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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.

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    $\begingroup$ 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. $\endgroup$
    – Moishe Kohan
    Commented Apr 14, 2022 at 18:20
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    $\begingroup$ @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. $\endgroup$
    – Lamda8
    Commented Apr 14, 2022 at 20:17

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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 $$\tilde M \times SU(2)/(x,g)\sim(\gamma \cdot x, \rho(\gamma)g)~.$$

Let me now answer your questions, by order of generality.

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).

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  • $\begingroup$ Why can a reducible connection not take values (up to conjugation) in the upper triangle matrices? Reducible need not be completely reducible. $\endgroup$
    – ThiKu
    Commented Apr 15, 2022 at 7:01
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    $\begingroup$ For SU(2)-connections, it does (the orthogonal of an invariant subundle is invariant) $\endgroup$
    – Nicolast
    Commented Apr 15, 2022 at 7:16
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    $\begingroup$ 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. $\endgroup$
    – Steven Sivek
    Commented Apr 15, 2022 at 9:06
  • $\begingroup$ 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? $\endgroup$
    – Nicolast
    Commented Apr 15, 2022 at 9:22
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    $\begingroup$ 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. $\endgroup$
    – Steven Sivek
    Commented Apr 15, 2022 at 9:39

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