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?



Holonomy group of Enriques surface
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 $J^2 = - I\_2$. 


