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Given the many-valued algebraic function $w =\sqrt{(z-1)(z-2)(z-3)}$, we can get a Riemann surface $S$ that is topologically equivalent to a torus. I am wondering whether the $1$-form $ \displaystyle\frac{dz}{w}$ is a holomorphic form on the torus $S$.

On the surface we know that

$$ \frac{dz}{w}=\frac{dz}{\sqrt{(z-1)(z-2)(z-3)}},$$

I think that $z=1,2,3$ are poles of the above 1-form, so it is a meromorphic 1-form on the $z-$plane. Is it a holomorphic 1-form on the torus $S$ determined by the algebraic function $w =\sqrt{(z-1)(z-2)(z-3)}$?

I just know a little about Riemann surface theory. I will appreciate any suggestions and comments. Thanks.

Jacob.Lee
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2 Answers2

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What Riemann surface did you obtain? $C:w^2=(z-1)(z-2)(z-3)$ or $C-(1,0),(2,0),(3,0)$ or the projective closure $E\simeq C\cup (\infty,\infty)$?

Anyway $dz/w$ is holomorphic on them all: near $(1,0)$ we have $z=f(w)w^2$ with $f$ analytic at $0$. At $(\infty,\infty)$ we have $z=(z/w)^{-2} g(z/w),w=(z/w)^{-3} h(z/w)$ with $g,h$ analytic at $0$.

Eventually $dz/w$ is a nowhere vanishing holomorphic 1-form and $p\to \int_{p_0}^p dz/w$ is an isomorphism $E \to \Bbb{C/(uZ+vZ)}$ where $uZ+vZ$ is the lattice obtained by integrating over the closed-loops.

reuns
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  • The Riemann surface I think is the first one, the torus $T$, which is obtained by Riemann's method from the algebraic function. $T\subset \Bbb{ C^2\times C^2}$. I do not understand why $z=f(w)w^2.$ – Jacob.Lee Jan 18 '21 at 01:06
  • And why $f$ is analytic at 0? – Jacob.Lee Jan 18 '21 at 01:07
  • What is $0$ ? The points on $C:w^2=(z-1)(z-2)(z-3)$ are pairs of complex numbers satisfying the equation. What do you mean with $T\subset \Bbb{ C^2\times C^2}$ ? – reuns Jan 18 '21 at 02:31
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    $F(z)= (z-1)(z-2)(z-3)$ is biholomorphic near $z=1$ thus $z=F^{-1}(w^2)=f(w)w^2$ with $f$ analytic at $0$. This holds only locally, and we'll have a different $f$ near $z=2$ and $z=3$. – reuns Jan 18 '21 at 02:35
  • Sorry for the mistake. Yes. It is $T\subset \Bbb{C\times C}.$ – Jacob.Lee Jan 18 '21 at 03:19
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    $w^2=(z-1)(z-2)(z-3)$ is not a complex torus, it is an (affine) elliptic curve, its projective closure is isomorphic to a complex torus ($\Bbb{C/(uZ+vZ)}$, the quotient of $\Bbb{C}$ by a lattice) – reuns Jan 18 '21 at 03:23
  • Yes.The algebraic function determines a torus $T, \dim_\Bbb{R}(T)=2$. – Jacob.Lee Jan 18 '21 at 03:32
  • If $z=F^{-1}(w^2)=f(w)w^2$, then $dz=[f'(w)w^2+2wf(w)]dw$. Hence we have $$ \frac{dz}{w}=[f'(w)w+2f(w)]dw$$ which is holomorphic 1-form. But how can $F^{-1}(w^2)=f(w)w^2$ be obtained? and why $f(w)$ is analytic at $w=0$? – Jacob.Lee Jan 18 '21 at 03:34
  • No it is not a torus, it is an elliptic curve, but it is isomorphic to a torus. – reuns Jan 18 '21 at 03:36
  • Yes, you are right. I mean it from topologist's view. – Jacob.Lee Jan 18 '21 at 03:39
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    And for it to be isomorphic to a torus, not a torus minus finitely many points, you need to consider the projective closure. – reuns Jan 18 '21 at 03:57
  • Yes. But I still do not understand the equation $F^{-1}(w^2)=f(w)w^2.$ Could you give me the details? – Jacob.Lee Jan 18 '21 at 04:22
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Given the algebraic function $w^2=(z-1)(z-2)(z-3)$, differentiate each side, we get $$2wdw=[(z-2)(z-3)+(z-1)(z-3)+(z-1)(z-2)]dz .$$ Therefore we have $$\frac{dz}{w}=\frac{2dw}{(z-2)(z-3)+(z-1)(z-3)+(z-1)(z-2)}.$$ It is easy to see that the RHS is holomorphic locally at (1,0), (2,0),(3,0) respectively.

Jacob.Lee
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