I'm trying to understand this question and I was hoping someone could help me. One thing in particular is confusing me.

Question: They are starting with a riemannian manifold $(M,g)$ and considering the "metric-induced" almost complex structure on $T^*M$. What exactly is the metric induced almost complex structure? Is there a nice way to visualize/think of it?

Thoughts: Does this come from first defining a metric $g_0$ (induced by $g$) on $T^*M$ so that we would then have the canonical symplectic form $\omega_0$ and and riemannian $g_0$ on $T^*M$, which would determine an almost complex structure via the compatible triples. If this is the case, how exactly is $g_0$ defined using $g$? (a reference for this construction will definitely suffice.)

  • $\begingroup$ I think as metric and symplectic forms are both non degenerate, they give maps TM to T^* M, now you can define J by composing one map with the inverse of other. $\endgroup$
    – Soham
    Mar 27, 2020 at 1:09

1 Answer 1


Daniel Huybrechts: "Complex Geometry: An Introduction," Springer Universitex, 2005, presents, on page 48, a very nice, clear picture of what is going on in your question. I find his explanation clear, and hope you too find it so.

With regard to your thoughts, Riemannian means isometric. That is, the metric of the target is pulled back to the source, so in the induced metric on the source, the two manifolds are isometric or Riemannian.

I hope this helps.

  • $\begingroup$ Thank you for the reference. I don't understand what you mean by metric on target though. The metric we start with is on $M$. We have the projection from $T^*M$ to $M$, but this doesn't pull back the metric. $\endgroup$
    – Ashley
    Jan 6, 2017 at 22:43
  • $\begingroup$ I don't know. In any game, there are two players, source and target. The way I'm playing the game is to declare M the target, and then at every point z=* of the target M to declare TM the tangent hyperplane to M at some point z=.. Perhaps I am playing the game backwards. $\endgroup$ Jan 7, 2017 at 4:41
  • $\begingroup$ I’m sorry but I did not find anything about induced metric on the cotangent bundle in your reference, in that page I could only find discussion about open subset of $\mathbb C^n$. $\endgroup$
    – Soham
    Mar 27, 2020 at 1:07

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