# T-Duality for closed bosonic string: what is the difference between a graviphoton and a B-vector?

Consider the closed bosonic string in 26 dimensions, of which one physical dimension, say $X^{24}$ is compactified into a circle of radius $R$. The lowest lying massless states are the graviton, the antisymmetric $B_{\mu\nu}$ field, the dilaton $\phi$, and two vectors'', the so called graviphoton and B-vector. They are described by

$\alpha^{i}_{-1} \tilde{\alpha}^{24}_{-1}|p\rangle = 0$ and $\alpha^{24}_{-1}\tilde{\alpha}^{i}_{-1}|p\rangle$, with $p^2 = 0$.

What is the difference between these two states? I see that they are defined differently, but the only difference is in which (antiholomorphic) creation operator appears first.

Is it correct to say that the symmetrized version is the graviphoton and the antisymmetrized version is the B vector field?

• I do not think the definition is correct. The B vector should carry two anti-symmetric indices. But here you only have one. However, B field is indeed originated from the same level with gravitons, with latter symmetric and traceless. – Kevin Ye Feb 24 '15 at 2:59
• So are you saying that there's a fundamental field and by considering the anti symmetric part one gets the anti symmetric B field and from the symmetric part one gets the photon? – leastaction Mar 2 '15 at 22:45

The graviphoton comes from the components $g^{\mu 24}=g^{24 \mu}$ of the metric tensor and the B-vectors from the components $B^{\mu 24}=-B^{24 \mu}$ of the anti-symmetric tensor $B^{\mu\nu}$. Now to obtain the states is trivial. The metric tensor is related to the state $(\alpha_{-1}^\mu\tilde\alpha_{-1}^{\nu} +\alpha_{-1}^\nu\tilde\alpha_{-1}^{\mu})|0;k\rangle$ and the anti-symmetric tensor to the state $(\alpha_{-1}^\mu\tilde\alpha_{-1}^{\nu}-\alpha_{-1}^\nu\tilde\alpha_{-1}^{\mu})|0;k\rangle$. So you are right in saying that is the symmetric and anti-symmetric combinations.