Trivial canonical bundle of a Ricci-flat, simplyconnected Kähler manifold

Hallo,

I have two questions where I do not really know how to deal with them. Let $(M,J,g)$ be a Kähler manifold, where $g$ is the Riemannian metric and denote by $\omega(\cdot , \cdot) = g(J \cdot , \cdot)$ the Kähler form. How can one show:

1. If $\pi_{1}(M)=0$ and $Ric(g)=0$ then the canonical bundle $K_{M} := \Lambda^{(n,0)}M$ is trivial.

2. If $Ric(g)=0$ and $K_{M}$ is trivial then $Hol(g) \subseteq SU(n)$.

I have no idea and I also dont find any proof. Do you know any reference where I can find these? Thanks

berhard

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The answer to the first question is that the Ricci tensor defines a (1,1) form (called the Ricci form) and this is the curvature of the connection on $K_M$ induced by the Levi-Civita connection on $T M$. So if the manifold is Ricci-flat, $K_M$ is flat and hence if $M$ is simply-connected it is trivial. Flatness says the restricted holonomy group acts trivially, simply-connectedness says the holonomy group coincides with the restricted holonomy group.
The answer to the second question is that the holonomy representation on $K_M$ is the determinant of the holonomy representation on the holomorphic tangent bundle $T^{(1,0)}M$. Hence if $K_M$ is trivial, the determinant representation is trivial: the holonomy around any loop has determinant $1$. Since the manifold is Kähler, the holonomy representation is in $U(n) \subset SO(2n)$, whence the unit determinant condition says it is actually in $SU(n) \subset SO(2n)$.