TITLE: Holonomy of a Kähler manifold
QUESTION [5 upvotes]: Hi,
I have the following question: Let $(M,J, \omega)$ be a Kähler manifold (not necessary compact). We know that the holonomy group is a subgroup of $U_{n}$. Let $\Omega$ be a constant ($\nabla \Omega = 0$) holomorphic non-vanishing (n,0)-form. Can one say that the holonomy group is now cotained in $SU_{n}$ ? Is it true ? I hope a lot of answers. Thanks in advance.
Mina

REPLY [5 votes]: The answer is yes.
The holonomy principle states that a given a riemannian manifold $(M,g)$ and a point $x\in M$, the datum of a parallel tensor field of a given type is equivalent to the datum of a tensor of the same type at the point $x$ which is invariant under the action of the holonomy group.
Now, $SU(n)$ is the subgroup of $U(n)$ of operators which preserve a (complex) non zero $n$-form in $\mathbb C^n$. Thus, if your manifold is Kähler, thanks to the holonomy principle, the holonomy group of your manifold is contained in $SU(n)$ if and only if there exists a non zero parallel $(n,0)$-form on $M$.
Such a from is closed, hence holomorphic.
In other words, the holonomy group is contained in $SU(n)$ if and only if there exists a holomorphic $n$-form on $M$, which is non zero and parallel.
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