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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.

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It seems to me that this is pretty immediate once you think in terms of holonomy representations. Have you thought in that direction? Maybe the bulk of the work is in relating the notion of reducible connection to reducibility of the holonomy representation. (Actually, I'm not completely sure I understand your definition of reducible: are you saying a connection is reducible if there is a non-trivial gauge transformation that fixes it? That doesn't sound right.) –  Dan Ramras Jan 26 '11 at 5:55
    
@ Xuanting - What do you mean by the "Product connection"? Do you mean the product of a $1$-form on $M$ and a map from $M$ to $\mathfrak{su}_2$ ? @ Dan - I believe the definition the OP stated is correct. At least it's the one that is used in Fukaya's notes and a couple of other places that I have seen. –  Somnath Basu Jan 26 '11 at 6:41
    
Xuanting, hope you don't mind that I edited your question (added Latex and a tag), and made the question the title (which is the convention on MO). –  Tim Perutz Jan 26 '11 at 14:15
    
Of Course. Thank you for your help. –  Xuanting Cai Jan 26 '11 at 23:29
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2 Answers

up vote 4 down vote accepted

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.

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Thanks for your explaination. –  Xuanting Cai Jan 26 '11 at 23:06
    
I am a knot theory student. Try to learn some 4 manifold thing. So maybe this is a dummy question for expert. Thank you. –  Xuanting Cai Jan 26 '11 at 23:28
    
You're welcome. I don't think it's a dumb question - if I did, I wouldn't have bothered to answer! –  Tim Perutz Jan 27 '11 at 3:33
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The following is probably well known. For example, one can find it in Fukaya's notes on Floer homology, $A_\infty$-categories and topological field theory.

Let $P$ be the trivial principal $SU(2)$-bundle on $M$ and let $\mathcal{A}(M)$ (resp. $\mathcal{A}^{\textrm{flat}}(M)$) denote the space of connections (resp. flat connections) on $P$. For $a\in\mathcal{A}(M)$ let $\mathcal{G}^a$ denote the stabilizer of $a$ under the action of the gauge group $\mathcal{G}(M)$. Then $\mathcal{G}^a=\pm 1, U(1)$ or $SU(2)$.

One can elaborate on a proof of this later. May be from this one can deduce that there are no flat connections with stabilizers $U(1)$ and the ones that have full stabilizers are the ones that you're looking for. On the other hand, following what Dan had said before, one has the following :

$\mathcal{A}^{\textrm{flat}}(M)/\mathcal{G}(M)\cong \textrm{Hom}(\pi_1(M),SU(2))/SU(2)$.

Here the right hand quotient is to be interpreted as all homomorphisms up to conjugation. Again, this may lead to some proof but it's been a while since I have seen these things. I'll put up something once I think it through a bit.

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$A^flat(M)/G(M)≅Hom(π_1(M),SU(2))/SU(2)$ is easy to prove. Just use geometric definition of connection. –  Xuanting Cai Apr 18 '11 at 16:07
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