User stefan waldmann - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T21:50:33Z http://mathoverflow.net/feeds/user/12482 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108733/convergence-for-a-family-of-poisson-structures/125700#125700 Answer by Stefan Waldmann for Convergence for a family of poisson structures Stefan Waldmann 2013-03-27T08:00:56Z 2013-03-27T08:00:56Z <p>In general, this seems not to be possible. Consider the case of a vanishing Poisson structure $\Pi = 0$. Then every bivector field $\Lambda$ is closed for the differential $\delta$, which is just the zero map $\delta = 0$. So the wanted $\Lambda(t)$ should satisfy the Jacobi identity $[\Lambda(t), \Lambda(t)] = 0$ and it should be smooth (or even analytic) in $t$ with first derivative at $t = 0$ given by $\Lambda$. Differentiating this at $t = 0$ twice gives the Jacobi identity $[\Lambda, \Lambda] = 0$ for the original $\Lambda$. So this is a real obstruction... So this only shows that the continuation of an infinitesimal deformation to a honest (either formal or even analytic) deformation is generally very difficult.</p> http://mathoverflow.net/questions/119991/is-the-quantum-algebra-unique-up-to-isomorphism-in-deformation-quantization/119999#119999 Answer by Stefan Waldmann for Is the quantum algebra unique (up to isomorphism) in deformation quantization ? Stefan Waldmann 2013-01-27T08:57:34Z 2013-01-30T07:43:53Z <p>In deformation quantization there is a full classification available: let us first focus on the symplectic case which is easier. If $(M, \omega)$ is a symplectic manifold (like the $\mathbb{R}^2$ in your example) then the equivalence classes of star products are classified by formal series in the second deRham cohomology of $M$, a purely topological quantity.</p> <p>Here equivalence means that $\star$ and $\star'$ are called equivalent if there is a formal series $T = \mathrm{id} + \sum_{n = 1}^\infty \hbar^r T_r$ of differential operators $T_r$ starting with the identity in zeroth order such that $$T(f \star g) = Tf \star'Tg$$ This implies clearly that the deformations yield isomorphic algebras. But it is sligtly stronger, as one requires $T$ to start with the identity in zeroth order. This means that $T$ is invisible in the classical limit. If you are interested in a physical interpretation, this is very important as the interpretation of the elements of $C^\infty(M)$ should not be changed under quantization: the same function should correspond to the same physical observable, both in the classical and the quantum world.</p> <p>There is a good reason that the above notion of equivalence can be interpreted as the freedom of having possible different ordering descriptions as they show up in other quantization schemes.</p> <p>So in your example, there is only one deformation up to equivalence since the second deRham cohomology of $\mathbb{R}^2$ is trivial.</p> <p>If $M$ admits a non-trivial deRham cohomology then the higher order terms in the classifying "characteristic class" $c(\star)$ of the star product have the physical interpretation of a magnetic field. Their non-triviality then gives a magnetic monopol. So from that point of view (which can be justified physically more striclty) quantization requires to be told what magnetic charges are around and then it is unique up to equivalence, which means physically, up to the choice of an ordering description.</p> <p>Names involved in this classification results are Bertelson-Cahen-Gutt, Nest-Tsygan, Fedosov, Weinstein-Xu, and many others.</p> <hr> <p>Let me add here the precise definition of the characteristic class. On a symplectic manifold, given a star product $\star$ there is an intrinsic way to construct a formal series $$c(\star) \in \frac{[\omega]}{i\hbar} + \mathrm{H}^2_{dR}(M, \mathbb{C})[[\hbar]]$$ of second de Rham cohomology classes in such a way that $\star$ is equivalent to $\star'$ iff their classes coincide. Moreover, for every such formal series there is also a star product $\star$ having this series as characteristic class. The choice of the origin in this affine space is by convention, and given by the class of the symplectic form itself. </p> <p>There are several ways to construct $c(\star)$, all leading to the same result. The easiest is probably the construction in Gutt-Rawnsley <a href="http://www.ams.org/msnmain?fn=130&amp;form=fullsearch&amp;Submit=Start+Search&amp;pg1=MR&amp;s1=MR1675581" rel="nofollow">link text</a>. As a result: there is always the way to choose the class $c(\star) = \frac{[\omega]}{i\hbar}$ without higher order contributions. This is the somehow canonical choice on a symplectic manifold. However, having this choice does of course not mean to have a unique quantization (up to equivalence): depending on the non-triviality of the deRham cohomology, there are hiogher orders possible and then one get's non-trivial equivalence classes of quantizations.</p> <hr> <p>Now for the Poisson case: here the situation is similar in so far as in the symplectic case you can interprete the higher order terms of the class as perturbation of the symplectic form. Kontsevich proved using his formality theorem that the possible star products are classified by formal deformations of the Poisson tensor $\pi = \hbar \pi_1 + \hbar^2 \pi_2 + \cdots$ up to the action of the formal diffeomorphism group. This is given by exponentiating the formal vector fields $X = \hbar X_1 + \hbar^2 X_2 + \cdots$.</p> <p>In the symplectic case, this of course matches with the "characteristic class" classification. One can even show that, under correct identification, the classes are equal (this was done by Bursztyn, Dolgushev, and myself).</p> <p>In particular, not all deformations will be isomorphic in general, this depends now really very much on the underlying manifold and the Poisson tensor.</p> <p>Surprisingly, and this point I don't really understand well, in the Poisson case, the classification, i.e. the identification of the class of $\star$ and the class of $\pi$ depends to some extend on the choice of the formality map (and on those, GT acts). This is funny, because in the symplectic case, the classification can be done quite intrinsically, not depending on any sort of choices. Here the construction of Deligne, explained later also in Gutt-Rawnsley, is very nice.</p> <p>Note added: I fear that for the refined question I have no answer. The situation is the following: one has two types of data needed to construct a quantization (in the sense of star products) of a Poisson manifold together with a meaningful classification:</p> <p>First is to choose a quantization machinery. In the generic Poisson case, the only one we know is a formality map, which can be obtained by e.g. Kontsevich's or Tamarkin's methods. These constructions require certain choices, like the ones mentioned by Alexander Chervov_ a propagator in Kontsevich's construction or an associator for Tamarkin. </p> <p>Second, having a formality we can choose a classical deformation $\pi = \hbar \pi_1 + \hbar^2 \pi_2 + \cdots$ of the given Poisson structure $\pi_1$ and plug it into the formality. This will give a star product and any two obtained this way will be equivalent iff the classical deformations $\pi$ and $\pi'$ are equivalent in the sense of the action of the formal diffeos.</p> <p>So the question, how the notion of equivalence depends on the choice of the formality seems to be much more subtle. I have no idea about that. ONe knows that there are really different formalities (in a sense of being non-homotopic) but whether and how they yield a really different classification scheme seems to be not known (please correct me). I had some discussions about that with Vasiliy Dolgushev some time ago, but we didn't come to a real conclusion...