5
$\begingroup$

NOTATION

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.

PROBLEM

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.

EDIT 1

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 $\phi^{[4]}$.

EDIT 2

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

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

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

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
2
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.