For any subgroup $G$ of $\text{Homeo}(S^1)$, we have the Euler class $\chi$ in the group cohomology $H^2(G;\mathbb{Z})$. One can think of this class as the pullback of the generator of $H^2(\mathrm{B}\text{Homeo}(S^1);\mathbb{Z})$. I have two related questions regarding the (non)vanishing of the powers of Euler class for different subgroups of $\text{Homeo}(S^1)$.

- Does there exist a torsion-free, finitely generated group $G\subset\operatorname{Homeo}(S^1)$ such that $\chi^k$ for all $k$ are nonzero in $H^{2k}(G;\mathbb{Z})$?

If we let the group $G$ be the group of piecewise linear homeomorphisms of the circle $\mathbb{R}/\mathbb{Z}$ which send $\mathbb{Q}_2/\mathbb{Z}$ onto itself and have singular points only on $\mathbb{Q}_2/\mathbb{Z}$, where $\mathbb{Q}_2$ is the ring of dyadic numbers. Ghys and Sergiescu proved, in this case, $\chi^k$ for all $k$ are nonzero in $H^{2k}(G;\mathbb{Z})$ but $G$, which is one of the variants of Thompson groups, has many torsion elements.

Another example related to this question is due to Solomon Jekel. Let $\Gamma_{g}^1$ be the mapping class group of a surface of genus $g$ with a marked point. It is a subgroup of $\text{Homeo}(S^1)$. Jekel showed that in this case $\chi^{g-1}$ is nonzero in $H^{2g-2}(\Gamma_{g}^1;\mathbb{Q})$. Now, one could let $G$ be a finite index torsion-free subgroup of $\Gamma_{g}^1$. But the power $g-1$ is a threshold for the non-vanishing of powers of the Euler class.

- In the direction of the second example, let $\text{Tor}_g^1$ be the Torelli group of the surface of genus $g$ and a marked point. What is the threshold $k$ for which $\chi^k$ is nonzero in $H^{2k}(\text{Tor}_g^1;\mathbb{Q})$? Is it less than $g-1$?