Quantization and noncommutative deformations - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T06:38:51Z http://mathoverflow.net/feeds/question/71484 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71484/quantization-and-noncommutative-deformations Quantization and noncommutative deformations amine 2011-07-28T12:43:29Z 2011-12-13T15:54:39Z <p>Hello, I would like to introduce myself to the theory of quantization and noncommutative deformations of Riemann Poisson structures. In fact, I am familiar with Riemannian and Poisson geometry, but I cannot grasp the principle of the theory above.</p> <p>As I understand, by reading some introductory texts on the subject, ideas come from physics. This involves replacing the coordinates of the phase space (the cotangent of $\mathbb{R}^3$ endowed with its canonical symplectic structure) by operators (on a Hilbert space) which do not commute with each other and is the description of quantum mechanics in a form similar to classical mechanics. Then there was conflict between the riemannian description (or pseudo-Riemannian) of space in general relativity and the space of quantum mechanics. On small scales, the innermost structure of space-time is broken: the position and velocity of a point do not make sense at the same time, if one is defined precisely the other will not defined (the Heisenberg uncertainty principle). This raises the question how to give a physical meaning to the concept of point?</p> <p>The Gelfand-Naimark theorem first brought a solution to this "internal conflict" of physics, by establishing a bridge between topology and algebra. So we look at a point $x$ in a space $X$ be fixed, and considering its "shadow" consists of the values $f(x)$ taken by all continuous functions $f$ on $X$. This leads to replace the space $X$ by commutative algebra $C(X)$.</p> <p>The fact that pseudo-Riemannian geometry is a sufficient description of space-time for most purposes, suggests that noncommutativity might be treated as the limit where Planck's constant $\hbar$ tends to $0$ then I understand it, how to become a constant variable and move towards $0$! hence the idea of deformation quantization, which is to construct noncommutative algebras $\mathcal{A}_\hbar$, by deformation of the Poisson algebra $\mathcal{A}=C^\infty(M)$ (formal series in $\hbar$ with coefficients in $\mathcal{A}$) that converges to $\mathcal{A}$ when $\hbar$ tends to $0$. The work most results in this direction is that of Kontsevich.</p> <p>1) How Heisenberg's uncertainty principle reviews the classical definition of a point?</p> <p>2) Why deformation always starts with a Poisson manifold? if it is to deform the phase space of hamiltonian mechanics it suffices to consider symplectic manifolds!</p> <p>3) Why the deformation of the algebra of functions $C^\infty(M)$ of a Poisson manifold is a way of quantization? in this case what is the Hilbert space, in which the observable $f$ are replaced by a bounded operators $\widehat{f}$?</p> <p>Alain Conne then directed the program algebraization of differential geometry, in order to then work on "noncommutative spaces". He considered riemannians manifolds with additional structure of Spin. Such structure is canonically attached to the triple $(C^\infty(M),L^2(M),D)$ ; what is $L^2(M)$? and $D$ is the Dirac operator, with some number of properties that can be generalized to spectral triples $(\mathcal{A}, H,D)$: $\mathcal{A}$ a noncommutative algebra, $H$ Hilbert space, and $D$ the Dirac operator.</p> <p>1) Why in physics we need to replace the Laplacian (which is of order $2$) by a differential operator of order $1$ the Dirac operator (which is the square root of the Laplacian)?</p> <p>3) How to define, precisely, a spectral triple and how to find the metric using the Dirac operator?</p> <p>4) Why we deform only Spin manifolds?</p> <p>On the other hand, in an article \url{http://arxiv.org/abs/math/0504232v2} Eli Hawkins gave some definitions that I not understand (not being familiar with the language of algebraic geometry). In particular, the definition 1.4 (page 4) and the definition of "metacurvature" (page 5). In particular, how an extension of the algebra of differential forms gives rise to a Poisson bracket!?