I'm a physicist studying higher-spin theory. In my research, we work with fully symmetric tensors using a notation which is implicit both in the dimension of space and the order of the tensor, i.e. an order-$s$ fully symmetric tensor is simply written as $$ \phi_{\mu_1 \cdots \mu_s} \equiv \phi. $$ The $n$-th gradient of $\phi$ is written as $\partial^n \phi$, the $n$-th divergence, as $\partial^n \cdot \phi$ and the $n$-th trace as $\phi^{[n]}$. Lower traces are simply written with a prime, e.g. $\phi''$ for the second trace. All indices are implicitly symmetrized, without weight factors, using the minimal number of terms.
For example, if $s=2$, $$ \partial^2 \phi \equiv \partial_\mu \partial_\nu \phi_{\sigma \rho} + \partial_\mu \partial_\sigma \phi_{\nu \rho} + \partial_\mu \partial_\rho \phi_{\sigma \nu} + \partial_\nu \partial_\sigma \phi_{\mu \rho} + \partial_\nu \partial_\rho \phi_{\sigma \mu} + \partial_\sigma \partial_\rho \phi_{\mu \nu} $$ $$ \partial (\partial \cdot \phi) \equiv \partial_\nu (\partial^\lambda \phi_{\mu \lambda}) + \partial_\mu (\partial^\lambda \phi_{\lambda \nu})$$ $$ \eta \partial^2 \cdot \phi \equiv \eta_{\mu\nu} \partial^\rho \partial^\sigma \phi_{\rho \sigma} $$ The formalism implies the following set of rules $$ ( \partial^p \phi )' = \Box \partial^{p-2} \phi + 2 \partial^{p-1} \left( \partial \cdot \phi \right) + \partial^p \phi' $$ $$ \partial \cdot (\partial^p \phi) = \Box \partial^{p-1} \phi + \partial^p \left( \partial \cdot \phi \right) $$ $$ \partial^p \partial^q = {{p+q}\choose{q}} \partial^{p+q} $$ which you can simply take as axioms of the game. By the way, $\Box = \partial \cdot \partial$, which we are allowed to invert to $\frac{1}{\Box}$, but don't worry about what this means, you can just use the three rules written above.


So, the thing I'm interested in is the recursion relation $$\mathcal{F}_{n+1} = \mathcal{F}_{n} - \frac{1}{n+1} \frac{\partial}{\Box} \left( \partial \cdot \mathcal{F}_{n} \right) + \frac{1}{(n+1)(2n+1)} \frac{\partial^2}{\Box} \mathcal{F}_{n}{}'$$ with the initial condition $$ \mathcal{F}_0 = \Box \phi.$$

I would like solve the recursion fully, i.e. to express $\mathcal{F}_{n+1}$ only in terms of $\phi$'s and not $\mathcal{F}_{k}'s$ with $k<n+1$.

It's easy to rewrite $\mathcal{F}_{n+1}$ as $$\mathcal{F}_{n+1} = \mathcal{F}_{n} - \frac{1}{n(2n+1)} \frac{\partial^2}{\Box} \mathcal{F}_{n}' + \frac{1+2n}{n} \frac{\partial^{2n+2}}{\Box^n} \phi^{[n+1]}$$ but I don't know how to proceed from there and I've been trying for days. I could really use some help.

Please do tell me if you want me to clarify anything.


By lower traces I simply mean a number of traces low enough to actually write out the primes explicitly, so, for example, $\phi''''$ could just be written as $\phi^{[4]}$.


I provide here the explicit form of first two $\mathcal{F}_n$ after $n=0$.

$$ \mathcal{F}_1 = \Box \phi - \partial (\partial \cdot \phi) + \partial^2 \phi'$$

$$ \mathcal{F}_2 = \Box \phi - \partial (\partial \cdot \phi) + \frac{2}{3 \Box} \partial^2 (\partial^2 \cdot \phi) + \frac{1}{3} \partial^2 \phi' - \frac{1}{\Box} \partial^3 (\partial \cdot \phi') + \frac{1}{\Box} \partial^4 \phi''$$

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    $\begingroup$ Just to check if I understand what you mean, is it true that $\phi^{[2]} = \phi''$? $\endgroup$ – Jules Lamers Oct 25 '17 at 10:08
  • $\begingroup$ Yes, that is correct. $\endgroup$ – Gwynbleidd Oct 25 '17 at 11:47
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    $\begingroup$ P.S. Perhaps it's worth mentioning that, physically, $\mathcal{F}_1$ is the Maxwell tensor if $s=1$ and it's the linearized Riemann tensor if $s=2$. Or it's simply the Fronsdal tensor if you happen to be familiar with higher-spin theory. $\endgroup$ – Gwynbleidd Oct 25 '17 at 16:33
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    $\begingroup$ If you define $\mathcal{G}_n:= \Box^{n-1} \mathcal{F}_n$, then by induction every term in the expression for $\mathcal{G}_n$ have the schematic form of $\partial^{2n} \phi$ with $n$ contractions/traces. By your commutation relationships you can always move the traces as far "right" as possible, so you have a total of $n(n+1)/2$ different types of terms. From this you get a recursive rule on their coefficients. I guess how this simplify is a matter of combinatorics. $\endgroup$ – Willie Wong Oct 25 '17 at 20:58
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    $\begingroup$ @IvanV. In that section I don't think there's anything to do with AdS, that comes later in the notes $\endgroup$ – Jules Lamers Oct 27 '17 at 7:14

To make my comment more concrete: if you define $$\mathcal{G}_0 = \phi $$ and $$ \mathcal{G}_{n+1} = \Box \mathcal{G}_n - \frac{1}{n+1} \partial (\partial\cdot \mathcal{G}_n) + \frac{1}{(n+1)(2n+1)} \partial^2 \mathcal{G}_n' $$ then $\mathcal{G}_n = \Box^{n-1} \mathcal{F}_n$.

The terms of $\mathcal{G}_n$ can always be written (by induction) as a sum

$$ \mathcal{G}_n = \sum_{i+j+k = n} \alpha^{(n)}_{(i,j,k)} \Box^i \partial^{j+2k} (\partial^j \cdot \phi^{[k]}) $$

by virtue of your commutation relations $$ \partial\cdot (\partial\phi) - \partial (\partial\cdot \phi) = \Box \phi $$ and $$ (\partial\phi)' - \partial (\phi') = \partial \cdot \phi $$ (and the ones that you didn't list: $$ (\partial\cdot\phi)' - \partial\cdot (\phi') = 0 $$ and $\Box$ commutes with everything.)

In principle this would allow you to write down a recurrence relation relating the coefficients $\alpha^{(n+1)}_{(i,j,k)}$ in terms of $\alpha^{(n)}_{(i,j,k)}$, and you are trying to solve for the case $\alpha^{(0)}_{(0.0.0)} = 1$. I would suggest just first writing down the recurrent relation for the $\alpha$s and show the relation to experts in combinatorics and see whether they look familiar.

(Since you are posting on MO anyway, maybe a first step is follow what I described and do the grunt work and copy down the recurrence relations for $\alpha$ into your question statement.)

  • $\begingroup$ Great, this could turn out to be useful, I'll see what I can do with it! $\endgroup$ – Gwynbleidd Oct 25 '17 at 22:02

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