</p> <p>The somehow surprising thing is that in the symplectic case, the classification is independent of any construction, and done by this characteristic class. Since the classifying space is the deRham cohomology $H_{dR}^2(M, \mathbb{C})[[\hbar]]$, we have a canonical choice for the trivial class. So in this sense, we really have a canonical quantization up to equivalence. Note however, that there are good reasons for considering also nontrivial classes in the symplectic case.</p> <p>From that point if view, the original question in the Poisson case can also be formulates as folows. given $\pi = \hbar \pi_1$ and a formality yielding $\star$, is there another formality such that $\star$ has a corresponding class $\pi' = \hbar \pi_1 + \hbar^2 \pi_2 + \cdots$ which is not equivalent to $\pi$ by means of the formal diffeos?</p> <p>OK, not much of a help, but just a refomulation of the problem.</p> http://mathoverflow.net/questions/114640/can-a-sphere-be-a-phase-space/114641#114641 Answer by Stefan Waldmann for Can a sphere be a phase space? Stefan Waldmann 2012-11-27T09:54:13Z 2012-11-27T09:54:13Z <p>Of course, the spheres are compact while cotangent bundles are noncompact (unless in dimension 0). Nevertheless, a bit more interesting is the question whether the even dimensional spheres can be phase spaces in the sense of symplectic manifolds. There the $\mathbb{S}^2$ is an example: the volume form is non-degenerate and a two-form. Closedness is for free in 2 dimensions. The higher dimensional spheres $\mathbb{S}^{2n}$ are never symplectic as on a compact symplectic manifold, the deRham cohomology has to be sufficiently non-trivial: the class of the symplectic form and all its $\wedge$-powers up to $n$ are non-trivial. For $\mathbb{S}^{2n}$ and $n \ge 2$ this is known to be not true: all cohomologies vanish except for the zeroth and the $2n$-th, which are both one-dimensional.</p> http://mathoverflow.net/questions/109957/what-is-the-definition-of-the-picard-group-of-a-non-necessarilly-commutative-ri/109980#109980 Answer by Stefan Waldmann for what is the definition of the Picard group of a (non necessarilly commutative) Ring? Stefan Waldmann 2012-10-18T06:31:00Z 2012-10-18T06:31:00Z <p>The definition I appreciate most comes from Morita theory: you consider unital (just for simplicity, something slightly weaker will also work) rings and the bimodules between them. Using the tensor product this makes almost a category: the tensor product of bimodules (over the ring in the middle) is associative up to a natural isomorphism and the rings themselves, viewed as bimodules over themselves with the mutliplication being the left/right module structures constitute the units, again up to a natural isomorphism. So either you prefer a bicategory setting, then you are done with what I said, or you prefer an honest category, then you have to pass to isomorphism classes of bimodules as morphisms between the rings. Of course, there are the usual set-theoretic issues that you should work in some universe etc.</p> <p>But now the definition of the Picard groupoid is very simple: it is the groupoid of invertible arrows in this category. The Picard group of a single ring is then just the isotropy group at this ring of the big Picard group. In other words: the Picard group of $R$ is the group of invertible bimodules with respect to the tensor product as multiplication and the ring $R$ as unit.</p> <p>It is then a famous theorem of Morita that characterizes the invertible bimodules $M$ between $R$ and $S$: there are various versions but a simple one is that they are finitely generated projective right modules over $R$ such that if you write $M = eR^n$ with an idempotent $e \in M_n(R)$ one has $ReR = R$, i.e. the two-sided ideal generated by the entries of the matrix $e$ is the whole ring $R$. Then the ring $S$ is given (via the left module structure) by the $R$-linear endomorphisms of $M$.</p> <p>You can find this in many algebra textbooks like e.g. Lam's book on Modules and Rings, or Bass' book on Algebraic $K$-Theory. On my homepage you can also find some (very preliminary) lecture notes covering this stuff and other versions of Morita theory for rings with additional structures.</p> http://mathoverflow.net/questions/53876/hilberts-17th-problem-for-smooth-functions Hilbert's 17th Problem for smooth functions Stefan Waldmann 2011-01-31T11:03:08Z 2012-10-02T11:28:53Z <p>Consider an open subset $U \subseteq \mathbb{R}^n$ and a smooth function $f\colon U \longrightarrow \mathbb{R}$ with $f(x) \ge 0$ for all $x \in U$.</p> <p>It is then known (if I remember correctly: by Michor?) that $f = g^2$ with a function $g$ which can be shown to be twice differentiable but not $C^2$ in general. In particular, a smooth square root does not exist in general.</p> <p>My question is whether $f$ can be represented as a sum of squares of smooth functions, i.e. the smooth version of Hilbert's problem 17, and if so, what is the minimal number of squares needed?</p> http://mathoverflow.net/questions/107678/reference-for-a-frechet-nuclear-space-is-montel/107682#107682 Answer by Stefan Waldmann for Reference for : a Fréchet nuclear space is Montel Stefan Waldmann 2012-09-20T12:57:04Z 2012-09-20T12:57:04Z <p>Maybe not in a single theorem, but you can go for Cor1 in Section 33 and Cor3 in Section 50 in Treves book.</p> http://mathoverflow.net/questions/61275/analysis-and-finitely-generated-groups Analysis and finitely generated groups Stefan Waldmann 2011-04-11T09:27:34Z 2012-04-08T02:19:56Z <p>Dear all, this is perhaps a bit a vague question, but some references would already be very helpfull.</p> <p>So let $G$ be a finitely generated group and choose some finite set of generators. This allows to define the length of a group element as usual, by the minimal number of generators needed to write it as a product of these generators. Using this length you can define all sort of analytic functions by replacing the "n" in various summation formulas by the $\mathrm{Length}(g)$ and the summation is now over the group elements $g$. To have a concrete example in mind, the "exponential series" is now e.g. $$\mathrm{lexp}(z) = \sum_{g \in G} \frac{z^{\mathrm{Length}(g)}}{\mathrm{Length}(g)!}.$$ Since for a given length there are at most exponentially many group elements, the length-exponential function is entire. For $G = \mathbb{Z}$ and $1$ as generator this reproduces the usual exponential series up to a factor $2$ as negative and positive $n \in \mathbb{Z}$ contribute with the same power of $z$.</p> <p>So my question is: what is known about the analytic features of such functions (depending on the choice of generators, depending on the group itself, etc). I guess there should be some literature on the market, but I'm really not in this buisness...</p> http://mathoverflow.net/questions/58951/toy-models-of-quantum-mechanics/59042#59042 Answer by Stefan Waldmann for Toy Models of Quantum Mechanics Stefan Waldmann 2011-03-21T08:03:43Z 2012-02-22T11:54:30Z <p>Though it is not a finite field the non-archimedian versions of quantum mechanics have indeed been used quite a bit in the recent years. Here I would like to draw attention to a different approach, not to the $p$-adic numbers which have been mentioned already in several answers/comments.</p> <p>One big flaw of the $p$-adics is that they are not ordered: there is no notion of positivity in the field of $p$-adics. Positivity on the other hand is crucial at many places in quantum physics (stability, probabilistic interpretations, ...). So one may wonder whether there are field extensions of $\mathbb{Q}$ which are <em>ordered</em> and still relevant in QM. Of course, if we stick to archimedian orders then the only field (which is also complete) are the reals, so this we all know. However, if we allow for non-archimedian orders then there are plenty more such fields. The most simple ones are perhaps the formal Laurent series $\mathbb{R}((\hbar))$ in a formal parameter $\hbar$. This is the quotient field of the (ordered) ring of formal power series $\mathbb{R}[[\hbar]]$.