</p> <p>If $\mathcal{A}_0$ is an algebra, that means an extension of the form $$0 \rightarrow\hbar\mathbb{A}\rightarrow\mathbb{A} \rightarrow\mathcal{A}_0$$ where $\hbar$ is a central multiplier $\mathbb{A}$, and for all $a\in\mathbb{A}$ $$\hbar^2a=0 \Longrightarrow a\in\hbar\mathbb{A}$$</p> http://mathoverflow.net/questions/71484/quantization-and-noncommutative-deformations/71508#71508 Answer by Theo Johnson-Freyd for Quantization and noncommutative deformations Theo Johnson-Freyd 2011-07-28T17:04:28Z 2011-07-28T23:27:49Z <p>There are a lot of questions here, and it would be much better if you broke them up into separate MO posts. (Aside: you may also want to break up your text into paragraphs, as it aids reading.) I will in this answer not touch the second half of your questions (about Spin manifolds and Connes' NCG), but I'll try to say something to the first few questions. Note also that the what and why of quantization have been discussed many times on MO.</p> <ol> <li><p><strong>How Heisenberg's uncertainty principle reviews the classical definition of a point?</strong><br><br> I'm not entirely sure I understand what you're trying to ask, but I'll make a stab. From a modern perspective, the uncertainty principle comes in two steps: (i) We allow, or even demand, noncommutativity in our algebras of observables. (ii) Fix a state $\phi$, normalized so that $\langle \phi | \phi \rangle = 1$. If $f$ is any observable, the <em>expectation value of $f$ in state $\phi$</em> is $\langle f \rangle = \langle \phi | f | \phi \rangle$. Note that we allow everything to be complex-valued; denote complex (i.e. Hermitian) conjugation by $\Box \mapsto \Box^\ast$. The <em>uncertainty</em> of $f$ in state $\phi$ is $\Delta f = \sqrt{ \langle f^\ast f \rangle - \langle f \rangle \langle f^\ast \rangle} = \sqrt{ \bigl\langle ( f - \langle f\rangle)^\ast (f - \langle f \rangle) \bigr\rangle }$. Let $f,g$ be any observables. The Cauchy-Schwartz inequality then says that: \begin{align} (\Delta f)^2(\Delta g)^2 &amp; = \bigl\langle (f - \langle f \rangle)\phi \big| (f - \langle f \rangle)\phi \bigr\rangle \bigl\langle (g - \langle g \rangle)\phi \big| (g - \langle g \rangle)\phi \bigr\rangle \\ &amp; \geq \bigl\langle (f - \langle f \rangle)\phi \big| (g - \langle g \rangle)\phi \bigr\rangle \bigl\langle (g - \langle g \rangle)\phi \big| (f - \langle f \rangle)\phi \bigr\rangle \\ &amp; = \bigl\langle \phi \big| (f - \langle f \rangle)^\ast (g - \langle g \rangle) \big| \phi \rangle \bigl\langle \phi \big| (f - \langle f \rangle)^\ast (g - \langle g \rangle) \big| \phi \rangle^\ast \\ (\Delta f) (\Delta g) &amp; \geq \bigl| \bigl\langle (f - \langle f\rangle)^\ast (g - \langle g \rangle) \bigr\rangle \bigr| \end{align} Note that $\Delta f = \Delta(f^\ast)$. Set $f' = f - \langle f \rangle$ and $g' = g - \langle g \rangle$. Then, switching $f \leadsto f^\ast$ in the above calculation, we've shown that $(\Delta f)(\Delta g) \geq | \langle f'g'\rangle|$, and also $\geq | -\langle g'f'\rangle|$, adding a minus sign just for fun. But then: \begin{align} (\Delta f)(\Delta g) &amp; \geq \frac12 \left( \bigl| \langle f'g'\rangle\bigr| + \bigl| -\langle g'f'\rangle\bigr|\right) \\ &amp; \geq \frac12 \left| \langle f'g'\rangle - \langle g'f'\rangle \right| \\ &amp; = \frac12 \left| \langle [f',g']\rangle \right| = \frac12 \left| \langle [f,g]\rangle \right| \end{align} by the triangle inequality. (Note that you can also get a lower bound of $\frac12 \bigl| \langle f'g' + g'f' \rangle\bigr|$, and thus by the triangle inequality you have the following sharpening by Schrodinger: $(\Delta f)(\Delta g) \geq \sqrt{ \left( \frac12 \bigl| \langle [f,g]\rangle \bigr| \right)^2 + \left( \frac12 \bigl| \langle f'g' + g'f' \rangle\bigr| \right)^2 }$. Note also that many accounts give the much weaker but still useful estimate that $(\Delta f)(\Delta g) \geq \frac12 \operatorname{Im}\langle [f,g]\rangle$.) <br><br> Anyway, the particular example is whenever $f,g$ are in canonical commutation, i.e. $[f,g]$ is a nonzero constant (usually fixed at Planck's constant, up to some $2\pi i$s). Then $\langle [f,g]\rangle$ is the expectation value of a constant, and so is