# Deriving Virasoro algebra question

I'm reading a book Lie groups, Lie algebras, cohomology and some applications in physics by Azcarraga and Izquierdo, and on page 347, when deriving the exact form of the central extension term I came to a bit which I don't understand how they got it.

The algebra is given by

$$[L_m,L_n]=(m-n)L_{m+n}+c_{m,n}$$

And using the Jacobi identity I get

$$(m-n)c_{m+n,p}+(n-p)c_{n+p,m}+(p-m)c_{m+p,n}=0$$

If we transform the generators of the algebra, by adding a constant $b(m)$, the algebra itself won't change: $L_m\to L_m'=L_m+b(m)$. After putting that in the first expression I get

$$[L_m',L_n']=(m-n)L_{m+n}'+c_{m,n}'$$

where $c_{m,n}'=c_{m,n}+c_{m,n}^{\textrm{cob}}$, and $c_{m,n}^{\textrm{cob}}=-b(m+n)(m-n)$ is the coboundary generated by $b(m)$.

Now starts the part that I don't know where they got it from. They say that taking this into the account we see that, with

$$b(m)=\frac{1}{m}c_{m,0},\quad m\neq 0$$ $$b(0)=\frac{1}{2}c_{1,-1}$$

we have

$$c_{0,n}'=c_{0,n}+\frac{1}{n}c_{n,0}\ n=0,\quad n\neq 0$$ $$c_{1,-1}'=c_{1,-1}-\frac{1}{2}c_{1,-1}2=0$$

I get why the last one is 0 (the factors cancel each other out), but why is the $c_{0,n}'=0$? And from where do they get the relations with $b(m)$ and $b(0)$?

Also, how does that come to play when they set $p=0$ in the equation for $c$ in the Jacobi identity, and getting the result

$$(n+m)c_{m,n}=0$$

I have the feeling like there are some steps that were ommited. I cannot reproduce this last equation from Jacobi identity one.

Any help will be appreciated.

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Perhaps math.stackexchange might be better suited for this problem? – Kyle Kanos Oct 13 '13 at 2:01
I posted it here because it's usually used in physical context. But if a mod can migrate it there than that'd be great :) – dingo_d Oct 13 '13 at 11:24

Let us consider the most general central-charge extension of the Witt algebra, given by $$\begin{split} \left[ L_m, L_n \right] = \left( m - n \right) L_{m+n} + c(m, n) \end{split}$$ Clearly, we must have $c(m,n) = -c(n,m)$. Naively, central extensions are parameterized by a function $c(m,n)$. However, we now show that by appropriately redefining the generators $L_m$, we can rewrite any general central charge extension into the standard form where the function $c(m,n)$ is parameterized by a single $c$-number. First note that the algebra takes the same form if we redefine $L_m \to L_m + a(m)$ where $a(m)$ is a $c$-number function. Under this redefinition, the function $c(m,n)$ changes to $$\begin{split} c(m, n) \to c(m,n) + \left( m - n \right) a(m + n) \end{split}$$ By choosing $a(m) = - \frac{1}{m} c(m,0)$ for $m \neq 0$ and $a(0) = - \frac{1}{2} c(1,-1)$, we find $$\begin{split} c(m,0) &\to c(m,0) + m a(m) = 0,~~ \forall~m \neq 0 \\ c(1,-1) &\to c(1,-1) + 2 a(0) = 0 \end{split}$$ We can use this redefinition to set $c(m,0) = c(0,m) = 0$ and $c(1,-1) = 0$. The conformal algebra then takes the form $$\begin{split} \left[ L_m, L_0 \right] &= m L_m,~~ \left[ L_1, L_{-1} \right] = 2 L_0 \end{split}$$ while the remaining generators satisfies the same old algebra. Consider now the following Jacobi Identity for $m,n\neq 0$ and $m+n\neq0$. $$\begin{split} 0 &= \left[ \left[ L_m, L_n \right], L_0 \right] + \left[ \left[ L_n, L_0 \right], L_m \right] + \left[ \left[ L_0, L_m \right], L_n \right] = - \left( m + n \right) c (m,n) \\ \end{split}$$ Since $m+n\neq0$, we must have $c(m,n) = 0$. Putting it all together, we find $c(m,n) = a(m) \delta_{m+n,0}$. We then find that the most general central extension of the Witt algebra takes the form $$\begin{split} \left[ L_m, L_n \right] = (m-n) L_{m+n} + a(m) \delta_{m+n,0} \end{split}$$ where $a(\pm 1) = 0$ (from $c(1,-1) = 0$) and $a(m) = a(-m)$. The functional form of $a(m)$ can be set by looking at a generic Jacobi Identity $$\begin{split} 0 &= \left[ \left[ L_m, L_n \right], L_p \right] + \left[ \left[ L_p, L_m \right], L_n \right] + \left[ \left[ L_n, L_p \right], L_m \right] \\ &= (m-n)\left[ L_{m+n} , L_p \right] + (p-m) \left[ L_{p+m} , L_n \right] + (n-p)\left[ L_{n+p}, L_m \right] \\ &= \left[ (m-n)a(m+n) - ( 2m + n ) a(n) + (m+2n)a(m) \right] \delta_{n+p+m,0} \\ \end{split}$$ We then get $$\begin{split} (m-n)a(m+n) - ( 2m + n ) a(n) + (m+2n)a(m) \end{split}$$ For $m=1$, we get $$\begin{split} (m-1)a(m+1) - (m+2)a(m) = 0 \implies \frac{a(m+1)}{a(m)} = \frac{m+2}{m-1} \end{split}$$ We then have $$\begin{split} a(m) &= \frac{a(m)}{a(m-1)} \frac{a(m-1)}{a(m-2)} \cdots \frac{a(3)}{a(2)} \\ &= \frac{m+1}{m-2} \frac{m}{m-3} \frac{m-1}{m - 4} \frac{m-2}{m-5} \cdots \frac{4}{3} \\ &= \frac{a(2)}{6} m(m^2-1) \end{split}$$ We define $a(2) = \frac{c}{2}$. The most general central extension of the Witt algebra then takes the form $$\begin{split} \boxed{ \left[ L_m, L_n \right] = \left( m - n \right) L_{m+n} + \frac{c}{12} m \left( m^2 - 1 \right) \delta_{m+n,0} } \end{split}$$ This algebra is called the Virasoro algebra and is denoted by ${\cal L}_c$. It is parameterized by a single number $c$, called the central charge of the theory.
Your derivation is somewhat different than from the book, but I think I know what I missed. You said it's obvious that $c(m,n)=-c(n,m)$, and that wasn't clear to me, but then I remembered that Lie bracket anticommute ($[L_m,L_n]=-[L_n,L_m]$), and then the obviousness is apparent :D Thank you :) That was the part that was missing for all the rest. When I went through the derivation in the book I could follow it, but this was the piece that was missing :D – dingo_d Oct 13 '13 at 20:45