I am going through Tong's lecture notes on String Theory and came across the following irrep decomposition (Chap 2, p.43) of the bosonic string first excited states:

$$\text{traceless symmetric} \oplus \text{anti-symmetric} \oplus \underbrace{\text{singlet}}_{=\text{trace}}$$

He then goes on and claims that the traceless symmetric tensor is the spin-2 graviton.

What is the reason behind that claim? Is there a relationship between degrees of freedom and the spin of a particle in any number of dimensions? I remember from the $SU(2)$ irrep decomposition that the $\ell=1$ irrep has 3 d.o.f. just like a massive spin-1 particle would have. But what about massless particles living in 26 dimensions?

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    $\begingroup$ Well, the graviton must couple to the totally symmetric rank-2 stress energy tensor, so it must be rank/spin-2, and an anti-symmetric part won't contribute when the Lorentz indexes are contracted with the symmetric stress energy tensor. $\endgroup$ – innisfree May 23 '13 at 18:11
  • $\begingroup$ @innisfree, the question is actually raised in the context of light-cone quantisation, so the indices are not really Lorentzian. $\endgroup$ – Peter Kravchuk May 23 '13 at 23:01
  • $\begingroup$ In fact, it is an interesting question. Naively, $SO(24)$ corresponds to $D_{12}$ simple Lie algebra, and should have a plethora of inequivalent representations, which cannot be labeled by a single number in an obvious way. Probably, all (corresponding to vector reps of the group, leave spinors alone) arise from tensoring the fundamental rep, hence the "spin-.." names. It would be wonderfull if someone could answer this properly. $\endgroup$ – Peter Kravchuk May 23 '13 at 23:11
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    $\begingroup$ In a $D$ dimensional space-time, a massless particle is a representation of $SO(D - 2)$. The graviton corresponds to the symmetric traceless representation of $SO(D - 2)$ which has dimension $\frac{D(D - 3)}{2}$ $\endgroup$ – Trimok May 25 '13 at 11:06

The spin $s$ of a particle characterizes how the rotation generators act on it. In $D$ dimensions, you represent the little group $SO(D-1)$ for massive particles and $SO(D-2)$ for massless ones. In fact, you really need to consider its universal cover $\textrm{Spin}(n)$ which happen to be just its double cover.

Now, you can define the spin to be the largest real $s$ such that $$ \mathrm{exp}\left(\frac{2i\pi}{s}J\right) = \rm{id} $$ where $J$ is any generator in your representation. This is basically the statement that, to return to identity, you need to do a $4\pi$ rotation for spin $\frac{1}{2}$, a $2\pi$ for spin 1, a $\pi$ for spin 2... Note that because the universal cover is the double cover, $s$ has to be a half-integer.

It is clear that Lorentz vectors have spin 1. Let's take a symmetric 2-tensor $T_{\mu\nu}$. It transforms as $T'_{\alpha\beta} = R^{\:\,\mu}_{\alpha}R^{\:\,\nu}_{\beta}T_{\mu\nu}$, where $R = \mathrm{exp}(i\theta J)$ are the usual $SO(n)$ matrices. Infinitesimally, \begin{align} \delta T_{\alpha\beta} & = i\theta(J^{\:\,\mu}_{\alpha}\delta^{\:\,\nu}_\beta + J^{\:\,\nu}_{\beta}\delta^{\:\,\mu}_\alpha)T_{\mu\nu} \\ & = 2 i\theta J^{\:\,\mu}_{\alpha}T_{\mu\beta} \\ \end{align} where the symmetry of the tensor is crucial. Now, because $\mathrm{exp}(2i\pi J) = 1$, you see that for $\theta = \pi$, you get back to identity. This is spin 2!

You can generalize this to show that traceless symmetric n-tensors are spin n. You need them to be traceless because you want irreducible representations. With this, it shouldn't be too hard to derive the degrees of freedom of a spin $s$ particle in $D$ dimensions e.g. for the graviton, it is $D(D-3)/2$.

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    $\begingroup$ Thanks for this answer. The field strength $F$ also transforms with $F'_{\alpha \beta}=R_\alpha^\mu R_\beta^\nu F_{\mu \nu}$, correct? So why does this train of logic not also mean that $F$ is spin-2? $\endgroup$ – Rococo Apr 18 '16 at 1:32
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    $\begingroup$ You are right for the transformation law. $F$ is not spin 2 because it is antisymmetric. The variation of the two indices don't add up as for symmetric tensors. Thus, $F$ has to be considered as a spin 1 field. $\endgroup$ – Vigod Apr 18 '16 at 7:15
  • $\begingroup$ Okay, so that lets you exchange the indices in $T$. Thank you! $\endgroup$ – Rococo Apr 18 '16 at 14:08

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