independent of the state $\phi$. This proves that as soon as you have such observables in your algebra, then there <em>cannot</em> be states for which all uncertainties vanish. This is in contrast with the commutative situation: a "point" might be defined as a state in which all uncertainties vanish, and in a (well-behaved) commutative algebra every state is a convex combination of points. <br><br> So I guess the answer to your first question is that Uncertainty requires that you abandon the notion of "point" in noncommutative spaces, in favor of a more spread-out notion of "state". In commutative land, it suffices to consider the "points", but in noncommutative land there generically are no points at all.</p></li> <li><p><strong>Why deformation always starts with a Poisson manifold? if it is to deform the phase space of hamiltonian mechanics it suffices to consider symplectic manifolds!</strong> <br><br> This question, I think, starts from a mistaken impression of physical systems. At first, students study physical systems that are symplectic &mdash; indeed, one begins only with cotangent bundles, being the ones that describe physical systems with a good notion of "configuration space" or "position space". Note that none of this complicated machinery of deformation quantization is necessary there: if you have a configuration space $N$ with a distinguished volume measure, then you can immediately write down the $L^2$ functions on $N$ as your Hilbert space, and the (bounded operators within the appropriate completion of) the differential operators as your algebra. The algebra of differential operators is naturally filtered, and its associated graded is canonically isomorphic to the algebra of functions on the cotangent bundle; so the Rees algebra construction without effort introduces Planck's constant into the game and lets you take classical limits, etc. (If $N$ does not have a measure, then you still have the algebra of differential operators, which acts on the vector space of functions, but that vector space does not have a Hilbert space structure; you also have the Hilbert space of $L^2$ half-densities on $N$, and its algebra of bounded operators, but you do not have a canonical "classical limit" of this algebra.) <br><br> But there are many more complicated physical systems, and they are not all symplectic. Indeed, Dirac recognized early on that constrained mechanical systems can be well-described by possibly-non-symplectic Poisson manifolds (maybe with singularities). He talks about "first class constraints" and "second class constraints", and I don't remember which is which, but one of them is constraints that preserve symplecticity and the other do not, if memory serves. There's probably a discussion of this in the book by Marsden and Weinstein, although I don't have it in front of me so I can't check.</p></li> <li><p><strong>Why the deformation of the algebra of functions $C^\infty(M)$ of a Poisson manifold is a way of quantization? in this case what is the Hilbert space, in which the observable $f$ are replaced by a bounded operators $\hat f$?</strong> <br><br> Ah, this I think is the most interesting of your questions. The answer is: there isn't one. The GNS construction assures that for noncommutative algebras with good analytic control, there always exists a Hilbert space on which the algebra is faithfully represented, although it's not (in most write-ups) at all canonical. (Deformation quantization does not produce algebras with sufficient analytic control, but I will sweep that issue under the rug.) This is well-illustrated by the example of cotangent bundles I discussed above. Let $N$ be a manifold and $M = \mathrm T^\ast N$ its cotangent bundle. A very good deformation quantization of $C^\infty(M)$ is the Rees algebra of differential operators on $N$: it is generated by smooth functions on $N$ and $\hbar\times$ vector fields on $N$, and then you may have to do some analytic work if you really want to be a deformation of $C^\infty$ and not just of the part that's polynomial in the $\mathrm T^\ast$-direction. This deformation quantization always exists, and always acts on $C^\infty(N)$. But $C^\infty(N)$ is not a Hilbert space without