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For a $n$-dim smooth projective complex algebraic variety $X$, we can form the complex line bundle $\Omega^n$ of holomorphic $n$-form on $X$. Let $K_X$ be the divisor class of $\Omega^n$, then $K_X$ is called the canonical class of $X$.

Question: Is homology class of $K_X$ in $H_{2n-2}(X)$ a topological invariant? If it's true, please tell me the idea of proof or some references. If not, please give me the counterexamples.

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Just to be sure, does "topological invariant" here mean "diffeomorphism invariant"? –  user5117 Jul 15 '11 at 12:25
It may seems that the case "homeomorphism" and "diffeomorphism" have many differences, but I don't know whether the differences will make the answer different. However, I want to know the answers both in "homemorphism" and "diffeomorphism" cases. –  Yuchen Liu Jul 15 '11 at 12:33
The answer is yes if the diffeomorphism is induced by a deformation equivalence; this is an easy consequence of Ehresmann theorem. Otherwise, I think that the answer is not at all obvious. For instance, using Seiberg Witten theory, one proves that any diffeomorphism $\phi \colon X \to X'$ between smooth $4$-manifolds (for instance, algebraic surfaces) maps $K_X$ either into $K_{X'}$ or into $-K_{X'}$, and the second case may occur. I do not know, however, if there are examples where the second case occurs and $X$, $Y'$ are both smooth and projective. –  Francesco Polizzi Jul 15 '11 at 13:23
For a simple counterexample to the "homeomorphism" version of the question, take, say, a K3 surface (which of course has canonical class zero) which elliptically fibers over $\mathbb{P}^1$, and then do a logarithmic transformation of multiplicity $p>1$ to one of the fibers. You won't have changed the homeomorphism type and the resulting manifold will still admit Kahler forms, but the result will now have nontrivial canonical class. (with divisibility $p-1$ in $H^2(M;\mathbb{Z})$). Of course these examples are all mutually nondiffeomorphic. –  Mike Usher Jul 15 '11 at 14:10
You are so kind! <3 hehe –  diverietti Jul 15 '11 at 15:18

3 Answers 3

up vote 13 down vote accepted

This answer is about the case of complex surfaces $X$ and their diffeomorphisms (all my diffeos are assumed to be orientation-preserving!).

(1) Examples of self-diffeomorphisms that reverse the sign of the canonical class.

Take $X=\mathbb{C}P^1\times \mathbb{C}P^1$. Let $\tau$ be reflection in the equator of $S^2=\mathbb{C}P^1$. Then $\tau \times \tau$ preserves orientation and acts as $-I$ on $H^2(X)$. It therefore sends $K_X$ to $-K_X$.

One can also realise the automorphism $-I$ of $H^2(X)$ by a diffeomorphism when $X$ is the blow-up of the projective plane at $k$ points, $k = 2,3,\dots,9$. This follows from a result of C.T.C. Wall from

Diffeomorphisms of 4-manifolds, J. London Math. Soc. 39 (1964) 131–140, MR0163323

Wall says that if $N$ is a simply connected, closed oriented 4-manifold with $b_2(N)<9$, and $X$ is the connected sum of $N$ with $S^2 \times S^2$, then all automorphisms of the intersection form of $X$ are realised by diffeos. To apply this, recall that the 1-point blow-up of $\mathbb{C}P^1\times \mathbb{C}P^1$ is the 2-point blow up of the projective plane. (Wall's strategy, by the way, is to factor the automorphism into reflections along hyperplanes, and to realise those.)

(2) Results from Seiberg-Witten theory.

These results tie complex geometry amazingly closely to differential topology. They say that the unsigned pair $\pm K_X$ is invariant under diffeomorphisms (Witten http://arxiv.org/abs/hep-th/9411102 and others); so too is the Kodaira dimension; so too are the plurigenera (Friedman-Morgan http://arxiv.org/abs/alg-geom/9502026).

In Kodaira dimension $<2$, one can take this further and prove that oriented-diffeomorphic surfaces are actually deformation-equivalent (to be safe, let me specify the simply connected case). But that's not the explanation in general: there are pairs of simply connected general-type surfaces that are diffeomorphic (by diffeos preserving the canonical class), which are not deformation-equivalent (Catanese-Wajnryb http://arxiv.org/abs/math/0405299).

(3) How it happens.

The Seiberg-Witten invariant (for an oriented 4-manifold with $b^+(X)>1$) is a map $$SW: Spin^c(X)\to\mathbb{Z}$$ defined on the $H^2(X)$-torsor of $Spin^c$-structures. The overall sign is equivalent to a "homology orientation". It's natural under diffeomorphisms. It's also invariant under "conjugation" $\mathfrak{s}\mapsto \bar{\mathfrak{s}}$ of $Spin^c$-structures.

