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Image from S. Cecotti "Introduction to String Theory". Introduction to generating functional method

Question 1: is it always possible to write the metric in that form? Is it sufficient the local conformally-flat form to obtain the volume?

Question 2: Is the volume form in (4.1) well-defined? Going in real coordinates one has $z=x+iy$ from which $ds^2=g(x,y)(dx^2+dy^2)$ and then the volume form (probably I miss a factor 2 from canonical definition) is $$dV=\sqrt{det(g)}dx \wedge dy=g(x,y) dx \wedge dy.$$ But $$dx\wedge dy=\frac{i}{2} dz\wedge \bar{z}$$ and thus I expect a volume form in complex coordinates as: $$dV=g(z,\bar{z}) \frac{i}{2} dz\wedge \bar{z}$$ but the book defines $d^2z$ without a factor 2. Where is the mistake? Is there something wrong in the changing of real to complex coordinates?

Question 3: why in the first integral of (4.2) there is no $g(z,\bar{z})$ term while in there is in the second integral? Is it related to the symmetries of Polyakov action?

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  1. Yes, any Euclidean 2D metric can always be locally written in the form $$ ds^2 = 2 g_{z{\bar z}} d z d{\bar z} . $$ Naively, you can think of it like this - a generic 2D metric has 3 independent components, and we have 2 diffeomorphisms. We can use the two diffeos to set 2 out of 3 components to zero.

  2. The volume form is $\epsilon = \sqrt{g} dz \wedge d{\bar z} = i g_{z{\bar z}} dz \wedge d{\bar z} = g_{z{\bar z}} d^2 z$.

  3. The first term in the action is $$ I_1 \sim \int ( d^2 x \sqrt{g} ) ( g^{ab} \partial_a X^\mu \partial_b X_\mu ) $$ Moving to complex coordinates, $$ I_1 \sim \int ( g_{z{\bar z}} d^2 z ) ( 2 g^{z {\bar z} } \partial_z X^\mu \partial_{\bar z} X_\mu ) = 2 \int d^2 z \partial_z X^\mu \partial_{\bar z} X_\mu . $$ All factors of $g_{z{\bar z}}$ cancel. This is related to the Weyl invariance of the 2D boson action.

The same cancellation does not occur in the second term (check for yourself).

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  • $\begingroup$ For the point 2: what is your $d^2 z?$? The book defines $d^2 z=i dz \wedge d\bar{z}$ not $d^2 z=2 i dz \wedge d\bar{z}$ (no with a 2 factor), there is something wrong in the third and last step. But there is always a different factor 2 with respect to the book. $\endgroup$ Commented Nov 28, 2025 at 8:31
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    $\begingroup$ ya, I screwed my factors of 2. Fixed it now. $\endgroup$ Commented Nov 28, 2025 at 9:49

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