How could a vanishing cycle be zero? I am studying the book "Etale Cohomology and the Weil Conjectures" by Reinhard and Kiehl. For the following, let's assume $\Lambda$ is a finite group of order prime to the characteristic of all schemes we are working with.
In chapter III, paragraph 4 is about a "local" Picard-Lefschetz formula which statement starts as following:

Let S be the spectrum of a strictly Henselian local ring $R$, $f: X \to S$ a proper flat morphism of odd fiber dimension, let $X_s$ be its special fiber and $X_\eta$ its generic fiber. Assume $f$ is such that it is singular on a single point $a$, then the strict henselization of $\mathcal{O}_{X,a}$ is $R$-isomorphic to $R[X_1,\ldots X_n]/\left(\sum\limits_{\nu = 0}^mX_\nu X_{\nu+m+1} + \lambda\right)$ for some $\lambda \in \mathfrak{m}(R)$.
Then $H^\nu(X_\eta,\Lambda) \cong H^{\nu}(X_s, \Lambda)$ for all $\nu \neq n, n+1$,
There is a "vanishing cycle" $\delta \in H^n(X_\eta, \Lambda)\otimes \Lambda(m)$ and a "covanishing cycle" $\delta^* \in H^{n + 1}(X_s, \Lambda(n-m))$...

The vanishing and covanishing cycle are then constructed as follow:
First recall the normal form of smooth quadrics: they are the zero sets in $\mathbb{P}^{n+1}$ of the $n+2$ variable quadratic forms:

*

*$Q = \sum\limits_{\nu=0}^mX_\nu X_{\nu+m+1} + x_{n+1}^2$ if $n = 2m+1 $

*$Q = \sum\limits_{\nu=0}^mX_\nu X_{\nu+m+1}$ if $n = 2m$.

Now for a quadric $Y$ in the case $n = 2m$, we consider the linear subspaces:

*

*$L_1: X_0 = \cdots = X_m = 0$

*$L_2: X_0 = \cdots = X_{m-1} = X_{2m+1} = 0$
Let's call $\theta_1$ and $\theta_2$ the classes in $H^n(Y, \Lambda(m))$ associated to these cycles. They are such that $H^n(Y, \Lambda(m)) = \Lambda.\theta_1 + \Lambda.\theta_2$.
Now in an odd-dimensional quadric $X$ in normal form, the hyperplane section $X_{n+1} = 0$ may be identified with $Y$, call $X_e$ the complement $X \setminus Y$, it is an affine quadric defined by $\sum\limits_{\nu = 0}^m X_\nu X_{\nu+m+1} + 1 $, one has a long exact sequence \begin{align*}
H_c^{\nu}(X_e, \Lambda(-)) \rightarrow H^{\nu}(X, \Lambda(-)) \rightarrow H^{\nu}(Y, \Lambda(-)) \overset{\partial}{\rightarrow} \cdots
\end{align*}
The image $\theta$ of either $\theta_1$ or $\theta_2$ by $\partial$ in $H^{n}_c(X_e, \Lambda(m))$ is then a generator of it.
Then, in the setting of the Picard-Lefschetz formula, the vanishing cycle is the image of $\theta$ in $H^n(X_\eta, \Lambda)$ (the morphism is induced by the same exact triangle but with $X_{\eta}$ instead of $X$.
The vanishing cycles are uniquely determined up to sign.
Now in a latter theorem, there are some cases handling the case of a null vanishing cycle. It comes as a surprise to me, because I don't really see in the construction where the cycle could become zero. $\theta_1$ and $\theta_2$ determine the cohomology of a quadric, and they can't be $zero$, same for $\theta$ I think (otherwise, by the long exact sequence, that would mean $\theta_1$ and $\theta_2$ are images of elements of $H^{\nu}(X, \Lambda())$, which seems impossible to me because of their definition, but I might be wrong here.
The last point where $\delta$ could become zero would be if the image of $\theta$ in $H^{n}(X_\eta, \Lambda(-))$ is zero, which would mean that $\theta$ comes from a cycle in $H^{n-1}(Y_\eta, \Lambda(-))$, which could be possible since this group is $\Lambda \oplus \Lambda$ ($n = 2m+1$ and $Y$ is a quadric which lands in the  case $2m$, so its $H^{n-1}$ is generated by two cycles).
My question is: I have no idea of

*

*Wether this is actually correct, i.e wether I have correctly identified the conditions for $\delta$ to be null or not.

*Whether the case I described can actually happens (I think it could since later theorems handle the case of a null vanishing cycle).

*What it would actually mean for a vanishing cycle to be zero, so far I only see an algebraic condition and I have no intuition on what it should mean, or what should cause it to happen besides "it could just happens sometimes".

I have read this MO question to help build my intuition of the Picard-Lefschetz formula, but the picture given is making me think even more that the vanishing cycle can not vanish.
So I would gladly accept any answer providing me with either comprehensive references on the vanishing cycles in this case (I tried to read the relevant part of SGA7, but the situation seems similar and I can not understand why a vanishing cycle could be zero as well), or explain to me how it could happen, what it would mean and what is known about this case (besides the consequences on the monodromy being "the singularity does not twist the monodromy action"). Or even better, provide an example in which the vanishing cycle is zero.
 A: I don't know if a vanishing cycle is always non zero in the case of an algebraic Lefschetz pencil, I recall something like that but I am not sure at all. If this is true, it will follow a posteriori from the cohomological study. At first glance, this is not obvious at all.
Also, recall that vanishing cycles can be defined in a non algebraic context where you just have a map $f:X\to\mathbb{C}$ such that $\mathrm{d}f$ vanishes in one point $p$ and the Hessian is non-degenerate. In this context, vanishing cycles can definitely be zero.
The Picard-Lefschetz formula is local in two ways :


*

*in the target : consider a singular value $0$ and a small disk around $0$. We then study the local monodromy

*in the source : consider a singular point $p\in X$. Then a neighbourhood of $p$ in $X_0$ looks like a singular quadric. A "nearby fiber in a neighbourhood of $p$" looks like a smooth quadric. By nearby fiber, I mean $X_t$ for $t$ close enough to $0$. 


So the situation is that locally around $p$ (and not $0$), the fiber looks like quadric with a cycle which is pinched as the fiber gets closer to the singular fiber $X_0$.
But this says nothing about the rest of the fiber : what is going on in the fiber $X_t$ when we are not close to $p$ ?
Take this example which is not algebraic : consider a sphere of radius $1$ and mark the equator. This will be the fiber over $1$. Over another point $t$ (with $|t|\leq 1$), just shrink the equator so that it is of radius $|t|$. When $t=0$, you will get two tangent spheres : the equator is then reduced to one point. In this example, the vanishing cycle is the equator. But this is definitively zero since $H^1(S^2)=0$. The point is that, locally around $p$, the equator is not zero in cohomology, but becomes $0$ when considering the whole fiber.