</p> <p>As my notion already suggest this is the field underlying quantization theory itself if one (as a first step into the right direction) treats quantization as a (at this stage still) infinitesimal deformation of classical physics with deformation parameter $\hbar$, commonly known as deformation quantization. The non-achimedian order simply means that $\hbar$ is positive but smaller than e.g. $\frac{1}{n}$ for all $n \in \mathbb{N}$. So physically speaking the quantum effects are still "very small" at this stage of the theory.</p> <p>It is now the positivity which allows to mimick many constructions from operator algebra also in this framework like e.g. a GNS construction, notions of pre Hilbert spaces, etc. Of course, for a honest quantum theory, $\hbar$ is not infinitesimally small, but has to be replaced by a real positive number. From this point of view the formal deformation quantization approach is only a first step, but still a very important one when it comes to the construction of quantum mechanical models for complicated classical theories.</p> <p>OK, I hope this is not too off-topic, but I was inspired by the nice answers on the $p$-adics ;)</p> http://mathoverflow.net/questions/88741/methods-for-determining-domains-of-influence/89107#89107 Answer by Stefan Waldmann for Methods for determining domains of influence Stefan Waldmann 2012-02-21T14:24:50Z 2012-02-21T14:24:50Z <p>I'm not quite sure if this is really the situation you are interested in, but in the book of Bär, Ginoux, and Pfäffle: Wave Equations on Lorentzian Manifolds and Quantization. ESI Lectures in Mathematics and Physics, European Mathematical Society, 2007, they discuss in quite some detail the Cauchy problem for hyperbolic linear wave equations on, and that is the catch, globally hyperbolic spacetimes. I guess that one should require something like that since otherwise you can at best hope for some local statements. But in their situation, I'm pretty sure to remember correctly, they have statements like the one you are looking for (don't they?). In any case, this is maybe a too special situation for you, but the book is nevertheless very nice. Unlike many other texts on hyperbolic PDE, it emphasizes the geometry very much.</p> http://mathoverflow.net/questions/80021/entire-calculus-and-clmc-algebras Entire calculus and clmc algebras Stefan Waldmann 2011-11-04T08:53:55Z 2012-02-10T16:42:15Z <p>If $\mathcal{A}$ is a complete locally convex (Hausdorff) associative unital algebra (over $\mathbb{C}$) one is interested in defining "transcendental" functions of a given algebra element $a \in \mathcal{A}$ like e.g. exponentials $\exp(a)$ by means of the power series expansion. This works fine for complete locally <em>multiplicatively</em> convex algebras. Recall that $\mathcal{A}$ is called lmc if there is a defining system of seminorms which are submultiplicative for the product. Equivalently, such an algebra is a (suitable) projective limit of Banach algebras. Then the (algebraic) polynomial calculus sending a polynomial $p \in \mathbb{C}[z]$ to the algebra element $p(a)$ extends by completion to an entire calculus $$\mathcal{O}(\mathbb{C}) \ni f \mapsto f(a) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} a^n \in \mathcal{A},$$ which is a continuous algebra homomorphism for a given $a$. Here $\mathcal{O}(\mathbb{C})$ is equipped with its usual Fréchet topology of locally uniform convergence. Equivalently and more convenient in this context, one can use the seminorms given by $p_R(f) = \sum_{n=0}^\infty \frac{|f^{(n)}(0)|}{n!} R^n$, from which one sees the continuity of the entire calculus on the nose.</p> <p>Now there are many lc algebras which are definitely not lmc like the Weyl algebra generated by the canonical commutation relations $[Q, P] = i\hbar \mathbb{1}$ (whatever lc topology you may put on it). </p> <p>My question is whether it is possible to have an entire calculus in the sense that there is a continuous algebra homomorphism extending the polynomial calculus to $\mathcal{O}(\mathbb{C})$ <em>without</em> having a lmc algebra but just a locally convex algebra. Can one give examples, reasonable conditions etc? </p> http://mathoverflow.net/questions/88056/is-there-a-sheaf-theoretical-characterization-of-a-differentiable-manifold/88089#88089 Answer by Stefan Waldmann for Is there a sheaf theoretical characterization of a differentiable manifold? Stefan Waldmann 2012-02-10T10:28:02Z 2012-02-10T10:28:02Z <p>This is more a long comment than an answer, but the comments concerning the question Hausdorff or non-Hausdorff triggered this...</p> <p>The second countable condition is certainly desirable for many reasons (embedding theorems etc) but there are prominent examples of non-Hausdorff manifolds in differential geometry. If you have a Lie algebroid, i.e. a vector bundle $E \longrightarrow M$ such that the sections of $E$ are equipped with a Lie bracket which satisfies a Leibniz rule along a bundle map $E \longrightarrow TM$ (the anchor) then there is a notion of a corresponding Lie groupoid integrating this. The obstructions for existence have been described in a beautiful paper by Crainic and Fernandes. Nevertheless, the resulting Lie groupoid is typically not Hausdorff but the non-Hausdorffness (funny word) is not so bad. The fibers of the source/target as well as the base of the Lie groupoid are all Hausdorff, it is only the way they are glued together which makes it non-Hausdorff.</p> <p>So this does not answer your question at all, but I think that an answer should also take care of these kind of situation as this is really important in many areas of differential geometry (Lie groupoids are everywhere...).</p> http://mathoverflow.net/questions/65678/lagrangian-submanifolds-in-deformation-quantization/65758#65758 Answer by Stefan Waldmann for Lagrangian Submanifolds in Deformation Quantization Stefan Waldmann 2011-05-23T11:49:44Z 2012-02-08T16:11:37Z <p>Well, in the symplectic case, the situation is somehow much simpler as in the general Poisson case where you only can speak about coisotropic (there is no good meaning of minimal coisotropic as the rank may vary). In the symplectic case you have a theorem of Weinstein which states that a there is a tubular neighbourhood of $L$ which is symplectomorphic to a neighbourhood of the zero section of $T^*L$. Thus the question of a module structure is reduced to the case of a cotangent bundle since star products are local. For cotangent bundles there is a good understanding whether you can have a module structure on the functions on the configuration space $L$: the characteristic class of $\star$ has to be trivial. In fact, together with Martin Bordemann and Nikolai Neumaier we constructed such module structures in a series of papers in the end of the nineties. Also Markus Pflaum has some papers on this. Thus the global statement is that on $L$ you have a module structure for $\star$ iff the characteristic class of $\star$ is trivial in an tubular neighbourhood of $L$.</p> <p>The module structures have (for particular star products) a very nice interpretation as global symbol calculus for differential operators on $L$. Moreover, if the char. class is not trivial but at least integral (up to some $2\pi$'s) then there is a module structure on the sections of some line bundle over $L$, coming quite close to the functions on $L$. Physically, this is important for the quantization of Dirac's magnetic monopole.</p> <p>As DamienC already said, the situation in the general Poisson case or even in the general coisotropic case on symplectic manifolds is much more involved. Here my answer to <a href="http://mathoverflow.net/questions/64452/in-the-dictionary-between-poisson-and-quantum-what-corresponds-to-coisotropic/64478#64478" rel="nofollow">http://mathoverflow.net/questions/64452/in-the-dictionary-between-poisson-and-quantum-what-corresponds-to-coisotropic/64478#64478</a> might also be of interest for you.