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Since every question has to be asked in a seperate topic, I'm asking a question refering to the following topic: Beginners questions concerning Conformal Field Theory In particula I'm refering to the subsection "Implementing a symmetry on operators" of the answer given by Lubos Motl.

It is clear to me that we obtain $\delta\phi=i\epsilon [L_{m},\phi]=i \epsilon z^{m+1} \partial_{z} \phi$ and that $z^{m+1} \partial_{z}$ are the generators of conformal symmetry. What confuses me though is that these generators fulfill the Witt-Algebra and not the Virasoro Algebra. Since we're in the quantum case and we want to show that the theory is invariant on a quantum level, shouldn't we get generators that fulfill the Virasoro Algebra?

Thanks in advance for the responses.

Best regards.

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3 Answers

The main answer to the question is that the full generator

$$L_n = L^{(\mathrm{m})}_n+L^{(\mathrm{g})}_n$$

in Bosonic string theory is a sum of a matter part with normal ordering constant $a=1$,

$$L^{(\mathrm{m})}_n=\frac{1}{2} \eta_{\mu\nu}\sum_m:\alpha_{n-m}^{\mu}\alpha_m^{\nu}:-\hbar a\delta_n^0,$$

and a ghost part

$$ L^{(\mathrm{g})}_m=\sum_n(m-n):b_{m+n}c_{-n}:.$$

The matter and ghost part satisfy (normal-order-shifted) Virosoro algebras with central charges $\pm 26$,

$$ [L^{(\mathrm{m})}_m,L^{(\mathrm{m})}_n]=\hbar(m-n)L^{(\mathrm{m})}_{m+n}+\hbar^2 \left(\frac{Dm(m^2-1)}{12}+2ma \right) \delta_{m+n}^0, $$

$$ [L^{(\mathrm{g})}_m,L^{(\mathrm{g})}_n]=\hbar(m-n)L^{(\mathrm{g})}_{m+n}+\hbar^2 \frac{m(1-13m^2)}{6} \delta_{m+n}^0, $$ respectively, while the full generator $L_n$ satisfies a Witt algebra

$$ [L_m,L_n]=\hbar(m-n)L_{m+n} +\hbar^2\frac{(D-26)m^3+(24a+2-D)m}{12}\delta_{m+n}^0 $$ $$=\hbar(m-n)L_{m+n}$$

in the critical dimension $D=26$ and $a=1$. To summarize, while the individual matter and ghost parts, and more generally, various subsectors of the theory, can have separately non-vanishing Virasoro quantum anomalies, there is no net Virasoro quantum anomaly in the full construction. To return to the question, the full generators $L_n$ act/have a representation/interpretation as conformal transformations $\sim z^{n+1}\partial/\partial z$ on the world sheet. In precisely the critical dimension $D=26$, there is no Weyl/conformal anomaly.

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I thought at first this question was more general involving the nature of the Virasoro algebra. As a result the first two paragraphs are boiler plate discussions on that. The actual question is addressed in the third paragraph.

The vector fields $T^m~=~z^{m+1}\partial z$ satisfy the Witt algebra or Virasoro algebra without central extension for each index value, $$ [T^m,~T^n]~=~(n~-~m) T^{m+n}. $$ The Virasoro algebra maybe extended by a one-dimensional center with the map $$ {m:Vir\oplus {\mathbb C}~\rightarrow~Vir,} $$ where this central extension is the kernel of the map. The extended Virasoro group is then, $$ [T^m,~T^n]~=~(n~-~m)T^{m+n}~+~c(m,~n). $$ This is imposed since with the case $m~+~n~=~0$ then $[T^m,~T^{-m}]~=~-2mT^{2m}$ and there are two infinite sums with the same result, one for $m~\ge~0$ and $m~\le~0$. To give a central extension for a specific index $m$ this is a c-number in the $\mathbb C$ in the domain of the above map, and $c(m,~n)~=~c(m)\delta_{m+n}$ The central extension may then be computed with the Jacobi identity $$ [T^k,~[T^m,~T^n]]~+~ [T^m,~[T^n,~T^k]]~+~ [T^n,~[T^k,~T^m]]~=~0. $$ which is equal to $$ (m~-~n)c(k)~+~(n~-~k)c(m)~+~(k~-~m)c(n)~=~0. $$ To get $c(m)$ we set $k~=~1$ and $n~=~-(m~=~1)$. This gives us $$ c(m~+~1)~=~\frac{(m~+~2)c(m)~-~(2m~+~1)c(1)}{m~-~1}. $$ This is a nice recursion relationship and we get $c(2)~=~3c(1)~-~3c(1)$ $=~0$ and the general recursion $$ c(m)~=~c(3)m^3~-~c(1)m $$

