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 ?


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.)

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

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.