</p> <p>Oh, I forgot: the first order term can be obtained as in the flat case, at least morally. On the configuration space $L$ you chose your favorite connection then the first order term of the module structure is something like "half the Poisson bracket" which means that $$f \bullet \psi =<br> \iota^* f \psi + i \hbar \iota^* \frac{\partial f}{\partial p_i} \frac{\partial \psi}{\partial q^i} + \cdots$$ (modulo some constants I forgot) where $q^i$ are coordinates on $L$ and $p_i$ are the corresponding fiber coordinates on the cotangent bundle. This has an intrinsic meaning. Here $\iota: L \longrightarrow T^*L$ is the zero section...</p> http://mathoverflow.net/questions/87819/is-there-dual-space-of-the-distributions-mathcaldr/87822#87822 Answer by Stefan Waldmann for Is there dual space of the distributions $\mathcal{D}'(R)$? Stefan Waldmann 2012-02-07T17:37:29Z 2012-02-07T17:37:29Z <p>Well, that depends on what topology you want to put on the space of distributions. The weak$^*$ is probably not really the one you would like to take. Instead, the strong dual might be more useful. The seminorms of this topology are given by $$p_B(\varphi) = \sup_{f \in B} |\varphi(f)|$$ where $B \subseteq \mathcal{D}(\mathbb{R})$ runs through the bounded subsets of the LF space $\mathcal{D}(\mathbb{R})$. The it is a theorem that (since the test functions are Montel etc) the dual with respect to this is again the space of test functions, i.e. the test functions are reflexive...</p> <p>I guess for the weak$^\ast$ topology this is not true and one gets a different dual of the dual. Your inclusion is correct, any test function gives a linear functional on the distribtions (by evaluation) which is continuous in the weak$^*$-topology. But you probably get more...</p> http://mathoverflow.net/questions/86738/which-functions-have-all-derivatives-everywhere-positive/86741#86741 Answer by Stefan Waldmann for Which functions have all derivatives everywhere positive? Stefan Waldmann 2012-01-26T17:51:14Z 2012-01-26T17:51:14Z <p>Well, there are certainly more. If you look at the chain rule then you see that the $n$-th derivative is a linear combinations of products of derivatives of the two functions you compose with positive coefficients. Thus if you have two functions with your property, then their composition will again have only positive derivatives. So you can go on...</p> http://mathoverflow.net/questions/49384/tools-for-long-distance-collaboration/86726#86726 Answer by Stefan Waldmann for Tools for long-distance collaboration Stefan Waldmann 2012-01-26T15:07:17Z 2012-01-26T15:07:17Z <p>It seems that there are still some tools missing in this long list. What I really enjoy more and more is a version control system. Personally, I prefer git over other more centralized solutions like subversion. It has several nice advantages when you're using different computers (say a desktop in your office and a laptop on the train or so) for which you do not have always a reliable internet connection. With the de-centralized approach of git, this is no big problem, you can commit changes locally and merge things back globally at a later time.</p> <p>I have by now made some nice experience with collaborators all over the globe using this...</p> http://mathoverflow.net/questions/86022/can-you-tell-the-volume-of-a-symplectic-manifold-from-the-poisson-brackets/86199#86199 Answer by Stefan Waldmann for Can you tell the volume of a symplectic manifold from the Poisson brackets? Stefan Waldmann 2012-01-20T11:38:02Z 2012-01-20T11:38:02Z <p>Let me add a few remarks on Theo's answer. For a compact connected symplectic manifold, it is known that the integration with respect to the Liouville volume form (whatever normaliyation you prefer) is the only linear functional on $C^\infty(M) = C^\infty_0(M)$ which vanishes on all Poisson brackets, unique up to rescaling of course. Remarkably, you don't even need to assume that the functional is continuous in any sense: this statement holds in the full algebraic dual of $C^\infty(M)$. There are various proofs of this available. Now on the other hand, you know that the integral of $1$ is the volume (in your prefered normalization) which shows that the constant functions on a compact symplectic manifold are <em>not</em> linear combinations of Poisson brackets (this is of course wrong in the noncompact case). From this you obtain that a function is a linear span of Poisson brackets iff its integral vanishes, a result due to Lichnerowicz (maybe???). This is the space Theo is talking about.</p> <p>In particular, I'm rather sceptic that one has a simple way of computing the volume in terms of Poisson brackets, as you can rescale the integration functional without destroying the above conclusions about the subspace of linear spans of Poisson brackets. Hmm...</p> http://mathoverflow.net/questions/85973/analysis-over-non-archimedean-ordered-fields/85974#85974 Answer by Stefan Waldmann for analysis over non-Archimedean ordered fields Stefan Waldmann 2012-01-18T09:46:38Z 2012-01-18T12:42:49Z <p>Well, there seems to be a lot of literature. I have encountered similar questions once when discussing problems in deformation quantization. here the ordered field is simply the field of formal Laurent series $\mathbb{R}((\hbar))$ with the ordering that $\hbar$ is positive (this fixed it uniquely). The idea was to transfer as much of stuff from $C^*$-algebra theory to the formal star products which are algebras of the above non-Archimedian field. Ultimately, we wanted to develop a spectral theory for these algebras, which utterly failed :) Nevertheless, you might be interested in examples of such fields and the "completed Newton Puiseux field" might be a funny one. First you take the algebraic or real closure of the formal Laurent series yielding the Newton-Puiseux series. This is not yet complete (in the sense of convergent Cauchy sequences) so you still have to complete it. The result is again field (either real, i.e. ordered) or algebraically closed, the smallest one containing the formal Laurent series.</p> <p>Let me probably clarify what I mean by completion here: since you have an ordered field, the naive $\epsilon$-definition of a Cauchy sequence makes sense with the main point that the $\epsilon$ is now a positive element of you field and not just a positive real number. So this defines a topology on your ordered field, but also a uniform structure. So one can speak about Cauchy sequences and completeness etc. It is then a standard argument to show that the completion (in the sense of uniform structures, i.e. taking Cauchy sequences module zero sequences) gives again an ordered field and the original one is included via constant sequences as usual. The slightly less trivial point is that if you start with an ordered field which is real closed then the completion is again real closed. This is needed to see that the completed Newton-Puiseux series are indeed both: real (or algebraically, in the case you started with $\mathbb{C}$ instead of $\mathbb{R}$) closed and (Cauchy) complete.</p> <p>A first reading may be</p> <p>Narici, N., Beckenstein, E., Bachman, G.: Functional Analysis and Valuation Theory. New York: Marcel Dekker, 1971.</p> <p>However, the main point is that analysis does not work too well. The reasons is that whenever you need suprema, you're on your own. And this happens quite often in analysis :) The other problem arising is that $1/n$ is no longer a zero sequence, a fact which is also used very often in analysis: you named already the intermediate value theorem...</p> <p>So my conclusion is that notions of positivity work well (needed a lot in formal DQ and representation theory of $^*$-algebras) but notions of calculus do not work well.