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 



Trivial canonical bundle of a Ricci-flat, simplyconnected Kähler manifold
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 $SU(n) \subset SO(2n)$.


This ought to be explained in Besse or in Joyce, to mention but two books. I don't have them handy, since I'm travelling, so cannot be more precise as to where in the books to find them.


**Edit**


I found a copy of Besse's *Einstein Manifolds*. You may wish to look at Chapter 2F in that book for the answer to the first question and in Chapter 10 (especially 10.28-10.30) for the answer to the second question.


