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Note: I'm learning complex analysis from Gamelin and Riemann surfaces appear there in the first chapter as prerequisites. I have no background in Topology/Complex Analysis so please don't use that while answering this.

I don't understand what is actually a Riemann surface (because it's definition involves terms which I don't know), but based on intuitive explanation given in Gamelin I get the following notion of a Riemann surface (I don't know how much is this correct and this may be very wrong):

Let $f(z)$ be a multivalued complex function.

Let $S \in \mathbb{C}$ be the set of points such that for any "sufficiently" small loop $\gamma$ near a point $P \in S$ it's not possible to define $f$ continiously on $\gamma$.

Now we make a bunch of slits $s_1, \cdots, s_k$ (ray/line segment) in $\mathbb{C}$ such that each point $P \in S$ is adjacent to atleast one slit. Now define $f$ continuously on $T:= \mathbb{C} - \{s_1 \cup s_2 \cup \cdots s_k \}$ as $f_1$. Now consider all such functions $f_1, \cdots, f_n$ and glue the slitted complex plane's ($T$) together to form a surface $S$ such that the value of $f$ on $S$ is continuous.

This $S$ is the required Riemann surface

It's not clear to me why the surface $S$ obtained should be independ of in which way you make the slits or how you define $f$ over the slitted complex plane and glue them up.

Consider $f(z) = \sqrt{z(1-z)}$. Here $S = \{0, 1 \}$. Now I think there're two ways to make a slit: either slit $[0,1]$ or slit both $(-\infty, 0]$ and $[1, \infty)$. Doesn't which way you make the slit changes the resulting Riemann surface ?

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Also suppose you make the only slit $[0,1]$. Now consider two small circles $\gamma_0, \gamma_1$ centered at $0$ and $1$ respectively. When you approach the slit $[0,1]$ through a variable point $p \in \gamma_0$ in the counterclockwise direction, the sign of $Re(p)$ either is postive or negative. Similar for $\gamma_2$, so total four ways (I guess) you can define $f$ continuously on $\mathbb{C}-[0,1]$ - i.e w. Among them, why it's not possible to define $f$ on $\mathbb{C} - [0,1]$ so that it's like the either of the cases in the right side ?

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  • $\begingroup$ BTW why such weird surfaces are useful and why should we care about them ? $\endgroup$ – cdt Apr 2 at 13:37
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    $\begingroup$ "I have no backgroud in topology and complex analysis" I would say that then it's unreasonnable for you to learn about Riemann surfaces. At the very least you should have a first course in complex analysis and differential geometry (so that you are familiar with notions of charts and manifolds). It would also be really good to have some notions about covering maps $\endgroup$ – Glougloubarbaki Apr 2 at 13:43
  • $\begingroup$ @Glougloubarbaki But then why it's given in the first chapter of Gamelin ? And I've heard Gamelin requires very little prior CA background and is good for first timers ? $\endgroup$ – cdt Apr 2 at 13:44
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    $\begingroup$ good question... I'm not familiar with this book, so I don't really know. I guess that the content on Riemann surfaces is either very light or very handwavy. In any case, if you want to really go in depth learning about Riemann surfaces you will need all I mentionned and more. My first course on Riemann surfaces was in master 2 (rough equivalent of 1st year of grad school in the US system) $\endgroup$ – Glougloubarbaki Apr 2 at 13:48
  • $\begingroup$ You have to read more in order to get an answer to this question (at the very least, you have to get through the section on covering spaces). Once you are there, the argument will go as follows: Suppose that $p: X\to {\mathbb C}, q: Y\to {\mathbb C}$ be two RS corresponding to the same multivalued function. Then you construct a multivalued function $X\to Y$ by composing $q^{-1}\circ p$. Then you have to work (by looking at the branch points) to prove that the composition is well-defined and is biholomorphic. $\endgroup$ – Moishe Kohan Apr 2 at 17:51

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