What is the spin of the Kalb-Ramond field?

Question

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.

Answer 1

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.

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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.