MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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.


share|cite|improve this question
up vote 4 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.