_{p^2}.

The fact that we can’t actually find a two-dimensional Q-vector space on which the quaternion algebra acts explains why the category of motives over a field of char. p doesn’t admit a fibre functor over Q (which is to say, I think, that it is not neutral).

14. Wow! The title of this blog entry sure got my attention! I’ve been fascinated by this subject for ages!

The physicist Stephen Adler wrote a big book on quaternionic quantum mechanics, which in my opinion was severely hurt by his failure to fully grapple with the following issue:

In quaternionic quantum mechanics, you’d like to use “quaternionic Hilbert spaces”. To combine physical systems you’d like to tensor such Hilbert spaces. But you have to think very hard before tensoring left Q-modules when your ring Q is noncommutative.
The same problem shows up when you try to define a “quaternionic C*-algebra”.

Toby Bartels and I proposed one obvious solution: use Q-Q bimodules. Better: use Q-Q bimodules for which the left and right actions of the copy of the reals inside the quaternions Q agree.

But there’s also a nice symmetric monoidal category in which the quaternions are a commutative monoid object. If we work inside this category, we can tensor Q-modules to our heart’s content. Is this good for something? I don’t know.

I see that Allen Knutson has already confirmed Ben Webster’s description of this symmetric monoidal category. The same trick works for any Clifford algebra: you just need more factors of Z/2 in your grading group.

It’s even more fun to see a symmetric monoidal category in which the *octonions* are a commutative monoid object. You can see a brief description of this near the bottom of this page.

I have no idea if either of these remarks helps answer David’s question. I’ll think about that question later… I just had to say this stuff.

15. John Mangual says:

David, I would love to hear a fleshed-out version of comment 9 which had some explanation of ideas like “endomorphism ring of an elliptic curve”, “quaternion order”, “Tate module” and “Dieudonne module”.

Reading about $\mathbb{Q}_l}$ in Wikipedia leads to a discussion on Etale cohomology and the Weil conjectures. I would really like to understand why motives arise in Number Theory. You could call it “motivating motives”.

Also, I am amused that by adding enough gradings you can make even the Octonions into the commutative object in some category. I think you guys have talked about this theme in other parts of this blog.

16. John Mangual says:

err $\mathbb{Q}_l$

17. What I’m coming to realize is that my vague understanding of Tanaka-Krein duality was wrong over a non-algebraically-closed ground field.

For example, here is a symmetric monodial category over $\mathbb{Q}$. There are three simple objects, $1$, $L$ and $L^{\vee}$. As the notation suggests, $L$ and $L^{\vee}$ are dual. Also, $L \otimes L = L^{\vee}$, $L \otimes L^{\vee} = L^{\vee} \otimes L = 1$ and $L^{\vee} \otimes L^{\vee} = L$. On any product of simple objects, the associator and commutor act by the identity.

This category clearly wants to be $\mathrm{Rep}(\mathbb{Z}/3)$. If I replaced $\mathbb{Q}$ with $\mathbb{C}$, it would be equivalent to $\mathrm{Rep}(\mathbb{Z}/3)$.

Before I started thinking about this stuff, I would have thought that I couldn’t define this category. The group $\mathbb{Z}/3$ doesn’t have any nontrivial characters over $\mathbb{Q}$, and I thought that the Tanaka-Krein theorem would therefore tell me that this category doesn’t exist.

Obviously, I was wrong …

18. John, I might write something up. But honestly, I think Milne did a great job. Have you thought much about why number theorists want a good notion of cohomology, and about why it’s frustrating that all the definitions give coefficients in $\mathbb{Q}_{\ell}$ or $\mathbb{Q}_{p}$, not in $\mathbb{Q}$ or $\mathbb{R}$?

19. Ben Wieland says:

David, it seems worth pointing out that the category in 17 is called Rep(mu_3).

just because no one’s yet written out an example of a non-neutral tannakian category in the comments, here’s an attempt.

let D be a quaternion algebra over a field k of char != 2, and fix an algebra isomorphism f:D -> D^op. then define C(D) to be semisimple abelian k-linear category generated by two objects: L and L(1) satisfying (a) End(L) = k, (b) End(L(1)) = D, (c) L and L(1) are mutually Hom-orthogonal. in addition, define a tensor structure on C(D) as follows: L is the unit, and L(1) (x) L(1) =~ L, with the map on endomorphism rings D (x) D -> k given by the reduced trace map (using the chosen isomorphism f).

if D is non-split, then i believe the above is an example of a non-neutral tannakian category. i was trying to write down the category QC(non-split mu_2-gerbe associated to (D,a)) explicitly, so i might have made a mistake.

