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In symplectic geometry, Darboux's theorem says that locally, any symplectic manifold of dimension $2n$ looks like symplectic Euclidean space (that is, there is some set of coordinates $(x_i, y_i)$ such that the symplectic form looks like $\sum dx_i \wedge dy_i$. Equivalently (I think), there are no local invariants of symplectic manifolds.

Riemannian geometry has an analogous theorem: that a manifold is Euclidean-flat iff the Riemannian curvature is everywhere 0.

Is there an analogous theorem for Kahler geometry? That is, is there some tensor $A$ determined by the Kahler form such that if $A$ is globally 0, then there is some set of coordinates $(x_i y_i)$ such that the Kahler form looks like $\sum dx_i \otimes dy_i$?

Also, are there any other generalizations of the two theorems above?

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  • $\begingroup$ You mean in your first sentence symplectic manifold, and not symmetric manifold. $\endgroup$
    – Malkoun
    Commented Nov 13, 2016 at 7:57
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    $\begingroup$ A Kahler metric can be thought of as consisting of 2 pieces, a metric $g$ and a complex structure $J$, with the two being compatible, in the sense that $J$ is $g$-orthogonal, and such that the Kahler form is closed. I will assume that your local coordinates $x_i$ and $y_i$ are the real and imaginary parts of some holomorphic local coordinates $z_i$. The vanishing of the Riemann curvature is a necessary condition for the Kahler form to be of that form (because $g$ is then locally Euclidean). It is also sufficient, because it forces the (local) holonomy to be trivial. $\endgroup$
    – Malkoun
    Commented Nov 13, 2016 at 8:38
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    $\begingroup$ No. It has to do with the fact that a Kahler form is more than a symplectic form. It also defines a metric. For a Kahler form to be locally canonic we need the associated metric to be locally Euclidean. And we know from the early work of Riemann that there is obstruction to that happening and it is called curvature. $\endgroup$
    – Liviu Nicolaescu
    Commented Nov 13, 2016 at 11:06
  • $\begingroup$ @Malkoun My apologies; I see that I forgot a key phrase. I've now edited to include it. Both of your answers seem to suggest that the Riemannian curvature is the tensor I'm looking for; would you mind expanding on why the vanishing of the Riemannian curvature allows for coordinates that also work for the symplectic form? $\endgroup$
    – user44191
    Commented Nov 14, 2016 at 20:19
  • $\begingroup$ @LiviuNicolaescu My apologies; I see that I forgot a key phrase. I've now edited to include it. Both of your answers seem to suggest that the Riemannian curvature is the tensor I'm looking for; would you mind expanding on why the vanishing of the Riemannian curvature allows for coordinates that also work for the symplectic form? (copied because I could only notify one user per comment) $\endgroup$
    – user44191
    Commented Nov 14, 2016 at 20:19

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Let $(M,g,J)$ be a Kahler manifold. This means that $J$ is a (integrable) complex structure, so that $(M,J)$ is a complex manifold, $g$ is a smooth metric such that $J$ is $g$-orthogonal, i.e. $g(JX,JY) = g(X,Y)$ for any smooth vector fields $X$, $Y$ on $M$ and, moreover, the Kahler form is closed.

Ok, now let us assume that given an arbitrary point $p \in M$, there exist local holomorphic coordinates $z_i$ around $p$ such that

$\omega = \frac{i}{2} \sum_k dz_k \wedge d\bar{z}_k$

with $\omega$ being the Kahler form. By the definition of the Kahler form, we have $g(X,Y) = \omega(X, JY)$, so that in the local coordinates $z_i$, one has therefore

$g = \frac{1}{2} \sum_k dz_k \odot d\bar{z}_k$

Hence $g$ is locally Euclidean, so that the Riemann curvature tensor vanishes identically. I think the OP is more interested in the converse though.

Conversely, suppose $(M,g,J)$ is a Kahler manifold with Riemann curvature vanishing identically. Being flat, given an arbitrary point $p \in M$, the local holonomy group $Hol_0(g,p)$ is trivial. Thus starting with a basis $\alpha_1,\cdots,\alpha_m$ of the space of $1$-forms at $p$ of type $(1,0)$ (with $m$ being the complex dimension of $M$), one can locally extend the $\alpha_i$ using parallel transport locally to some parallel $\alpha_i$ defined in some neighborhood of $p$. In particular, these $\alpha_i$ form a local holomorphic coframe, with each $\alpha_i$ being further closed. Then using Dolbeault cohomology, it follows that there exist local holomorphic coordinates $z_i$ such that $dz_i = \alpha_i$.

Moreover, if at $p$, the $\alpha_i$ were chosen to be unitary (at $p$), it follows that their extensions to a neighborhood are also unitary (and as we have seen, parallel too) so that the metric $g$ can be written locally as:

$ g = \frac{1}{2} \sum_k \alpha_k \odot \bar{\alpha}_k = \frac{1}{2} \sum_k dz_k \odot d\bar{z}_k$

Thus the tensor you are looking for is the Riemann curvature tensor.

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