# Moduli space of flat connection over homology 3-sphere

I'm trying to understand the space of flat connections of the trivial $$\mathrm{SU}(2)$$-bundle over a closed, oriented homology three-sphere (for the purpose of understanding the instanton Floer homology of it).

From now on I will just define the space of gauge equivalent classes of flat connections on it to be $$R$$.

Are the following facts correct and why?

1. The trivial connection is the only reducible connection.

2. $$R$$ is isolated generically (according to the perturbation of Chern-Simons functional)

3. $$R$$ is compact.

• Cross-posed at MSE, here: As a general rule, if you post on one site, you should wait for several days before posting on another (if no satisfactory answer emerges). In any case, you should add a note on cross-posting in order to eliminate duplication of efforts. Commented Apr 14, 2022 at 18:20
• @MoisheKohan Thanks for letting me know this! I wasn't sure where to post this question so did it on both sites. Sorry for the confusion. Commented Apr 14, 2022 at 20:17

Let $$M$$ be your homology $$3$$-sphere. First, (as suggested by @MoisheKohan on MSE), note that your space is the quotient of $$\mathrm{Hom}(\pi_1(M), SU(2))$$ by the conjugation action of $$SU(2)$$. Indeed, every flat connection gives a holonomy representation, defined up to conjugation, which is invariant by Gauge transformation, from every representation $$\rho$$ you can build the flat $$SU(2)$$-bundle $$\tilde M \times SU(2)/(x,g)\sim(\gamma \cdot x, \rho(\gamma)g)~.$$

Question 3: Yes. This is simply because $$\pi_1(M)$$ is finitely generated, and $$\mathrm{Hom}(\pi_1(M), SU(2))$$ embeds in $$SU(2)^S$$ where $$S$$ is a generating set.

Question 1: Yes. Indeed, the holonomy of a reducible connection takes values (up to conjugation) in $$U(1)\times U(1)$$ which is abelian. It is thus trivial, since $$H_1(X) = \{0\}$$.

Question 2: I don't understand what you mean by "generically" but I think the answer is No.

The trivial representation is isolated because its algebraic tangent space in the character variety is $$H^1(\pi_1(X),\mathfrak{su}(2)) = \{0\}$$ since $$X$$ is a homology sphere.

But other representations might not be: take $$X_1$$ and $$X_2$$ two homology spheres admitting non-trivial flat $$SU(2)$$-bundles. You have irreducible representations $$\rho_i: \pi_1(X_i) \to SU(2)$$. Consider now the homology sphere $$X= X_1 \sharp X_2$$ (connected sum). We have $$\pi_1(X) = \pi_1(X_1) \star \pi_1(X_2)$$. For every $$g\in SU(2)$$, there is a representation $$\rho_g: \pi_1(X) \to SU(2)$$ such that $${\rho_g}_{\vert \pi_1(X_1)} = \rho_1$$ and $${\rho_g}_{\vert \pi_1(X_2)} = g\rho_2 g^{-1}$$.

These form a pairwise non-conjugate $$SU(2)$$-family of representations of $$\pi_1(X)$$.

I wonder if $$SU(2)$$ representations of irreducible homology spheres are rigid, in which case the above construction would completely describe deformations of flat bundles on homology spheres. The only explicit examples I can think of are Brieskorn spheres, the $$SU(2)$$-representations of which are rigid (I think).

• Why can a reducible connection not take values (up to conjugation) in the upper triangle matrices? Reducible need not be completely reducible. Commented Apr 15, 2022 at 7:01
• For SU(2)-connections, it does (the orthogonal of an invariant subundle is invariant) Commented Apr 15, 2022 at 7:16
• Rigidity can fail even for an irreducible homology sphere $Y$. Say $Y$ splits along an incompressible torus $T$ as $Y=Y_1 \cup_T Y_2$ (example: glue two nontrivial knot complements, meridian to longitude and vice versa), and $\rho:\pi_1(Y)\to SU(2)$ is irreducible on each $Y_i$ separately. Then $\rho$ is reducible on $T$, with image in a $U(1)$ subgroup, and you can "bend" it by the same trick as for connected sums: keep $\rho|_{\pi_1(Y_1)}$ fixed and replace $\rho|_{\pi_1(Y_2)}$ with $g\rho g^{-1}$ for elements $g$ in that subgroup. This gives a $U(1)$ family of non-conjugate representations. Commented Apr 15, 2022 at 9:06
• Nice! One needs to be a bit careful however : not every representation of $pi_1(T)$ extends to $pi_1(Y_i)$ and we need a representation that extends on both sides. Do you have an argument for that? Commented Apr 15, 2022 at 9:22
• I don't claim that one always exists, but that this construction applies when it does. E.g. if you splice together two copies of the same knot complement $E_K$ in this way, then it suffices to find an irreducible representation $\rho: \pi_1(E_K) \to SU(2)$ with $\rho(\mu)=\rho(\lambda)$ and use this $\rho$ on each copy of $E_K$. But this is the same as an irreducible representation of the (-1)-surgery group $\pi_1(S^3_{-1}(K)) \cong \pi_1(E_K)/\langle\mu\lambda^{-1}\rangle$, and Kronheimer and Mrowka's proof of property P showed that for $\pm1$-surgery, such representations always exist. Commented Apr 15, 2022 at 9:39