From the level matching condition of the physical state $$(L_0-\bar L_0)|\phi\rangle=0$$ and thus $$h-\bar h=0$$. However, $h-\bar h$ was also the quantity that defined the spin. This seemed to imply that there's no spin none zero states.But it seemed to be counter intuitive since the graviton was said to be the spin 2 particle.
My advisor mentioned that this was specific to the string theory and connected to the topic of the "gauge gravity", where in the "?reduced sector?" $h=\bar h\neq 0$ could happen and in the "?internal inseparate space?" $h\neq \bar h$ could happen.
On a second thought the physical state condition $$(L_0-a)|\phi\rangle$$ but since $a=1$ was solved for the $D=26$ does this mean the only highest weight representation was of $h=\bar h=1$?(This was for the string theory, not CFT.)
Why in the gauge gravity the spin state was zero?
Related Post: Does String Theory explain spin?
