Let $G$ be the mapping class group of a closed surface $S\_{g}$. Bestvina-Bromberg-Fujuwara <https://arxiv.org/abs/1006.1939> recently constructed a finite index subgroup $B$ of $G$ such that for every essential closed simple curve $\gamma$ on $S\_{g}$ and every $h\in B$ either $h\gamma=\gamma$ or $h\gamma$ intersects $\gamma$ (everything is modulo isotopy of $S\_{g}$). The construction is not difficult: $B$ is just the subgroup of $G$ fixing $\pi\_1(S\_{g})/N$ for some characteristic subgroup $N$ of the $\pi\_1$ of finite index. The $N$ can be found by intersecting kernels of the first homology mod $6$ with kernels of the first homology mod 2 of all index 2 subgroups of the $\pi\_1$. Question: can one find another finite index subgroup $B'$ of $G$ with the same property but with a smaller index. Most probably $B'$ cannot be above Torelli subgroup, but can it be "not far from Torelli". I am interested in $B'$ that does not act non-trivially on a simplicial tree. The motivation is here: <https://arxiv.org/abs/1005.5056>.


 **Update:**  I read somewhere (I do not remember where now) that if $g$ is an element of the Torelli subgroup, $\gamma$ is a simple closed curve, then at least two of the three curves $\gamma, g\gamma, g^2\gamma$ must intersect. From this I deduced that two curves are not enough and that a Bestvina-Bromberg-Fujiwara subgroup cannot contain the Torelli subgroup. Now the question is whether it may contain the Johnson kernel. For this one needs first to answer the following question: if $g$ is the Dehn twist about a separating curve, $\gamma$ is a simple closed curve, is it true that $g\gamma$ intersects $\gamma$ or $g\gamma=\gamma$?



A finite index subgroup of the Mapping Class Group
Regarding the new question, you're correct that $f\in \mathcal{I}$ doesn't imply $f(\gamma)$ meets $\gamma$: for example, if $f$ is a product of disjoint bounding pairs and $\gamma$ is a (nonseparating) curve meeting each curve-defining-$f$ in at most 1 point, then $f\in\mathcal{I}$ but $f(\gamma)$ is disjoint from $\gamma$.


For the rest of the question, the following theorems are taken from Farb-Leininger-Margalit, "The lower central series and pseudo-Anosov dilatations", American Jour. Math, 130 (2008), no. 3, 799--827. ([PDF](http://www.math.uchicago.edu/~farb/papers/dil.pdf))


Lemma 2.2: if $f\in \mathcal{I}$, $\gamma$ is separating, and $f(\gamma)\neq\gamma$, then $i\big(f(\gamma),\gamma\big)\geq 4$.


Lemma 2.3: if $f\in \mathcal{I}$ and $\gamma$ is nonseparating, and $f(\gamma)\neq\gamma$ then $i\big(f(\gamma),\gamma\big)\geq 2$ or $i\big(f^2(\gamma),\gamma\big)\geq 2$.


Proposition 3.2: if $f\in\mathcal{K}$ and $f(\gamma)\neq \gamma$, then $i\big(f(\gamma),\gamma\big)\geq 4$.


In particular 3.2 answers your question about a separating twist. The proof of the proposition is as follows: $\mathcal{K}$ contains no nonseparating bounding pairs (use the Johnson homomorphism). By Lemma 2.2 we can assume $\gamma$ is nonseparating. But if $f(\gamma)$ is disjoint frmo $\gamma$, then $[f,T\_{\gamma}]=f T\_{\gamma}f^{-1} T\_{\gamma}^{-1}=T\_{f(\gamma)}T\_{\gamma}^{-1}$ is a bounding pair, which lies in $\mathcal{K}$ if $f$ does, yielding a contradiction.




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I do appreciate the nice calming green background, but I'm not sure we're done. If $f\in\mathcal{K}$, and $g\in \mathcal{BBF}$ (the group defined by Bestvina-Bromberg-Fujiwara), how do you know that $fg$ has the desired property?


Here is a sketch of a possible construction of a BBF-subgroup which properly contains the Johnson kernel. Let $G$ be the subgroup generated by the Johnson kernel $\mathcal{K}$ together with ($\mathcal{BBF}\cap\mathcal{I}$). Since $G$ is contained in $\mathcal{I}$, separating curves are taken care of by Lemma 2.2 above. If $\gamma$ is a nonseparating curve such that $f(\gamma)$ is disjoint from $\gamma$, then $[f,T\_\gamma]$ is a nonseparating bounding pair; $[f,T\_\gamma]$ lies in $G$ if $f$ does, since $G$ is normal. 


Now the idea is to show that $G$ contains no nonseparating bounding pairs. Certainly $\mathcal{BBF}$ doesn't, but we need to show that this is still true after throwing in $\mathcal{K}$. Since $G$ contains $\mathcal{K}=\ker \big(\tau\colon\mathcal{I}\to \bigwedge^3 H/H\big)$, this is the same as showing that $\tau(\mathcal{BBF}\cap \mathcal{I})$ does not contain the image under $\tau$ of a genus $k$ bounding pair. We know the image of such a BP, and you should be able to write down a condition on $\tau(f)$ for $f\in\mathcal{BBF}$ coming from the triviality of the action on the homology of the double cover. I have the feeling that this condition will rule out such a bounding pair, which would conclude the proof. I don't know how hard this approach would be if you wanted to actually carry it out, though.


