What does "quantization is not a functor" really mean? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T22:17:49Z http://mathoverflow.net/feeds/question/8606 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/8606/what-does-quantization-is-not-a-functor-really-mean What does "quantization is not a functor" really mean? Theo Johnson-Freyd 2009-12-11T21:23:26Z 2010-02-27T16:41:53Z <p>The answers to <a href="http://mathoverflow.net/questions/6200/what-is-to-quantize-something" rel="nofollow">this question</a> do a good job of exploring, at a heuristic level, what "quantization" should be. From my perspective, quantization involves replacing a (commutative) Poisson algebra by some related noncommutative associative algebra. Poisson algebras arise naturally especially as algebras of functions in geometry and physics. Noncommutative algebras arise naturally as algebras of operators on linear spaces.</p> <p>I've often heard it said that "quantization is not a functor". I'm wondering what a precise statement of this is.</p> <p>For example, I could imagine statements of the following form.</p> <ol> <li>There is no functor from the category of Poisson manifolds (and Poisson maps?) to the (opposite of the) category of associative algebras satisfying some nice property.</li> <li>There is no functor from the category of symplectic manifolds (and Poisson maps?) to the (opposite of the) category of associative algebras satisfying some nice property.</li> <li>Recall that for any smooth manifold, its cotangent bundle is naturally symplectic. There is no functor from the category of smooth manifolds to the category of associative algebras that quantizes the cotangent bundle.</li> <li>Recall that the dual to the universal enveloping algebra of a Lie bialgebra is naturally Poisson Hopf. There is no functor from the category of Lie bialgebras to the category of Hopf algebras satisfying some nice property.</li> </ol> <p>Actually, 4. is false. Indeed, Etingof and Khazdan constructed a functor from bialgebras to Hopf algebras satisfying a host of properties, and Enriquez classified all the ones with nice properties. Note that Kontsevich does give a quantization of any Poisson manifold, but perhaps his isn't functorial?</p> http://mathoverflow.net/questions/8606/what-does-quantization-is-not-a-functor-really-mean/8607#8607 Answer by JosÃ© Figueroa-O'Farrill for What does "quantization is not a functor" really mean? JosÃ© Figueroa-O'Farrill 2009-12-11T21:52:57Z 2009-12-11T22:51:53Z <p>I believe that the phrase "quantization is not a functor" originated in the days before deformation quantization. Originally, based perhaps on the very few examples of quantum systems that people had at their disposal, the hope was that one could quantize a classical hamiltonian system by the simple procedure of replacing (up to $i\hbar$) the Poisson bracket of functions by the commutator of operators. In other words, suppose that $P$ is a classical phase space and $H \in C^\infty(P)$ a hamiltonian function, with time evolution given by the hamiltonian vector field $\lbrace H,-\rbrace$. Quantization would then be a map $f \mapsto O_f$ from $C^\infty(P)$ (the <em>classical observables</em>) to self-adjoint operators in a Hilbert space in such a way that $$[O_f,O_g] = i\hbar O_{\lbrace f,g\rbrace}.$$</p> <p>It was soon realised that this could not work for all classical observables. The first result of this kind is the so-called Grönewald/Van Hove's theorem, which shows that ordering ambiguities force the above equation to be true only up to terms of higher order in $\hbar$.</p> <p>A discussion of this is given in Section 5.4 of Abraham and Marsden's <a href="http://caltechbook.library.caltech.edu/103/" rel="nofollow">Foundations of Mechanics</a>.</p> <p>In general, it is accepted that one has to make choices in quantization. Even in the context of deformation quantization, the deformation may not be unique. Moyal quantization, for example, corresponds to a choice of ordering prescription (Weyl's symmetric ordering).</p> <p>There are also subtler effects. For example, it is possible to cook up one-dimensional quantum mechanical systems for which the hamiltonian is not self-adjoint, but admits inequivalent self-adjoint extensions, each one giving rise to different physical predictions.</p> <p>I realise that this does not actually answer your question, which as far as I understand it asks between which categories does quantization fail to be a functor. My point is that quantization may not even be a map!