Clifford algebra as an adjunction? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T16:11:20Z http://mathoverflow.net/feeds/question/7687 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/7687/clifford-algebra-as-an-adjunction Clifford algebra as an adjunction? José Figueroa-O'Farrill 2009-12-03T17:14:24Z 2012-07-03T05:34:50Z <p><strong>Background</strong></p> <p>For definiteness (even though this is a categorical question!) let's agree that a <em>vector space</em> is a finite-dimensional real vector space and that an <em>associative algebra</em> is a finite-dimensional real unital associative algebra.</p> <p>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 <strong>quadratic vector space</strong>.</p> <p>Let $A$ be an associative algebra and let's say that a linear map $\phi:V \to A$ is <strong>Clifford</strong> if $$\phi(x)^2 = - Q(x) 1_A,$$ where $1_A$ is the unit in $A$.</p> <p>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 <strong>Clifford algebra of $(V,Q)$</strong> 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 $\Phi \circ i = \phi$.</p> <p>(This is the usual definition one can find, say, in the <a href="http://ncatlab.org/nlab/show/Clifford+algebra" rel="nofollow">nLab</a>.)</p> <p><strong>Question</strong></p> <p>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</p> <p>$$\mathrm{hom}_{\mathbf{Assoc}}(Cl(V,Q), A) \cong \mathrm{cl-hom}(V,A)$$</p> <p>where the left-hand side are the associative algebra morphisms and the right-hand side are the Clifford morphisms.</p> <p>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 $F$.</p> <p>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.</p> http://mathoverflow.net/questions/7687/clifford-algebra-as-an-adjunction/7694#7694 Answer by sdcvvc for Clifford algebra as an adjunction? sdcvvc 2009-12-03T18:08:30Z 2009-12-03T18:44:42Z <p>If I understand the definitions correctly:</p> <p>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.</p> <p>Let $D$ be the category of unital algebras over the field. Morphisms are linear maps preserving multiplication and identity.</p> <p>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.</p> <p>(I'm inexperienced, so this might be plain wrong. But surely an adjoint functor is hiding here.)</p> http://mathoverflow.net/questions/7687/clifford-algebra-as-an-adjunction/7697#7697 Answer by Alberto García-Raboso for Clifford algebra as an adjunction? Alberto García-Raboso 2009-12-03T18:27:36Z 2009-12-03T18:57:54Z <p><strong>UPDATE</strong>: the following argument is wrong, see the comments.</p> <p>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 $$\mathcal{C}l(V \oplus V', Q \oplus Q') \cong \mathcal{C}l(V, Q) \otimes_{\mathbb{R}} \mathcal{C}l(V', Q')$$</p> <p>Here's a counterexample: take $V = V' = \mathbb{R}$ with $Q = Q' = -1$. By the <a href="http://en.wikipedia.org/wiki/Classification%5Fof%5FClifford%5Falgebras" rel="nofollow">classification of Clifford algebras</a>, 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 $$\mathbb{H} \not\cong \mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$$</p> http://mathoverflow.net/questions/7687/clifford-algebra-as-an-adjunction/7717#7717 Answer by Alberto García-Raboso for Clifford algebra as an adjunction? Alberto García-Raboso 2009-12-03T21:26:06Z 2009-12-04T13:40:23Z <p>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).</p> <p>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 $\mathcal{C}l: \mathcal{Q} \to \mathcal{A}ss$.</p> <p>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,</p> <p>$$\mathrm{Hom}_{\mathcal{A}}(\Lambda^\ast V, A) \cong \lbrace \phi: V \to A \; | \; \phi(v)^2 = 0 \rbrace$$</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/7687/clifford-algebra-as-an-adjunction/7769#7769 Answer by Andrew Stacey for Clifford algebra as an adjunction? Andrew Stacey 2009-12-04T10:48:42Z 2009-12-04T10:48:42Z <p>Disqualifier: this isn't a complete answer.</p> <p>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 <strong>not</strong> 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 <em>out</em> of the vector space, not into it, so isn't an "operation".</p> <p>However, you may get some mileage if you work with <strong>pointed</strong> 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!).</p> <p>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.</p> <p>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 <em>pointed</em> sets) and the identity $\eta \epsilon b = b$ to ensure that $b$ really lands up in the $\mathbb{R}$-component of $V$ (plus symmetry).</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/7687/clifford-algebra-as-an-adjunction/101209#101209 Answer by Joe Hannon for Clifford algebra as an adjunction? Joe Hannon 2012-07-03T05:05:40Z 2012-07-03T05:34:50Z <p>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.</p> <p>We know that a universal construction, if it exists for every object in the source category, always gives an adjunction between categories.</p> <p>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.</p> <p>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 $a_1\cdot a_2\mapsto a_1a_2$.</p> <p>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?</p> <p>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?</p> <p>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?</p> <p>And do we really need to consider, as Andrew Stacy suggests, some kind of pointed vector spaces? If so, why?</p> <p>I wanted to post my questions as comments, not an answer, but I guess I don't have enough rep. Please forgive me.</p>