more choices: namely, you must choose a measure on $N$ to be able to define $\langle f,g\rangle = \int_N f^\ast g$. A choice of measure essentially amounts to a choice of state for the GNS construction applied to the algebra of differential operators. <br><br> Note furthermore that a priori, there also is not a canonical map $C^\infty(M) \to$ differential operators, at least not without choices. Kontsevich's result that you alluded to above picks out such a map, but does require a choice (that said, I get myself confused about how much of a choice it requires). <br><br> The point is that for many questions, just having a noncommutative algebra with some parameter $\hbar$, whose $\hbar \to 0$ limit is your commutative algebra (and that's maybe "flat" over the space of values of $\hbar$) is enough for many applications. The early quantum mechanists (and still all textbooks) were simply asking too much to have a Hilbert space and a quantization map and all that &mdash; at least, that's my opinion about it. The buzzwords in physics books are "Schrodinger picture" and "Heisenberg picture": in "Schrodinger picture" the Hilbert space is actually a physical, existing thing, whereas in "Heisenberg picture" basically all you have is the algebra of observables. The Heisenberg picture is much more philosophically reasonable: as scientists, we make observations and measurements; we don't directly manipulate vectors in a Hilbert space. If possible, it's best to set up a mathematical framework that's as minimal and faithful-to-what-people-actually-do (not what they say they do) as possible.</p></li> </ol> http://mathoverflow.net/questions/71484/quantization-and-noncommutative-deformations/71559#71559 Answer by Stefan Waldmann for Quantization and noncommutative deformations Stefan Waldmann 2011-07-29T07:57:49Z 2011-12-13T15:54:39Z <p>Well, a lot of questions, some of which Theo already answered in a very nice way. Let me just give some additional remarks and hints how I think about DQ and Poisson geometry in relation to quantum physics.</p> <p>Concerning the first question:</p> <p>the good replacement (in view of Gel'fand duality) of a point on phase space is a (pure) state on the quantum algebra. While for $C^\ast$-algebras this is standard lore, in formal DQ things are slightly more tricky: of course you can argue that a formal star product yields not yet a quantum observable algebra as $\hbar$ does not have a "value" (say $1$ in your favorite unit system), so you should postpone the question of states till when you reach a "convergent/strict" DQ. This is often possible but in general completely unknown. Surprisingly, there is a good notion of states already for formal star products: essentially the same definition applies, take <em>positive</em> functionals of the algebra $C^\infty(M)[[\hbar]]$ which are $\mathbb{C}[[\hbar]]$-linear and take values in $\mathbb{C}[[\hbar]]$. To define positivity you make use of the fact that $\mathbb{R}[[\hbar]]$ is an ordered ring. Then many techniques of $C^*$-algebra theory can be carried over to this entirely algebraic framework. In fact, we have worked out many things like the GNS construction of representations etc.</p> <p>Now the point is that a classically positive functional $\omega_0\colon C^\infty(M) \longrightarrow \mathbb{C}$ (which is a positive Borel measure with compact support by a smooth version of Riesz' Theorem) may no longer be positive with respect to a given star product $\star$. Thus you need to add higher order corrections $\omega = \omega_0 + \hbar\omega_1 + \cdots$ in order to gain positivity. It is a (quite non-trivial) theorem that this is always possible, even in a "differential" sense that all the higher orders are of the form $\omega_0 \circ D_r$ with some differential operator $D_r$.</p> <p>You can apply this now to your favorite classical state, the delta-functional at a given point. The corresponding (non-unique) deformation is then the quantum analog of what a point can be, in some sense the best thing you can get.</p> <p>The uncertainty principle can be understood as the reason why positivity fails for $\omega_0$ itself and why higher orders are necessary...