</p> <p><strong>Update</strong></p> <p>Having thought about the question some more, I think that perhaps it is meant in the following sense. Quantization, whatever it is, should be a process by which one goes from a classical hamiltonian system to a quantum mechanical system. The former is given by a Poisson manifold $P$ and a choice of hamiltonian, the latter by a Hilbert space $\mathcal{H}$ and a self-adjoint operator. Quantization should relate classical states and classical observables to quantum states and quantum observables. In particular, it should relate points in $P$ with rays in $\mathcal{H}$. Hence one possible hope would be that quantization be a functor from the category of Poisson manifolds to the category of (projectivised?) Hilbert spaces. Perhaps it is in this sense that the phrase in your question is meant. I was trying to remember where it was that I read that for the first time, but so far nothing. It was a long time ago...</p> http://mathoverflow.net/questions/8606/what-does-quantization-is-not-a-functor-really-mean/8611#8611 Answer by David Jordan for What does "quantization is not a functor" really mean? David Jordan 2009-12-11T22:20:19Z 2010-02-27T16:41:53Z <p>A prop P is a symmetric monoidal category whose objects are the non-negative integers, with tensor product defined as addition of integers (so morally, it's a single object V, the unit object, and various powers of V; so it's the morphisms that make it interesting). A representation of prop P in a symmetric tensor category C is a symmetric tensor functor from P to C.</p> <p>A prop is a way of encoding the morphisms defining a structure you are interested in, in a universal way. A representation of a prop P in a category C is then the same thing as a gadget modelled by P, in the category C. For instance see the prop Alg below; it's representations in C are the same thing as algebras in C. One nice thing about Props is that they can be given by generators and relations, and also they express via Schur functors $\mathbb{S}_\lambda$ defined in any symmetric tensor category similarly to the definition in $GL_n$ (if someone wants I can elaborate, but a search for Schur functors will probably yield enough results).</p> <p>Examples include the prop "Alg": it has a morphism $u:1\to V$, and a morphism $\mu: V\otimes V\to V$, and a relation $\mu\circ(\mu\otimes id)=\mu\circ(id\otimes \mu)$, and a relation $\mu(u\otimes id)=\mu(id\otimes u)=id$, etc. So you formally take the symmetric tensor category on objects the non-negative integers, which admits maps like $\mu$ and $u$, and you quotient by those relations. One can similarly define Lie-Alg, Lie-Bialg, Bi-alg, Hopf, quasi-Hopf,... you can go all day.</p> <p>A key point is that morphisms between props induce pullback functors between their representations in any symmetric tensor category, but in the reverse order (meaning, as usual $\rho:P\to S$ induces $\rho^*:S-mod\to P-mod$. For instance, there is a morphism of props from Lie-Alg to Alg, sending the bracket $[,] \mapsto m - \tau\circ m$. This induces the familiar forgetful functor from Alg to Lie-alg.</p> <p>In situations where there is a quantum structure that is a quantization of a classical structure, you get two props defined over $k[[\hbar]]$. For example in the case of Lie-Bialg, and Hopf alg, both of these make sense over k[[\hbar]]. The classical limit functor is induced by a prop map from Bialg to Hopf alg. It turns out to have a section Hopf alg to Bialg, which is the sort of thing you can check by generators and relations. Of course it doesn't have a unique section, there are many choices. However, the fact that you make those choices ONCE and FOR ALL on this particular prop, then implies that you have a functor of quantization between these two structures in any symmetric tensor category. I think a moral is that asking some construction in symmetric tensor categories be functorial, which is in general a tricky business, can in certain instances be reduced to giving a single (not even necessarily canonical!) map of props.</p> <p>In the case of Kontsevich quantization, The problem seems to be that quantizations are classified (up to isomorphism!) by HH^2(M) (if it's a Poisson manifold), and HH^2(A,A) of the algebra in general. Obstructions to quantization appear in HH^3. If, for instance HH^3(A,A) vanishes, it means you can quantize your Poisson algebra step-by-step. You basically start with your two-cocycle, it gives you a first order deformation, you try and see if your new algebra structure is associative so far (meaning write down associativity identity, expand in power of $\hbar$, and examine the first place where it isn't trivially zero); this will yield a certain very explicit 3 co-cycle which you need to vanish. If that 3-cocycle vanishes in HH^3, then you get it as d(w) for some 2 co-cycle. This then gives you the next higher order deformation of the multiplication, on and on ad infinitum.</p> <p>The problem here is that at each step you are making a choice of cocycle w such that dw=your previous obstruction. The results you get will be unique up to some non-canonical isomorphism, but that isn't good enough for functoriality. In particular, there is not a prop morphism between non-commutative algebras and Poisson algebras which would unify the approach for all examples.