# Derivation for the commutation relation of $\alpha_m^\mu$ and $\alpha_n^\nu$

I have been trying to derive the commutation relation of $$\alpha_m^\mu$$ and $$\alpha_n^\nu$$ in a closed-string mode expansion, but I found an extra factor of $$2$$ that ruins things out:

Given $$\dot X = \dot X^\mu_R (\tau-\sigma)+ \dot X^\mu_L(\tau+\sigma)$$ and only $$X_R$$ involves $$\alpha$$ while only $$X_L$$ involves $$\tilde \alpha$$, I will focus on $$X_R$$ here.

So $$\left[\dot X_R^\mu (\sigma, \tau),X_R^\nu(\sigma',\tau)\right] = \left(\frac i 2\right)^2l_s^2\sum_{n\neq 0}\sum_{m\neq0}\frac 1 n \frac 1 m (-2in) e^{-2in(\tau-\sigma)-2im(\tau-\sigma')}[\alpha_n^\mu,\alpha_m^\nu]$$ $$= \left(\frac i 2\right)l_s^2\sum_{n\neq 0}\sum_{m\neq0}\frac 1 m e^{-2i(m+n)\tau-2im\sigma'-2in\sigma}[\alpha_n^\mu,\alpha_m^\nu]$$ $$=i\alpha'\sum_{n\neq 0}\sum_{m\neq0}\frac 1 m e^{-2i(m+n)\tau-2im\sigma'-2in\sigma}[\alpha_n^\mu,\alpha_m^\nu]$$ where we have used the relation $$\frac {l_s^2} 2 = \alpha'$$.

Now we compare this to the relation $$\left[\dot X_R^\mu (\sigma, \tau),X_R^\nu(\sigma',\tau)\right]=\frac 1T\eta^{\mu\nu}\delta(\sigma-\sigma') = 2\pi\alpha'\eta^{\mu\nu}\frac 1\pi\sum_{n\in\mathbb{Z}}e^{-2in(\sigma-\sigma')} = 2\alpha'\eta^{\mu\nu}\sum_{n\in\mathbb{Z}}e^{-2in(\sigma-\sigma')}$$

This holds when $$\begin{split} i\alpha'\frac 1m[\alpha_n^\mu,\alpha_m^\nu] &= 2\alpha'\eta^{\mu\nu} \delta_{m+n,0}\\ [\alpha_n^\mu,\alpha_m^\nu]&=-2im\eta^{\mu\nu} \delta_{m+n,0} = 2in\eta^{\mu\nu} \delta_{m+n,0} \end{split}$$ with an additional factor of $$2$$. Which part am I wrong?