</p> 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/56304/looking-for-interesting-actions-that-are-not-representations/56311#56311 Answer by Stefan Waldmann for Looking for interesting actions that are not representations Stefan Waldmann 2011-02-22T19:24:53Z 2011-12-13T15:34:45Z <p>In addition to what has been said already: I think that everywhere in Mathematics when you speak of symmetries you mean "group plus action" and not just the group itself.</p> <p>The thoughts about symmetry are probably of geometric nature: asking for symmetry means asking for the symmetry of a geometric object (we have already the examples of Riemannian manifolds, but there are many more) In differential geometry you can ask for "symmetries" of all kind of structures: metric, but also symplectic forms or Poisson tensors. In this case you enter the realm of dynamical systems with symmetries. The symmetries usually help to simplify the dynamical system by using "conserved quantities" to eliminate degrees of freedom. You may remember this from your first mechanics courses when dealing with the Kepler problem...</p> <p>But symmetries in crystals might yet give another example, not related to Lie groups and some inherited action from a linear action: treating a crystal as an abstract lattice with colored edges and vertices one may well ask for its symmetries and arrives at discrete groups acting in a much more combinatorial way. The original possibility that the lattice can be embedded into some Euclidean space is no longer relevant.</p> <p>In addition, symmetries arise in much more abstract concepts that these geometric ones. A prominent example is perhaps the question of solving polynomial equations. Here the symmetries of the polynomial might allow for general formulas or not. This is the beginning of Galois theory in field theory, where not Lie groups but discrete groups are acting.</p> <p>From my own field a statement which I would like to understand better: the Grothendieck-Teichmueller group acts on the set of Drinfeld associators. Not a linear action at all :(</p> <p>On the other hand: One reason why linear actions are so omnipresent is perhaps that (beside being the simplest type of actions) all types of geometric actions dualize to a linear action on the spaces of reasonable functions on the geometric spaces. Hence even a group action on some geometric object (manifold, lattice, ...) can by studied by means of representation theory when one looks for the induced action (via pull-back) on the functions on it. However, this is typically quite complicated as the representation spaces typically are infinite-dimensional.</p> http://mathoverflow.net/questions/57025/down-to-earth-uses-of-de-rham-cohomology-to-convince-a-wide-audience-of-its-usefu/57092#57092 Answer by Stefan Waldmann for Down-To-Earth Uses of de Rham Cohomology to Convince a Wide Audience of its Usefulness Stefan Waldmann 2011-03-02T08:38:26Z 2011-12-13T15:31:32Z <p>Perhaps even simpler than the examples from electromagnetism in $\mathbb{R}^3$ minus some points is the following:</p> <p>The angle "function" $\varphi\colon S^1 \longrightarrow \mathbb{R}$ is not really globally defined as turning aroung one time gives a discontinuity. It jumps by $2\pi$. Neverhtheless, the differential $\mathrm{d}\varphi$ is a perfectly global one-form on $S^1$. It is the usual volume form, not being exact but closed for dimensional reasons. So the non-trivial first deRham cohomology of $S^1$ is responsible for counting angles and the fact that $0 \ne 2\pi$ ;)</p> <p>This can be upgraded to the more interesting statement that on a orientable compact manifold without boundary you have a non-trivial top-degree deRham cohomology: again, the reason is that we can integrate a volume form resulting in a non-zero volume. Thus (by Stokes theorem) the volume form can not be exact. It is closed without thinking about it, simply for dimensional reasons.</p> http://mathoverflow.net/questions/83234/monomials-in-the-universal-enveloping-of-a-lie-algebra-in-terms-of-the-symmetric/83240#83240 Answer by Stefan Waldmann for monomials in the universal enveloping of a Lie algebra in terms of the symmetric basis Stefan Waldmann 2011-12-12T13:42:06Z 2011-12-13T13:49:10Z <p>This will not be an honest answer but just a long remark: it reminds me a bit on the relation between Weyl quantization and standard ordered quantization. Here one has the polynomials in $q$ and $p$ with their canonical Poisson bracket $\lbrace q, p\rbrace = 1$ which should be quantized into operators as usual. For higher polynomials one has to choose an ordering, e.g. Weyl or standard or many more... Pulling back the operator product gives then a star product, depending on the choice of the ordering. All of them are isomorphic by explicit isomorphism. In case of Weyl/standard the isomorphism is given by the exponential of the indefinite Laplacian $\frac{\partial^2}{\partial q \partial p}$. This is very explicit and allows for many nice formulas and computations.</p> <p>Now in your situation it seems to me that you would like to have some similar isomorphism between the two quantizations of the Polynomials on the dual $\mathfrak{g}^*$ which you obtain by total symmetrization (= Weyl) and a standard ordering with respect to the choice of a basis of $\mathfrak{g}$. The point is that the standard ordering you are considering is much less canonical as in the case of canonical quantization: you really have to specify a basis.</p> <p>I don't think that this has been worked out in detail, but maybe the old work of Simone Gutt from 1983 in Lett. Math. Phys. might be inspiring.</p> <p>EDIT: Maybe I have misunderstood the question in the first place. So if your intention is to find an explicit formula for the Weyl ordered case alone, then one indeed has an "explicit" formula: I'm not sure where this showed up first, but the formula goes as follows: given $\xi, \eta \in \mathfrak{g}$ one considers the (formal) BCH series $BCH(\hbar \xi, \hbar\eta)$ with $\hbar$ being a formal parameter. Then the Weyl product of the formal exponentials $\exp(\hbar\xi)$ and $\exp(\hbar\eta)$ is given by $$\exp(\hbar\xi) \star \exp(\hbar\eta) =\exp (\hbar^{-1} BCH(\hbar \xi, \hbar\eta))$$</p> <p>Well, from this one get's the formula for monomials by differentiating and polarization. But you see, you won't get what you want: this is not at all explicit for two reasons. First, the differentation and polarization has to be done, very ugly. But second, and this is the more severe point, you have to know the BCH series.</p> <p>Now on the other hand, this formula shows that you probabaly can not expect to get a simpler formula <em>without</em> using BCH. I fear, one can not go beyond this... :(</p> http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/53463#53463 Answer by Stefan Waldmann for Why should one still teach Riemann integration? Stefan Waldmann 2011-01-27T08:19:05Z 2011-12-07T14:43:00Z <p>Maybe yet another argument in favor for the Riemann integral: it is fairly easy to generalize it to integration of vector-valued functions. Here the Lebesgue theory becomes more tricky as soon as the target space is something beyond a Banach space. In many situations you want to integrate a not too badly behaved function, say a continuous one, with values in a rather complicated topological vector space (not only Fréchet, but perhaps dual of Fréchet, only sequentially complete, or something like that if you think of distributions). OK, I admit that this is not what you teach in first year calculus, but just be prepared :)</p> http://mathoverflow.net/questions/60201/what-is-a-lagrangian-submanifold-intuitively/60220#60220 Answer by Stefan Waldmann for What is a Lagrangian submanifold intuitively? Stefan Waldmann 2011-03-31T18:18:43Z 2011-12-07T14:07:21Z <p>Everything is a lagrangian submanifold (A. Weinstein's lagrangian creed...) Indeed, every manifold is the lagrangian zero section of its cotangent bundle ;)</p> <p>But more serious: Weinstein's tubular neighbourhood theorem states that every lagrangian submanifold in a symplectic manifold has a neighbourhood <em>symplectomorphic</em> to a neighbourhood of the zero section of the cotangent bundle.