A confusion about vanishing cycles and the Picard Lefschetz formula 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 ? 
 A: 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 $H_1(E_t)$.
