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Ok so this might be a very stupid and trivial question but I have spent a couple of hours on this little problem.

I am trying to derive a simple formula in a paper. We have a real commuting spinorial variable $y_\alpha$ ($\alpha = 1,2$) ($y^\alpha = \epsilon^{\alpha\beta}y_\beta$, $\epsilon^{\alpha\beta}=-\epsilon^{\beta\alpha}$, $\epsilon^{12}=\epsilon_{12}=1$).

The linear space of functions of the variables $y_\alpha$ is endowed with the structure of the algebra with the following $*$ product law:

$$(f*g)(y) = \int_{\mathbb{R}^4} \! d^2u~d^2v~ \exp \left[i u_{\alpha} \epsilon^{\alpha\beta} v_{\beta}\right] f(y+u)g(y+v).$$

I am trying to compute the $*$ commutator

$$[y_\alpha,y_\beta]_{*} := y_\alpha * y_\beta - y_\beta * y_\alpha = 2i\epsilon_{\alpha\beta},$$

but when I try to use the formula for lets say computing $y_1*y_2$, I get a divergent integral

$$y_1*y_2 = \int d^2u\,d^2v \,\,\, (y_1+u_1)(y_2+v_2)\exp[i (u_1v_2-u_2v_1)].$$

I feel like I am applying the formula wrong but I can't understand where. Can someone help me?

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  • $\begingroup$ Btw which paper? $\endgroup$ – Qmechanic Sep 6 '13 at 20:43
  • $\begingroup$ Right now I am studying "Unfolded representation for relativistic equations in 2+1 anti-de-sitter space" by M.A. Vasiliev, Class. Quant. Grav. 11, 649, 1994. But this is really a starting point for any higher spin theory paper so I feel it must be something very trivial I am failing to understand. $\endgroup$ – Bilentor Sep 6 '13 at 20:45
  • $\begingroup$ The version of the paper I looked at doesn't have an imaginary unit in the exponential in the definition of the product law. Is this an error? $\endgroup$ – Brian Moths Sep 6 '13 at 21:37
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I worked a lot with this kind of formulas :) It would be useful to recover the imaginary unit in $\exp$ and think of integration variables as taking real values. Then, upon appropriate normalization $$ \int\int \exp (i u_\alpha v^\alpha) d^2u d^2v=\int \delta^2(u) d^2u=1$$ Once you agree with this formula, which is a standard representation for the delta function (that the variables have some spinorial meaning is irrelevant, they are usual commuting variables) the rest is easy.

For example, let us compute $y_\alpha \star g(y)$ where $g(y)$ is any function $$\int (y+u)_\alpha\, g(y+v) \exp (iu_\alpha v^\alpha)=\\\int y_\alpha\, g(y+v) \exp (iu_\alpha v^\alpha) +\int u_\alpha\, g(y+v) \exp (iu_\alpha v^\alpha)=A+B$$ A is easy, one can integrate over $u$ to get delta function $$A=\int \delta^2(v) y_\alpha g(y+v) d^2 v=y_\alpha g(y)$$ To compute B we integrate by parts $$ B=\int g(y+v)(-i\frac{\partial}{\partial v^\alpha}\exp (iu_\alpha v^\alpha))=\\ =\int (i\frac{\partial}{\partial v^\alpha}g(y+v))\exp (iu_\alpha v^\alpha)=\\ =i\frac{\partial}{\partial y^\alpha}\int g(y+v)\exp (iu_\alpha v^\alpha)=\\ =i\frac{\partial}{\partial y^\alpha} g(y)$$ Alltogether we derived, $$y_\alpha \star g(y)=(y_\alpha+i \partial_\alpha)g(y)$$ which for example gives $$y_\alpha \star y_\beta=y_\alpha y_\beta+i \partial_\alpha y_\beta=y_\alpha y_\beta+i\epsilon_{\alpha\beta}$$ The integral formula for star product is a very useful one if you are to compute star products of something more complicated than polynomials. This is what happens with Vasiliev theory since the functions encode higher-spin fields together with all derivatives thereof, so only the anti-de Sitter solution (vacuum) is of type $yy$ while perturbations are analytic functions of $y$. For example, the bulk-to-boundary propagator is somethins like $\exp(yy+y)$.

That the integral formula is more useful can be seen in trying to compute the star product of two gaussians $\exp(yy)\star \exp(yy)$, which is just a gaussian integral with the help of the integral formula and I have no idea how to compute this using $\exp(\partial\partial)$ formula

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  • $\begingroup$ Thank you. That makes a lot of sense. I am attending a school where Vasiliev gave some lectures on HS theories and he did mention that integral formula is more useful because the differential version doesn't make sense for non-polynomial functions such as distributions like $\delta(y)$ and these operators come up as Klein operators. $\endgroup$ – Bilentor Sep 8 '13 at 12:07
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The integral formula for the Moyal/Groenewold $*$ product in Ref. 1 reads

$$\tag{1} (f * g)(y)~=~\int_{\mathbb{R}^4} \! d^2u~d^2v~ \exp \left[i u_{\alpha} \epsilon^{\alpha\beta} v_{\beta}\right] f(y+u)g(y+v).$$

Remark: The argument $u_{\alpha} \epsilon^{\alpha\beta} v_{\beta}$ inside the exponential of the integral formula (1) has an interesting geometric interpretation as (twice) a signed area of a triangle in phase space $\mathbb{R}^2$, cf. Ref. 2.

