7
$\begingroup$

Compact Kahler manifolds have the property that surjective maps induce injections on cohomology with coefficents in $\mathbb{Q}$ (That is, if $X,Y$ compact Kahler, then a surjective map $\phi: X \rightarrow Y$ induces injections $\phi^*: H^i(Y, \mathbb{Q}) \rightarrow H^i(X, \mathbb{Q})$ for all $i$, [Voisen, Hodge Theory I, p 177]).

Question: I'm wondering if I should think of this as a property of compact Kahler manifolds, or as an instance of something more general. For example, can the Kahler condition be replaced with a more general class of manifolds (not dropping the compactness hypothesis). Perhaps one that includes not just complex manifolds but maybe a few manifolds of odd (topological) dimension? I know that if we require that $\dim X = \dim Y$ then the fact above is true more generally just for compact oriented manifolds, for formal reasons.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
15
$\begingroup$

In order to have this property it is sufficient to require that $\phi^{-1}(y)$ is a non-zero cycle in $H_*(X,\mathbb Q)$, where $y$ is a generic point in $Y$. This holds indeed when $X$ and $Y$ are Kahler.

Let me give an example showing that this does not work when $X$ and $Y$ are just complex.

Example. Let $X$ be a Hopf surface $X=(\mathbb C^2\setminus 0)/\mathbb Z$ where $\mathbb Z$ is acting on $\mathbb C^2$ by multiplication by (say) $2$. Then there is a fibration $\phi\colon X\to \mathbb CP^1=Y$. This fibration comes from the standard $\mathbb C^*$-action on $\mathbb C^2$. Now $\phi^{-1}(y)$ is null-homlogous.

Improve this answer
$\endgroup$
8
  • $\begingroup$ Nice criteria! Is it easy to see that $\phi^{-1}(y)$ is a non-zero cycle in $H_*(X,\mathbb{Q})$ in the Kahler case? $\endgroup$
    – LMN
    Commented Oct 1, 2012 at 2:57
  • $\begingroup$ Sure, the preimage of a point is a complex submanifold and the appropriate power of Kahler form of $X$ is a volume form on it. $\endgroup$
    – Dmitri Panov
    Commented Oct 1, 2012 at 3:00
  • $\begingroup$ Great! Can you give me a hint (or reference) of how you would prove this criteria implies that the maps on cohomology with $\mathbb{Q}$ coefficents are injective? $\endgroup$
    – LMN
    Commented Oct 1, 2012 at 3:04
  • 2
    $\begingroup$ This follows form Poincare duality. So let $Z\in Y$ be a non-trivial cycle (i.e $[Z]\ne 0 \in H_*(Y,\mathbb Q)$). Let us prove that $\phi^{-1}(Z)$ is a non-trivial cycle in $X$. Chose $Z'$ in $Y$ Poincare dual to $Z$. And let us make so that $Z$ intersects $Z'$ transversally. Then $\phi^{-1}Z\cap \phi^{-1}\Z'$ is a collection of fibers. So the intersection is non-zero in homology of $X$, so both preimages in non-zero in homology of $X$. $\endgroup$
    – Dmitri Panov
    Commented Oct 1, 2012 at 3:13
  • 2
    $\begingroup$ Remark: your condition is necessary and sufficient, since it is equivalent to $\phi^\ast:H^n(Y)\hookrightarrow H^n(X)$, where $n=dim Y$. $\endgroup$
    – Ian Agol
    Commented Oct 1, 2012 at 14:13
3
$\begingroup$

Suppose one has compact symplectic manifolds $(X,\omega), (Y,\sigma)$, and a map $f:X\to Y$ such that $f^\ast\sigma=\omega$ (if $f$ is also a diffeomorphism, then this map is a symplectomorphism, but I'm not sure the terminology in this case). Then $f^\ast: H^*(Y;\mathbb{Q})\to H^\ast(X;\mathbb{Q})$ will be injective. If $dim Y=2k$, then $\sigma^k\neq 0\in H^{2k}(Y)$ is a fundamental class, and $dim X=2n \geq 2k$, then $\omega^n\neq 0\in H^{2n}(X)$. So $f^\ast(\sigma^k)= \omega^k \neq 0 \in H^{2k}(X)$.

Now, consider $\alpha \in H^j(Y)$. By Poincare duality, there exists $\beta\in H^{2k-j}(Y)$ such that $\alpha\cup \beta = \sigma^k$. Then $f^\ast(\alpha\cup \beta)=\omega^k\neq 0$, so $f^\ast \alpha\neq 0$.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ In fact $Y$ is not required to be symplectic. It suffices that $f$ is surjective and $X$ has a closed $2$ form which is symplectic on a generic fiber (so that dimensions have equal parity). When $f$ is a submersion, it is called a symplectic fibration, but there are important examples with critical points, the so-called (symplectic) Lefschetz fibrations. $\endgroup$
    – BS.
    Commented Oct 1, 2012 at 11:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .