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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?

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    $\begingroup$ 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 $\pi_1(B\setminus\{0\},b)$. $\endgroup$
    – Jason Starr
    Commented Jun 26, 2013 at 14:47
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    $\begingroup$ 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 $c\ne 0$. $\endgroup$
    – Serge Lvovski
    Commented Jun 26, 2013 at 15:06
  • $\begingroup$ yes, both of you are right, i'm sorry, i will edit the question $\endgroup$
    – michael waltz
    Commented Jun 26, 2013 at 16:13

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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 $$\cup: R^2\phi_* \mathbb{Z}\otimes R^2\phi_* \mathbb{Z}\to R^4\phi_* \mathbb{Z}\cong \mathbb{Z}$$

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