# Flat connections on 3-manifold with boundary

Suppose $Y$ is a 3-manifold and the boundary $\Sigma:=\partial Y$ is non-empty. Let $G$ be a Lie group with trivial center. Let $\overline {\mathcal A}_{flat}(\Sigma)$ and $\overline {\mathcal A}_{flat}(Y)$ denote the gauge equivalence classes of flat connections on the trivial bundles $\Sigma \times G$ and $Y \times G$ respectively. The space $\overline {\mathcal A}_{flat}(\Sigma)$ is a finite dimensional symplectic manifold with singularitites. I am trying to understand why the image of the restriction map $\overline {\mathcal A}_{flat}(Y) \to \overline {\mathcal A}_{flat}(\Sigma)$ is a Lagrangian submanifold. I know why the image is an isotropic subspace, but I am having a problem with the dimensions. In particular, the space $$\overline {\mathcal A}_{flat}(\Sigma)\simeq \operatorname{Hom}(\pi_1(\Sigma),G)/G$$ has dimension $(2\operatorname{genus}(\Sigma)-2)\dim(G)$, but I can not see why the space $\operatorname{Hom}(\pi_1(Y),G)/G$ has half that dimension.

For completeness, I include a proof of why the image of $\overline {\mathcal A}_{flat}(Y)$ in $\overline {\mathcal A}_{flat}(\Sigma)$ under the restriction map is isotropic. Let $\mathcal A(Y)$ be the space of connections on $Y \times G$. There is a one-form $\mathcal F$ on $\mathcal A(Y)$ defined as $${\mathcal F}_A(\alpha):=\int_Y \langle \alpha \wedge F_A\rangle,$$ where $A \in \mathcal A(Y)$ is a connection, $F_A \in \Omega^2(Y,\mathfrak g)$ is its curvature and $\alpha \in T_A\mathcal A(Y)$. We can see by some calculations that $$d\mathcal F_A(\alpha,\beta)=\int_\Sigma \langle \alpha \wedge \beta \rangle,$$ where $\alpha, \beta \in T_A\mathcal A(Y)$, which coincides with the symplectic form on $\mathcal A(\Sigma)$ the space of connections on $\Sigma \times G$. When restricted to $\mathcal A_{flat}(Y)$, $\mathcal F$ vansihes, and so does $d\mathcal F$. Therefore the restriction map takes $\mathcal A_{flat}(Y)$ to an isotropic submanifold of $\mathcal A_{flat}(\Sigma)$. By gauge invariance, the restriction map takes $\overline {\mathcal A}_{flat}(Y)$ to an isotropic submanifold of $\overline {\mathcal A}_{flat}(\Sigma)$.

• See arxiv.org/pdf/0902.2589.pdf. The image of the restriction map need not be smooth even in the smooth part of the surface group character variety. If you treat the image as a variety and not a scheme then it is unknown if the image of half-dimensional. However, if you work on the scheme-theoretic level then indeed the image is Lagrangian. – Misha May 12 '16 at 6:27
• Thanks @misha. The paper you refer to takes care of the case when $G$ is complex reductive. Can we say something when $G$ is compact, especially when it is $SU(2)$? In that case, the result seems to be taken for granted by many authors. – Anon May 12 '16 at 8:04
• If you work with schemes, this is the same computation. – Misha May 12 '16 at 20:53

It is an easy exercise with Poincare duality to see that the image of the map in cohomology (so at the level of "Zaraski" tangent spaces) $$H^1(Y;ad_\rho) \to H^1(\Sigma;ad_\rho)$$ is half dimensional. Indeed this map appears in the long exact sequence of the pair $(Y,\Sigma)$ $$H^1(Y;ad_\rho) \to H^1(\Sigma;ad_\rho)\to H^2(Y, \Sigma;ad_\rho)$$ and the two arrows are dual by PD so have the same rank. The exactness tells you the sum of these ranks is $dim(H^1(\Sigma;ad_\rho))$.
Here $\rho$ is a representation of $\pi_1(Y)\to G$ corresponding to the given flat connection and $ad_\rho$ is the local system with fiber $\mathfrak{g}$ corresponding to the adjoint action of $\rho$ on $\mathfrak{g}$.
• It might be helpful to comment here that since $G$ is assumed to be semisimple, the adjoint representation is self-dual, and hence Poincaré duality applies. – Ian Agol May 16 '16 at 15:37
For the case of compact $G$, see Proposition 3.27 in: Freed, Daniel S. Classical Chern-Simons theory. I. Adv. Math. 113 (1995), no. 2, 237–303.
For the case of complex reductive $G$, see Theorem 61 in: Sikora, Adam S. Character varieties. Trans. Amer. Math. Soc. 364 (2012), no. 10, 5173–5208.