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In section 2.2 of David Tong's String Theory lecture notes, he claims that conformal transformations on the flat worldsheet are such that $$\sigma^\pm \to \tilde{\sigma}^\pm(\sigma^\pm).\tag{2.10}$$ I'm trying to develop this step-by-step so we consider a line element in the worldsheet with the conformal gauge $h_{ab}=\eta_{ab}$, where $h_{ab}$ is the worldsheet metric in Polyakov action. We have that

\begin{equation} \begin{aligned} ds^2&= \eta_{ab}d\sigma^a d\sigma^b\\ &= -d\tau^2 + d \sigma^2\\ &= -d\sigma^+ d\sigma ^-, \qquad \sigma^\pm = \tau \pm \sigma. \end{aligned} \end{equation} By performing a conformal transformation on the coordinate system $(\sigma^+,\sigma^-)$ we have new coordinates $(\tilde{\sigma}^+,\tilde{\sigma}^-)$ such that

$$ds^2 = - e^{\omega(\tilde{\sigma}^+,\tilde{\sigma}^-)} d\tilde{\sigma}^+d\tilde{\sigma}^- = -d\sigma^+ d\sigma ^-.$$ But we have that

\begin{equation} \begin{aligned} d\tilde{\sigma}^+d\tilde{\sigma}^-&=\left(\partial_+ \tilde{\sigma}^+ d \sigma^+ + \partial_- \tilde{\sigma}^+ d\sigma^- \right)\\ &\times \left(\partial_+ \tilde{\sigma}^- d \sigma^+ + \partial_- \tilde{\sigma}^- d\sigma^- \right)\\ &\propto d \sigma^+ d\sigma^-. \end{aligned} \end{equation}

One of the possible choices and the one chosen by Tong is to consider

$$\partial_+ \tilde{\sigma}^- = \partial_- \tilde{\sigma}^+=0. \tag1$$

However, I could choose instead

$$\partial_+ \tilde{\sigma}^+ = \partial_- \tilde{\sigma}^-=0. \tag2 $$

What kind of argument do I use to eliminate the possibility of $(2)$ and obtain the same result as Tong?

Qmechanic
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Генивалдо
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    you can eliminate (2) by restricting to orientation-preserving maps – Wakabaloola Feb 05 '22 at 12:00
  • @Wakabaloola: You're talking about $J = +\partial_+ \tilde{\sigma}^+ \partial_-\tilde{\sigma}^- - \partial_+ \tilde{\sigma}^- \partial_- \tilde{\sigma}^+>0$? If so, $(2)$ would implies that $ - \partial_+ \tilde{\sigma}^- \partial_- \tilde{\sigma}^+>0$. Whats the problem with this? – Генивалдо Feb 05 '22 at 13:37
  • if your question is why we might prefer to consider orientable manifolds (rather than non-orientable manifolds, such as möbius strips, Klein bottles etc), then in the bosonic string there’s no good reason that i’m aware of, but in the superstring you can’t put fermions consistently on non-orientable manifolds as far as i know (technically because they have a non-trivial Stiefel-Witney class). – Wakabaloola Feb 05 '22 at 15:01
  • As mentioned in @Qmechanic's answer, you can think of the other transformation as the usual one composed with $\sigma^\pm \to \sigma^\mp$. But, this is equivalent to parity reversals $\sigma \to - \sigma$ and since this is not connected to the identity, it is not always a symmetry of the theory. – Prahar Feb 07 '22 at 11:42

1 Answers1

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OP has a point. Tong only considers conformal transformations in the 1+1D Minkowski plane $\mathbb{R}^{1,1}$ of the form $\tilde{\sigma}^{\pm}= \tilde{\sigma}^{\pm}(\sigma^{\pm})$ in terms of light cone coordinates. OP's alternative case (2) leads to conformal transformations of the form $\tilde{\sigma}^{\pm}= \tilde{\sigma}^{\pm}(\sigma^{\mp})$, which in principle should also be taken into account. However, a transformation of type (2) is just a transformation of type (1) composed with the map $(\sigma^+,\sigma^-)\mapsto (\sigma^-,\sigma^+)$.

Concerning the global structure of the group(oid) of conformal transformations, see e.g. this Phys.SE post.

Qmechanic
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  • $\sigma^\pm \to \sigma^\mp$ is equivalent to parity $\sigma \to - \sigma$ and since this is not connected to the identity, it is not always a symmetry of the theory. While interesting, it should definitely be considered separately from all the other conformal transformations. – Prahar Feb 07 '22 at 11:43
  • $\uparrow$ Right. – Qmechanic Feb 07 '22 at 13:04