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The Weyl invariance symmetry of the Polyakov action is said to be considered as the invariance of the theory under a local change of scale which preserves the angles between all lines.

However, why does the Weyl transformation $g_{\alpha \beta} (\sigma) \to \Lambda ^2 (\sigma) g_{\alpha \beta} (\sigma)$, where $\sigma$ is the worldsheet coordinate, preserve the angles? Could somebody prove it mathematically?

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    $\begingroup$ See e.g. here (conformal transformations are closely related to Weyl symmetry). Also take a look at this related question. $\endgroup$ – Danu Sep 10 '15 at 8:36
  • $\begingroup$ Locally, the conformal transformation is isotropic in the sense that the change to a vector's length is independent of the vector's direction. This must mean that all angles are preserved. $\endgroup$ – Robin Ekman Sep 10 '15 at 14:43
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    $\begingroup$ @Danu In fact, I want an answer more explicit in mathematics, and it is below. Thanks for your advice! $\endgroup$ – Simon Sep 10 '15 at 14:45
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Consider two vector fields $a^\mu(x)$ and $b^\mu(x)$ in a space-time. The local angle between the two vector fields is given by $$ \cos\theta(x) = \frac{ a(x) \cdot b(x) }{ \|a(x)\| \|b(x)\| } = \frac{ g_{\mu\nu}(x) a^\mu(x) b^\nu(x) }{ \left| g_{\alpha\beta} a^\alpha(x) a^\beta(x) \right|^{\frac{1}{2}} \left| g_{\rho\sigma} (x) b^\rho(x) b^\sigma(x) \right|^{\frac{1}{2} } } $$ Note that this is a direct analog of what the definition of angle between vectors is in flat ${\mathbb R}^3$. It should be clear from the formula above that $\theta(x)$ is invariant under a Weyl transformation $g_{\mu\nu}(x) \to \Lambda^2(x) g_{\mu\nu}(x)$.

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