I have a question concerning the complexification of a real analytic Riemannian manifold. Let $(M,g)$ be a compact Riemannian manifold. It is a well known fact that in a neighbourhood $U$ of the zero section of the cotangent bundle $T^{*}M$, $M$ admits a complexification. By Stenzel and Guillemin this complexification is called the adapted complex structure. It is even Kähler. In this complexification, the phase function $\varphi$ solves the homogenous Monge-Ampere equation. My question is now: are there Kähler structures on a neighbourhood $U$ of the zero section in the cotangent bundle of $M$, which turn $M$ into a Lagrangian submanifold and such that $M$ is isometrically embedded in $U$, BUT differ from the adapted complex structure (up to biholomorphism)? Do all such structures come from an adapted complex structure? Is this already known or can it be derived easily?

  • $\begingroup$ Try the paper "Symplectic geometry and the uniqueness of Grauert tubes" by D. Burns and R. Hind, deepblue.lib.umich.edu/bitstream/2027.42/41842/1/… $\endgroup$ Dec 7 '12 at 13:45
  • $\begingroup$ I dont see how this answers my question? I am interested in: are there Kälhler structures in a neighbourhood of the zero section of some Riemannian manifold that are not adapted in the sense of Guillemin and Stenzel? $\endgroup$
    – Dmitri
    Dec 7 '12 at 16:51

Suppose that $M \subset Z$ is a compact Lagrangian submanifold of a Kaehler manifold $Z$ with Kaehler form $\omega$. Take a function $f$ so that $\partial \bar\partial f=0$ at every point of $M$. Then for a small enough neighborhood of $M$ in $Z$, $\omega+i\partial\bar\partial f$ is a Kaehler form in which $M$ is a Lagrangian manifold with the same induced metric. In your case, take $Z=T^*M$ with the Stenzel metric.

  • $\begingroup$ Is it clear that the new metric is not biholomorphic to the old one? $\endgroup$ Mar 28 '16 at 9:24
  • 1
    $\begingroup$ Biholomorphisms of a complex $n$-dimensional manifold depend on $2n$ functions of $n$ variables, while our function $f$ is one essentially arbitrary function of $2n$ variables, so it seems unlikely that a biholomorphism can identify such metrics. Of course, this isn't a rigorous argument. But $f$ can have compact support here, so it is easy to make a rigorous argument. $\endgroup$
    – Ben McKay
    Mar 28 '16 at 9:33

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.