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For Cylindrical $Y\times\mathbb R$, where $Y$ is a closed oriented 3-manifold.

If it is necessary, we could consider the $b_1(Y)=0$ case.

Fix a Line bundle $L\to Y\times \mathbb R$ and a Hermitian connection $A_0$, we know that all other Hermtian connections, denoted by $\mathcal C$, can be regarded as the 1 forms on $Y\times\mathbb R$, i.e. $\mathcal C\overset{\sim}{=}\sqrt{-1}\Omega^1(Y\times\mathbb R)$.

And the $U(1)$-gauge group $\mathcal G:=Map(Y\times\mathbb R,U(1))$, the action is given by $$(g,A_0+a)\mapsto (A_0,a-g^{-1}dg).$$

Question: Is there a method to find a subspace $V\subset \mathcal C$ such that $$V\overset{\sim}{=}\mathcal{C/G}.$$

PS:

We know that for the temporal gauge, $\{a\in \mathcal C~|~ a \mbox{ contains no } dt \mbox{ part and } d^*_Ya=0 \}$, is surjective to the quotient space.

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    $\begingroup$ The reduced gauge group $\mathcal G_0$ s the kernel of the evaluation map $\mathcal G \to S^1$ at some point $p \in Y \times \Bbb R$. It acts freely on $\mathcal C$ with the same quotient, and is homotopy equivalent to $H^1(Y;\Bbb Z)$. If the quotient map has a section, then necessarily $b_1(Y) = 0$. When this is the case, then the subspace of connections in temporal gauge is such a section. $\endgroup$
    – mme
    Commented Jul 24, 2016 at 5:51
  • $\begingroup$ Yes, $b_1=0$ gives the global slicing. But it is still surjective. Of course if anyone could find a method for $b_1=0$ case, it is also a great job. $\endgroup$
    – DLIN
    Commented Jul 24, 2016 at 6:38

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