# Question about NS ground state of superstring in the RNS formalism

We know that in the RNS formalism of the superstring, there is a unique ground state corresponding to the NS sector. This is defined by $\varphi_{r}^{\mu} | 0 \rangle_{NS} = 0$ for all $r > 0$. Here $r$ runs over the half-integers. The states with $r < 0$ then act as raising operators.

In all books it is claimed that, because there is a unique ground state in the NS sector, this state corresponds to a spin $0$ state in spacetime. By acting with raising operators, one then obtains spacetime bosons. I don't understand why the ground state should be spin zero? What is the reason for this?

I read somewhere that this can be checked by applying the Lorentz generators $$\Sigma^{\mu \nu} = - \frac{i}{2} \sum_{r \in \mathbb{Z} + \nu} [ \varphi_r^{\mu}, \varphi_{-r}^{\nu} ]$$ to the ground state. But I don't really understand this argument. Can someone elaborate please?

One can verify that for $\mu\neq \nu$, $$\Sigma^{\mu\nu}|0\rangle_{NS}=-\frac{i}{2}\sum_{r\in \mathbb{Z+\frac{1}{2}}}[\varphi_r^{\mu}, \varphi_{-r}^{\nu}]|0\rangle_{NS}=-i\sum_{r\in \mathbb{Z+\frac{1}{2}}}\varphi_r^{\mu}\varphi_{-r}^{\nu}|0\rangle_{NS}=i\sum_{r\geq \frac{1}{2}}\varphi_{-r}^{\nu}\varphi_r^{\mu}|0\rangle_{NS}-i\sum_{r\leq \frac{1}{2}}\varphi_r^{\mu}\varphi_{-r}^{\nu}|0\rangle_{NS}=0,$$ where in the second step one uses $\{\varphi_r^{\mu}, \varphi_{-s}^{\nu}\}=\eta^{\mu\nu}\delta_{r,s}$, and in the last step one uses the definition of $|0\rangle_{NS}$.
Since $\Sigma^{\mu\nu}$ is the spacetime Lorentz generator, it indicates that $|0\rangle_{NS}$ transforms trivially under Lorentz transformation and thus a scalar.