(more generally, if D is an non-split Azumaya algebra with D^-1 = D^op, then \oplus_{i in Z} Mod(D^i) should be a non-neutral tannakian category, realised as QC(G_m-gerbe of splittings of D)).

whoops, slight correction. in the preceeding example, i think L(1) (x) L(1) should be D (viewed as an object of = Vect_k), and the map on endomorphism rings is simply D (x) D =~ D (x) D^op =~ End(D). not sure if this works!

22. John Mangual says:

David, I have not done many homology calculations, so I have never gotten frustrated when the base ring is wrong. :)

It seems like the representation theory of G is equivalent to studying the category C[G]-mod and which behaves like a ring w/ resp to tensor products and direct sums. And now you are asking what happens if we replace C with the quaternions. All of these generalize the theory of vector spaces when G = {1}.

23. Gentlemen,
I have been using quaternions for 20+ years for spacecraft attitude and appendage control. After stumbling across this blog I feel that I have missed a lot of recent developments concerning quaternions. Much of the nomenclature I see here is unfamiliar to me. Are there references that would help me come up to speed so that, if I follow this blog, I might have a better understanding of the discussions?

24. Noel, I’m not an expert in control theory, but as far as I can tell, the arithmetic quaternion theory we’re discussing doesn’t have a strong relationship with rotations in three dimensional space. I should also mention that it isn’t particularly new, but there probably wasn’t a good reason for you to have heard of it. Deuring wrote about quaternion algebras as endomorphism rings of elliptic curves in 1941, and the language of tensor categories and motives was introduced in the 1960s. You can read about elliptic curves over F_p in chapter 5 of Silverman’s The Arithmetic of Elliptic Curves, and you can find some information about tensor categories and motives in chapter 2 of some book by Deligne, Milne, Ogus, and Shih (Springer Lecture Notes in Mathematics 900). I think if you search for “James Milne”, you’ll find a scan on his web page.

25. Noel, Scott is right. Yes, quaternions are important in control theory, because trajectories in the group of quaternions are a very good way to model rotation of objects. This technique is useful in computer graphics as well as for spacecraft attitude.

However, what David Speyer is talking about is not important in control theory. This use of quaternions is to your use of quaternions is what Renoir is to cartoon animation. That is, they are painting, they are difficult, and there are certain connections too, but don’t feel pressed to be expert at both of them.

26. noel hughes says:

Thank you for the condescending and, frankly, insulting reply to my honest inquiry for information. If this is typical of the thinking of people in this group, I do not want to be involved.

27. Noel-

Don’t let greg’s ill-chosen analogy bother you too much. It’s just that you make a request that’s rather hard to fill, especially since none of us know your background, and a lot of diverse stuff came up in the discussions above: did you want to know about the tensor categories or the crystalline cohomology? This is all stuff that’s Greek to most mathaticians, so it’s pretty difficult to come up with appropriate suggestions for some one of unspecified mathematical background.

28. Peter L. Griffiths says:

Hamilton in his letter of 17 October 1843 to John Graves seems to be very confused about the relationship between i, j, +1, and -1. He asks what are we to do with ij when i and j are the unequal roots of a common square. In fact there is no law of arithmetic which makes ij equal to anything but +1. It is these doubts of Hamilton which are the source of his fallacious theory of the non-commutative properties of the multiplication of imaginary numbers. All multiplication whether of real or imaginary numbers is commutative.

29. Peter L. Griffiths says:

Hamilton’s Quaternions Equation i^2=j^2=k^2=ijk=-1 is incorrect because -1, like all real imaginary and complex numbers, only has two square roots, so that -1=k^2 is wrong unless k can equal either i or j. Likewise all real, imaginary and complex numbers can only have three cube roots, four fourth roots and five fifth roots etc.