# How can one calculate the central term of the conformal field theory algebra (and show it's really the virasoro algebra)?

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 $$[\alpha_m^\mu, \alpha_n^\nu] = m\delta_{m+n}\eta^{\mu\nu}$$ 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: $$L_n = \frac{1}{2}\sum\limits_{m=-\infty}^\infty{\alpha_{n-m}\cdot\alpha_m}$$ 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 $$[L_m,L_n]$$, 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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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.

References:

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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