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So I'm following Szabo's book "An Introduction to String Theory and D-brane Dynamics (2nd ed, 2011); still on the canonical treatment in chapter 3. After doing a mode expansion, we get (up to a constant) a set of nice ladder operators \begin{equation}[\alpha_m^\mu, \alpha_n^\nu] = m\delta_{m+n}\eta^{\mu\nu}\end{equation} where $\delta_{m+n}$ is unity if $m=-n$, zero otherwise; $\eta$ is the Minkowski metric in D dimensions, $m$ and $n$ are integer labels and the $\alpha$s are Fourier coefficients. I'm omitting the LM/RM sets and concentrating on the open string.

We use these to define these composite operators: \begin{equation} L_n = \frac{1}{2}\sum\limits_{m=-\infty}^\infty{\alpha_{n-m}\cdot\alpha_m}\end{equation} which we want to use for generate conformal transformations. Cool.

So we need some Witt-like algebra, and because we're quantising we're expecting some weirdness, and it turns out you get the Virasoro algebra. But I just can't get the central term. For a start, it doesn't look right: each pair of $\alpha$s can give me a factor of m, so having \begin{equation}[L_m,L_n]\end{equation}, we have four $\alpha$s and hence we'd expect to be able to extract some central term of order $n^2$. But the Virasoro central term is $n(n^2-1)$ - so where does the extra n come from? There are no conveniently-placed sums that might give rise to this. Every time I attempt to plug through the algebra I end up with something silly, for example $\sum\limits_{p=-\infty}^\infty p$. So no central term at all!

So the root of my question is: where does the extra n come from? If I can understand that, hopefully I'll be able to do the rest of the derivation myself (don't take away my fun:)).

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2 Answers 2

The derivation of the Virasoro algebra is obviously a very important calculation in ST/CFT. For starters, OP (v1) does not mention explicitly the normal ordering ": :" in the definition of the Virasoro generators

$$ L_n ~=~ \frac{1}{2}\sum_{k\in\mathbb{Z}} :\alpha^I_{n-k}\alpha^I_k: $$

Normal ordering is important for $L_0$. Here the index $I$ runs over transversal dimensions. We recommend to divide the calculation into two cases: $n+m\neq 0$ and $n+m=0$. We will not further spoil the fun for OP, but just mention that a pedagogic derivation can e.g. be found in Ref. 1.


  1. B. Zwiebach, A first course in ST, 2nd edition, 2009, p. 256. (Note that the presentation in 2nd edition is improved as compared to 1st edition.)
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The solution to this problem, which Szabo gives in the appendix, is somewhat misleading since it invokes the regularization of the $\zeta$ function, whereas the central term arises from ordinary sums as follows.