</p> <p>The second question: of course, for hard physical applications you only need $\mathbb{R}^{2n}$, maybe a cotangent bundle but that's it. Even a generic symplectic manifold is hard to motivate from this point of view.</p> <p>But there are also reasons from physics why one should take care of DQ of more general Poisson manifolds:</p> <p>a) Symmetries: whenever you have a classical symmetry encoded by a momentum map, then $\mathfrak{g}^\ast$ is a Poisson manifold. Quantizing a symmetry then amounts to quantize the momentum map in an appropriate way. There are several competing definitions but essentially all involve a DQ of the linear Poisson structure on $\mathfrak{g}^*$.</p> <p>b) Aesthetics: to have a general framework in which you can discuss your relevant examples might be useful and open your view, even though the examples might be very very special inside this bigger framework.</p> <p>c) Applications in NCG: many models of noncommutative space-times require more general Poisson structures to be quantized than just symplectic ones. It is even not clear that space-time allows for a symplectic structure at all, but it certainly carries interesting Poisson structures. In serious models of NC space-times, the Poisson structure itself should be treated as a <em>dynamical</em> quantity, i.e. a field. Then there is no reason why it should be non-degenerate everywhere. These models are of course all still very speculative...</p> <p>d) Toy models: one can view complicated Poisson/symplectic manifolds as toy models for the infinite-dimensional phase spaces of classical field theories with gauge symmetries. Here the true phase spaces are sort of Marsden-Weinstein quotients (in ugly infinite dimensions) which can have quite generic geometry. So one tries to learn something about their quantization by looking at finite-dim models having at least also a complicated geometry.</p> <p>Third question: Where is the Hilbert space...</p> <p>After what I said for the first question, this is now pretty clear and follows the same line of argument as in AQFT: Having the algebra of observables, one takes a look at <em>all</em> $^*$-representations on, say pre-Hilbert spaces, by means of a GNS construction. The notion of pre Hilbert space works very much the same for ordered rings like $\mathbb{R}[[\hbar]]$. This has been worked out in detail in many places and gives indeed physically interesting results. The main advantage is now that one can take a look at <em>different</em> representations which can encode different physical situations...</p> <p>OK, sorry for such a long blurp. I hope it gives some inspiration.</p> http://mathoverflow.net/questions/71484/quantization-and-noncommutative-deformations/71599#71599 Answer by Nicola Ciccoli for Quantization and noncommutative deformations Nicola Ciccoli 2011-07-29T18:35:13Z 2011-07-29T18:35:13Z <p>Hi Amine, nice seeing you here. I would like to add some comments on the "metric" side of your questions, which was left untouched by previous comments.</p> <p>About 3, well, I guess that there are no short-cuts. The list of axioms about spectral triples was clarified by Connes, is quite long to repeat here, and quite obscure unless one goes deeply into the subtleties of the reconstruction theorem, the theorem that answers precisely your question, proved finally by Connes himself in <em>On the spectral characterization of manifolds, arXiv:0810.2088</em>. In the compact case, the metric can be reconstructed explicitely from the Dirac operator $D$ via the formula $d(p,q)=sup|f(p)-f(q)|$ where $f\in{\cal C}(M)$ is such that the norm of $[D,f]$ is less than 1. But much work goes into proving that the axioms of the spectral triple allow to reconstruct the smotth manifold at first.