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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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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. – innisfree May 23 '13 at 18:11
@innisfree, the question is actually raised in the context of light-cone quantisation, so the indices are not really Lorentzian. – Peter Kravchuk May 23 '13 at 23:01
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. – Peter Kravchuk May 23 '13 at 23:11
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 $\frac{D(D - 3)}{2}$ – Trimok May 25 '13 at 11:06

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