The star product of two smooth functions $f,g$ on $\mathbb R^{2n}$ can be defined as $$ f\star g = \exp\left(-\omega^{ij} \frac{\partial}{\partial y^i} \frac{\partial}{\partial z^j}\right) f(y)g(z) \vert_{z=y}. $$

where $\omega^{ij}$ are the components of a symplectic form. There is a similar formula for the Clifford product (where instead of derivatives there are interior products) when translated to the exterior algebra.

In both cases the product is associative and I've seen many references say that it is easy to check that the Moyal product is associative (directly, without appealing to a symbol map).

Unfortunately I do not see an easy way to check associativity. I was also wondering if this sort of product is a standard thing, considering that I've seen it in two contexts (though I guess they are very much related as the Clifford algebra is a deformation of the exterior algebra and the Moyal product gives a deformation quantization of $\mathbb R^{2n}$).


  • $\begingroup$ For a graphic geometrical "behold!" proof, see this. $\endgroup$ – Cosmas Zachos Jun 14 '18 at 14:05

$\def\dd#1{\tfrac{\partial}{\partial #1}}$Observe first that $$\dd{x}\Big(f(x,y)|_{x=y}\Big) = \Bigg(\Big(\dd x+\dd y\Big)f(x,y)\Bigg)\Bigg|_{x=y}$$

Let us write $E(\dd{y}, \dd{z})=\exp(-\omega^{i,j}\dd{y^i}\dd{z^j})$, so that $$ (f\star g)(x)=\Big(E(\dd{x},\dd{y})\cdot\big(f(x)g(y)\big)\Big)\Big|_{x=y}$$ and consequently \begin{align} ((f\star g)\star h)(x)&=\Bigg[E(\dd x,\dd z)\cdot\Bigg(\Big(E(\dd{x},\dd{y})\cdot\big(f(x)g(y)\big)\Big)\Big|_{x=y} h(z)\Bigg)\Bigg]\Bigg|_{x=z}\\ &=\Bigg[E(\dd x+\dd y,\dd z)E(\dd x,\dd y)\cdot f(x)g(y)h(z)\Bigg]\Bigg|_{x=y=z} \\ &=\Bigg[E(\dd x,\dd z)E(\dd y,\dd z)E(\dd x,\dd y)\cdot f(x)g(y)h(z)\Bigg]\Bigg|_{x=y=z}\end{align}

Computing the product in the other order, we get a similar expression, which only differs in the differential operator involved: it is a product of the same three factors in a different order. Since these factors commute, this is not a problem.

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    $\begingroup$ Ah, thanks a lot for the answer, Mariano! That crucial step in the last line of using the properties of $\exp$ is what I kept missing. $\endgroup$ – Eric O. Korman Oct 1 '11 at 2:59
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    $\begingroup$ Am I right to say that it is crucial for this proof that $\omega^{ij}$ be constants? Otherwise, $\exp(-\omega^{i,j}\dd{y^i}\dd{z^j})$ would also involve derivatives of $\omega^{ij}$, a fact which seems to ruin this proof. I ask because the OP had requested a proof for arbitrary symplectic structures (in fact, he never uses the non-degeneracy of $\omega$, a Poisson structure would seem to do as well). $\endgroup$ – Alex M. Mar 26 '16 at 9:54
  • $\begingroup$ @Alex M, For arbitrary ω the associative product had to wait for Kontsevich to construct it, no? $\endgroup$ – Cosmas Zachos Jun 13 '18 at 22:06
  • $\begingroup$ @CosmasZachos: Indeed. For $\omega$ of constant rank, Fedosov had obtained the same result just a few years before Kontsevich. $\endgroup$ – Alex M. Jun 14 '18 at 6:46
  • $\begingroup$ @CosmasZachos: Oh, only now I think I see what you mean. Notice that the OP used the words "$\omega^{ij}$ are the components of a symplectic form": to me, this means that the OP is using the term "Moyal product" for what he intends to be a $\star$-product on some open subset of $\mathbb R^{2n}$ with respect to some arbitrary symplectic form $\omega$. Hence my comment. $\endgroup$ – Alex M. Jun 14 '18 at 8:24

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