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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 $M$ ?

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You should tell us what your definition of Calabi-Yau manifold is. –  diverietti Mar 20 '12 at 9:20

1 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.

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Whether compact or noncompact, a Riemannian manifold with parallel complex structure and parallel holomorphic volume form either (1) has holonomy $SU(n)$, or else (2) has a parallel splitting of the tangent bundle or else (3) has a parallel holomorphic symplectic form. This follows from Berger's classification of holonomy groups. –  Ben McKay Mar 20 '12 at 8:47
    
I am not interested in this results. just in the holomony group. How could one calculate it ? –  pascal Mar 20 '12 at 9:10
    
@ Ben McKay: yes and what when te manifold is calabi-yau??? –  pascal Mar 20 '12 at 9:11
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Henri, diverietti, and Ben are all trying to explain to you that there are MANY definitions in the literature of the term "Calabi-Yau". These definitions are NOT equivalent. There is no standard definition that is universally accepted. In fact, there is one definition that says precisely that Calabi-Yau means SU(n) holonomy. So the answer to your question depends on what your definition of the term "Calabi-Yau" is. –  Spiro Karigiannis Mar 20 '12 at 16:05
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By the way- the holonomy group is in some sense a "global" invariant. It is not easy to "calculate" it in general. One usually does need to check exactly what Ben McKay is telling you- that there are nontrivial parallel forms. –  Spiro Karigiannis Mar 20 '12 at 16:06

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