There are a lot of different definitions of a Calabi-Yau manifold. Roughly, we can divide them in two sets, see Wikipedia https://en.wikipedia.org/wiki/Calabi%E2%80%93Yau_manifold . I will refer to this link for notations and everything else.

Wikipedia says that when $M$ is simply-connected, these two definitions are equivalent. But if $M$ is only connected, then we have that the first one implies the second one but not the vice versa. I agree with both the statements.

My problem is the following. Suppose that $M$ is not simply-connected and the holonomy group $Hol(M)$ is contained in $SU(n)$ (here $n$ is the complex dimension of $M$). By the holomy principle there exists a parallel $n$-form, i.e. a parallel section of the canonical bundle $K_M=\Omega_M^n$. In particular this implies that $K_M$ is flat and since the curvature of $K_M$ is proportional to the Ricci tensor of $M$, we get that $M$ is Ricci-flat. According to Wikipedia, $M$ is Ricci-flat if and only if $c_1(M)=0$. So we are leading to say that $Hol(M)\subset SU(n)$ if and only if $c_1(M)=0$. But Wikipedia itself says that this double implication is not true when $M$ is not simply-connected. I am stuck in front of this absurd.

Can someone please point out where my argument fails and why?

Thank you in advance!

EDIT: I think to have solved my problem(s) now.

1) first of all I was wrong because Wikipedia talks about local holonomy and not restricted holonomy (for which the holonomy principle doesn't hold);

2) if $Hol(M)\subset SU(n)$ I conclude that $M$ is Ricci-flat, but the vice versa is not true in general because of the following: if $K_M$ is flat I claimed that there exists a parallel section; but what happens if the bundle has no sections? We cannot go the other direction. This is for example the case of an Enrique surface.

3) in the non simply-connected case, I can run my argument on the universal cover of $M$, which is simply-connected and hence everything works well. After that I can push-down the result and see what happens. In the case the holonomy is contained inside $SU(n)$ we can still state that the canonical bundle is trivial, but it can have non trivial curvature.

  • $\begingroup$ First of all, $M$ has a Ricci flow metric iff $c_1(M)_{\mathbb R} = 0$, where $c_1(M)_\mathbb{R}$ is the image of the (integral) first Chern class $c_1(M) \in H^2(M, \mathbb Z) \to H^2(M, \mathbb R)$. So having a Ricci flat metric is a weaker condition than $c_1(M) = 0$. But according to wiki, even $c_1(M) = 0$ is a weaker condition. So I guess there might be cases that $K_X$ is trivial as a (smooth) complex line bundle but not trivial holomorphically? $\endgroup$ – user99914 Oct 8 '15 at 5:41
  • $\begingroup$ Dear @John, thank you for your reply. Yes I agree with you and an example is an hyperelleptic surface: it has trivial integral first chern class but non-trivial canonical bundle. What I cannot see is that with my argument I say that $Hol(M)\subset SU(n)$ iff $M$ is Ricci-flat, which by Wikipedia is a 'iff' with having the restricted holonomy group $Hol^0(M)\subset SU(n)$. So for sure my argument fails in some point, but I cannot see where... $\endgroup$ – A. Prufrock Oct 8 '15 at 14:00
  • $\begingroup$ Related: mathoverflow.net/questions/15003/… $\endgroup$ – user99914 Oct 10 '15 at 2:46

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