# Deriving the critical dimension of bosonic string theory

I am going through the lecture note by Gleb Arutyunov on the derivation of critical dimension for bosonic string theory. I was able to reproduce all the results till the last step given on page 62.

For $[S^{i-},S^{j-}]$, the term that should vanish and give us the critical dimension is

$$\sum_{n=1}^{\infty}\left[\left(\frac{4\pi T}{(p^{+})^{2}} 2(n-1)-\frac{f(n)}{n^{2}}\right)\alpha_{n}^{[i}\alpha_{n}^{j]}\right]$$

where

$$f(n)=\frac{4\pi T}{(p^{+})^{2}}\left[\frac{n(n^{2}-1)}{12}(D-2)+2an\right]$$

I realize that I might be missing something obvious but at this moment I don't see how the above expression can vanish for all $n$ when $D=26$ and $a=1$. Thanks!

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As far as I see, it doesn't vanish, but $f \propto 2nn^2$, so all that remains is $-\sum \frac{4\pi T}{(p^+)^2} \alpha_n^{[i}\alpha_n^{j]}$. – jdm Jul 22 '13 at 15:58
The $-1$ term is gone in the next step ("we thus find..."), and then it's easy to see $D=26$. I'm not sure how he gets there exactly, because the PDF doesn't render correctly for me. But if you collect all the $\alpha_0^{-}$, you can probably get each of the three rows in the next step. Specifically, you can get $\alpha_{-n}^{[i}\alpha_n^{j]}$, which cancels the $-1$. I don't know where some of the $\alpha_0^{-}$ go, or the $p^i, p^j$ come from though. Maybe I'm missing a font, maybe the alphas have a funky commutator :-) – jdm Jul 22 '13 at 16:18
It doesn't vanish, I think you're looking at the wrong equation. You want the commutator of the {i,-} components to vanish for the Lorentz generators J, not the spin generators S. – David M Jul 22 '13 at 16:18
Thanks for all the comments. Just to set the record straight, the lecture note is correct. It turns out I overlooked something very simple. Just as jdm said, by collecting the $\alpha_{0}^{-}$ appropriately, the -1 term in the above expressions cancels and that give us the desired result :) – user27365 Jul 26 '13 at 0:08