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I have the following question: Let $(M,I,\omega)$ be a not necessary compact Kaehler manifold of complex dimension $n$. Assume that there exists a nowhere vanishing holomorphic $(n,0)$-form $\omega$ such that $\omega^{n} = K(n) \Omega \wedge \overline{\Omega}$, where $K(n)$ is a constant depending only on the dimension $n$. From this it obviously follows that $M$ is Ricci-flat. Can one say something about the holonomy of the Levi-Civita connection with respect to the metric? Is the holonomy contained as a subgroup of $SU_{n}$ ? Or is it the whole $SU_{n}$ (well I think not, since I am not assuming something like: $M$ to be simply connected)? I would be very thankfull for a lot of answers.

Greetings Mina

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does anyone have an idea ? – Mina Oct 25 '12 at 17:27
The holonomy doesn't have to be the whole of $SU(n)$, because flat complex Euclidean space has holonomy $\{1\}$ as does any flat complex torus or a product of complex Euclidean space with a flat complex torus. On the other hand, the holonomy could be $SU(n)$. – Ben McKay Jul 5 '13 at 13:51

Yes, the holonomy of this manifold is in $SU(n)$. Indeed, the Chern connection on the canonical bundle is flat and its holonomy preserves $\Omega$, because its curvature is $\partial\bar\partial |\Omega|^2=0$. However, the Chern connection on canonical bundle is induced by the Levi-Civita connection on $M$, hence the Levi-Civita connection also satisfies $\nabla\Omega=0$.

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