Despite the apparent simplicity of the following question I couldn't find the answer so far.


I am looking to construct an elliptic fibration $X \to \mathbb{P}^1$ with $X$ smooth, and exactly two reduced singular fibres $X\_0, X\_1$ such that the monodromy action around each of the fibres on the first cohomology of the nearby fibre is of the form
$$
\left(\begin{array}{cc}1 & a\\ 0 & 1\\ \end{array}\right)
$$


Does such a fibration exist?



elliptic fibration over $\mathbb{P}^1$ with exactly two fibres with monodromy of unipotency rank 1
If you look at the global monodromy action $\pi\_1 ( \mathbb P^1 -\{0,1\}) \to SL\_2(\mathbb Z)$, you see that $\pi\_1 ( \mathbb P^1 -\{0,1\}) = \mathbb Z$ so the image is abelian and therefore has infinite index. On the other hand, if we consider the $j$ map $\mathbb P^1 \to X(1)$, the image of the fundamental group has finite index in $\pi\_1 ( X(1) -\{0, 1728, \infty\} )$ unless the $j$ map is constant. This is a contradiction, so the $j$ map is constant.


An alternate way to see that the $j$ map is constant is to observe that, if we pull back to the universal cover of $\mathbb P^1 - \{0,1\}$, this map lifts to the upper half plane. But the universal cover is the complex plane, and any map to the upper half plane is constant by Liouville's theorem.


But unipotent local monodromy can only occur when there is a pole of the $j$ invariant, so the $j$ invariant also cannot be constant. So no such fibration exists.