</p> http://mathoverflow.net/questions/8606/what-does-quantization-is-not-a-functor-really-mean/8624#8624 Answer by YBL for What does "quantization is not a functor" really mean? YBL 2009-12-11T23:08:11Z 2009-12-11T23:20:49Z <p>There is no good definition of what quantization is. It usually means that we start with a commutative structure $A$, a 1st order (non commutative) deformation $A\otimes k[h]/(h^2)$ of this structure and we want to lift it to an actual (or formal) deformation $A_h$ so that $A_h/h^2A_h = A\otimes_k k[h]/(h^2)$. </p> <p>In some cases we know that the lifting exists but it is never unique. Instead, in the formal setting, there is a group that acts simply transitively on the set of quantizations. So the set of "quantizations" is a torsor under this group.</p> <p>For example, quantum groups theory is a theory of quantization of semi-simple groups or more precisely of their Hopf algebra. Drinfeld showed that, given the enveloping algebra $A = Ug$ of a Lie algebra $g$ (it is co-commutative Hopf algebra) over a field of caracteristic 0, and an ad-invariant symetric tensor $t\in Sym^2(g)^g$ (corresponding to a first order deformation), there exists a quantization of $Ug$, namely a quasi-triangular quasi-Hopf algebra $A_h \simeq Ug[[h]]$ reducing to these data mod $h^2$. The set of (universal) quantizations is a pro-algebraic variety $Assoc$ of "associators" and it is a torsor under the Grothendieck-Teichmuller group $GT$ which is an extension of $G_m$ (because we have an action $\varphi:h\mapsto \lambda h$) by a pro-unipotent group (because we can filter it by $\varphi \equiv id \mod h^n$). </p> <p>The theorem of Kontsevich on deformation quantization of Poisson varieties shows a similar pattern. Here the structure is associative algebra (with some additional properties). A 1st order deformation of a commutative algebra $A$ is a Poisson bracket. And we try to lift it to a formal deformation $A_h$. </p> <p>In both cases, I think the choice of an associator gives you a functor. But a) we only consider "universal" quantizations. If we consider a single object (Hopf algebra or Poisson algebra) it may have some deformations that are not given by the universal recipe. b) in both cases, we just ask for an isorphism $A_h/h^2A_h = A\otimes k[h]/(h^2)$. We do not ask for a section $a\mapsto \hat a_h$ (a quantization rule like symetric ordering). I think such a requirement would always break functoriality.</p> http://mathoverflow.net/questions/8606/what-does-quantization-is-not-a-functor-really-mean/9463#9463 Answer by Greg Kuperberg for What does "quantization is not a functor" really mean? Greg Kuperberg 2009-12-21T07:42:18Z 2009-12-21T07:42:18Z <p>Certainly if you're not just looking at deformation quantization, there is no reasonable way to make a functor that creates every quantum object from a classical object. I can't think of a rigorous obstruction offhand, but for instance there are a great variety of <code>$C^*$</code>-algebras, which are quantum compact Hausdorff spaces, that don't look anything like classical Hausdorff spaces. To give another example, it is technically true that there is a functor from finite sets to the quantum version, finite-dimensional matrix algebras. But this functor is so far away from an isomorphism of categories, using either unitary operators or completely positive maps in the quantum case, that it misses the point. One famous discovery is that the matrix algebras have more computational power (quantum computation) than finite sets or finite probabilistic spaces (classical computation). If a quantization functor is surjective on objects but far from surjective on morphisms, then that is a second level at which "quantization is not a functor".</p> <p>Here is a related point that may be of some interest. In some important cases, a quantum object is not the same as a classical object defined in the quantum category. In particular, a quantum group, defined properly as a Hopf algebra, is not a group object. A commutative Hopf algebra really is a group object in the reverse category of commutative algebras. In other words, a group object in the category of affine schemes is an affine scheme whose coordinate ring is a Hopf ring. The same is true of graded-commutative Hopf algebras. But a general Hopf algebra is a different beast, because the tensor product of noncommutative algebras is not a coproduct in the sense of category theory.</p> <p>Nonetheless the tensor product is the orthodox quantum equivalent of a coproduct, and a Hopf algebra is the orthodox quantum equivalent of a group. In free probability, you replace the tensor product with a free product, and that is perfectly interesting, but it is a modification of standard quantum probaibility. I am not sure if there is a good way to explain the difference between the orthodox quantum product and the categorical coproduct. (Maybe there isn't really anything to explain.) Well, one important feature is that the Hopf algebra axioms are self-dual.