This is my attempt to prove something (I'm not even sure if it's true to begin with) using somewhat loose arguments. I present here all the steps and ideas and I would be extremely grateful if someone could confirm or refute it, explaining why I'm wrong (if I'm wrong) or explaining why I'm right (if I'm right for the wrong reasons or if I don't seem to be aware of some important subtleties). Thank you!

Assumptions and definitions

  • let $\mathcal{P}$ be a $2n$-dimensional phase space, i.e. a symplectic manifold with a linear symplectic form given by a $2n \times 2n$ matrix$$\omega = \begin{bmatrix} 0 & I_n \\ -I_n & 0 \end{bmatrix}$$
  • functions on $\mathcal{P}$ are equipped with the Moyal $\star$-product, explicitly given by $$f(x,p)\star g(x,p)=f(x,p)\exp \left[ \frac{i}{2}\left( \overleftarrow{\partial_x} \cdot \overrightarrow{\partial_p} - \overrightarrow{\partial_x} \cdot \overleftarrow{\partial_p}\right) \right] g(x,p)$$
  • the Moyal $\star$-product induces a Lie algebra through the Moyal bracket$$[f(x,p) \overset{\star}{,} g(x,p) ] = f(x,p) \star g(x,p) - g(x,p) \star f(x,p)$$
  • In my calculations, there appears an object $\Omega_{ab}$ closely resembling the Maurer-Cartan form $$\Omega_{ab} = \partial_af_b - \partial_b f_a + i [f_a \overset{\star}{,} f_b]$$ which becomes more clear soon.
  • we define framelike functions as functions of the form $f_a(x,p)$ where $a$ is a spacetime (pseudo-Euclidean) index such that, using Einstein summation convention $$f_a(x,p) \star g^a(x,p) = \eta^{ab} f_a(x,p) \star g_b(x,p)$$ where $\eta=\mathrm{diag}(-1,\underbrace{+1,\dots,+1}_{n-1})$


$$\Omega_{ab} = \partial_af_b - \partial_b f_a + i [f_a \overset{\star}{,} f_b] = 0 \Longleftrightarrow \exists \, \varepsilon(x,p) \, \text{s.t.} \, f_a=\partial_a \varepsilon$$

($f_a \equiv f_a(x,p)$, for notational clarity)


  • Since Moyal product is isomorphic to the matrix product in $M_\infty$, we can write $f \star g$ as the operator product $\hat{f} \hat{g}$. Equivalently, we can view $f$ and $g$ as functions of non-commutative coordinates $\hat{x}$ and $\hat{p}$ represented as operators $\hat{x} = x + \frac{i}{2}\partial_p$ and $\hat{p} = p - \frac{i}{2}\partial_x$. In symbols, $$f(x,p) \star g(x,p) = \hat{f}(x,p) \hat{g}(x,p) = f(\hat{x},\hat{p}) g(\hat{x},\hat{p}) = $$ $$= f\left(x + \frac{i}{2}\partial_p, p - \frac{i}{2}\partial_x\right) g\left(x + \frac{i}{2}\partial_p, p - \frac{i}{2}\partial_x\right)$$
  • In any case, we can write framelike functions $f_a(x,p)$ as Lie algebra-valued $1$-forms $$\mathbf{f}=\hat{f}_a \mathbf{dx}^a \, ,$$ which allows us to write $\Omega_{a,b}$ as a 2-form $$\mathbf{\Omega} = \mathbf{d} \mathbf{f} + \frac{i}{2} \mathbf{f} \wedge \mathbf{f}.$$ This looks like the Maurer-Cartan form, except for the $i$ multiplying the second term.
  • The main statement becomes $$ \mathbf{\Omega} = 0 \Longleftrightarrow \mathbf{f} \text{ is exact}$$

$\Longleftarrow$ proof

$$\mathbf{f} \, \text{exact} \Longleftrightarrow \exists \varepsilon \, : \, \mathbf{f} = \mathbf{d} \varepsilon$$ $$\implies \mathbf{d} \mathbf{f} = \mathbf{d}^2 \varepsilon = 0 $$ Now, $$0=\mathbf{d}^2 ( \varepsilon \wedge \varepsilon ) = \mathbf{d}^2 \varepsilon \wedge \varepsilon + \varepsilon \wedge \mathbf{d}^2 \varepsilon + 2 \mathbf{d} \varepsilon \wedge \mathbf{d} \varepsilon = 2 \mathbf{d} \varepsilon \wedge \mathbf{d} \varepsilon $$ $$\implies \mathbf{d} \varepsilon \wedge \mathbf{d} \varepsilon = 0$$ $$\implies \mathbf{\Omega}=0 \tag*{$\blacksquare$} $$

$\Longrightarrow$ proof ?

Does this implication even hold? Why? How would one prove it?

What does $\mathbf{d} \mathbf{f} = - \frac{i}{2} \mathbf{f} \wedge \mathbf{f}$ even mean for $\mathbf{f}$?

  • $\begingroup$ i) Are framelike functions possibly something else than 1-forms? ii) Is the product of framelike functions supposed to involve the Moyal product? iii) Could you specify what is the definition of the bracket $[ - \overset{\star}{,} - ]$? iv) Could you recall the definition of $\wedge$ between operator-valued (OV) forms? v) Why does $d\epsilon \wedge d\epsilon = 0$ for OV forms? vi) Why does $df \wedge f = f \wedge df$ for OV forms? vii) Why does $f \wedge df = 0$ implies $df=0$? (This is false for the real-valued 1-form $f = xdy - ydx$, hence for some 'diagonal' OV forms too.) $\endgroup$ – Jordan Payette Aug 3 '18 at 14:51
  • $\begingroup$ @JordanPayette Reasonable and helpful questions, thank you! I'll append the answers to my post as an edit. $\endgroup$ – PhysSE is Cancer Aug 3 '18 at 16:24
  • 1
    $\begingroup$ Just to give possible clues: for the Lie algebra $\mathfrak{g}$ of a Lie group $G$, a $\mathfrak{g}$-valued 1-form $\omega$ (defined over some manifold $M$) satisfies the Maurer-Cartan equation $d \omega + \frac{1}{2}[ \omega \overset{\wedge}{,} \omega ] = 0$ if it is '$G$-exact', that is if it is the Darboux derivative of a $G$-valued function defined over $M$ c.f. en.wikipedia.org/wiki/Darboux_derivative . In particular for $G = Gl(n, \mathbb{R})$ and $\mathfrak{g} = M(n, \mathbb{R})$, the MC eqn can be written $d\omega + \omega \wedge \omega = 0$, which ressembles your context. $\endgroup$ – Jordan Payette Aug 7 '18 at 12:26
  • $\begingroup$ @JordanPayette Hmm, I just arrived at similar conclusions myself, I'm glad to see that we're on the same page. I updated the original question to reflect what I've figured out so far, I think it should be much more clear now, do take a look :) $\endgroup$ – PhysSE is Cancer Aug 8 '18 at 20:21

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