Definition: One definition of a Calabi-Yau manifold is that of a Kahler manifold $X$ of dimension $n$, with a non-vanishing holomorphic $(n,0)$-form $\Omega$ satisfying $$ \omega^n/n! = (-1)^{n(n-1)/2}(i/2)^n \Omega \wedge \bar{\Omega}, $$ where $\omega$ is the symplectic form on $X$.

For any symplectic form we can take Darboux coordinates $(x,y)=(x_1,\dots,x_n, y_1,\dots,y_n)$ such that $$\omega = \sum_i dx_i \wedge dy_i.$$

(Note: In general these coordinates are not holomorphic, i.e. $z= x+iy$ are not complex coordinates.)

Question: In these coordinates, does $\Omega$ look like $\Omega = dx_1 +idy_1 \wedge \dots \wedge dx_n +idy_n$?

Thoughts: This seems like a lot to expect since the condition only requires the wedge product to be non-trivial as we are in the top degree and the coefficients can be adjusted. So I guess this must be a highly underdetermined equation.

On the other hand, I don't see that it is has to be false. The Calab-Yau condition would definitely hold and it seems like it is true when $n=1$. In the $X=\mathbb C^n$ setting it is also true (however, in that case $(x,y)$ do induce complex coordinates). Another reason I'm thinking this might be true up to some function is that a Kahler manifold with SU$(n)$ holonomy admits such an $(n,0)$-form $\Omega$ and it is unique up to change of phase (see Lemma 4.4). So maybe it is not as underdetermined as it looks?


Considering the generality of your question, the answer is 'no'.

Your insight towards an affirmative answer relies on the complete Kahler structure of the manifold, not just the associated symplectic form. However, a Darboux chart only 'sees' the symplectic form and is indifferent to the specific (compatible) complex structure you consider. So it shall be possible to come up with a Darboux chart such that $\Omega$ does not have the simple expression you conjectured.

In order to prove this possibility, suppose that you have a Darboux chart $\phi : p \mapsto (x_1(p), y_1(p), \dots, x_n(p), y_n(p))$ such that $\Omega = dz_1 \wedge \dots \wedge dz_n$. Let $U(n) \subset Sp(2n)$ be the unitary subgroup associated to the given complex structure; It is the set of symplectic transformations which are complexe linear. Then given $\Psi \in Sp(2n) \setminus U(n)$, the chart $\Psi \circ \phi$ is still Darboux, say with coordinates $(X_1, Y_1, \dots, X_n, Y_n)$; But as $\Psi$ does not preserve the complex structure, there is no reason for $(\Psi^{-1})^{\ast}(dz_1 \wedge \dots \wedge dz_n)$ (which is the expression of $\Omega$ in the new chart) to have the simple form $dZ_1 \wedge \dots \wedge dZ_n$.

For instance, let $X = \mathbb{R}^{2n}$ with its standard Kahler structure; for simplicity, take $n=1$. The matrix

$$\Psi = \left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right) \mbox{ with inverse } \Psi^{-1} = \left( \begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array} \right)$$

is symplectic (as it has determinant 1) but not unitary (as it is not orthogonal), and we compute $(\Psi^{-1})^{\ast}dz_1 = d(X_1 - Y_1) + i dY_1 = dZ_1 - dY_1 \neq dZ_1$. For $n > 1$, the $n$-fold product of the counter-example yields a counter-example for $X = \mathbb{R}^{2n}$.

Given any point $p \in X$, it is always possible to find a chart around $p$ such that the Kahler structure is standard at that point. The chart can even be chosen such that the components of the three tensors $\omega$, $g$ and $J$ are almost standard up to second order in the geodesic distance from $p$. However, one cannot expect in general to be able to find a chart which 'standardize' the three tensors, since the (vanishing of the) Riemann curvature tensor is precisely the obstruction to finding such 'standard' coordinates for the metric $g$. (Since $\mathbb{C}^n$ is flat, you were able to find such a global chart, but it is not any Darboux chart.)

It also means that a Darboux chart is not usually a complex chart: in Darboux coordinates $(x_1, \dots, y_n)$, the function $z_k = x_k + i y_k$ is not necessarily holomorphic (i.e. one does not necessarily have $i dz_k = dz_k \circ J$). Consequently, it is not clear how your conjectured expression for $\Omega$ can be meaningfully linked to the Kahler structure on the manifold.

Observe that in any Darboux chart $(x_1, \dots, y_n)$, the form $dz_1 \wedge \dots \wedge dz_n$ satisfies $\omega^n/n! = (-1)^{n(n-1)/2}(i/2)^n \Omega \wedge \bar{\Omega}$. However, (i) this form $\Omega$ is not clearly the restriction to this chart of a global form which satisfies this equality, and (ii) even in the chart, for reasons explained above, this form $\Omega$ is not necessarily holomorphic, so it is not automatically related to the whole Kahler structure. Conversely, given the 'Calabi-Yau' form $\Omega$, it is far from clear that it could be expressed in this way in some Darboux chart.

Of course, since a Calabi-Yau is Ricci-flat, one might expect to find charts which get close to 'standardize' the three tensors. It might be close enough so that $\Omega$ indeed has this simple expression in such coordinates, but I doubt it.

Even in the 'easy' case when the holonomy takes values in $SU(n)$, Lemma 4.4 does not claim that $\Omega$ has this simple expression (up to phase) in some Darboux chart; It rather implies that in any chart which standardize the Kahler structure at any fixed point, then $\Omega$ has this simple expression (up to phase) at that point.

Incidentally, if you could come up with a Calabi-Yau which has non-vanishing Riemann curvature and which has $SU(n)$-holonomy, you might be able to prove (precisely because of this lemma) that no Darboux chart exists for which $\Omega$ is expressed everywhere in this simple manner. Indeed, you would have to perform a symplectic and non-unitary change of basis at any point in order to standardize the Kahler structure, but this change of basis would have little chance to preserve (up to phase) the conjectured simple expression (preservation implied by your conjecture and the lemma). But I cannot definitely tell; Food for thoughts.

  • $\begingroup$ This is a really nice answer, thank you for explaining it so well. $\endgroup$ – Ashley Jan 29 '17 at 17:14
  • $\begingroup$ Thank you for your appreciation. I am not familiar enough with Calabi-Yau manifolds to exhibit a proof of or a counter-example to your 'complete' conjecture, but I would be interested to know if you (or anyone else reading this) could come with such a thing. $\endgroup$ – Jordan Payette Jan 29 '17 at 17:46

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