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Let $f : E \to D$ a family of elliptic curve, such that $E_t$ is smooth and $E_0$ is a nodal curve. Let $\delta \in H_1(E_t) := H$ ($t \neq 0$ is fixed) the vanishing cycle and $\gamma$ a cycle so that $\{\delta, \gamma\}$ is a symplectic basis of $H$.

I don't understand how to connect these two facts :

1) We should have $I \cap V = 0$ where $I \subset H$ is the set of invariant cycles and $V \subset H$ the set of vanishing cycles.

2) Let $m \in End(H)$ be the monodromy operator, then the Picard Lefschetz formula says that $m(\delta) = \delta$ and $m(\gamma) = \gamma + \delta$.

So we should have $I = V = \Bbb Z \langle \delta \rangle$. Where is my mistake ?

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  • $\begingroup$ Can you give your references ? There is often some confusions with $\delta$ being the "covanishing" cycle (the dual of the vanishing cycle). $\endgroup$ – Roland May 8 '18 at 19:22
  • $\begingroup$ @Roland : Thanks for your interest ! The statement $I \cap V = 0$ is written here page 29, the Picard-Lefschetz formula is page 40. It seems both statements are written for homology, but I don't understand the arguments very well so I might have done a mistake. $\endgroup$ – student May 8 '18 at 20:13
  • $\begingroup$ Weird, I have worked with the Picard-Lefschetz formula with another (cohomological) presentation, but not with this one. So what seems weird is the proof of the monodromy theorem (theorem 7.3 p45) since it says that $\pi_1$ acts non trivially on $V:=\mathbb{Z}\delta$. I'll think about it. $\endgroup$ – Roland May 8 '18 at 21:18
  • $\begingroup$ Ok I think the problem is just the situation : if $D$ is a small disk around $0$, then this fail $I\cap V\neq 0$ : the invariant homology is too big. On the other hand, if $D=\mathbb{P}^1$, then the Picard-Lefschetz formula only gives the local monodromy. Monodromy is a weird thing. It is locally trivial, though it is local and not necessarily trivial... $\endgroup$ – Roland May 9 '18 at 10:11
  • $\begingroup$ @Roland : I see, thanks a lot ! So in the case of the family $E_t$ given by $y^2 = x(x-1)(x-t)$ we have $I = 0$ and $V = H_1(E_t)$. $\endgroup$ – student May 9 '18 at 12:00
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The local situation is actually easy to picture. Imagine a torus got by revolving a circle $\delta$. Then make $\delta$ smaller and smaller until $\delta$ becomes a point, and the torus becomes a pinched torus. This description is the degeneration of $E_t$ into $E_0$.

The circle $\delta$ is a vanishing cycle since it becomes trivial in $E_0$. The other cycle $\gamma$ of the torus does not change and stay a cycle in $E_0$. From the topology of the torus, we have $\delta.\delta=0$, $\gamma.\gamma=0$ and $\delta.\gamma=\pm1$ (depending on the chosen orientations for $\delta$ and $\gamma$, so choose the orientation so that this product is $1$).

So the vanishing homology is $V=\mathbb{Z}.\delta$.

Now, we look at the monodromy : it is harder to visualize, but you have when $t$ moves around $0$, then $\delta$ does not change (in fact, up to a constant, $\delta$ is the only cycle which vanish, so the monodromy can only act as $m(\delta)=\pm\delta$, it turns out that in odd relative dimension $m(\delta)=\delta$). On the other hand, $\gamma$ does change, and you have the Picard-Lefschetz formula $m(\gamma)=\gamma+\delta$.


If $E_t$ the elliptic curve defined by $y^2=x(x-1)(x-t)$. When $t=0$ it degenerates into a pinched torus. It also degenerates at $t=1$ (also into a pinched torus) and at $t=\infty$, but this case the singularity is worse so it does not fit in the general study.

I don't have a nice example at hand on families of elliptic curves with quadratic singularities. You can take your favorite cubic surface and take a Lefschetz pencil, but it will have quite a lot of critical points... Computation are already quite hard. So I won't do them here. Instead I will try to give the picture.

So say there are $n$ critical values $v_1,v_2,...,v_n$ and let $t\neq v_1,...,v_n$. For each critical value $v_i$, there is an associated vanishing cycle $\delta_i\in H_1(E_t)$, an associated "dual" $\gamma_i\in H_1(E_t)$ and a monodromy operator $(m_i)_*$ (a rotation around the critival value $v_i$). The Picard-Lefschetz formula says that $(m_i)_*\delta_i=\delta_i$ and $(m_i)_*\gamma_i=\gamma_i+\delta_i$. But we have no idea what $(m_j)_*\delta_i$ and $(m_j)_*\gamma_i$ are.

Note also that $\delta_i,\gamma_i\in H_1(E_t)$ which is of dimension 2. So there are a lot of linear relations between our $2n$ elements.

In this case, we will have $V=H_1(E_t)$ and $I=0$. Indeed, $V$ is spanned by the vanishing cycles which are not zero (in this case). So take any vanishing cycle $\delta_i$. On has $\mathbb{Z}\delta_i\subset V$. But the intersection pairing is trivial on $\mathbb{Z}\delta_i$ whereas by Hard Lefschetz it is not degenerate on $V$. This implies that the rank of $V$ is greater that $1$ and thus is the whole $H_1(E_t)$.

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  • $\begingroup$ Thanks a lot, this is very useful as usual. A last question : how do you know that there is $6$ critical points ? $\endgroup$ – student May 9 '18 at 14:41
  • $\begingroup$ @student I did some calculation for that but got confused while rechecking. So it is probably wrong, I'll remove that and make a generic $n$. The point is $n$ is usually quite big... $\endgroup$ – Roland May 9 '18 at 16:54
  • $\begingroup$ a cubic surface $X$ has $\chi(X) = 9$ and for a smooth elliptic fibration $E$ we have $\chi(E) = 0$. I'm not sure but I think that every singular fiber with a single double points add +1 to the Euler characteristic so we should have $9$ singular fibers, what do you think ? $\endgroup$ – student May 9 '18 at 17:16
  • $\begingroup$ Well, I did this kind of computation, but I think we should first blow up the base of the pencil, leading to $\chi(\tilde{X})=12$. But I am not sure anymore, I got confused... $\endgroup$ – Roland May 9 '18 at 17:40
  • $\begingroup$ I'm not sure how to do it with a cubic surface, but if we take a pencil of cubics in $\Bbb P^2$ so there are 9 intersections points, and blowing up 9 times gives $\tilde X$ with $\chi(\tilde X) = 12$, which mean that there is $12$ singular fibers, assuming that there are only ordinary double points (I guess this is what happens if the pencil is generic enough). Thanks again for your wonderful answer and comments ! $\endgroup$ – student May 9 '18 at 20:27

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