Usually the Moyal/Groenewold $*$ product is defined as

$$\tag{2} (f * g)(y)~:=~ f(y) \exp \left[\stackrel{\leftarrow}{\frac{\partial}{\partial y_{\alpha}}} i\epsilon_{\alpha\beta} \stackrel{\rightarrow}{\frac{\partial}{\partial y_{\beta}}} \right] g(y). $$

Let us for completeness prove$^1$ the integral formula (1) from the definition (2) for a pair of sufficiently well-behaved functions $f$ and $g$:

$$(f * g)(y)~\stackrel{(2)}{:=}~ f(y) \exp \left[\stackrel{\leftarrow}{\frac{\partial}{\partial y_{\alpha}}} i\epsilon_{\alpha\beta} \stackrel{\rightarrow}{\frac{\partial}{\partial y_{\beta}}} \right] g(y) $$ $$ ~=~\int_{\mathbb{R}^4} \! d^2u~d^2v ~\delta^2(u)~\delta^2(v) \exp \left[\frac{\partial}{\partial u_{\alpha}} i\epsilon_{\alpha\beta} \frac{\partial}{\partial v_{\beta}}\right] f(y+u)g(y+v)$$ $$ ~=~\int_{\mathbb{R}^8} \! d^2u~d^2v~\frac{d^2p}{(2\pi)^2}\frac{d^2q}{(2\pi)^2} \exp i\left[p\cdot u+q\cdot v+\frac{\partial}{\partial u_{\alpha}} \epsilon_{\alpha\beta} \frac{\partial}{\partial v_{\beta}}\right] f(y+u)g(y+v)$$ $$ ~\stackrel{\text{int. by part}}{=}~\int_{\mathbb{R}^8} \! d^2u~d^2v~\frac{d^2p}{(2\pi)^2} \frac{d^2q}{(2\pi)^2} \exp i\left[p\cdot u+q\cdot v-p^{\alpha} \epsilon_{\alpha\beta} q^{\beta}\right] f(y+u)g(y+v)$$ $$ ~=~\int_{\mathbb{R}^6} \! d^2u~d^2v~\frac{d^2q}{(2\pi)^2} ~\delta^2(u_{\alpha}-\epsilon_{\alpha\beta} q^{\beta}) ~e^{i q\cdot v} f(y+u)g(y+v)$$ $$ ~\stackrel{(6)}{=}~\int_{\mathbb{R}^6} \! d^2u~d^2v~\frac{d^2q}{(2\pi)^2} ~\delta^2( q^{\beta}-u_{\alpha}\epsilon^{\alpha\beta} ) ~e^{i q\cdot v} f(y+u)g(y+v)$$ $$\tag{3} ~=~\int_{\mathbb{R}^4} \! d^2u~d^2v~ \exp \left[i u_{\alpha} \epsilon^{\alpha\beta} v_{\beta}\right] f(y+u)g(y+v).$$

The functions $f$ and $g$ should be sufficiently well-behaved in order for the above integrals and manipulations (3) to make mathematical sense. Let us here give a necessary and sufficient condition for the integrand

$$\tag{4} h(u,v) ~:=~\exp \left[i u_{\alpha} \epsilon^{\alpha\beta} v_{\beta}\right] f(y+u)g(y+v), \qquad u,v,y ~\in ~\mathbb{R}^2, $$

of the integral formula (1) to be integrable. In general an integrand $h$ is integrable (i.e belongs to $L^1$) if and only if (i) the integrand $h$ is Lebesgue measurable, and (ii) the absolute value $|h|$ of the integrand has a finite integral $\int |h| dm<\infty$. Note that $|h(u,v)|=|f(y+u)|~|g(y+v)|$ factorizes in a $u$- and a $v$-dependent factor. Assuming that both functions $f$, $g$ (and therefore $h$) are Lebesgue measurable, this means (via Tonelli's and Fubini's theorems) that the integrand (4) is integrable $h\in L^1(\mathbb{R}^4)$ if and only if (i) both $f,g\in L^1(\mathbb{R}^2)$ are integrable, or (ii) at least one of the two functions $f$ and $g$ vanishes almost everywhere.

For instance, inserting the two non-integrable first-order polynomials $f(y)=y_{\alpha}$ and $g(y)=y_{\beta}$ inside the integral formula (1) is ill-defined as OP also mentions.

On the other hand, the definition (2) does not involve integrals. So we can put $f(y)=y_{\alpha}$ and $g(y)=y_{\beta}$ in eq. (2) to derive the sought-for formula

$$\tag{5} y_{\alpha} * y_{\beta}~=~y_{\alpha} y_{\beta}+ i\epsilon_{\alpha\beta}. $$

References:

  1. M.A. Vasiliev, Unfolded representation for relativistic equations in 2 + 1 anti-de Sitter space, Class. Quant. Grav. 11, (1994) 649. Note that there is an imaginary unit $i$ missing in the published version of formula (1). A preprint version from KEK Preprint Library has the $i$ factor so it seems that the $i$ was lost during the publishing phase.

  2. C. Zachos, Geometrical Evaluation of Star Products, J. Math. Phys. 41 (2000) 5129, arXiv:hep-th/9912238.

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$^1$ Note that Ref. 1 uses the convention

$$\tag{6} \epsilon_{\alpha\beta} \epsilon^{\beta\gamma}~=~-\delta_{\alpha}^{\gamma}. $$

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  • $\begingroup$ Thanks a lot for the detailed and explicit derivation. I had already used that formula to get the answer but I was wondering if I was using the integral formula wrong as I wasn't getting the answer. It just seemed odd to me that the paper will have the integral version and right below it the star bracket without any mention of how it should be used or not. $\endgroup$ – Bilentor Sep 6 '13 at 22:10

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