For a start, we can re-write the Virasoro operators by making the normal ordering explicit: $$ L_n=\frac{1}{2}:\!\left( \sum_{k\in\mathbb N}\alpha_{n-k} \cdot \alpha_k \right)\! := \frac{1}{2}\sum_{k\ge n/2}\alpha_{n-k} \cdot \alpha_k +\frac{1}{2}\sum_{k< n/2}\alpha_{k} \cdot \alpha_{n-k} $$ $$ L_0=\frac{1}{2}\alpha_0^2+\sum_{k\ge 1}\alpha_{-k}\cdot \alpha_k. $$ Using the commutation relations $$ [\alpha^\nu_n,\alpha^\rho_m]=n\delta_{n+m}\eta^{\nu\rho} $$ one immediately obtains $$ [L_n,\alpha_m^\mu]=-m \alpha_{m+n}^\mu; $$ using this relation, we have $$ [L_n,L_m]=\frac{1}{2}\left[ L_n, \sum_{l\ge m/2}\alpha_{m-l} \cdot \alpha_l +\sum_{l< n/2}\alpha_{l} \cdot \alpha_{m-l} \right]=\\ = \frac{1}{2}\left[ \sum_{l\ge m/2}(-m+l)\alpha_{n+m-l}\cdot \alpha_l+ \sum_{l\ge m/2}(-l)\alpha_{m-l}\cdot \alpha_{n+l}+\\ \sum_{l<m/2}(-l)\alpha_{n+l}\cdot \alpha_{m-l}+ \sum_{l<m/2}(-m+l)\alpha_l\cdot \alpha_{n+m-l} \right]. $$ Now, in this expression if $m+n\not=0$ each pair of $\alpha$ operators commutes and therefore the r.h.s. can be rearranged by letting $l=l'-n$ in the second and third term $$ (n-m)\frac{1}{2}\sum_{l\in \mathbb Z} \alpha_{n+m-l}\cdot \alpha_l = (n-m)L_{n+m}. $$ Consider now the case $m=-n$ and let $n>0$ $$ [L_n,L_{-n}]=\\= \frac{1}{2}\left[ \sum_{l\ge -n/2}(n+l)\alpha_{-l}\cdot \alpha_l+ \sum_{l\ge -n/2}(-l)\alpha_{-n-l}\cdot \alpha_{n+l}+\\ \sum_{l<-n/2}(-l)\alpha_{n+l}\cdot \alpha_{-n-l}+ \sum_{l<-n/2}(n+l)\alpha_l\cdot \alpha_{-l} \right] $$ changing the sign of summation in the last two sums $$ \frac{1}{2}\left[ \sum_{l\ge -n/2}(n+l)\alpha_{-l}\cdot \alpha_l+ \sum_{l\ge -n/2}(-l)\alpha_{-n-l}\cdot \alpha_{n+l}+\\ \sum_{l>n/2}(l)\alpha_{n-l}\cdot \alpha_{-n+l}+ \sum_{l>n/2}(n-l)\alpha_{-l}\cdot \alpha_{l} \right] $$ shifting the two middle sums and adding similar terms $$ \frac{1}{2}\left[ (2n) \sum_{l>n/2}\alpha_{-l}\cdot \alpha_l + \sum_{-n/2\le l \le n/2}(n+l)\alpha_{-l}\cdot \alpha_l+ (2n)\sum_{l\ge n/2} \alpha_{-l}\cdot \alpha_l + \sum_{-n/2<l<n/2}(n+l)\alpha_{-l}\cdot \alpha_l \right]. $$ Among these sums, the only terms we need to commute to achieve normal ordering are the lower halves of the second and fourth sums: $$ \sum_{-n/2\le l \le-1}(n+l)\alpha_{-l}\cdot \alpha_l+ \sum_{-n/2<l\le -1}(n+l)\alpha_{-l}\cdot \alpha_l=\\= \sum_{1\le l\le n/2}(n-l)\alpha_{l}\cdot \alpha_{-l}+ \sum_{1\le l < n/2}(n-l)\alpha_{l}\cdot \alpha_{-l}=\\= \sum_{1\le l\le n/2}(n-l)\alpha_{-l}\cdot \alpha_{l}+D\sum_{1\le l \le n/2}l(n-l)+ \sum_{1\le l < n/2}(n-l)\alpha_{-l}\cdot \alpha_{l}+D\sum_{1\le l<n/2}l(n-l) $$ where $D=d+1$ is the number of space-time dimensions ($d$ space dimension); now every operator term is normal ordered and putting everything together we have $$ (2n)\left[\sum_{l\ge 0}\alpha_{-l}\cdot \alpha_l+\frac{1}{2}\alpha_0^2\right]+ D\sum_{1\le l \le n/2}l(n-l)+ D\sum_{1\le l < n/2}l(n-l) $$ the c-number can be computed using $$ \sum_{i=1}^n i = \frac{n(n+1)}{2}\qquad \sum_{i=1}^ni^2 =\frac{n(n+1)(2n+1)}{6};$$ indeed, taking for example $n$ to be even, we have $$ D\left[ n\frac{n(n+2)}{8}-n\frac{(n+2)(n+1)}{24} + n\frac{n(n-2)}{8}-n\frac{(n-2)(n-1)}{24} \right]=D\frac{n}{12}(n^2-1). $$ To sum up $$ [L_n, L_{-n}]=(2n)L_0 + D\frac{n}{12}(n^2-1), $$ hence $$ [L_n,L_m]=(n-m)L_{n+m}+\delta_{n+m}D\frac{n}{12}(n^2-1). $$

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