</p> <p>Occurences of lagrangian submanifolds are indeed manifold: they arise as semiclassical support for certain FIO's and can also be thought of as semiclassical version of states in quantum mechanics via the WKB expansion. This point of view is exemplified a lot in the nice booklet of Bates and Weinstein.</p> <p>Another occurence of coisotropics is in constraint mechanics: In Dirac's theory of constraints a coisotropic submanifold is what he calls a first class constraint. They arise very often in geometric mechanics with degenerate Lagrangians etc.</p> <p>They are also the natural starting point for reduction: this is perhaps the more modern "coisotropic creed" of Lu (everything is a...)</p> <p>Finally, to make the deformation quantization aspects a bit more precise: if you are looking for a submanifold $C \subseteq M$ in a Poisson manifold with star product $\star$ which allows for a (say) left module structure on $C^\infty(C)[[\hbar]]$ in such a way that the zeroth order of the module structure is the usual multiplication by the restriction, then you can show quite easily that $C$ has to be coisotropic. Martin Bordemann has a nice point of view how this relates to a theory of quantizing reduciton etc. in his (french!) big preprint :) In particular, the classical vanishing ideal becomes deformed into a left ideal for the star product (this was, I guess, essentially Lu's suggestion)</p> <p>Note however, that there are other left ideals not of this form, e.g. the Gel'fand ideals of positive functionals, which can be much smaller.</p> <p>The role of the lagrangian submanifolds $L$ in this context is that the corresponding representation on $C^\infty(L)[[\hbar]]$ becomes "irreducible" in a meaningful way (trivial commutant in the local operators). However, this only makes sense in the symplectic surrounding. In a truly Poisson manifold, only coisotropic makes sense.</p> http://mathoverflow.net/questions/54661/borel-lemma-for-vector-valued-functions Borel Lemma for vector-valued functions Stefan Waldmann 2011-02-07T16:46:47Z 2011-12-06T22:24:40Z <p>The classical Borel Lemma states that for an arbitrary sequence $(v_n)_{n \in \mathbb{N}_0}$ of complex numbers there is a smooth function $f\colon \mathbb{R} \longrightarrow \mathbb{C}$ with Taylor coefficients at $0$ given by the $v_n$. Some generalizations work for functions of $d$ variables and also for values taken in an arbitrary Fréchet space $V$ instead of the complex numbers.</p> <p>The proofs I know use essentially the fact that $V$ has a <em>countable</em> set of seminorms defining the topology. On the other hand, taking the space of compactly supported smooth functions with its usual LF topology as $V$ and $v_n$ with increasing support gives easily a counter-example that for this sequence we can not have a smooth $f\colon \mathbb{R} \longrightarrow C^\infty_0(\mathbb{R})$ with $v_n$ being the Taylor coefficients, unless the $v_n$ are all in same $C^\infty_0(K)$ for a <em>fixed</em> compact subset $K$. </p> <p>So my question is to which lcs one actually can extend the Borel lemma? Are Fréchet spaces the end of the story?</p> <p>EDIT: There is of course a stupid way to extend it beyond Fréchet: whenever you have a coarser lc topology on $V$ then every smooth function with respect to the orignal one is also smooth with respect to the coarser one. So if $V$ is Fréchet then every coarser topology on $V$ will also have a valid Borel Lemma. Examples are e.g. the operator topologies on the bounded operators on a Hilbert space (soooorry for overlooking this in the first try).</p> <p>So the refined question is: are there other lcs with topologies not dominated by a Fréchet one for which the Borel Lemma holds?</p> http://mathoverflow.net/questions/79907/what-is-a-complex-inner-product-space-really/79908#79908 Answer by Stefan Waldmann for What is a complex inner product space "really"? Stefan Waldmann 2011-11-03T09:13:53Z 2011-11-04T08:24:34Z <p>Suppose first that you're not interested in the complex situation at all and just consider real inner product spaces (with positive definite inner product for simplicity). Then you have the notion of transposed operators with respect to the inner product and you get a notion of normal operators this way. Now the spectral theory for normal operators is very nice provided they are even symmetric ones: they can be diagonalized. Now there are other normal operators as e.g. the rotations where $D D^T = 1 = D^T D$. However, they do not even need to have an eigenvector (depending on the dimension...) So their normal form is much more complicated and cumbersome for many reasons.</p> <p>The reason is that the characteristic polynomial needs not to have real roots. But as we know, over the complex numbers we have the fundamental theorem of algebra telling that it decomposes into linear factors. From this one gets the idea that <em>complexifing</em> the real vector space to a complex one might be a good idea for spectral purposes. Indeed, doing so allows now to diagonalize rotations, they have eigenvalues on the unit circle.</p> <p>Of course, one still has to see how "normal" is translated into the complex situation: you also have to complexify the inner product leading to a sesquilinear inner product (not a complex bilinear one, this is sort of useless here).</p> <p>Then of course, you loose the interpretation of the inner product encoding an "angle" between vectors, but you still have a Cauchy-Schwarz inequality. The only thing which remains after this complexification is that zero remains zero: so it still makes sense to talk about orthogonal vectors.</p> <p>Well, this might be a motivation why complex Hilbert spaces arise quite naturally, even if you're only interested in real ones, supporting your point 3.)</p> http://mathoverflow.net/questions/63749/dimensional-analysis-in-mathematics/63786#63786 Answer by Stefan Waldmann for Dimensional Analysis in Mathematics Stefan Waldmann 2011-05-03T09:25:29Z 2011-10-26T21:56:57Z <p>Dimensional analysis can be viewed as the study of graded objects in algebra. The grading then corresponds to "counting the units" in a precise way. There are of course many examples and I believe that morally, every graded (say $\mathbb{Z}$-graded) algebra/vector space/etc can be viewed as collection of objects having an intrinsic dimension. If you have a graded algebra then this just means that the dimensions mutliply correctly. If you have a grading on a vector space and a bilinear map (product, Lie bracket...) which is still homogeneous but not respecting the grading additively then the bilinear map itself carries a fixed dimension:</p> <p>On example is the polynomial algebra $\mathbb{C}[x,y]$ with its usual $\mathbb{Z}$-grading by the total polynomial degree. Then the canonical Poisson bracket is determined by $\lbrace x, y \rbrace = 1$ and hence has dimension $-2$ times the units of the generators. Relabelling then into $q$ and $p$, you have the Poisson bracket of classical mechanics with its usual dimension being that of an "inverse action".</p> <p>More sophisticated examples can be found e.g. in differential geometry where you have zillions of graded algebras/spaces arising naturally. Staying in the realm of Poisson geometry, the canonical Poisson bracket on a cotangent bundle has "momentum degree $-1$": Mathematically, this means that one first has an Euler vector field $E = \sum_i p_i \frac{\partial}{\partial p_i}$ on $T^*M$ as on every vector bundle (where the $p_i$ are fiber coordinates). Heuristically, this vector field "counts" how many $p$'s you have. Then the Poisson tensor is something like $\pi = \sum_i \frac{\partial}{\partial q^i} \wedge \frac{\partial}{\partial p_i}$ which satisfies $\mathcal{L}_E \pi = - \pi$, making the above statement precise.