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Background

For definiteness (even though this is a categorical question!) let's agree that a vector space is a finite-dimensional real vector space and that an associative algebra is a finite-dimensional real unital associative algebra.

Let $V$ be a vector space with a nondegenerate symmetric bilinear form $B$ and let $Q(x) = B(x,x)$ be the associated quadratic form. Let's call the pair $(V,Q)$ a quadratic vector space.

Let $A$ be an associative algebra and let's say that a linear map $\phi:V \to A$ is Clifford if $$\phi(x)^2 = - Q(x) 1_A,$$ where $1_A$ is the unit in $A$.

One way to define the Clifford algebra associated to $(V,Q)$ is to say that it is universal for Clifford maps from $(V,Q)$. Categorically, one defines a category whose objects are pairs $(\phi,A)$ consisting of an associative algebra $A$ and a Clifford map $\phi: V \to A$ and whose arrows $$h:(\phi,A)\to (\phi',A')$$ are morphisms $h: A \to A'$ of associative algebras such that the obvious triangle commutes: $$h \circ \phi = \phi'.$$ Then the Clifford algebra of $(V,Q)$ is the universal initial object in this category. In other words, it is a pair $(i,Cl(V,Q))$ where $Cl(V,Q)$ is an associative algebra and $i:V \to Cl(V,Q)$ is a Clifford map, such that for every Clifford map $\phi:V \to A$, there is a unique morphism $$\Phi: Cl(V,Q) \to A$$ extending $\phi$; that is, such that $\Phi \circ i = \phi$.

(This is the usual definition one can find, say, in the nLab.)

Question

I would like to view the construction of the Clifford algebra as a functor from the category of quadratic vector spaces to the category of associative algebras. The universal property says that if $(V,Q)$ is a quadratic vector space and $A$ is an associative algebra, then there is a bijection of hom-sets

$$\mathrm{hom}_{\mathbf{Assoc}}(Cl(V,Q), A) \cong \mathrm{cl-hom}(V,A)$$

where the left-hand side are the associative algebra morphisms and the right-hand side are the Clifford morphisms.

My question is whether I can view $Cl$ as an adjoint functor in some way. In other words, is there some category $\mathbf{C}$ such that the right-side is $$\mathrm{hom}_{\mathbf{C}}((V,Q), F(A))$$ for some functor $F$ from associative algebras to $\mathbf{C}$. Naively I'd say $\mathbf{C}$ ought to be the category of quadratic vector spaces, but I cannot think of a suitable $F$.

I apologise if this question is a little vague. I'm not a very categorical person, but I'm preparing some notes for a graduate course on spin geometry next semester and the question arose in my mind.

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I was just wondering this a week ago! I'd also be very interested in the answer to this question. –  Qiaochu Yuan Dec 3 '09 at 18:04
    
Are categorical questions not definite? :P –  Mariano Suárez-Alvarez Dec 3 '09 at 18:14
    
I(and likely others) would appreciate it if you posted these notes for download. There are some references that I know of that address relationships between Clifford algebras and other branches of modern physics(Girard, Iordănescu), but I would like to see what perspective you are coming from on this. Thanks for the question though, it is interesting! –  B. Bischof Dec 3 '09 at 22:17
    
Certainly. I put all my notes online as a matter of principle. I've only just started preparing them, though... so it may be a while. Thanks for your interest! –  José Figueroa-O'Farrill Dec 4 '09 at 7:06

5 Answers 5

up vote 7 down vote accepted

Disqualifier: this isn't a complete answer.

There's a basic "chalk and cheese" problem here. The "categories" that you are comparing are of two different types, although they do seem similar on the surface. On the one hand you have an honest algebraic category: that of associative algebras. But the other category (which, admittedly, is not precisely defined) is "vector spaces plus quadratic forms". This is not algebraic (over Set). There's no "free vector space with a non-degenerate quadratic form" and there'll (probably) be lots of other things that don't quite work in the way one would expect for algebraic categories. For example, as you require non-degeneracy, all morphisms have to be injective linear maps which severely limits them. You could add degenerate quadratic forms (which means, as AGR hints, that you regard exterior algebras as a sort of degenerate Clifford algebra - not a bad idea, though!) but this still doesn't get algebraicity: the problem is that the quadratic form goes out of the vector space, not into it, so isn't an "operation".

However, you may get some mileage if you work with pointed objects. I'm not sure of my terminology here, but I mean that we have a category $\mathcal{C}$ and some distinguished object $C_0$ and consider the category $(C,\eta,\epsilon)$ where $\eta : C_0 \to C$, $\epsilon : C \to C_0$ are such that $\epsilon \eta = I_{C_0}$. In Set, we take $C_0$ as a one-point set. In an algebraic category, we take $C_0$ as the free thing on one object. Then the corresponding pointed algebraic category is algebraic over the category of pointed sets (I think!).

