Several physicists consider non-Poisson bivectors but still apply Kontsevich formality in order to get deformation quantization type results: see e.g. Szabos's review An introduction to nonassociative physics.

The result they get is a `non-associative star product’.

What is the algebraic structure that they get, or, how to state correctly the deformation quantization problem in this non-associative and non-Poisson context?

  • 2
    $\begingroup$ Is the question "what goes in the blank?"? Also, it would be nice to see more detailed references to several physicists and Maxim's morphisms. $\endgroup$
    – LSpice
    Jul 2, 2019 at 23:33
  • 2
    $\begingroup$ good overview of where nonassociativity can be found in physics. arxiv.org/pdf/1903.05673.pdf $\endgroup$ Jul 4, 2019 at 14:49
  • 1
    $\begingroup$ @DamienC thanks for your efforts to improve the question $\endgroup$
    – YCor
    Feb 28, 2020 at 16:35
  • 2
    $\begingroup$ Dear Jim, it is a little bit unfortunate that what you describe as a good review of where nonassociativity can be found in physics is written in a very careless way. On page 4, it is claimed that the Jordan identity is equivalent to power-associativity, and the identity (1.12) is called the alternative property (whereas this is what is called the flexible identity). People who wish to use nonassociative algebras should really start with getting their definitions and statements right. $\endgroup$ Feb 28, 2020 at 19:13
  • 2
    $\begingroup$ @VladimirDotsenko: you are right that there is room for improvement, but this survey is actually a very nice one, despite a few inaccuracies. I like very much how the physics that is behing all this is introduced. $\endgroup$
    – DamienC
    Feb 28, 2020 at 23:13

2 Answers 2


This is probably not really an answer to this question, but there are two different context I know where deformation quantization produces something not exactly associative, but associative in a larger sense. This doesn't encompass all cases covered by the question, but might give some hints.

a) First, there are so-called quasi-Poisson $G$-manifolds for which the bivector doesn't satisfy the Jacobi identity. Here is the setup. A quasi-Poisson $G$-manifold ($G$ being a Lie group) is a $G$-manifold $M$ together with a bivector $\pi$ and an element $Z\in\wedge^3(\mathfrak{g})^G$ ($\mathfrak{g}$ being the Lie algebra of $G$) and such that $[\pi,\pi]=\vec{Z}$ (here $x\mapsto\vec{x}$ send an element of $\mathfrak{g}$ to the associated fundamental vector field of the action). In other words, the default of the Jacobi identity is determined by $Z$.

In this context one can chose an associator $\Phi=1+\hbar Z+\cdots\in(U(\mathfrak{g})^{\otimes3})^G[[\hbar]]$, and consider the monoidal category $\mathcal C_\hbar$ of $\mathfrak{g}[[\hbar]]$-modules whith non-trivial natural associativity isomorphism being given by $\Phi$. A quantization of $\pi$ is then a product on $C^\infty(M)[[\hbar]]$ deforming the usual product of fonctions as an associative monoid in $\mathcal C_\hbar$.

Since the forgetful functor $\mathcal C_\hbar\to Vect_\hbar$ is not monoidal, then the quantization doesn't seem associative. But secretely, it is.

b) The second situation I have in mind in which one gets non-associative deformation is probably more in the spirit of the what is alluded to in the question. It is the one of twisted Poisson manifolds, where the default of satisfying the Jacobi identity is rather controlled by a closed $3$-form $H$ on $M$ (sometimes refered to as a magnetic charge in more physics papers). Here is what happends:

  • locally, this $3$-form is exact, and thus the twisted Poisson structure is isomorphic to a genuine non-twisted one.

  • on intersections, one can prove that we do have compatibilities... which unfortunately don't glue well (i.e. they don't satisfy the expected cocycle identity).

  • it turns out that things do glue to higher order (getting a gerby version of a sheaf of Poisson algebras). The glueing data is determined by a Cech $3$-cocycle, which represents the same cohomology class in $H^3(M,\mathbb{R})$ as $H$.

These gadgets can be naturally deformed to so-called algebroid stacks, which are a generalization of sheaves of algebras, satisfying a higher cocycle condition (sometimes called tetrahedron equation). They can also be seen as linear versions of Giraud's gerbes. A lot of people worked on that in the algebaic and holomorphic contexts: Kontsevich, Yekutieli, Van den Bergh, Halbout and myself, Bressler--Gorokhovsky--Nest--Tsygan, Kashiwara--Schapira, etc... In these contexts these gadgets are necessary for proving existence theorems for deformation quantization of algebraic/holomorphic Poisson structures (indeed, glueing property are known to be harder to handle in these contexts than in the differentiable case).

This approach is less known in the differentiable context, though there are references (I guess it would be fruitful to revisit this paper in the light of all the huge work that has been done in the algebraic and holomorphic cases).

I'm probably biaised, but it seems to me that all the non-associative structures appearing in deformation quantization are secretely associative in a way: either associativity is satisfied up to homotopy (like in the case of deformation quantization with branes, after Cattaneo--Felder), or it is satisfied in an appropriate monoidal category (like for quasi-Poisson manifolds), or it is satisfied locally but glue in a non-trivial way (like in the twisted Poisson situation), etc...

I don't pretend at all that this is all understood. But I would be tempted to say that a good strategy could be to search for hidden associativity whenever one encounters non-associative products of observables in deformation quantization problems.

  • $\begingroup$ Damien, I also am so tempted. I'm especially interested in ``gluing up to homotopy', $\endgroup$ Feb 29, 2020 at 15:14

I figured out that in full generality this problem has no chance of leading to a different algebraic structure for which the given one is a quasi-classical limit (like it is for associative/Poisson): see https://arxiv.org/abs/2306.08351 for a proof of that.


Your Answer

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