# Homotopy group of space of gauge fields modulo gauge equivalence on T^4

Singer observed in 1978 (Comm.Math.Phys. 60, 7-12) that the homotopy group of the space of gauge fields modulo gauge equivalence with gauge group $G$ on $S^4$ is given by

$\pi_n({\cal A}/{\cal G}) = \pi_{n-1}{\cal G} = \pi_{n+3} G$

Does anyone know what the corresponding expression is if the base manifold $S^4$ is replaced by $T^4$?

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Doesn't the first equality follow from the long exact sequence of homotopy sequence associated to the fibration $\mathcal{G} \to \mathcal{A} \to \mathcal{A}/\mathcal{A}$ together with the fact that $\mathcal{A}$ is contractible, so it is really the second equality you're interested in? – skupers Aug 9 '10 at 11:56
I meant $\mathcal{A}/\mathcal{G}$... – skupers Aug 9 '10 at 11:56
@skupers: this is no fibration! – Orbicular Aug 9 '10 at 12:20
In fact, I. Singer has $G=SU(n)$, or, which is more or less the same, $G=PSU(n)=SU(n)/\mathbb{Z}_n$, and observes that the subspace of $\mathcal{A}$ where the $\mathcal{G}$-stabilizer is non trivial (i.e. reducible holonomy) is "infinite codimensional" in $\mathcal{A}$. Hence the subspace $\mathcal{B}$ with irreducible holonomy has trivial homotopy groups, and $\mathcal{A}/\mathcal{G}$, $\mathcal{B}/\mathcal{G}$ have the same homotopy groups. Since now $\mathcal{G}\to\mathcal{B}\to\mathcal{B}/\mathcal{G})$ is a fibration, you obtain the claim. – BS. Aug 9 '10 at 13:36
But how specific are these statements to $S^4$? I'm only worried about $T^4$ because for example the space of flat connections on $T^4$ are totally different from $S^4$ (because $T^4$ is not simply connected). So unless somebody argues why this is the case I'm hesitant to generalize the above to $T^4$ from $S^4$. – Daniel Aug 9 '10 at 15:03