The point (ha ha) of this is that in the category of pointed associative algebras one does have a "trace" map: $\operatorname{tr} : A \to \mathbb{R}$ given by $(a,b) \mapsto \epsilon(a \cdot b)$. Thus one should work in the category of pointed associative $\mathbb{Z}/2$-graded algebras whose trace map is graded symmetric.

In the category of pointed vector spaces, one can similarly define quadratic forms as operations. You need a binary operation $b : |V| \times |V| \to |V|$ (only these products are of pointed sets) and the identity $\eta \epsilon b = b$ to ensure that $b$ really lands up in the $\mathbb{R}$-component of $V$ (plus symmetry).

Whilst adding the pointed condition is non-trivial for algebras, it is effectively trivial for vector spaces since there's an obvious functor from vector spaces to pointed vector spaces, $V \mapsto V \oplus \mathbb{R}$ that is an equivalence of categories.

Assuming that all the $\imath$s can be crossed and all the $l$s dotted, the functor that you want is now the forgetful functor from pointed associative algebras to pointed quadratic vector spaces.

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I'm accepting this answer; although to be honest I'm still not close to understanding the "pointed" point of view. I'm learning category theory in earnest, so perhaps I can come back to this a little later. Thanks! –  José Figueroa-O'Farrill Mar 3 '10 at 0:02
    
Gosh, I'd forgotten about this completely! Well done for going back through your questions and sorting them out. If, at some point, you want to work through the details then I'd like to see the workings (even help if that's allowed). You should feel free to do this on the nlab if you wanted! –  Loop Space Mar 3 '10 at 9:20

If I understand the definitions correctly:

Let $C$ be the category of pairs (V,q) where V is a vector space on a fixed field and q is a quadratic form. A morphism $f: (V,q) \rightarrow (V',q')$ is a linear map $V \to V'$ preserving the quadratic form.

Let $D$ be the category of unital algebras over the field. Morphisms are linear maps preserving multiplication and identity.

We've got a forgetful functor $D \rightarrow C$ that maps an algebra V to the quadratic vector space $ (V,q)$ where $q(x)=(x \cdot x) \cdot 1$. This functor has as left adjoint the Clifford algebra construction.

(I'm inexperienced, so this might be plain wrong. But surely an adjoint functor is hiding here.)

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q(x) isn't a quadratic form. But I think if e denotes the identity and e* denotes its dual then defining q(x) = e* x^2 works. –  Qiaochu Yuan Dec 3 '09 at 18:11
    
Also, I'm not sure if it matters or not whether you want algebra homomorphisms to preserve the identity. –  Qiaochu Yuan Dec 3 '09 at 18:19
    
Thanks, corrected. –  sdcvvc Dec 3 '09 at 18:45
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"its dual"? What is the dual to the identity? –  Theo Johnson-Freyd Dec 3 '09 at 19:39
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I am not sure if this is whole answer, but it seems to be in the right direction. First, $q(x) = - e^* x^2$ seems better, the way I have defined Clifford maps. My main concern is that $e^*$ is not canonically defined. Perhaps one has to add more structure to the algebras... –  José Figueroa-O'Farrill Dec 3 '09 at 19:41

This answer builds on sdcvvc's answer and the comments below it, and in particular concerns the (non)existence of a canonical quadratic form $q$ (in sdcvvc's notation).

Let me denote by $\mathcal{Q}$ the category of quadratic real vector spaces (where the symmetric bilinear form is not necessarily nondegenerate), and by $\mathcal{A}$ some subcategory of the category $\mathcal{A}ss$ of finite-dimensional real unital associative algebras that contains the image of the Clifford functor $\mathcal{C}l: \mathcal{Q} \to \mathcal{A}ss$.

Notice that $\mathcal{Q}$ contains $\mathrm{\mathbf{Vect}}_\mathbb{R}$ as the full subcategory whose objects of the form $(V, 0)$, and that the restriction of the functor $\mathcal{C}l: \mathcal{Q} \to \mathcal{A}$ to this subcategory is the exterior algebra functor $V \mapsto \Lambda^{\ast}V$. Then,

$$\mathrm{Hom}_{\mathcal{A}}(\Lambda^\ast V, A) \cong \lbrace \phi: V \to A \; | \; \phi(v)^2 = 0 \rbrace$$

You can make $\Lambda^{\ast}(-)$ into a left adjoint by restricting $\mathcal{A}$ to be the category of $\mathbb{Z}_2$-graded supercommutative algebras (maybe you can take a bigger subcategory?). The right adjoint should then be the functor taking such an algebra to its odd-degree part considered as a vector space. This makes the Clifford condition $\phi(v)^2 = 0$ trivially true.

It is the latter observation the one that allows us to cook up such an $\mathcal{A}$. However, in the general case the Clifford condition does involve the quadratic form on the vector space that is the domain, and so it doesn't seem possible to me that we could do something like the above universally.

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UPDATE: the following argument is wrong, see the comments.

