Let $M$ be a homology sphere. Suppose $P=M\times SU(2)$ is the trivial $SU(2)$ principal bundle. Let $R$ be all reducible connections on $P$. Here $A$ in $R$ is reducible if the gauge transformation group acting on $A$ has nontrivial stable subgroup. I want to see that the only flat connection in $R$ is the product connection.



Why must a reducible flat SU(2)-connection over a homology sphere be trivial?
I think the statement is that in a principal $SU(2)$-bundle $P$ over a connected manifold $M$ with $H\_1(M;\mathbb{Z})=0$, every reducible flat connection has trivial holonomy and therefore trivializes $P$.


For a Lie group $G$, gauge transformations $u$ that stabilize a $G$-connection $A$ are covariant-constant, and therefore determined by their value $u(x)$ at a point $x\in M$. Since $u$ commutes with the parallel transport defined by $A$, it centralizes the holonomy group $H\_A\subset G$ (the image of the holonomy representation $\rho\colon \pi\_1(M,x)\to G$). This sets up an isomorphism between the stabilizer of $A$ and $C\_G(H\_A))/Z(G)$, where $C\_G(H\_A)$ is the centralizer of $H\_A$ in $G$ and $Z(G)$ the center ($\pm I$ for $G=SU(2)$). See Donaldson-Kronheimer's book.


The centralizer of a non-central element in $SU(2)$ is a circle $\mathbb{T}$ (to see this, observe that any element is conjugate to a diagonal element, or else consider rotations in $SO(3)$). The centralizer of any non-abelian subgroup of $SU(2)$ lies in the intersection of two different circle-subgroups, and is therefore just the center.


When $H\_1(M)=0$, every non-trivial homomorphism $\pi\_1(M)\to SU(2)$ has non-abelian image, since the abelian ones factor through $H\_1$. For any flat connection $A$ with non-trivial holonomy, the centralizer therefore equals the center, which makes $A$ irreducible.


