1
$\begingroup$

For what type of manifolds $M$,

1) $TM$ does not admit a holomorphic structure?

2)$TM$ is not diffeomorphic to an algebraic variete in some $\mathbb{C}^n$?

$\endgroup$
7
  • $\begingroup$ $TM$ always has an almost complex structure; it is quite likely that such a structure can be chosen to be integrable. As for the 2nd question: sometimes yes, sometimes no, it is unclear what you are asking here. $\endgroup$ Jan 22, 2018 at 0:47
  • $\begingroup$ @MoisheCohen what is that almost complex structure? The second question ask for obstructions agains this (possible) holomorphic manifold $TM$ to be bioholomorphic to an algebraic curve. I was motivated by $TS^n$ which is diffeomorphic to $\sum x_i ^2=1$ and was motivated by holomorphic manifolds which can not be embeded in a complex projective space. $\endgroup$ Jan 23, 2018 at 8:17
  • 1
    $\begingroup$ $TM$ has a standard symplectic structure, hence, a compatible almost complex structure. $\endgroup$ Jan 23, 2018 at 14:36
  • $\begingroup$ @MoisheCohen Ah yes it is defined via $\omega(X, J(Y))=<X,Y>$. I forget it. But what about integrabiluty, is it an obvious auestion? $\endgroup$ Jan 23, 2018 at 14:43
  • 1
    $\begingroup$ @AnonymousCoward: I was thinking about a Riemannian manifold. I also did not say that the structure is canonical, only that it is standard. $\endgroup$ Jan 23, 2018 at 20:26

2 Answers 2

1
+50
$\begingroup$
  1. There is always an integrable almost complex structure on $TM$, see Theorem 2.2 in

R. Szöke, Complex structures on tangent bundles of Riemannian manifolds. Mathematische Annalen, 291, (1991) 3, page 409-428.

  1. I know of two types of obstructions to the existence of a structure of a smooth complex quasiprojective variety on $TM$:

a. Each quasiprojective variety is tame (admits a compactification as a manifold with boundary), while $TM$ need not be tame, e.g. it can have non-finitely generated fundamental group (for instance, take $M$ to be a surface of infinite genus) or infinite Betti numbers.

b. Assume, therefore, that $M$ is compact. There are some known restrictions on fundamental groups of smooth complex quasiprojective varieties coming from the Rational Homotopy Theory:

J. Morgan, The algebraic topology of smooth algebraic varieties. Inst. Hautes Études Sci. Publ. Math. No. 48 (1978), 137–204.

Since every finitely presented group can be realized as the fundamental group of some smooth closed 4-dimensional manifold, we obtain, therefore, more examples.

$\endgroup$
1
$\begingroup$

I would like to point out that the fact that $TM$ admits a complex structure follows from an earlier result than the one mentioned in Moishe Cohen's answer.

Landweber proved the following result in his $1974$ paper Complex Structures on Open Manifolds:

Let $X$ be an open manifold of dimension $2q$. If $H^i(X; \mathbb{Z}) = 0$ for $i > q$, then each almost complex structure on $X$ is homotopic to an integrable almost complex structure.

If $M$ is a $q$-dimensional manifold without boundary, then $TM$ is a $2q$-dimensional open manifold. Moreover, $TM$ is homotopy equivalent to $M$ and hence for $i > q$, $H^i(TM; \mathbb{Z}) \cong H^i(M; \mathbb{Z}) = 0$. As pointed out in the comments, $TM$ always admits an almost complex structure, so by the result above, it admits an integrable almost complex structure, i.e. a complex structure.

$\endgroup$
1
  • $\begingroup$ Thank you very much for your answer. I am sorry that I can not accept two answers. $\endgroup$ Jan 26, 2018 at 18:38

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.