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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?

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    $\begingroup$ Do you want all fibers other than $X_0, X_1$ to be smooth? If so, it is impossible. $\endgroup$ – Will Sawin Apr 3 at 11:38
  • $\begingroup$ I do. Could you tell my why this is impossible? $\endgroup$ – Dima Sustretov Apr 3 at 12:21
  • $\begingroup$ the monodromy should be semi-simple, but also unipotent, hence the identity ... $\endgroup$ – Geordie Williamson Apr 3 at 13:18
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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.

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    $\begingroup$ why must the $j$ function avoid 0 and 1728 in the first version of the argument? $\endgroup$ – Dima Sustretov Apr 3 at 14:16
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    $\begingroup$ @DimaSustretov For no reason. I just removed those points so that the map from $\pi_1( (X(1) - \{0, 1728, \infty\})$ to $SL_2(\mathbb Z)$ is well-defined. So for this part of the argument you need to remove the inverse image of $\mathbb 0$ and $1728$ from $\mathbb P^1$. Or you can work with Galois groups, that works too. $\endgroup$ – Will Sawin Apr 3 at 14:19
  • $\begingroup$ Alternatively, one can observe that such configuration of singular fibres contradicts the Shioda-Tate formula $rk\,E(k(x)) = rk\,NS(S) - 2 - \sum (r_i-1)$ where $E$ is the generic fibre of the elliptic surface $S$ over $P^1$ regarded as a variety over $k(x)$, $rk\,E(k(x))$ is the rank of the grop of rational sections and $r_i$ are the numbers of irreducible components in singular fibres. If there are two $I_6$ fibres, then the Picard number is 10 by Euler characteristic computation, and so the rhs is negative, which is impossible. $\endgroup$ – Dima Sustretov Apr 8 at 9:39
  • $\begingroup$ Similarly, this formula rules out the configuration of three fibres of $I_4$ type $\endgroup$ – Dima Sustretov Apr 8 at 9:39

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