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I am trying to understand what a holomorphic quadratic differential is, i have read a local definition on two books: Jürgen-Jost-"Compact Riemman Surfaces" and Kurt Strebel-"Quadratic-Differentials". The definition that they use is local:

Definition: Let ( M , g) be a Riemann surface with a conformal metric and $z$ a local conformal coordinate, we say that $\varphi dz^2$ is a holomorphic quadratic differential if $\varphi$ is holomorphic.

That is the definition in Jürgen-Jost book, in Kurt Strebel "Quadratic-Differentials" They only add that a transformation rule for other coordinates is needed.

I would like to understand a global definition in terms of sections, in wikipedia i found this:

"a quadratic differential on a Riemann surface is a section of the symmetric square of the holomorphic cotangent bundle"

I guess holomorphic cotangent bundle means the bundle of holomorphic 1-forms. I don't understand the "symmetric square part". I would like to have a global definition and a local coordinate representation. Of course I don't trust at all about a wikipedia link. So I would like to ask, is this definition right? Is there a good reference i should read ?

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Since the holomorphic cotangent bundle $T^\star_X$ is of complex rank one, we can get rid of the word "symmetric". A quadratic differential is simply a section of the tensor product $T^\star_X \otimes T^\star_X$.

From a practical point of view, a section of this bundle looks like $\varphi(z) dz \otimes dz$ in a given coordinate patch parametrised by a coordinate $z$. But if you go to a different coordinate patch parametrised by a new coordinate $w$, then the same section should be written as $\varphi(z(w)) \left(\frac{dz}{dw} \right)^2dw \otimes dw$.

Since you asked for a reference, I searched on Google and the first thing I found was this blog post. I know it's not a textbook, but I can assure you the person who wrote it know what he's talking about! (I used to share an office with him.)

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  • $\begingroup$ I see! the tensor product commutes in dimension 1! Thank u! $\endgroup$ – Alicia Basilio May 14 '17 at 3:30

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