Edit: According to the answers to the linked MSE question and the comment of Holonomia, I understand that the answer to the second question is that " Every tangent bundles is a complex manifold".

Let $M$ be a compact real manifold. Assume that the tangent bundle $TM$ carries a holomorphic structure, namely it is equipped with a holomorphic atlas. We fix a holomorphic structure for $TM$.

We say that a map $f:M \to M$ has a holomorphic derivative if $Df:TM \to TM$ is a holomorphic map.

The first Question: Is it true to say that the space of all diffeomorphism of $M$ with holomorphic derivative admits a structure of a finite dimensional Lie group?

The Second Question: What is an example of a manifold whose tangent bundle, as a manifold, does not admit a holomorphic atlas?

I asked the latter question on MSE but I did not get any answer.


Further questions:

Added: How can one decide that a given map has holomorphic derivative? In this investigation, and motivated by CR equations, what kind of differential operators would appear?

As a particular example, we consider the Hopf map $p: S^3 \to S^2$. Is it a map with holomorphic derivative? Of course this question is meaningless if we do not fix a holomorphic structure for $TS^2$ and $TS^3$. So it is natural to ask: What is a precise holomorphic structure for these space? Can the holomorphic structure of the tangent bundle of a Riemann surface or a parallelizable manifold be determined explicitly?

  • 2
    $\begingroup$ The tangent bundle $TM$ of any differentiable manifold carries a complex manifold structure: link.springer.com/chapter/10.1007/978-1-4757-9771-8_8 $\endgroup$
    – Holonomia
    Jan 25, 2018 at 18:54
  • $\begingroup$ @holonomia very great comment and link. Thank you so much for that. $\endgroup$ Jan 25, 2018 at 19:08
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    $\begingroup$ I think it would be interesting to consider more generally: suppose that $X$ is a connected complex manifold, and that $Y \subset X$ is a totally real compact submanifold without boundary, of real dimension equal to the complex dimension of $X$. ("Totally real" meaning: no complex line in any tangent space of $X$ is tangent to $Y$.) Is the group of biholomorphisms of $X$ preserving $Y$ always a finite dimensional Lie group acting smoothly and faithfully on $X$? In your problem, $Y$ is the zero section of $TM$ and $X=TM$. $\endgroup$
    – Ben McKay
    Jan 27, 2018 at 16:43
  • $\begingroup$ @BenMcKay Very interesting generalization for complex manifolds and totally real submanifolds. Thank you! $\endgroup$ Jan 28, 2018 at 9:27
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    $\begingroup$ @AliTaghavi: the group of biholomorphisms of $\mathbb{C}^2$ is not finite dimensional, containing (for example) the maps $(z,w)\mapsto(z,w+f(z))$ for any holomorphic entire function $f(z)$. $\endgroup$
    – Ben McKay
    Jan 28, 2018 at 12:34

1 Answer 1


I guess that you mean: $TM$ carries an almost complex structure $J:TM\to TM$ with $J^2=-1$. It is integrable to complex structure iff the Froelicher-Nijenhuis bracket $[J,J]$ vanishes.

First question: The Lie algebra of your group consists of all vector fields $X$ with $\mathcal L_XJ = [X,J]=0$. Since $J$ is invertible, $[X,J]=0$ is an elliptic equation of 1. order for $X$, so the solution space is finite dimensional on a compact manifold.

Second question: If $J$ exists, the manifold must be even dimensional and orientable.


So let us assume that $TM$ carries a complex structure. $TM$ is not compact, so the group of biholomorphic automorphismsm need not be finite dimensional.

First question: $\phi\mapsto T\phi=D\phi$ is an injective group homomorphism into the group of all those biholomophic diffeomorphisms of $TM$ which are also vector bundle homomorphisms; i.e., linear between fibers. Its Lie algebra consists of all vector fields which are holomorphic and linear when restricted to any fiber. To be holomorphic is an elliptic equation, and the linearity along fibers should imply that the corresponding solution space is finite dimensional.

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    $\begingroup$ I think the OP had in mind an arbitrary complex structure on $TM$, not one coming from a complex structure on $M$. $\endgroup$
    – abx
    Jan 20, 2018 at 12:56
  • $\begingroup$ @abx Yes, Exactly. I was motivated by $TS^n$ which is a regular algebraic curve so it is a holomorphic manifold. Thanks for your comment. $\endgroup$ Jan 20, 2018 at 19:33
  • $\begingroup$ So $M$ can be an odd dimensional manifold. $\endgroup$ Jan 20, 2018 at 19:36
  • $\begingroup$ @PeterMichor Dear Prof. Michor I consider $TM$ as a holomorphic manifold. I am not thinking to a particular structure on its fiber. Thank you for your attention to my question. $\endgroup$ Jan 20, 2018 at 19:41

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