14
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Piotr Achinger Feb 19 '16 at 17:45
  • 1
    $\begingroup$ 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). $\endgroup$ – Geordie Williamson Feb 19 '16 at 17:53
  • 1
    $\begingroup$ @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 Feb 22 '16 at 14:40
  • 1
    $\begingroup$ @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 Feb 22 '16 at 14:43
4
$\begingroup$

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.

cycles

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...

monodromy

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.)

$\endgroup$
3
$\begingroup$

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...

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Geordie Williamson Feb 19 '16 at 21:51
  • 1
    $\begingroup$ What happens if we take passerby's example and identify the two sections instead of removing them? $\endgroup$ – Piotr Achinger Feb 20 '16 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 Feb 20 '16 at 17:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.