</p> <p>In general, associated to a grading one can attach the grading "derivation" (or better just operator) which satisfies $deg = k id$ on the homogeneous components $V^k$ of the graded space $V$. In the above case, $deg$ is just the Euler vector field and the graded spaces are the tensor fields on $T^*M$...</p> <p>I hope that these examples have convinced you that a "dimensional analysis" happens quite often and naturally in many areas of mathematics. The associated grading operators usually play an important role and help a lot. The least is certainly a "self-correcting" aspect, that handling dimensions correctly avoids stupid errors.</p> http://mathoverflow.net/questions/75976/symbol-of-pseudodiff-operator/75983#75983 Answer by Stefan Waldmann for Symbol of pseudodiff operator Stefan Waldmann 2011-09-20T18:15:36Z 2011-09-20T18:15:36Z <p>There are several global approaches to pseudo-diffops. All of them seem to need some additional geometric objects, a connection. So suppose that you want to have $\Psi$DO's on some vector bundle $E \longrightarrow M$ then you need a linear connection on $E$ as well as one on the tangent bundle. Out of this you can build via adaptions of the usual integral formulas for the Weyl quantization on flat $\mathbb{R}^n$ a symbol calculus, intrinically global and also allowing for a total symbol and not a leading one only. You can find these kind of approaches in the works of Widom in the 80's if I remember correctly. There are more recent approaches by Pflaum as well as by Safarov. If you're interested in the relation to star products and quantization of cotangent bundles (which is essentially the pullback of the operator product to the symbols) then you may want to take a look at the work of Bordemann, Neumaier, Pflaum and myself :)</p> http://mathoverflow.net/questions/75793/open-symplectic-embeddings-and-deformation-quantization/75824#75824 Answer by Stefan Waldmann for Open symplectic embeddings and deformation quantization Stefan Waldmann 2011-09-19T06:54:04Z 2011-09-19T06:54:04Z <p>Hi Igor,</p> <p>there is a quite elementary way to see that star products restrict to open subsets: it's essentially part of the definition of a star product. Here, I will focus on the case of smooth (symplectic/Poisson) manifolds, where you do not have to rely too much on sheaf-theoretic notions. Now, $\star$ being a star product on $M$ means that it consists in each order of $\hbar$ of <em>bidifferential</em> operators if you consider a differential star product, or of local operators if you consider a local star product. Local operators are, by a celebrated theorem of Preetre locally differential, i.e. on a sufficiently small open subset the restrict to differential operators. Globally, the only problem may be that their order is infinite (take the real line and a bump function with supp in the unit interval. Now translate the bump function to a bump function $\chi_n$ having supp in $[n,n+1]$ Then taking the sum of all $\chi_n \frac{\partial^n}{\partial x^n}$ is a local but not a differential operator). In any case, bidifferential/local operators restrict to open subsets. </p> <p>THis induced for every open $U \subseteq M$ a new product $\star_U$ for $C^\infty(U)[[\hbar]]$. The point is now that associativity can be check locally, as in each order of $\hbar$ it is an equation between (multi-)differential operators which localize! So you indeed get a star product.</p> <p>Unfortunately, with the advent of Kontsevich's formality these things and techniques were mostly forgotten. It was kind of standard arguments and it's also present in all the earlier constructions of/with star products in the 80's... It might be that it is not spelled out here explicitly in the literature, but you just have to look at the early papers of Bayen etc as well as Gutt, Cahen, deWilde, Lecomte and so on, and scan for the words "local operator" or Peetre Theorem ;)</p> <p>Now for the second question about functoriality. This is of course much more subtle. The naive answer is that there is no thing like a quantization functor (with appropriate continuity properties) as one has the no-go theorem of Gronewold and van Hove. I'm sure you know this. So one has to refine things a bit: the crucial question seems to be what the domain of this functor should be: symplectic manifolds per se is not a good choice. A better choice will be symplectic manifolds with a symplectic connection (this is a huge choice to make, as there are zillions of symplectic connections...) </p> <p>In this case, one indeed gets a functorial quantization by means of the Fedosov construction of a star product (say for trivial characteristic class) This you can find at many places in the literatur (if you're crazy enough to read german, you can take a look at my book ;) As Theo say, one can modify the construction using a series of closed two-forms, so it might be better to chose this as a domain of you functor...</p> <p>Simillar things hold in the Poisson case as well: you have to choose a formality for $\mathbb{R}^n$ once and for all, say the Kontsevich one, subject to certain invariance properties ($\mathrm{GL}(n)$-invariant should do the job). Then you can globalize this formality by Dolgushev's construction. This gives a functorial construction of a formality for the price of choosing a connection on each manifold first (this is needed in Dolgushev). The again, one get's functorial star products... This is sort of implicit in a paper by Kontsevich. If I remember correctly, in an appendix... BUt more details on it are in the papers by Dolguushev.</p> http://mathoverflow.net/questions/74837/frechet-manifolds-book/74838#74838 Answer by Stefan Waldmann for frechet manifolds book Stefan Waldmann 2011-09-08T07:41:09Z 2011-09-08T07:41:09Z <p>There is the book by Kriegl and Michor called "Convenient setting of global analysis" published by the AMS. It goes much beyond Fréchet and really gives a big panorama. However, it is not easy reading and requires really some work. But I guess that is due to the subject...</p> http://mathoverflow.net/questions/64452/in-the-dictionary-between-poisson-and-quantum-what-corresponds-to-coisotropic/64478#64478 Answer by Stefan Waldmann for In the dictionary between Poisson and Quantum, what corresponds to Coisotropic? Stefan Waldmann 2011-05-10T08:32:45Z 2011-08-03T19:57:38Z <p>Concerning your first question I have a couple of suggestions: first coisotropic is in some sense the best we can have in a truely Poisson situation: there is nothing like lagrangian (unfortunately).</p> <p>As you already said, lagrangian sometimes is associated to pure states in the quantum regime, here the main argument is coming from the WKB approximation in physics, which has some very interesting mathematical formulations: in the booklet of Bates&amp;Weinstein (you probably know) you can find a lot of these ideas.</p> <p>However, I would like to draw your attention to some other point: Having a coisotropic $I$ in some Poisson algebra $A$ (so we left geometry, purely algebraic setting) then the thing you can do classically is <em>reduction</em>: You take the idealizer $B \subseteq A$ of $I$ with respect to the Poisson bracket and this turns out to be the largest Poisson subalgebra of $A$ having $I$ as Poisson ideal. So the quotient $B/I$ is again a Poisson algebra. In your favorite geometric setting with nice assumption this corresponds precisely to the Poisson algebra of functions of the (Marsden-Weinstein) quotient: But we see something more:</p> <p>$A/I$ is a $A$-left module (sure) and it becomes also a $B/I$ right module (just check that things are well-defined). In fact, a little exercise shows that if $A$ has a unit (let's assume that) then $B/I = End_A(A/I)^{op}$. So we are in some sense even very close to a Morita context (it is not, though...)