I have just learned from reconsidering my demystified book, that when conformally maping the worldsheet of a closed string to the complex plain by using the transformation $z = e^{\tau + i\sigma}$ and $\bar{z} = e^{\tau - i\sigma}$ the Virasoro generators can be calculated as

$$ L_m = \frac{1}{2\pi i} \oint \frac{dz}{z}z^{m+2}T_{zz}(z) $$


$$ \bar{L}_m = \frac{1}{2\pi i} \oint \frac{d\bar{z}}{z}\bar{z}^{m+2}T_{\bar{z}\bar{z}}(\bar{z}) $$

with $T_{zz}(z)$ and $T_{\bar{z}\bar{z}}(\bar{z})$ being the holomorphic and antiholomorphic component of the energy-momentum tensor.

By the deformation of path theorem, the contour of integration can be shrinked and expanded in the complex plain, which means that the Virasoro generators are invariant under a time translation because the mapping to the complex plain has transformed the time coordinate into the radial coordinate. So these Virasoro operators should commute with the Hamiltonian and they have conserved charges related to them.

My question now is: What are these conserved charges and what is their physical meaning?


The generator of translations along $\tau$, the long coordinate of the cylinder, is $L_0$ (or perhaps $L_0-a$ for some constant $a$ in the old covariant quantization, but I won't discuss old covariant quantization here). This generator is reinterpreted as the generator of scaling of the radial direction, $r\to r\cdot \exp(\Delta \tau)$.

Now, charges conserved under these $L_0$-generated translations of $\tau$ are those that commute with $L_0$. Clearly, $L_0$ is the only generator of the Virasoro algebra that is conserved in this sense because it commutes with $L_0$. Others obey $$[L_m,L_0] = m L_m $$ The commutator is nonzero so they don't commute. These $L_m$ operators are "eigenstates" under $L_0$ with the nonzero eigenvalue $m$ which guarantees that no combination of them may commute with $L_0$. $L_0$ measuring the total energy on the cylinder – or the dimension of the operator inserted at $r=0$ in the radial quantization – is conserved because it commutes with itself.

Of course, there may be symmetry generators outside the Virasoro algebra that commute with $L_0$, for example generators of spacetime isometries, heterotic $SO(32)$ or $E_8\times E_8$, etc.

If you expected more charges within the Virasoro algebra that should be conserved, well, it's not quite clear why you did. ;-) But if it has something to do with the fact that an arbitrary shape of the contour of integrals yields the same result, that's fine but that's the holomorphicity that is already "fully priced-in" into the Virasoro algebra.

The independence of the contour integral on the detailed shape of the contour is equivalent to the fact that the integral over a contour that encircles the origin at $r=0$ zero times vanishes. So all the integrals/charges that correspond to such contours avoiding the origin (for example things like $\oint T(z) dz$ over such contours) are not only conserved; they're zero. So they're not independent operators. Note that the zero vector (or operator) isn't independent of itself. ;-) We don't talk about "them" (all those zeroes) at all. If they can only be zero, they don't exist. They don't store any information: their measurement is guaranteed to yield $0$ so it's not needed to measure them. Only operators that have some nonzero eigenvalues are talked about and $L_0$ is the only conserved one in the Virasoro algebra.

The generator $L_0$ is equivalent to the integral of $T(z)z$ over a contour that does include the origin and that's the right power of $z$ that picks the residue. The other generators are integrals of $T_z$ times other powers of $z$ so they effectively pick the coefficients of other singularities $1/z^n$ near the origin. So all these charges are linked to some singular behavior near $z=0$. If all things are regular across the plane including $z=0$, then all the charges are just zero.

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    $\begingroup$ Thanks for this very nice helpful explanation and clarification Lumo, it is exactly what I needed and at an appropriate "Demystified" level such that I can perfectly understand it :-). Some of your nice answers you gave yesterday to other questions helped me too in better understanding what David Mc Mohan has written in the chapter I was just reconsidering :-D. $\endgroup$ – Dilaton Mar 27 '13 at 11:40

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