This is maybe a silly question (but hopefully not). I am trying to understand the local monodromy of elliptic fibrations, and I am running into a seeming contradiction I do not know how to resolve.
Let $\pi:X\to\Delta$ be a complex elliptic fibration with a single singular fiber above $0\in\Delta=\{c\in\mathbb C:|c|<1\}$. If one looks up tables of the matrices for the local monodromies around each type of singular fiber, then you see that some of these matrices are 0 along the diagonal. However, it is my understanding that the Picard-Lefschetz formula says the monodromy around a point should be given by
$$T(\alpha)=\alpha-(\alpha,\beta)\beta$$
where $\alpha\in H_1(X_s)$ ($s\neq0$ and $X_s=\pi^{-1}(s)$) and $\beta\in H_1(X_s)$ is the (unique up to signs, I think) vanishing cycle. In my mind, the vanishing cycle can always be taken to be one of the generators of $H_1(X_s)$, so I would expect that the matrix for $T$ is always nonzero in one of the diagonals (the one corresponding to the other generator), but this is apparently not the case.
To make things more concrete, Miranda's Basic Theory of Elliptic Surfaces suggests (lecture VI) that the monodromy around the singular fibre (of type III in Kodaira's classification) of the surface $$X=\left\{([z_0:z_1:z_2],s)\in\mathbb P^2\times\Delta:z_2z_1^2=z_0^3+sz_0^2z_2\right\}$$ is given by $$T=\begin{pmatrix}0&1\\-1&0\end{pmatrix}.$$ How does one compute this, and what is wrong in my reasoning that I should not expect the diagonal to be only 0's (I imagine the answer to this second part would be already contained in an answer to the first)?