I expect that the holonomy group of an Enriques surface $S$ is $SU(2)\times C_2$. I think this can be proven by the fact that its double cover, which is a K3 surface, has the full $SU(2)$ holonomy, but I failed proving it. The holonomy group should be either $SU(2)$ or its $\pi_1(S)=C_2$-extension $SU(2)\times C_2$. Could someone help me prove or disprove this?
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$\begingroup$ $U(2)$ is not a $C_2$ extension of $SU(2)$, though; unless you and I mean something different by the word extension. $SU(2)$ is the kernel of the determinant map $U(2) \to U(1)$. $\endgroup$– José Figueroa-O'FarrillCommented Dec 15, 2013 at 10:22
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$\begingroup$ You are right. $U(2)$ should have been $SU(2)\times C_2$. Thanks for pointing out the mistake. $\endgroup$– David K.Commented Dec 15, 2013 at 10:25
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Assume that you have endowed an Enriques surface $S$ with a Ricci-flat Kähler metric $g$. The holonomy $H$ of $g$ cannot be contained in $\mathrm{SU}(2)$ because the canonical bundle of $S$ is not trivial (though its square is trivial). Meanwhile, the identity component of $H$ has to be equal to $\mathrm{SU}(2)$ because this is the holonomy of the (simply-connected, non-product) K3 surface that is the double cover of $S$ endowed with one of its Ricci-flat Kähler metrics. The fundamental group of $S$ is $\mathbb{Z}_2$, so $H$ must be a $\mathbb{Z}_2$-extension of $\mathrm{SU}(2)$ that lies inside $\mathrm{U}(2)$. There is only one of these, namely the group consisting of those matrices in $\mathrm{U}(2)$ with determinant $\pm1$, so this must be $H$.
N.B.: $H$ is not isomorphic to $\mathrm{SU}(2)\times\mathbb{Z}_2$. In the latter, the set of elements that satisfy $a^2=1$ has $4$ members: $(I_2,1),(-I_2,1),(I_2,-1),(-I_2,-1)$, while, in the former, the set of matrices satisfying $A^2=I_2$ consists of two points, $\pm I_2$, together with the $2$-sphere consisting of the elements of the form $iJ$ where $J$ lies in $\mathrm{SU}(2)$ and satisfies $J^2 = - I_2$.
answered Dec 15, 2013 at 11:13
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$\begingroup$ @Dr. Bryant Thanks you very much for the answer. I do not understand how you conclude that the identity component of the holonomy is $SU(2)$. As to $H$, you are totally right. $\endgroup$– David K.Commented Dec 15, 2013 at 11:58
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$\begingroup$ @DavidK.: Oh. I am relying on the Berger classification of holonomy for this: We know that the identity component of the holonomy must be a subgroup of $\mathrm{SU}(2)$ and, by Berger, we also know that the only connected proper subgroup of $\mathrm{SU}(2)$ that can be the holonomy of a $4$-manifold is the identity subgroup, which can only happen if the metric $g$ is flat, which is clearly not possible. $\endgroup$ Commented Dec 15, 2013 at 12:08
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$\begingroup$ @Dr. Bryant Thank you very much for the clarification. I am not familiar with the Berger classification. I will take a close look at it. I appreciate your kind reply. $\endgroup$– David K.Commented Dec 15, 2013 at 12:14