For algebraic surfaces, there's a canonical spin-c structure $\mathfrak{s}$, so $Spin^c(X)$ is identified with $H^2(X)$. Witten (http://arxiv.org/abs/hep-th/9411102) observed that the elliptic equations that define $SW$ simplify drastically in the algebraic case; in evaluating $SW$ on a cohomology class represented by a complex line bundle $L\to X$, you're led to consider a moduli space of pairs consisting of a holomorphic structure on the line bundle and a holomorphic section of it, with an obstruction bundle on the moduli space. Conjugation-invariance becomes Serre duality.

For general type surfaces, $\pm SW(\mathfrak{s}) = \pm SW(\bar{\mathfrak{s}}) = \pm 1$; all other spin-c structures have vanishing invariant. Since $c_1(\mathfrak{s})=-c_1(\bar{\mathfrak{s}})=-K$, one deduces diffeomorphism-invariance of $\pm K$. For lower Kodaira dimension, a more complicated analysis is needed.

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Dear Tim, may I ask you to elaborate a little on (1)? More precisely, are you saying that Wall's theorem allows one to realise any automorphism of the lattice bz a diffeomorphism? (I don't have access to MathSciNet where I am, so I can't check this for myself.) I guess not, but that seems to be one interpretation. –  user5117 Jul 15 '11 at 19:28
"bz" -> "by" (ah, the joys of switching between German and English keyboard layout twice a day...) –  user5117 Jul 15 '11 at 19:30
Artie, yes: Wall says that if $N$ is a simply connected, closed oriented 4-manifold which is either indefinite, or has $b_2(N)<9$, and $M= N \# (S^2×S^2)$, then all automorphisms of the intersection form of $M$ are realised by diffeos. If I take $N= \mathbb{C}P^2$ then $M$ is the 2-fold blow up of $\mathbb{C}P^2$. –  Tim Perutz Jul 15 '11 at 20:05
Thanks for the details! –  user5117 Jul 15 '11 at 21:04
(Artie's question, and my response to it, are now covered by the edit to my answer.) –  Tim Perutz Jul 16 '11 at 16:39

It is well known that in dimension $3$ and higher there exist complex structures on diffeomerphic manifolds with totally different Chern classes (and Chern numbers).

For the case of complex manifolds you can check

Can one bound the todd class of a 3-dimensional variety polynomially in c_3

For the case of complex projective manifolds the reference given in the same answer:


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Nice. Do you know whether there exist examples of orientation reversing diffeomorphic algebraic surfaces? Or, more generally, if the conjecture in my answer has been settled? –  Francesco Polizzi Jul 15 '11 at 14:12
Francesco, thanks. I never saw this conjecture of Kostchik, it sounds interesting. Though since this conjectured such a long time ago... I would try to get a counterexample by taking ramified covers of $\mathbb CP^2$ in a real arrangement. Is it obvious this will not work? –  Dmitri Jul 15 '11 at 14:22
Dear Dmitri, I do not know. I have to think about it. Anyway, if it works it would be nice! –  Francesco Polizzi Jul 15 '11 at 14:28

For the question about homeomorphisms the answer is no, even if $X$ and $X'$ are algebraic surfaces.

In fact, in his paper [Orientation reversing homeomorphisms in surface geography, Math. Ann. 292 (1992)], D. Kotschick proves the following result:

Theorem. There exist infinitely many pairs of simply connected algebraic surfaces of general type which are orientation-reversing homeomorphic (with respect to their complex orientations), but not diffeomorphic.

He also makes a conjecture about orientation-reversing diffeomorphic algebraic surfaces. As I said in my comments before, by using Seiberg-Witten theory one proves that, given any diffeomorphism $\phi \colon X \to X'$ between two smooth $4$-manifolds, one has either $\phi(K_X)=K_{X'}$ or $\phi(K_X)=-K_{X'}$.

Kotschick's conjecture is therefore the following:

Conjecture. If two algebraic surface with finite fundamental group are orientation-reversing diffeomorphic, then they are homeomorphic to a geometrically ruled rational surface. In particular, they are simply connected.

I do not know the current state of this conjecture.

Added On February 29, 2012. D. Kotschick kindly informed me that he actually proved this conjecture in his paper Orientations and geometrizations of compact complex surfaces, Bulletin of the London Mathematical Society 29 (1997), 145-149.

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Interesting conjecture indeed! –  Dmitri Jul 15 '11 at 14:23
Francesco, sorry, I don't see any more how to get a counter-example -- the real conjugation of $\mathbb CP^2$ that I wanted to use, of course, preserves orientation :) . This makes the conjecture even more interesting. –  Dmitri Jul 15 '11 at 14:57

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