If $\mathcal{C}l$ admits a right adjoint then it preserves colimits, and coproducts in particular. Now, in your category of quadratic vector spaces, the coproduct of $(V, Q)$ and $(V', Q')$ is $(V \oplus V', Q \oplus Q')$; for associative algebras $A$ and $A'$, its coproduct is given by tensor product over $\mathbb{R}$. Hence, it is necessary that $$\mathcal{C}l(V \oplus V', Q \oplus Q') \cong \mathcal{C}l(V, Q) \otimes_{\mathbb{R}} \mathcal{C}l(V', Q')$$

Here's a counterexample: take $V = V' = \mathbb{R}$ with $Q = Q' = -1$. By the classification of Clifford algebras, we know that $\mathcal{C}l(\mathbb{R}, -1) \cong \mathbb{C}$ and $\mathcal{C}l(\mathbb{R}^2, \mathrm{diag}(-1,-1)) \cong \mathbb{H}$. It is now enough to observe that $$\mathbb{H} \not\cong \mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} $$

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The coproduct in the category of associative algebras is not the tensor product: two maps $f:A\to C$ and $f:B\to C$ give a map $A\otimes B\to C$ only if the images of $f$ and of $g$ commute. –  Mariano Suárez-Alvarez Dec 3 '09 at 18:42
    
Curiously, since Clifford algebras are $\mathbb{Z}_2$-graded, if you were to take the $\mathbb{Z}_2$-graded tensor product in your first displayed equation, then this would be a true statement. This perhaps suggestst that I have to consider the category of $\mathbb{Z}_2$-graded associative algebras. –  José Figueroa-O'Farrill Dec 3 '09 at 18:53
    
My bad: Clifford algebras are Z_2-graded, and hence the coproduct is the Z_2-graded tensor product. Proposition 1.5 on page 11 of Lawson and Michelsohn's "Spin Geometry" asserts that Cl indeed preserves coproducts. –  Alberto García-Raboso Dec 3 '09 at 18:54
    
In any case, the Clifford algebra of an orthogonal direct sum of quadratic spaces is isomorphic to the twisted tensor product of the corresponding tensor algebras: this is the "super" tensor product, the one which introduces signs by the parity graduation of the Clifford algebras. –  Mariano Suárez-Alvarez Dec 3 '09 at 18:55
    
Yes, but there are useful ways of being wrong :) –  José Figueroa-O'Farrill Dec 3 '09 at 19:44

I'm hoping for a second opinion on this question. The same question occurred to me, and google led me to this thread. At first glance, the consensus answer here (there is no right-adjoint to $Cl$) seems a plausibly argued. But after some thought, I'm not convinced.

We know that a universal construction, if it exists for every object in the source category, always gives an adjunction between categories.

An object satisfying the universal property for a Clifford algebra can be explicitly constructed from any vector space with quadratic form as a quotient of the tensor algebra. So an object satisfying the universal property always exists, therefore it is a left-adjoint. And what should the right-adjoint functor to the Clifford functor? Why nothing other than the underlying map from associative algebras to quadratic spaces, with quadratic form $q(x)=x^2$. This is the only possible quadratic form on the underlying vector spaces which will make the stipulation in the universal construction about the linear maps into morphisms in the category of quadratic vector spaces.

I should conclude that the right-adjoint of $Cl$ is a forgetful functor $k\text{-Alg}\to k\text{-Quad}$ which takes an associative algebra and forgets multiplication but remembers how to square vectors. The unit of this adjunction is the Clifford algebra structure map, and the counit is the map from the Clifford algebra on the quadratic vector space underlying any algebra $A$ to $A$ which takes $a_1\cdot a_2\mapsto a_1a_2$.

This is of course exactly the unaccepted answer that sdcvvc gives above, though without much detail. Qiaochu Yuan says that the claimed quadratic form $q(x)=x^2$ on the underlying vector space of an associative algebra is not actually quadratic. I cannot see why not. Why is sdcvvc's answer incorrect?

Alberto García-Raboso gives an answer as well, where in the discussion it is settled that $Cl$ preserves finite coproducts. If we can also show that it preserves cokernels then we know that it must have a right-adjoint, by Freyd adjoint functor theorem, right?

And have I misunderstood the relationships between universal morphisms and adjunctions? Is it not the case that we can simply read off the adjoint functor out of the universal property?

And do we really need to consider, as Andrew Stacy suggests, some kind of pointed vector spaces? If so, why?

I wanted to post my questions as comments, not an answer, but I guess I don't have enough rep. Please forgive me.

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$q(x) = x^2$ doesn't take values in the underlying field! –  Qiaochu Yuan Jul 3 '12 at 5:44
    
Right, duh. Thank you, Qiaochu, for explaining to the slow kid. So how do we reconcile the fact that the Clifford functor isn't a left-adjoint with the fact that every universal property determines an adjunction? I guess we conclude that the universal property which characterizes the Clifford algebra doesn't actually meet the technical definition of a universal morphism, in the sense that there is no functor for which the Clifford construction is initial in the slice over it? Are there other examples of universal properties which are not universal morphisms like this? –  Joe Hannon Jul 3 '12 at 14:32

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