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I am reading the book "String theory demystified" by David McMahon.

On page 149, the author discusses the "critical dimension" for superstrings.

the number of spacetime dimensions is easily extracted. One obtains a relation for the Lorentz generators $M^{-i}$: $$ \left[M^{-i}, M^{-j}\right]=-\frac{1}{\left(p^{+}\right)^{2}} \sum_{n=1}^{\infty}\left(\alpha_{-n}^{i} \alpha_{n}^{j}-\alpha_{-n}^{j} \alpha_{n}^{i}\right)\left(\Delta_{n}-n\right)\tag{7.70}$$ where $$\Delta_{n}=n\left(\frac{D-2}{8}\right)+\frac{1}{n}\left(2 a_{N S}-\frac{D-2}{8}\right).\tag{7.71}$$ In order to maintain Lorentz invariance, we must have $\left[M^{-i}, M^{-j}\right]=0$. This can only be true if the first term on the right-hand side of Eq. (7.71) is $n$ and the second term vanishes. This implies that$$\begin{aligned} &\frac{D-2}{8}=1 \\ &\Rightarrow D=10 \end{aligned}\tag{7.72}$$

My questions

  1. How to obtain the first equation? which seems like a kind of Lie algebra relation.

  2. How does the requirement of Lorentz invariance translates to the RHS being zero?

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  • $\begingroup$ $ P_{\pm}=\frac{1}{2}\left(1 \pm \Gamma_{11}\right) $ and $ \Gamma^{11}=\Gamma^{0} \Gamma^{1} \cdots \Gamma^{9} $ is a 10-dimensional chirality operator. It acts on spinors $\psi$ according to $ \Gamma^{11} \psi=\pm \psi $ That is, states have positive or negative chirality. Weyl spinors are states with definite chirality, and states can be projected into spinors with opposite space-time chirality using the operator $\endgroup$
    – Eden Zane
    Commented Jun 4, 2022 at 7:41
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    $\begingroup$ Hi @Eden Zane. Do you know how to do the corresponding calculation of $\left[M^{-i}, M^{-j}\right]$ in bosonic string theory? $\endgroup$
    – Qmechanic ♦
    Commented Jun 4, 2022 at 8:19
  • $\begingroup$ @Qmechanic In bosonic string theory, the left hand side is just $m^2$ I think. Does $[M^{-i},M^{-j}]$ reduce to $m^2$ in the bosonic case? $\endgroup$
    – Eden Zane
    Commented Jun 4, 2022 at 9:05
  • $\begingroup$ I figured it out myself. It was all due to a confusion in the notation. I found the same equation( and the same discussion too) in the "superstring theory" by Greens Schwarz and Witten. Refer p.213 $\endgroup$
    – Eden Zane
    Commented Jun 5, 2022 at 4:05
  • $\begingroup$ Should I delete this question or, let it remain here for those who come up with the same confusion later? $\endgroup$
    – Eden Zane
    Commented Jun 5, 2022 at 4:10

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