The $T^m$ are expanded according to string modes $$ T^m~=~\frac{1}{2}\sum_{n=-\infty}^\infty\alpha_{m-n}\alpha_n. $$ Form here the coefficients $c(3)$ and $c(1)$ can be determined. The $T^0,~T^{\pm 1}$ generate a closed $SL(2,{\mathbb R}$ subalgebra. The $T^0~=~\frac{1}{2}\sum_{n=-\infty}^\infty\alpha_{-n}\alpha_n$ annihilates the vacuum state as does $L^{\pm}$. For $m~=~2$ one can show by computing the vacuum expectation $\langle 0|T^{-2}T^2|0\rangle$ $$ c(m)~=~\frac{D}{12}(m^3~-~m) $$ with some additional work, which is a bit lengthy, you can show $D~=~26$ and work up the bosonic string.

Any field $\phi$ transforms under the conformal algebra as its commutator with a $T^m$ element of is $iz^{m+1}\partial_z\phi$. The $SL(2,{\mathbb R})$ is the symmetries of conformal quantum mechanics, and the Witt algebra or Virasoro algebra is then its generalization as the set of diffeomorphisms of the circle. Suppose a function or field that transforms under these diffeomorphisms requires this central extension. For instance the Hamiltonian is $H~=~T^0$ and suppose the field is $\phi(\tau)~=~\phi(0)e^{iH\tau}$, then $$ \delta\phi~=~\delta\phi(0)e^{iT^0\tau}~+~i\phi(0)\epsilon\sum_m[T^m,~T^0]~+~O(\tau^2). $$ It is then clear that the central extension is $c(0)~=~0$. It is not hard to see that a similar situation hold if the field is dependent on $T^{\pm}$. However, if the field depends upon $L^m$, for $|m|~>~1$ then care needs to be taken. The $L_m$ are the Fourier modes of the stress-energy tensor in the string $1~+~1$ dimensional spacetime. So it is unclear whether a field is of this nature. I would be tempted at this point to say no, except for the case of the $T^0$, which is the Hamiltonian. However, in that case the central or anomaly term is zero. The only reason one would need to be concerned with the central term is if $\phi$ is an element of the circle diffeomorphism, or is some function of it.

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Thanks for your response. Why is it in the last paragraph obvious that the central extension is c(0)=0? –  mr_conf Apr 17 '11 at 7:46
It just follows from $c(m) = D/12(m^3~-~m)$ –  Lawrence B. Crowell Apr 19 '11 at 11:51
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The point is that the adjoint representation may have vanishing central term. As an example of this situation consider operators $R(L_n)$ and $R(C)$ on a Hilbert space $H$ which represent the Virasoro algebra $V$ i.e. it holds $$[R(L_n),R(L_m)]=(n-m)R(L_{m+n}) +\frac{1}{12}\delta_{m+n,0}(n^3-n)R(C).$$ $$[R(C), R(L_n)]=0.$$ Note that $R(C)$ is the operator on $H$ which represents the central generator of the Virasoro algebra and we consider the case when $R(C)$ is just the multiple of the identity operator $I$ : $R(C)=cI$.

The representation $R$ naturally induces the "adjoint" representation $S$ of the Virasoro algebra $V$ on the space $L(H)$, the elements of which are the linear operators on $H$. This means that $S(L_n)$ acts on an operator $O$ as $$S(L_n)O\equiv [R(L_n),O], \quad S(C)O\equiv [R(C),O]=0.$$
In consequence $S(C)\equiv 0$ and it holds $$[S(L_n),S(L_m)]=(n-m)S(L_{m+n})$$. This means that $S$ is in fact the representation of the de Witt algebra.

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