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

Let $\phi: \mathcal{X}\rightarrow B$ be a family of complex manifolds i.e. $\phi$ is a proper submersive holomorphic morphism, i write $X_b$ for the fiber of $b\in B$.

Suppose $dim_{\mathbb{C}}X_b=2$ and $X_b$ is kahler. Suppose that $R^2\phi_*\mathbb{Z}$ is a local system, then i can consider the monodromy action

$\pi_1(B,b)\rightarrow Aut(H^2(X_b,\mathbb{Z}))$

I have read that the monodromy action actually is a subgroup of $O(H^2(X_0,\mathbb{Z}))$ i.e. it preserves the intersection pairing. Why is that?

share|improve this question
1  
I am not sure I understand the question, and I wonder if there might be a mistake in the formulation. What do you mean by "family of complex manifolds". Are you assuming that the morphism $\phi$ is proper and submersive (i.e., "smooth" in the language of algebraic geometry)? If so, and if $B$ is contractible, then the monodromy action is trivial, just as you say. However, if the morphism is only smooth over $B\setminus \{0\}$, then there can be a nontrivial action of $\pi_1(B\setminus\{0\},b)$. –  Jason Starr Jun 26 '13 at 14:47
1  
To what Jason said, I would add that monodromy does not act on the cohomology of $X_0$ (that is, fiber over zero, which contains critical points of $f$). It acts on the cohomology of $X_c=f^{-1}(c)$, where $c\ne 0$. –  Serge Lvovski Jun 26 '13 at 15:06
    
yes, both of you are right, i'm sorry, i will edit the question –  michael waltz Jun 26 '13 at 16:13
add comment

1 Answer

The intersection pairing is the composition of cup product followed by evaluation on the fundamental class. Both these operations are preserved by monodromy. Indeed, we have a pairing of local systems $$\cup: R^2\phi_* \mathbb{Z}\otimes R^2\phi_* \mathbb{Z}\to R^4\phi_* \mathbb{Z}\cong \mathbb{Z}$$

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.