</p> <p>Usually I don't like to do that: but to make a little advertisement for some own work, I have a quite detailed paper with Simone Gutt on the above reduction proceedure where we investigate the relations of the representation theories of the big algebra $A$ and the reduced algebra $B/I$ :)</p> <p>Now the remarkable thing is that this (still classical) bimodule structure might have good chances to survive quantization. In fact (and here one should quote Martin Bordemann's long french preprint as well as the works of Cattaneo&amp;Felder) under certain geometric conditions deformation quantization gives indeed such a quantization.</p> <p>So the noncommutative version is a <em>left</em> ideal $I$ and then the above quotient proceedure just works the same on the algebraic level. Of course, in DQ the trickey question is whether $B/I$ is still something like the quantized functions on the classical Marsden-Weinstein quotient and even more trickey: whether the classical coisotropic ideal can be quantized into a left ideal at all. For this there are obstructions, even local ones, in the Poisson setting, while it works locally in the symplectic setting by taking an adapted Darboux chart. Globally, it is also trickey in the symplectic setting: Martin Bordemann discusses this in detail...</p> <p>OK: the conclusion is something like coisotropic ideals are used for reduction and their quantization will be left ideals used in the same way. Both lead to the above <em>bimodule</em> structures on $A/I$ which is geometrically the coisotropic submanifold itself.</p> <p>As a small warning: it is <em>not</em> true in deformation quantization that all modules (of interest) arise this way. There are other modules which have their support say on points: one can use $\delta$-functionals as positive functional (after some correction terms) and get a GNS like construction also in formal deformation quantization. Then in this case, the module has sort of support on that point...</p> <p>Ah, the second question: never thought about that in detail, but perhaps the above picture gives some ideas on "reverse engineering"..?</p> http://mathoverflow.net/questions/119991/is-the-quantum-algebra-unique-up-to-isomorphism-in-deformation-quantization/119999#119999 Comment by Stefan Waldmann Stefan Waldmann 2013-01-30T07:45:44Z 2013-01-30T07:45:44Z @Theo Johnson-Freyd: I added a link to a paper of Gutt and Rawnsley, which I personally like best ;) I also added a little explanation of how the class works in the symplectic case. I think the real problem is that the classification in the Poisson case depends on the formality (how?) and it is not clear whether the statement &quot;trivial class&quot; (ie given by the Poisson tensor itself) is independent on the choice of the formality... http://mathoverflow.net/questions/119991/is-the-quantum-algebra-unique-up-to-isomorphism-in-deformation-quantization/119999#119999 Comment by Stefan Waldmann Stefan Waldmann 2013-01-28T07:36:45Z 2013-01-28T07:36:45Z @Alexander Chervov: yeah, it's funny to meet this way. I added a couple of remarks (too long for a comment) Essentially, I have no idea about you refined question in the Poisson case :( http://mathoverflow.net/questions/107678/reference-for-a-frechet-nuclear-space-is-montel/107682#107682 Comment by Stefan Waldmann Stefan Waldmann 2012-09-20T13:05:42Z 2012-09-20T13:05:42Z and he was a couple of second faster :) http://mathoverflow.net/questions/7957/books-well-motivated-with-explicit-examples/53969#53969 Comment by Stefan Waldmann Stefan Waldmann 2012-07-16T12:21:38Z 2012-07-16T12:21:38Z @Andrew Yepp, of course! it does not go very far and you still need some other topology books, if you really want to get some deeper insight. But for a start, I really like it :) http://mathoverflow.net/questions/91104/are-the-smooth-vectors-of-a-frechet-space-dense/91145#91145 Comment by Stefan Waldmann Stefan Waldmann 2012-03-14T11:42:01Z 2012-03-14T11:42:01Z +1 for the paper of Dixmier and Malliavin, a pearl! This is (unfortunately) not well-known at all but a really nice statement. http://mathoverflow.net/questions/88741/methods-for-determining-domains-of-influence/89107#89107 Comment by Stefan Waldmann Stefan Waldmann 2012-02-22T08:37:14Z 2012-02-22T08:37:14Z Hi Igor, maybe you're right and it is only about this inclusion. So the point would be to construct a particular wave equation which extremizes this support of its solutions. So I fear that my reference does not answer your question :( sorry http://mathoverflow.net/questions/88624/is-there-a-coordinate-free-proof-of-the-hamiltonian-character-of-the-geodesic-flo/88649#88649 Comment by Stefan Waldmann Stefan Waldmann 2012-02-16T20:17:21Z 2012-02-16T20:17:21Z @Deane: you might like this (even legal) way to take a look <a href="http://www.cds.caltech.edu/~marsden/books/Foundations_of_Mechanics.html" rel="nofollow">cds.caltech.edu/~marsden/books/&hellip;</a> http://mathoverflow.net/questions/80021/entire-calculus-and-clmc-algebras/88116#88116 Comment by Stefan Waldmann Stefan Waldmann 2012-02-13T21:30:00Z 2012-02-13T21:30:00Z Dear Jochen, thank you very much. I will try to get a copy of this and have a look. http://mathoverflow.net/questions/87819/is-there-dual-space-of-the-distributions-mathcaldr/87822#87822 Comment by Stefan Waldmann Stefan Waldmann 2012-02-08T09:32:00Z 2012-02-08T09:32:00Z Thanks for the clarification, Jochen! http://mathoverflow.net/questions/87819/is-there-dual-space-of-the-distributions-mathcaldr/87822#87822 Comment by Stefan Waldmann Stefan Waldmann 2012-02-07T18:35:24Z 2012-02-07T18:35:24Z Hi Johannes: Ok, that I didn't know. But the weak$^*$ topology is still different from the strong one, right? I should be strictly weaker... http://mathoverflow.net/questions/49384/tools-for-long-distance-collaboration/86726#86726 Comment by Stefan Waldmann Stefan Waldmann 2012-01-26T16:09:08Z 2012-01-26T16:09:08Z @Willie: sorry, you may be right. Nevertheless, I also use that for collaboration along, sending notes back and forth and so on. Only in the final stage, this ends up with writing a paper (hopefully). I guess, I will leave the answer here, mainly since I don't know how to move it efficiently :) http://mathoverflow.net/questions/86022/can-you-tell-the-volume-of-a-symplectic-manifold-from-the-poisson-brackets/86199#86199 Comment by Stefan Waldmann Stefan Waldmann 2012-01-23T17:50:05Z 2012-01-23T17:50:05Z Hi Deane. Of course, this is true What I mean is that there seems to be not a simple computational formula. If you first recover $\omega$ from the Poisson bracket then, of course, you're done. But it does not seem to be easy to just write down a simple formula (is it?) in terms of Poisson brackets of some functions, in order to get the volume... http://mathoverflow.net/questions/86028/chern-classes-of-a-trivial-bundle Comment by Stefan Waldmann Stefan Waldmann 2012-01-18T19:25:07Z 2012-01-18T19:25:07Z either follow Mariano's suggestion or use a flat connection :) http://mathoverflow.net/questions/85973/analysis-over-non-archimedean-ordered-fields/85982#85982 Comment by Stefan Waldmann Stefan Waldmann 2012-01-18T12:36:06Z 2012-01-18T12:36:06Z @Rob: OK, maybe there is a confusion about &quot;completeness&quot;. I was referring to the question whether Cauchy sequences do converge or not. In a non-Archimedian field e.g. $1/n$ is not a Cauchy sequence any more. However, e.g. in the field of formal Laurent series with formal parameter $\hbar$ the sequence $\hbar^n$ is Cauchy and converges to $0$. Of course, bounded sets do not have a sup ind inf in general. http://mathoverflow.net/questions/80021/entire-calculus-and-clmc-algebras Comment by Stefan Waldmann Stefan Waldmann 2011-11-08T08:19:33Z 2011-11-08T08:19:33Z Oh, that <i>is</i> misunderstanding. Of course, I want $\mathcal{A}$ to be a locally convex algebra... I'll fix it. Thanks...