# Explicit description of a subgroup of the braid group $\mathsf{B}_2(C_2)$

This is related to my previous MathOverflow question Fundamental group of $\mathrm{Sym}^2(C_g)$ minus the diagonal.

Let $C_2$ be a smooth curve of genus $2$ and $X:=\mathrm{Sym}^2(C_2)$ its second symmetric product. If $\delta \subset X$ is the diagonal, then the topological fundamental group $\pi_1(X-\delta)$ is isomorphic to the braid group $\mathsf{B}_2(C_2)$ on two strands on $C_2$.

Such a group is generated by five elements $a_1, \, a_2, \, b_1, \, b_2, \, \sigma$ subject to the following set of relations:

\begin{equation*} \begin{split} (R2) \quad & \sigma^{-1} a_1 \sigma^{-1} a_1= a_1 \sigma^{-1} a_1 \sigma^{-1} \\ & \sigma^{-1} a_2 \sigma^{-1} a_2= a_2 \sigma^{-1} a_2 \sigma^{-1} \\ & \sigma^{-1} b_1 \sigma^{-1} b_1 = b_1 \sigma^{-1} b_1 \sigma^{-1} \\ & \sigma^{-1} b_2 \sigma^{-1} b_2 = b_2 \sigma^{-1} b_2 \sigma^{-1}\\ & \\ (R3) \quad & \sigma^{-1} a_1 \sigma a_2 = a_2 \sigma^{-1} a_1 \sigma \\ & \sigma^{-1} b_1 \sigma b_2 = b_2 \sigma^{-1} b_1 \sigma \\ & \sigma^{-1} a_1 \sigma b_2 = b_2 \sigma^{-1} a_1 \sigma \\ & \sigma^{-1} b_1 \sigma a_2 = a_2 \sigma^{-1} b_1 \sigma \\ & \\ (R4) \quad & \sigma^{-1} a_1 \sigma^{-1} b_1 = b_1 \sigma^{-1} a_1 \sigma \\ & \sigma^{-1} a_2 \sigma^{-1} b_2 = b_2 \sigma^{-1} a_2 \sigma \\ & \\ (TR) \quad & [a_1, \, b_1^{-1}] [a_2, \, b_2^{-1}]= \sigma^2. \end{split} \end{equation*} The geometric interpretation for the above generators of $\mathsf {B}_2(C_2)$ is the following. The $a_i$ and the $b_i$ are the braids coming from the representation of the topological surface associated with $C_2$ as a polygon of $8$ sides with the standard identification of the edges, whereas $\sigma$ is the classical braid generator on the disk. In terms of the isomorphism with $\pi_1(X-\delta)$, the element $\sigma$ corresponds to the homotopy class in $\textrm{Sym}^2(C_2)-\delta$ of a topological loop that "winds once around $\delta$". For more details see P. Bellingeri's paper

On presentations of surface braid groups, Journal of Algebra 274 (2004), 543-563.

Now, the surface $X$ contains a $(-1)$-curve $E$, namely the smooth rational curve given by the graph of the hyperelliptic involution on $C_2$. This curve intersect $\delta$ at six points $p_1, \ldots, p_6$ corresponding to the six Weierstrass points of $C_2$.

Thus, if $i \colon E-\{p_1, \ldots, p_6\} \to X - \delta$ is the inclusion, we have an induced map of fundamental groups $$i_* \colon \, \pi_1( E-\{p_1, \ldots, p_6\}) \to \pi_1(X-\delta) \simeq \mathsf{B}_2(C_2).$$

Question. What is the image of $i_*$ in terms of the above presentation for $\mathsf{B}_2(C_2)$?

• Your surface of genus two comes from the octagon with the boundary labelling $a_1 b_1 A_1 B_1 a_2 b_2 A_2 B_2$? (Here capital letters stand for inverses.) I think that the boundary labelling $a b c d A B C D$ will work better for this problem. Using this labelling, the hyperelliptic acts on the octagon by 180 degree rotation. This makes it "clear" that $\sigma$ is in the image of $i_*$. – Sam Nead Mar 28 '17 at 8:20
• I think that something like $[a, \sigma]$ (and thus also $[b, \sigma]$, $[c, \sigma]$, $[d, \sigma]$) is in the image. – Sam Nead Mar 28 '17 at 9:24