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Let $C_g$ be a compact Riemann surface of genus $g$, and let $X:=\mathrm{Sym}^2(C_g)$ be its double symmetric product and $\Delta \subset X$ the diagonal.

Question. What is the topological fundamental group $\pi_1(X - \Delta)?$

I'm primarily interested to the case $g=2$, but I would like to know the general answer. I'm aware that this is probably standard material for the experts, so any reference will be greatly appreciated.

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    $\begingroup$ The group you're asking about is also known as the braid group of $C_g$ on two strands. Maybe the keyword "surface braid groups" alone already gets you somewhere. Otherwise, you should be more precisely about what you want to know. If you're interested in presentations, for example, you can take a look at Paolo Bellingeri's papers and some references given therein. Or are you looking for a relation to the fundamental group of $C_g$? As I said, a more precise question would help. $\endgroup$ Commented Aug 31, 2016 at 17:01
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    $\begingroup$ @Stefan Behrens: Braid groups, this was the magical word! Thanks a lot, I actually wanted a presentation and looking at the papers of Bellingeri I found it. You answered my question, if you turn your comment into an answer I will be glad to accept it. $\endgroup$ Commented Sep 1, 2016 at 11:58
  • $\begingroup$ No need, let's leave it at the comment. Glad I could help you. $\endgroup$ Commented Sep 1, 2016 at 13:56

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$\newcommand{\Sym}{\operatorname{Sym}}$The answer is above, in the comment by Stefan Behrens -- I am just adding this to make some robot happy.

The $k$-fold symmetric product $\Sym^k$ of a space $C$ (minus the big diagonal $\Delta$) is the "configuration space" of distinct unlabelled $k$-tuples of points in $C$. After fixing a basepoint $(c_i)$, a loop in $\Sym^k(C)$ is a one-parameter family of $k$-tuples - so it is a union of $k$ non-intersecting arcs in $C$ that start and end at the $c_i$.

Accordingly we call $\pi_1(\Sym^k(C) - \Delta)$ the $k$-strand braid group in $C$. Note that there is a homomorphism from $\pi_1(\Sym^k(C) - \Delta)$ to the symmetric group on the indices $\{1, 2, \ldots, k\}$.

If $C$ is a surface, then sometimes the phrase surface braid group is used. If $C$ is the complex plane, then we obtain the standard braid group.

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  • $\begingroup$ Thanks. By any chance, do you have any clue about my subsequent question mathoverflow.net/questions/254151/…? $\endgroup$ Commented Mar 27, 2017 at 9:20
  • $\begingroup$ It was that question that brought me here. I have thought about it, and I certainly do not "see" an answer. It is a very good question, of course. $\endgroup$
    – Sam Nead
    Commented Mar 28, 2017 at 7:58

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