2
$\begingroup$

(skip disclaimer)

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)?

$\endgroup$
3
$\begingroup$

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:

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

$\endgroup$
  • $\begingroup$ 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 $a=b=z$ $\endgroup$ – user26143 Jul 23 '13 at 19:13
  • $\begingroup$ or, is Eq. (2.4.4) just the definition of $T_{zz}$? $\endgroup$ – user26143 Jul 23 '13 at 19:40
  • $\begingroup$ 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 $T_{\bar z \bar z}$ $\endgroup$ – Trimok Jul 24 '13 at 10:01
  • $\begingroup$ 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? $\endgroup$ – user26143 Jul 24 '13 at 13:05
  • $\begingroup$ 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 $g_{zz}=0$ $\endgroup$ – Trimok Jul 24 '13 at 13:37
4
$\begingroup$

(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).

$\endgroup$
  • $\begingroup$ 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)? $\endgroup$ – user26143 Jul 23 '13 at 17:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.