I am looking for an example of a (edit: projective) family

$f : X \to Y$

of complex algebraic varieties which is a topologically locally trivial fibration in (singular) varieties and such that there exists an $q$ such that the monodromy representation

$\pi_1(Y,y) \to GL(H^q(F))$

is not semi-simple, where $F := f^{-1}(Y)$.

I guess that such examples should be plentiful but I don't know any.

By Deligne's theorem (the decomposition theorem for smooth families) such behaviour is impossible if $f$ is smooth. Hence the requirement that the fibres should be singular.

I am actually interested in the following: A variety $X$ and a semi-simple local system $\mathcal{L}$ on a Zariski open subset such that some cohomology sheaf of $IC(X, \mathcal{L})$ has non-semi-simple monodromy.

Bonus points: Is there an example with $\mathcal{L}$ trivial?

EDIT (following Piotr's relevant comment below): Note that it is important that $Y$ be an algebraic variety. Indeed, if $Y = \mathbb{A}^1 \setminus \{ x_1, \dots, x_n \}$ and $f$ is smooth then the monodromy around any $x_i$ will be quasi-unipotent. However the representation of the free group will still be semi-simple.

  • $\begingroup$ I'm confused. In case $Y=\mathbb{A}^1\setminus\{0\}$ (or just the germ of a smooth curve), $f$ smooth and proper, with semistable reduction at $0$, the monodromy action is unipotent, and hence unlikely to be semi-simple. What am I missing? $\endgroup$ Commented Feb 19, 2016 at 17:45
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    $\begingroup$ The monodromy will be unipotent around 0 but there will be other singular fibres (e.g. the Weiestrass family $y^2 = x(x-1)(x-\lambda)$ over $\lambda \in \mathbb{A}^1$ the monodromy is unipotent around 0 and 1, but in total one gets a semi-simple local system). $\endgroup$ Commented Feb 19, 2016 at 17:53
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    $\begingroup$ @PiotrAchinger - the argument for this is beautiful! By spreading-out one reduces to characteristic $p$ and then one measures the eigenvalues of Frobenius on this local system. Recall that a local system is pure of weight $w$ if every eigenvalue of $\operatorname{Frob}_q$ has absolute value $q^{w/2}$. $\endgroup$
    – Will Sawin
    Commented Feb 22, 2016 at 14:40
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    $\begingroup$ @PiotrAchinger By Deligne's Weil II theorem the $i$th cohomology of a smooth proper family is pure of weight $i$, hence any two irreducible factors have the same weight and thus by Weil II again the Ext group between them is the first cohomology of a local system of weight $i-i=0$, hence has weight $1$, which means the Frobenius eigenvalues are not $1$ so no extension is defined over the base field. So one is able to show that the global algebraic geometry is completely different from the local / analytic situation using arithmetic and counting, because these only make sense globally. $\endgroup$
    – Will Sawin
    Commented Feb 22, 2016 at 14:43

2 Answers 2


So it seems passerby's example can be modified to give a projective example.

(Thanks for de Cataldo and Migliorini for some of the following. All mistakes are mine.)

curve and desingularization

Fix $E$ an elliptic curve and consider a family over $B = E - \{ id \}$ where the fibre over $s$ is $E$ with $id$ joined to $s$. Let us call this singular elliptic curve E_s. (I have not checked that such a family exists, but I guess it isn't difficult.)

We have an exact sequence

$0 \to H_1(E) \to H_1(E_s) \to Z[c] \to 0$

where c is any cycle that passes through the singular point of E_s. In Deligne's theory $H_1(E)$ is of weight -1 and $Z[c]$ is of weight 0.


Now $\pi_1(B)$ is a free group on 2 generators. Let $p : \pi_1(B) -> H_1(E)$ denote the canonical map (the abelianization). Then I think that if $\gamma \in \pi_1(B)$ then $\gamma$ acts on $[c]$ by

$\gamma(c) = c + p(\gamma).$

This picture might help...


Anyway, this means that the representation of $\pi_1(B)$ is certainly not-semi-simple. (It "mixes weight 0 with weight -1".)

In this context BBD, Proposition 6.2.3 is useful: the weight filtration for a topologically locally trivial family is by locally constant subsystems.

Now suppose that $f : X \to Y$ is some family of stable curves of genus 2 such that the fibres are generically smooth and such there exists some subvariety $E - \{ id \} \in Y$ such that over this subvariety the family is the above example. Then applying the decomposition for $f_* \mathbb{Q}_X$ one gets the non-semi-simple local system above occurring.

I am not sure if such a family exists. But in any case the above seems to suggest that considerations of stable curves should give many such examples of non-semi-simplicity.

(It is nice in this example to imagine the genus 2 curve degenerating and deducing the mixed Hodge structure on the IC from the limit mixed Hodge structure.)


Won't even an open variety do? Take an elliptic curve $E$ (maybe over a base $B$) and remove the zero-section and another section $s$.

The fiber over $b \in B$ is a twice-punctured complex torus and thus has $3$-dimensional $H^1$; that $H^1$ has a $2$-dimensional subspace, coming from $H^1(E_b)$, which is invariant by monodromy.

So we get an extension of the local system $b \mapsto H^1(E_b)$ by the trivial local system. The resulting extension class in $H^1(B, H^1(E))$) should be something fairly close to the cycle class of $s$. So if you take something like an elliptic surface and take $s$ to be nontorsion in the Mordell-Weil group, you should get a non-split extension.

Apologies if I misunderstood the question...

  • $\begingroup$ Thank you for this example. I really want a projective example (I realise I didn't specify this...) I will add an adjective to the question. $\endgroup$ Commented Feb 19, 2016 at 21:51
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    $\begingroup$ What happens if we take passerby's example and identify the two sections instead of removing them? $\endgroup$ Commented Feb 20, 2016 at 10:40
  • $\begingroup$ I definitely think some such way of "singularizing" an open variety should work. But I didn't think about the details. Also, I don't know if that would answer your question of making the intersection cohomology non-semisimple. $\endgroup$
    – passerby
    Commented Feb 20, 2016 at 17:57

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