Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In the paper On the complex projective spaces, Hirzebruch and Kodaira prove the following:

If $X$ is compact Kähler manifold diffeomorphic to $\mathbb{CP}^n$, then $X$ is biholomorphic to $\mathbb{CP}^n$ if $n$ is odd or $n$ is even but $c_1 \neq -(n+1)g$.

Here, $c_1$ is the first Chern class of $X$, and $g$ is a generator of $H^2(X,\mathbb{Z})$ with the same sign as the Kähler class.

My question is: Has the case $X$ is diffeomorphic but not biholomorphic to $\mathbb{CP}^n$ with $n$ even and $c_1=-(n+1)g$ been ruled out in the following years? Or do we have some constructions of manifolds of this type?

share|improve this question
add comment

1 Answer 1

up vote 20 down vote accepted

The case when $n$ is even has been famously ruled out by Yau as a consequence of his proof of the Calabi Conjecture, see his original paper or these notes.

In fact, thanks to results of Novikov, you can replace diffeomorphism by homeomorphism and the statement remains true.

When $n=2$ in the same paper Yau proved that you can also drop the Kähler condition and replace homeomorphism by homotopy equivalence.

When $n>2$ the statement for Kähler manifolds but assuming only homotopy equivalence is true up to $n=6$ thanks to Libgober and Wood (and true in general if you assume furthermore that the Pontryagin classes are the same as $\mathbb{CP}^n$), see also this question. The case of higher $n$ is probably open.

Finally, it is still an open problem to decide whether a compact complex manifold diffeomorphic to $\mathbb{CP}^n$ is biholomorphic to it (when $n\geq 3)$. If this were true, it would imply that the differentiable manifold $S^6$ does not admit a complex structure (otherwise, blowing up a point you'd get a complex manifold diffeomorphic but not biholomorphic to $\mathbb{CP}^3$), see for example here.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.