Question

Let $(M,J,\omega, \Omega)$ be a calabi-yau manifold (not necessary compact). Does it follow that the holonomy group of $M$ is $SU_{n}$, where $n$ is the complex dimension of $M$ ?

Answer 1

Well, it depends on what you call a Calabi-Yau manifold (there are several possible terminologies indeed).

First of all, a compact Kähler manifold with trivial canonical class does not necessarily have holonomy group $SU_n$ (with respect to some Ricci-flat metric). More precisely, the holonomy group of some compact Kähler manifold $(X,\omega)$ is included in $SU_n$ iff there exists a non-zero parallel holomorphic $n$-form. As a consequence the restricted holonomy group $H_0$ is included in $SU_n$ iff $(X,\omega)$ is Ricci-flat.

Now the holonomy groups of a Ricci flat compact Kähler manifold can be smaller than $SU_n$: think about any torus (the holonomy is trivial) or any holomorphic symplectic variety (the holonomy is $SP(n/2)$.

However, there is a result, which was maybe what you had in mind:

Theorem. Let $(X,\omega)$ be a compact Kähler manifold of dimension $n\geq 3$ with holonomy group $SU_n$. Then $X$ is projective and $H^0(X, \Omega_X^p)=0$ for every $0 < p < n$ and $\chi(\mathcal O_X)=1+(-1)^n$.

A manifold with such properties is sometimes called Calabi-Yau, indeed. For a reference, see Beauville's article "Variétés Kähleriennes à première classe de Chern nulle".

As for the non-compact case, I don't know if such a results holds.