</p> http://mathoverflow.net/questions/8606/what-does-quantization-is-not-a-functor-really-mean/9788#9788 Answer by Harold Williams for What does "quantization is not a functor" really mean? Harold Williams 2009-12-26T08:37:30Z 2009-12-26T08:37:30Z <p>Here is one precise statement of how quantization is not a functor:</p> <p>5) There is no functor from the classical category $\mathcal C$ of Poisson manifolds and Poisson maps to the quantum category $\mathcal Q$ of Hilbert spaces and unitary operators that is consistent with the cotangent bundle/$\frac12$-density relation (explained below).</p> <p>The result is due to Van Hove, in "Sur le probleme des relations entre les transformations unitaires de la mecanique quantique et les transformations canoniques de la mecaniques classique." This is an old paper and I can't find a link for it, but the reference I found it in is Weinstein's "Lectures on Symplectic Manifolds."</p> <p>By "cotangent bundle/$\frac12$-density relation" I mean the following: if $\mathcal M$ is the category of smooth manifolds and diffeomorphisms, we have a cotangent functor $\mathcal M \to \mathcal C$. This assigns to each manifold its cotangent bundle with the canonical symplectic structure, and to each diffeomorphism the induced symplectomorphism of cotangent bundles. </p> <p>We also have a natural functor $\mathcal M \to \mathcal Q$. For any smooth manifold $X$ consider the bundle of complex $\frac12$-densities on $X$. (What is the bundle of complex $s$-densities? Well, the fiber over a point $x \in X$ is the set of functions $\delta_x: \bigwedge^{top} T_xX \to \mathbb{C}$ such that $\delta(cv) = |c|^{s}\delta(v)$.) If $\delta^1$ and $\delta^2$ are smooth compactly-supported $\frac12$-densities, their pointwise product $\delta^1 \bar{\delta^2}$ is a compactly supported 1-density which we can integrate to get a complex number. This turns the space of all such sections into a pre-Hilbert space, the completion of which is what our functor assigns to the manifold $X$. As we would hope for, the canonical nature of the construction lets us assign unitary operators between Hilbert spaces to diffeomorphisms between smooth manifolds, hence is functorial. </p> <p>(Note: If we choose a volume form on $X$, the above procedure produces something isomorphic with the space of $L^2$ functions on $X$ with respect to this form, but to get something functorial we want a canonical construction.)</p> <p>From this pair of functors $\mathcal M \to \mathcal C$ and $\mathcal M \to \mathcal Q$ we get a product functor $\mathcal M \to \mathcal{C} \times \mathcal{Q}$. The image of this functor is a subcategory of $\mathcal C \times \mathcal Q$ which we will call the "cotangent bundle/$\frac12$-density relation." (The word relation is meant in the same sense that an ordinary relation between two sets is a subset of their product).</p> <p>Now we can clarify just what is meant by our original statement: there is no functor $\mathcal C \to \mathcal Q$ whose graph contains the cotangent bundle/$\frac12$-density relation. The reasons why this is a desirable condition come from physics and are beyond me, but roughly speaking I think the point is that there exists a good idea of what a quantization functor is supposed to do to cotangent bundles.</p> http://mathoverflow.net/questions/8606/what-does-quantization-is-not-a-functor-really-mean/13924#13924 Answer by Pavel Etingof for What does "quantization is not a functor" really mean? Pavel Etingof 2010-02-03T03:52:56Z 2010-02-03T03:52:56Z <p>This is meant to explain a bit more some of the things that were already mentioned before. </p> <p>Quantization of Lie bialgebras is indeed a functor, as was shown in my work with Kazhdan. However, the Konstevich or Fedosov deformation quantization are not functors in this sense. In fact, there is not even a functorial deformation quantization of the symplectic plane, which is sometimes referred to as "Groenewald-van Hove theorem", mentioned in one of the previous answers. The essence of this theorem is that there is no way to quantize the symplectic plane preserving all its symmetries, i.e. symplectic diffeomorphisms. Speaking algebraically, the Lie algebra of infinitesimal symmetries of the symplectic plane is $L_0=C^\infty(\Bbb R^2)/\Bbb R$, and the Lie algebra of symmetries of the (say, Moyal) quantization is the quotient of the quantized algebra by its center, $L=C^\infty(\Bbb R^2)[[h]]_\ast/\Bbb R[[h]]$, with bracket $[a,b]:=(a\ast b-b\ast a)/h$. Now, $L$ is a flat deformation of $L_0$, but the key point is that it is ${\bf nontrivial}$. This nontriviality (which is the gist of the Groenewald-van Hove theorem) is the obstruction to existence of a quantization functor. </p> <p>However, as was mentioned in a previous answer, a Drinfeld associator does, according to Tamarkin, give a map between sets of isomorphism classes of Poisson structures and star products (which is not a functor, however, since we have modded out by automorphisms). </p>