</p> <p>About Hawkins work: it is reasonable (to some extent...) to expect that a noncommutative deformation of a manifold should also have some non commutative differential calculus. Since differential forms carry the algebraic structure of commutative graded differential algebra he just drops the assumption on commutativity stating that a non commutative calculus should be a non commutative DGA, having the commutative DGA as limit $\hbar\to 0$. Just as in the ungraded case, then, one can define a graded Poisson bracket as first order approximation of this non commutative deformation. </p> <p>In particular, inside this algebra, one can tak the bracket of a 0-form $f$, a function, with a 1-form $\alpha$, having as a result a 1-form. This basically defines what is called a flat contravariant connection on the underlying Poisson manifold via $[f,\alpha ]=\nabla_{df}\alpha$. The Jacobi identity for a 0-form and two 1-form may be seen as a vanishing of a certain tensor which Hawkins calls <em>metacurvature</em>. What he proves is that basically the whole non commutative DGA structure on forms is fixed once one has a flat contravariant connection with 0-metacurvature, and that this is all one needs to build the whole bracket. </p> <p>Therefore one can say that a non commutative deformation on a Poisson manifold extends to a non commutative deformation of the DGA of differential forms once a flat contravariant connection with metacurvature is given; you can see this as a geometric condition for a "nice" NC differential calculus.</p> http://mathoverflow.net/questions/71484/quantization-and-noncommutative-deformations/71720#71720 Answer by amine for Quantization and noncommutative deformations amine 2011-07-31T09:40:57Z 2011-07-31T09:40:57Z <p>Many thanks, Theo Stefan and Nicola (Nice to see you here) for your very interesting responses. After a Master in PDE's, I defended a Ph.D thesis in Poisson geometry, based on the work by Eli Hawkins <a href="http://arxiv.org/abs/0906.2887" rel="nofollow">link text</a> <a href="http://www.heldermann.de/JLT/JLT19/JLT193/jlt19024.htm" rel="nofollow">link text</a> without a clear idea of what it is quantization and noncommutative deformations. Actually, I have never taken a course in geometry, and I am totally ignorant in physics. Added to that, there is no geometer in my university! I strive to achieve a real understanding, not just writing a Ph.D thesis. I will back with more interesting comments.</p> http://mathoverflow.net/questions/71484/quantization-and-noncommutative-deformations/72610#72610 Answer by Daniel Pomerleano for Quantization and noncommutative deformations Daniel Pomerleano 2011-08-10T19:00:39Z 2011-08-10T19:00:39Z <p>Why do deformations always start with a Poisson manifold? This maybe slightly imprecise since I am transporting the story from a knowledge of the deformation theory of algebraic varieties and any corrections are welcome. Here is a small piece of the puzzle that is interesting to me as someone interested in deformation theory. Well there is a general fact in algebra that deformations of a ring or algebra R are governed by the Hochschild cohomology, $HH^*(R)$. That is to say that deformations to first order correspond exactly to classes in $HH^2(R)$ Hochschild cohomology, and deformations to all orders can be understood roughly as placing additional conditions on that class that the deformation extend to all orders, that is to say over $\mathbb{R}[[t]]$. Here the ring in question is $C^\infty(M)$, but it could be the coordinate ring of an algebraic variety as well. Technically speaking we need to think of $C^\infty(M)$ as a topological algebra but let's supress that. Anyways, what's true is that for a $C^\infty(M)$, $HH^2(C^\infty(M)) \cong \Gamma(\wedge^2(T_X))$, that is exactly the bivector fields. The additional condition is exactly that the bivector field be Poisson. Thus Poisson structures are exactly what gives rise to deformations. There is a relation on Poisson structures called gauge equivalence which determines when two deformations are deemed to coincide.</p> <p>It might be worth noting that this situation is specific to smooth manifolds. In the case of a smooth algebraic variety the situation can be more complicated depending upon what you are trying to deform. There one can as before deform the (sheaf of) functions by algebraic bivector fields, but one can also deform the complex structure of the variety as well. In the algebraic case, we care also about modules over our variety, and there is a third class of deformations of the category of modules, called the gerby deformations. So if you are bored with Poisson structures you can have a look at those! </p>