Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Deformation quantization of Poisson bracket without star-product

+ 5 like - 0 dislike
1395 views

Kontsevich's formality theorem implies in particular that star-products on a $C^\infty$-manifold $M$, $$f\star g = fg + \sum_{k\geq1} \hbar^k B_k(f,g),\qquad f,g\in C^\infty(M),$$ where $B_k$ are bidifferential operators of degree at most $k$, are classified, up to the gauge equivalence $$\star \sim \star' \iff \exists\ T=I+\sum_{l\geq1}\hbar^lT_l\ ,\quad T(f\star g)=T(f)\star' T(g)$$ where $T_l$ are differential operators of order at most $l$, by Poisson bivectors depending formally on $\hbar$ $$\Pi(\hbar) = \Pi_0+\sum_{k\geq0}\hbar^{k}\Pi_k \in C^\infty(M,\wedge^2 TM)[[\hbar]],\qquad [\Pi(\hbar),\Pi(\hbar)]_{\mathrm{SN}}=0$$ (where $[\cdot,\cdot]_{\mathrm{SN}}$ is the Schouten-Nijenhuis bracket) up to formal paths in the groups of diffeomorphisms of $M$ starting at the identity diffeomorphism.

The star product commutator $[f,g] := \frac{1}{\hbar} (f\star g - g\star f) = \{f,g\}+\sum_{k\geq0}\hbar^kC_k(f,g)$ starts with the Poisson bracket associated to the Poisson tensor $\Pi_0$. So a star product, which is an associative deformation $(C^\infty(M)[[\hbar]],\star)$ of the associative commutative algebra $(C^\infty(M),\cdot)$, induces in particular a Lie algebra deformation $(C^\infty(M)[[\hbar]],[\cdot,\cdot])$ of the Poisson algebra $(C^\infty(M),\{\cdot,\cdot\})$.

1) Is there a sensible notion of "deformation quantization" of the Poisson algebra $(C^\infty(M),\{\cdot,\cdot\})$ as a Lie algebra deformation $(C^\infty(M)[[\hbar]],[\cdot,\cdot])$ which does not require referring to (or the existence of) a star-product, i.e. a notion of quantum commutator without the corresponding star-product?

If yes,

2a) does any such special Lie algebra deformation come from a star-product anyway?

2b) is there a classification analogous to Kontsevich's one?

Motivation: in field theory one often faces the problem that, while commutators of local functionals can be defined as local functionals themselves, star-products (or even just classical products for that matter) of local functionals are not local functionals (they are sometimes defined only in some completion of the tensor algebra of local functionals). Can one do without star-products and consider the classification problem for commutators instead?

This post imported from StackExchange MathOverflow at 2016-05-29 11:29 (UTC), posted by SE-user issoroloap
asked May 24, 2016 in Theoretical Physics by issoroloap (25 points) [ no revision ]
retagged May 29, 2016
A note about your motivation. You were not specific about what you mean by local functional. If one takes it to mean "spacetime local", like $A[\phi] = \int_M f(x) a(\phi,\partial\phi,\ldots)$ with $f(x)$ having compact support on the spacetime $M$, then your insistence on local functionals is moot. The Poisson bracket of two local functionals, given by the Peierls formula $\{A,B\} = \int_{M\times M} \frac{\delta A}{\delta\phi(x)} G(x,y) \frac{\delta B}{\delta\phi(y)} dx \, dy$, is already only bi-local since the causal Green function $G(x,y)$ only vanishes when $x,y$ are spacelike separated.

This post imported from StackExchange MathOverflow at 2016-05-29 11:29 (UTC), posted by SE-user Igor Khavkine
@Igor Yes I was not very specific in the motivation, I wanted just to to give an idea. What I had in mind was more the non-relativistic hydrodynamic Poisson bracket $\{\overline{f},\overline{g}\} = \int \frac{\delta \overline{f}}{\delta u } K( \frac{\delta \overline{g}}{\delta u} )dx$, with $K = \sum K_i \partial_x^i$ a differential operator with coefficients $K_i(u,u_x,u_{xx},\ldots)$ that are diff. polynomials and $\overline{f}=\int f(u,u_x,u_{xx},\ldots)dx$, with $x\in S^1$. People consider quantization of such systems (KdV, etc) all the time and I have been wandering about its unicity.

This post imported from StackExchange MathOverflow at 2016-05-29 11:29 (UTC), posted by SE-user issoroloap

1 Answer

+ 2 like - 0 dislike

Maybe Rieffel quantization = strict deformation quantization is what you want. In place of expansion into a formal infinite series one maps into a $C^*$-algebra (often realized by operators on a suitable Hilbert space). To my knowledge, the Rieffel quantization of arbitrary finite-dimensional Poisson manifolds is still open.

Your examples in the comment are for infinite-dimensional manifolds, while Kontsevich assumes finite dimensions. In infinite dimensions very little is known. For 2-dimensional integrable field theories (e.g., KdV) one can quantize using the inverse scattering transform.

answered May 29, 2016 by Arnold Neumaier (15,787 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOve$\varnothing$flow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...