As explained in [Ash and Yasaki, Theorem. 2.8] the Steinberg module $St_{n}$ is the dualizing module for any finite index subgroup $\Gamma\le\mathrm{GL}_{n}(\mathbb{Z})$.
Following the notation of [Brown, Chapter 8] this means that $D:=H^{d}(\Gamma, \mathbb{Z}\Gamma)\cong St_{n}$, and by Theorem 10.1 of [Brown, Chapter 8] it follows that there are natural isomorphisms $H^{i}(\Gamma,-)\cong H_{d-i}(\Gamma, D\otimes -)$ for all $i$, where $d=cd(\Gamma)$ is the cohomological dimension of $\Gamma$ (i.e. the cohomological dimension of the trivial $\mathbb{Z}\Gamma$-module $\mathbb{Z}$).
The same theorem also implies that $D=H^{d}(\Gamma, \mathbb{Z}\Gamma)\cong St_{n}$ is torsion-free as an abelian group.
The particular case I am most interested in is when $n=2$ (and hence $d=1$). But this is where I am getting confused: in Milnor's notes on $K$-theory [Milnor, Theorem 10.5] he proves that $St_{2}$ is the universal central extension of $SL_{2}(\mathbb{Z})$, and then goes on to show that in fact $St_{2}$ is isomorphic to the tre-foil knot group (i.e. the fundamental group $\pi_{1}(S^{3}\setminus K)$ where $K$ is the tre-foil knot). Furthermore, it is well known that this knot group is actually the braid group on three strands $B_{3}=\langle\sigma_{1},\sigma_{2}\mid \sigma_{1}\sigma_{2}\sigma_{1} = \sigma_{2}\sigma_{1}\sigma_{2}\rangle$ which is not even an abelian group (in fact the abelianisation of $B_{3}$ is just $\mathbb{Z}$).
So my main questions are:
- Am I interpreting this notion of duality correctly, i.e. is $St_{2}$ as constructed by Ash and Yasaki really the dualizing module for $SL_{2}(\mathbb{Z})$ (and hence for any finite index subgroup $\Gamma\le SL_{2}(\mathbb{Z})$)?
- Is $St_{2}$ really isomorphic to $B_{3}$?
And as a further question: is there a way to explicitly calculate the dualizing module $D=H^{1}(\Gamma, \mathbb{Z}\Gamma)$ for any finite index subgroup $\Gamma\le\mathrm{SL}_{2}(\mathbb{Z})$, and if so can one also produce a projective resolution of $D$ as a $\mathbb{Z}\Gamma$-module?
