In bosonic string theory the massless states of the closed string are given by a rank 2 tensor, which is divided into its three irreducible spherical tensors: symmetric traceless, antisymmetric and scalar (trace).
The symmetric traceless tensor of rank 2 represents a spin 2 particle which is identified with the graviton. The scalar one has, of course, spin 0 and is called dilaton. The antisymmetric tensor is identified with the so called Kalb-Ramond field, a generalisation of the Maxwell field for strings, but what is its spin?
I have read that this decomposition is somehow related to the decomposition of spin representations:
$$ \textbf{1}\otimes \textbf{1}=\textbf{2}\oplus \textbf{1}\oplus \textbf{0} $$
so its spin would be 1, but I'm not sure. I've tried to check how the antisymmetric tensor transforms under the little group rotations but I'm not sure either.
Best Answer
Here, a spin should be understood as a quantum number you get after doing dimensional reduction of a higher dimensional theory to a $(1+3)$-dimensional one. If you admit this assumption, you find the spin of Kalb-Ramond field is $1$. I think most people have this in mind, when they say a spin in higher dimensions.
In general, you get more quantum numbers than you get in $(1+3)$-dimensions to represent how particles are transformed under rotations. So strictly speaking, it does not mean anything to say a single spin quantum number in higher dimensions.
For the representation, I think more correct expression is $$ (2,2)\otimes(2,2)=\underbrace{(3,3)_\text{S}}_{\text{graviton}}\oplus\underbrace{(3,1)_\text{A}\oplus(1,3)_\text{A}}_{\text{gauge boson}}\oplus\underbrace{(1,1)_\text{S}}_\text{scalar}\,. $$ Here the numbers inside parentheses mean dimensions of representations. Subscripts S and A represent symmetric and antisymmetric, respectively.
If you want to know more details about connecting a representation $(3,1)_\text{A}\oplus(1,3)_\text{A}$ to gauge boson, I recommend you to read a chapter 34 of Srednicki, http://web.physics.ucsb.edu/~mark/qft.html or https://amzn.com/0521864496.
As a response to the comment by Xavier:
Denote $10$-dimensional indices by $M,N=0,1,\dotsc,9$, $4$-dimensional ones by $\mu,\nu=0,1,2,3$ and internal $6$-dimensional ones by $i,j,=4,\dotsc,9$. Then the Kalb-Ramond field $B_{MN}$ could be decomposed as \begin{equation} B_{MN}=\begin{cases} B_{\mu\nu} &: \text{1 two-form field,} \\ B_{\mu i} &: \text{$6$ vectors,} \\ B_{ij} &: \text{$15$ scalars.} \end{cases} \end{equation} The next step is to dualize the two-form field to scalar fields through Hodge dual so that we could regard degrees of freedom of two-form field as degrees of freedom of $6$ scalar fields. This can be confirmed easily.
As a final result, you get $6$ vectors and $21$ scalars in $4$-dimension. So it contains spin $1$ and spin $0$ fields in this multiplet. I think people simply call the spin of Kalb-Ramond field as the top spin of this multiplet. And same to other higher dimensional fields.