**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](http://ncatlab.org/nlab/show/Clifford+algebra).)


**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.



Clifford algebra as an adjunction?
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.


