1
$\begingroup$

An $n$-dimensional complex manifold $M$, endowed with an Hermitian metric $g$, is Kähler if and only if the holonomy group of $g$ is contained in $U(n)$. If $g$ is Hermitian but not Kähler do we necessarily get a reduction of the holonomy group.

EDIT: To clarify, I mean holonomy group with respect to the Chern connection, i.e. the second connection specified by Robert below.

A manifold is called hypercomplex if it admits three integrable complex structures $I,J,$ and $K$ which together give a representation of the quaternions $\mathbb{H}$. Each complex structure will admit three Hermitian metrics. Can we conclude any reduction of their holonomy groups?

$\endgroup$
  • 3
    $\begingroup$ Any complex manifold is oriented, so a Hermitian metric must have holonomy in the special orthogonal group. I think that is the holonomy of the generic Hermitian metric on any complex manifold. $\endgroup$ – Ben McKay Apr 30 at 19:32
2
$\begingroup$

This is really a comment, but it's far too long to go into a comment window.

Your questions need to be made more precise.

First, if $(M,J)$ is complex and $g$ is a metric on $M$ that is $J$-Hermitian, there are (at least) two possible connections that you can associate with $(M,J,g)$ and the different connections can have different holonomies. There is the Levi-Civita connection $\nabla^g$ of $g$ (which may not have $\nabla^g J = 0$), and, second, there is a canonical connection $\nabla^{J,g}$ for which $J$ and $g$ are parallel but has torsion of type (1,1). These are equal iff $(M,J,g)$ is Kähler.

For which of these are you asking about the holonomy? (There is actually a whole line of connections joining the two when $(M,J,g)$ is not Kähler, as the space of connections on the tangent bundle is an affine space, so you could be asking about some connection that is a combination of these. Actually, there are even more than that associated to the triple $(M,J,g)$, but let's not go there.)

Second, I don't know what you mean, in the hypercomplex case, by the statement that "Each complex structure will admit three Hermitian metrics." Where are these metrics coming from? They can't naturally arise from any construction using the three anti-commuting complex structures alone. For example, consider $M= \mathbb{H}^n$ with its natural hypercomplex structure given by multiplication on the right by $i$, $j$, and $k$. The group $\mathrm{GL}(n,\mathbb{H})$ acts on the left (by matrix multiplication preserving the hypercomplex structure), but it does not preserve any metric.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I guess I just mean that there exists an Hermitian metric for each $I,J$, and $K$. Nothing more than this trivial observation. $\endgroup$ – Fofi Konstantopoulou May 1 at 13:52
  • $\begingroup$ @FofiKonstantopoulou: I see. Of course, there will also exist metrics that are Hermitian for all three. $\endgroup$ – Robert Bryant May 1 at 16:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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