Given any smooth manifold $M$, its cotangent bundle $T^*M$ is a symplectic manifold, with the canonical symplectic form.

If $M$ is a complex manifold then $T^*M$ is also a complex manifold.

Thus, $T^*M$ is a complex symplectic manifold.

Does it follow that the canonical symplectic form is holomorphic?

If not, what condition can be placed on $M$ so that it is? For example this question on Overflow says that if $M$ is Kahler then the symplectic form is holomorphic.


One should not expect such a thing.

On the one hand, the symplectic form on $T^*M$ is canonically constructible for any smooth manifold $M$, regardless of any more structure on $M$. On the other hand, no canonical complex structure exists on $T^*M$ without some specific structure on $M$ (such as a Riemannian metric or a complex structure). The specification of any such structure has a priori no reason to lead to a complex structure on $T^*M$ which is naturally related to the canonical symplectic structure on $T^*M$ since the latter does not depend on the specified structure on $M$.

Even when $M$ is Kahler, the induced symplectic (in fact, Kahler) form on $T^*M$ has a priori no reason to coincide with the canonical symplectic form. In fact, I doubt that they coincide in general.

To be more specific, note that if you put coordinates $(x^1, \dots, x^m)$ on $M$, you induce Darboux coordinates $(x^1, p_1, \dots, x^m, p_m)$ on $T^*M$. Incidentally, to speak loosely, the symplectic structure intertwines 'base' vectors (i.e. $\partial /\partial x^k$) with 'fiber' vectors (i.e. $\partial/ \partial p_k$).

If you put holomorphic coordinates $(x^1+iy^1, \dots, x^m+iy^m)$ on a complex manifold $M$, you induce holomorphic coordinates $(x^1+iy^1, p_m+iq_m, \dots, x^m+iy^m, p_m + iq_m)$ on $T^*M$. Incidentally, to speak loosely, the complex structure intertwines real 'base' vectors (i.e. $\partial/\partial x^k$) with other ones (i.e. $\partial/\partial y^k$) and real 'fiber' vectors (i.e. $\partial/\partial p_k$) with other ones (i.e. $\partial/\partial q_k$).

However, if the symplectic form were to be holomorphic with respect to the complex structure, it should be expressible (at least to first order in the coordinates) as

$$i/2 \sum_{k=1}^m \left( d(x^k + iy^k) \wedge d(x^k - i y^k) + d(p_k + i q_k) \wedge d(p_k - i q_k) \right) = \sum_{k=1}^m (dx^k \wedge dy^k + dp_k \wedge dq_k)$$

which suggests that the symplectic form would intertwine 'base' vectors with other ones and 'fiber' vectors with other ones. This is not the case of the canonical symplectic form.

All these considerations can be explicitly checked in the simplest case when $M = \mathbb{C}$ equipped with its standard Kahler structure, as they are then precise. $M$ is Lagrangian when seen as the zero section of $T^*M$ with its canonical symplectic form, but it is a symplectic subspace of $T^*M \simeq (\mathbb{C}^2, i)$ with its induced Kahler structure. But when an almost complex structure is compatible with a symplectic form, any complex subspace is then symplectic, which shows that the canonical symplectic structure on $T^*M$ is not compatible with the induced complex structure from $M$.


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