5
$\begingroup$

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?

$\endgroup$
4
  • 2
    $\begingroup$ 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. $\endgroup$ May 25, 2016 at 1:23
  • $\begingroup$ @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. $\endgroup$
    – issoroloap
    May 25, 2016 at 6:43
  • 1
    $\begingroup$ See also Arnold Neumaier's answer to the same question which says that Rieffel quantization could be relevant. $\endgroup$
    – Dilaton
    May 29, 2016 at 21:16
  • 1
    $\begingroup$ thank you @Dilaton, i'm going to reply to that answer. here i just remark that, while i agree with Arnold and found his reference very useful, it doesn't quite solve my problem, yet. $\endgroup$
    – issoroloap
    May 30, 2016 at 12:43

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.