Question

Apparently 2 smooth complex hypersurface in complex projective space that have the same degree are diffeomorphic. Does anyone know where the proof of this can be found ? Is there a counter example for symplectic manifolds?

Answer 1

A more general version of the question made by the OP is the following

Question. Let $H_1$ and $H_2$ be two smooth connect hypersurfaces in a projective manifold $X$. Suppose $H_1$ is homologous to $H_2$, or equivalently that the Chern classes of $\mathcal O_X(H_1)$ and $\mathcal O_X(H_2)$ coincide. Is it true that $H_1$ is diffeomorphic to $H_2$ ?

A similar question was proposed by Fulton: can we determine the Betti numbers of a smooth divisor as a function of $X$ and of its Chern class ? See Totaro's "The topology of smooth divisors and the arithmetic of abelian varieties" for a thorough discussion.

Positive answer when $H^1(X, \mathcal O_X)=0$. The argument sketched by Jack Huizenga in the comments shows that the answer is yes if we further assume that $H^1(X, \mathcal O_X)=0$. In this case, the line-bundles $\mathcal O_X(H_1)$ and $\mathcal O_X(H_2)$ coincide since the exponential sequence $$ 0 \to \mathcal O_X \to \mathcal O_X^* \to \mathbb Z \to 0 $$ implies the Chern class morphism $c: H^1(X,\mathcal O_X^*) \to H^2(X, \mathbb Z)$ is injective.

Thus $H_1$ and $H_2$ are both members of the same linear system, and we can consider the incidence variety $$ Z = \lbrace (x,[ \sigma ] ) \in X \times \mathbb P H^0(X, \mathcal O_X(H_1)) ; \sigma(x)=0 \rbrace $$ which comes with a natural morphism $\pi : Z \to \mathbb PH^0(X, \mathcal O_X(H_1))$. The subset $U \subset \mathbb P H^0(X,\mathcal O_X(H_1))$ corresponding to sections with smooth zeros is clearly open in the Zariski topology, and non-empty since $H_1$ and $H_2$ are smooth. Consequently $U$ is also connected.

Take a path $\gamma : [0 ,1] \to U$ connecting the sections defining $H_1$ and $H_2$. The real variety $Y = \pi^{-1} ( \gamma [0,1])$ now has as boundaries $H_1$ and $H_2$ and comes with a submersion $\pi:Y \to [0,1]$. If we consider the gradient of $\pi$ ( for any Riemmanian metric on $Y$ ) then its flow will define a diffeomorphism between any two fibers of $\pi$.

Counter-example. If $H^1(X,\mathcal O_X) \neq 0$ then $H_1$ and $H_2$ are not necessarily diffeomorphic. The following example appears in Totaro's paper. Let $C_1$ and $C_2$ be two smooth curves of genus $>1$. Let $B_1 \to C_1$ and $B_2 \to C_2$ be two non-trivial double coverings. The group $\Gamma = (\mathbb Z / 2 \mathbb Z)^2$ acts freely on $B_1 \times B_2$ with quotient $C_1 \times C_2$. Let $\Gamma$ act on $\mathbb P^1$ through the automorphism $x \mapsto -x$ and $x \mapsto 1/x$, and take $X$ as the quotient of $B_1 \times B_2 \times \mathbb P^1$ by $\Gamma$.

Clearly $X$ is smooth ( since the action is free ) and comes with a fibration $\pi: X \to \mathbb P^1$. This fibration has exactly two non-reduced fibers and each has multiplicity two. One has support equal to $C_1 \times B_2$ while the support of the other is $B_1 \times C_2$.

It can be verified that $H^2(X,\mathbb Z)$ is torsion free. Consequently the support of the two non-reduced fibers have the same Chern classes. For a general choice of $C_1$ and $C_2$ they are not diffeomorphic.

Another question. The example above suggests the following conjecture made by Totaro in the very same paper.

Conjecture. Let $H_1$ and $H_2$ be two smooth connect hypersurfaces in a projective manifold $X$. Suppose $H_1$ is homologous to $H_2$, or equivalently that the Chern classes of $\mathcal O_X(H_1)$ and $\mathcal O_X(H_2)$ coincide. Then there exists an étale cyclic covering of $H_1$ which is deformation equivalent to an étale cyclic covering of $H_2$.

Evidence toward this conjecture is also presented by Totaro. If the Picard variety of $X$ is isogenous to a product of elliptic curves then there exists étale coverings of $H_1$ and $H_2$ which deformation equivalent via passage to some characteristic $p>0$.

Another evidence toward this conjecture is presented here where an analogous statement for divisors in compact Kahler manifolds is proved after replacing "deformation equivalent" by "diffeomorphic".

Answer 2

Here is a proof of the much stronger result that, given any two nonsingular degree d hypersurfaces in $CP^{n+1}$, there exists a diffeomorphism $CP^{n+1} \rightarrow CP^{n+1}$ isotopic to the identity that restricts to a diffeomorphism of the two hypersurfaces. This is taken from the paper Topology of Nonsingular Complex Hypersurfaces by Kulkarni and Wood.

Let $X$ (resp. $Y$) be defined by the polynomial $p(z)$ (resp. $q(z)$). Then the polynomial $f(t,z) = t_0p(z) + t_1q(z)$ of homogeneous bidegree $(1,d)$ defines a hypersurface $F$ in $CP^1 \times CP^{n+1}$. The set $S \subset CP^1 \times CP^{n+1}$ of points $[t,z]$ at which $F \cap [t] \times CP^{n+1}$ is singular is a closed algebraic set. So the projection $\pi(S)$ of $S$ onto $CP^1$ is also a closed algebraic set and since evidently $\pi(S) \ne CP^1$, $\pi(S)$ is zero dimensional hence a finite set of points. Let $I$ be a smooth arc in $CP^1$ from $[1,0]$ to $[0,1]$ in the complement of $\pi(S)$. Then $\pi^{-1}(I)= I \times CP^{n+1}$ contains the smooth submanifold $M = \pi^{-1}(1) \cap F$ of real codimension $2$ such that $\pi: M \rightarrow I$ is a product bundle: $\phi: I \times X \rightarrow M$. Thus $M$ may be regarded as the graph of an isotopy of $X$ in $CP^{n+1}$. Let $\partial/\partial t$ be the vector field on $I \times X$ tangential to the first factor. Then $\phi_{*}(\partial/\partial t)$, which is tangent to $M$, extends to a vector field $V$ on $I \times CP^{n+1}$. The integral flow of this vector field gives the desired ambient isotopy.