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I have a question in Polchinski's string theory book volume 1 p54, related to the Virasoro algebra.

Introducing complex coordinates $$w=\sigma^1 + i \sigma^2 $$ $$z=\exp (-i \omega) =\exp( -i \sigma^1 + \sigma^2) $$ and $$T_m=L_m - \delta_{m0} \frac{c}{24}$$ $$ \tilde{T}_m=\tilde{L}_m - \delta_{m0} \frac{\tilde{c}}{24}$$

It is said

The Hamiltonian H of time translation in the $w=\sigma^1+i\sigma^2$ frame is $$H=\int_0^{2 \pi} \frac{ d \sigma^1}{2 \pi} T_{22} = L_0 + \tilde{L}_0 - \frac{ c +\tilde{c}}{24} (2.6.10) $$

My questions are (sorry ask two questions in a thread, since they are closely related)

(i) Does $T_{22}$ mean $T_{zz}$ or $T_{\sigma^2 \sigma^2}$?

(ii) How the anti holomorphic operator $\tilde{L}_0$ are obtained in Eq. (2.6.10)?

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up vote 2 down vote accepted

For (ii) :

Using $(2.1.3)$,$(2.3.15b)$ , you get :

$$T_{22} = -(T_{ww} + T_{\bar w \bar w} )$$

Now, looking at the expansion $(2.6.7a)$ and $(2.6.7b)$ ,we see that if we integrate $T_{ww}$ or $T_{\bar w \bar w}$ with $\sigma_1$ going from $0$ to $2\pi$, the only non-null terms are for $m=0$, so, finally :

$H = \frac{1}{2\pi} \int d\sigma^1 T_{22} = -\frac{1}{2\pi}\int d\sigma^1(T_{ww} + T_{\bar w \bar w} )=(T_0+\tilde T_0)$

By using $2.6.8$, which itself comes from $(2.6.9)$(that is $(2.4.26)$ with $z=e^{-i\omega}$), and the comparison of the expansions $2.6.5$ and $2.6.7 a b$, we have finally:

$$H = L_0 + \tilde L_0 -\frac{c+\tilde c}{24}$$

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Thank you very much! Your solution reminds me a basic question that I have overlooked at (or I forgot my solution). Namely in (2.3.15b) $T_{ab}= - \frac{1}{\alpha'} :( \partial_a X^{\mu} \partial_b X_{\mu} - \frac{1}{2} \delta_{ab} \partial_c X^{\mu} \partial^c X_{\mu}):$. But in (2.4.3) and (2.4.4) $T(z)=T_{zz}(z)=-\frac{1}{\alpha'} : \partial X^{\mu} \partial X_{\mu}:$ ,why the $ - \frac{1}{2} \delta_{ab} \partial_c X^{\mu} \partial^c X_{\mu} $ term disappears? Presumably here $a=b=z$ – user26143 Jul 23 '13 at 19:13
or, is Eq. (2.4.4) just the definition of $T_{zz}$? – user26143 Jul 23 '13 at 19:40
Look at $(2.1.6)$: $g_{..}$ is a tensor, so it transforms like a tensor. For instance : $g_{zz} = (\frac{\partial \sigma^1}{\partial z} )^2 g_{11} + (\frac{\partial \sigma^1}{\partial z} )(\frac{\partial \sigma^2}{\partial z} ) g_{12}+(\frac{\partial \sigma^2}{\partial z} )(\frac{\partial \sigma^1}{\partial z} ) g_{21} +(\frac{\partial \sigma^2}{\partial z} )^2 g_{22}$. And, when you calculate it, you find that $g_{zz} = g_{\bar z \bar z}=0$, so the second term in $(2.3.15.b)$ disappears for $T_{zz}$ and $T_{\bar z \bar z}$ – Trimok Jul 24 '13 at 10:01
Sorry I don't get it. We write the second term in (2.3.15b) for $T_{zz}$ as $$ \frac{1}{2\alpha'} : \delta_{zz} \partial_c X^{\mu} \partial^c X_{\mu} : = \frac{1}{2\alpha'} : g_{cd} \partial^c X^{\mu} \partial^d X_{\mu}: $$ why it vanishes? – user26143 Jul 24 '13 at 13:05
No, writing $T_{zz}$, you have a term: $\frac{1}{2\alpha'} g_{zz} \partial_c X^{\mu} \partial^c X_{\mu}$ (with $c=z, \bar z)$, which is zero because $g_{zz}=0$ – Trimok Jul 24 '13 at 13:37

(1) The symbol $T_{22}$ obviously means what you call – unusually – $T_{\sigma_2\sigma_2}$. The letter $z$ has no relationship with the number $2$, except that they look similar so one could make a typo if his handwriting were bad – and the expression where $T_{22}$ appears uses $\sigma_1$ so it is obvious that these are the coordinates used in the expression.

(2) Pretty much all equations in section 2.6 are written twice so everything that is done for $L_n$ is also done for $\tilde L_n$. $\tilde L_0$ itself is introduced by nothing else that eqn (2.6.10) or, more generally for all generators $\tilde L_n$, by (2.6.5). The only equation that isn't doubled is (2.6.6); the formula for $\tilde L_n$ just has tildes and bars at the obvious places and changed sign of $i$ (or, equivalently, the direction of the contour integration).

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Thanks you very much... I thought from $w,z$ coordinate, $z$ corresponds to the second, as the subscript 2... But how to derive Eq. (2.6.10)? – user26143 Jul 23 '13 at 17:10

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