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In Ch. III, section 11 of Barth-Peters-Van de Ven (I only have the old edition), Proposition 11.4, there is a formula that relates the topological Euler characteristic of a fibered surface with the characteristic of the fiberse and of the base. In the case of a K3 with an elliptic fibration it gives: $24=\sum_{s\in T} e(F_s)$, where $T\subset {\mathbb P}^1$ is the set of critical values of the fibration, and $e(F_s)$ is the Euler characteristic of the fiber $F_s$ over $s\in T$. So if all the singular fibers are of type $I_2$, you have 12 singular fibers.

The Hirzebruch K3 (call it $S$) that you describe has at least 5 elliptic fibrations. If $P_1,P_2,P_3,P_4$ are the 4 vertices of the quadrilateral, these pencils are obtained by pulling back the pencil of lines through any of the $P_i$ and the pencil of conics through the points $P_1,\dots P_4$. I suspect that they can all be transformed one into the other: besides the $S_4$ symmetry that you point out, one can perfom a standard Cremona transformation centered, say, at $P_1,P_2,P_3$ and fixing $P_4$. This exchanges the pencil of lines through $P_4$ with the pencil conics through the 4 points.

Consider the pencil $|F_1|$ of lines through $P_1$: one can check that the preimage in $S$ of a general element of $|F_1|$ is the disjoint union of $4$ smooth elliptic curves. The singular fibers correspond to the lines $P_1P_i$, $i=2,3,4$. It seems to me that the preimage of each of these is the disjoint union of 4 double fibers of type $2I_2$. As I remarked above, the configuration of singular fibers should be the same for all the 5 fibrations by symmetry.

In Ch. III, section 11 of Barth-Peters-Van de Ven (I only have the old edition), Proposition 11.4, there is a formula that relates the topological Euler characteristic of a fibered surface with the characteristic of the fiberse and of the base. In the case of a K3 with an elliptic fibration it gives: $24=\sum_{s\in T} e(F_s)$, where $T\subset {\mathbb P}^1$ is the set of critical values of the fibration, and $e(F_s)$ is the Euler characteristic of the fiber $F_s$ over $s\in T$. So if all the singular fibers are of type $I_2$, you have 12 singular fibers.

The Hirzebruch K3 (call it $S$) that you describe has at least 5 elliptic fibrations. If $P_1,P_2,P_3,P_4$ are the 4 vertices of the quadrilateral, these pencils are obtained by pulling back the pencil of lines through any of the $P_i$ and the pencil of conics through the points $P_1,\dots P_4$. I suspect that they can all be transformed one into the other: besides the $S_4$ symmetry that you point out, one can perfom a standard Cremona transformation centered, say, at $P_1,P_2,P_3$ and fixing $P_4$. This exchanges the pencil of lines through $P_4$ with the pencil conics through the 4 points.

Consider the pencil $|F_1|$ of lines through $P_1$: one can check that the preimage in $S$ of a general element of $|F_1|$ is the disjoint union of $4$ smooth elliptic curves. The singular fibers correspond to the lines $P_1P_i$, $i=2,3,4$. It seems to me that the preimage of each of these is the disjoint union of 4 double fibers of type $2I_2$.

EDIT: in fact I was wrong, the preimage of $P_1P_i$ is the disjoint union of two fibers of type $I_4$.

As I remarked above, the configuration of singular fibers should be the same for all the 5 fibrations by symmetry.

In Ch. III, section 11 of Barth-Peters-Van de Ven (I only have the old edition), Proposition 11.4, there is a formula that relates the topological Euler characteristic of a fibered surface with the characteristic of the fiberse and of the base. In the case of a K3 with an elliptic fibration it gives: $24=\sum_{s\in T} e(F_s)$, where $T\subset {\mathbb P}^1$ is the set of critical values of the fibration, and $e(F_s)$ is the Euler characteristic of the fiber $F_s$ over $s\in T$. So if all the singular fibers are of type $I_2$, you have 12 singular fibers.

The Hirzebruch K3 (call it $S$) that you describe has at least 5 elliptic fibrations. If $P_1,P_2,P_3,P_4$ are the 4 vertices of the quadrilateral, these pencils are obtained by pulling back the pencil of lines through any of the $P_i$ and the pencil of conics through the $P_1,\dots P_4$. I suspect that they can all be transformed one into the other: besides the $S_4$ symmetry that you point out, one can perfom a standard Cremona transformation centered, say, at $P_1,P_2,P_3$ and fixing $P_4$. This exchanges the pencil of lines through $P_4$ with the pencil conics through the 4 points.

Consider the pencil $|F_1|$ of lines through $P_1$: one can check that the preimage in $S$ of a general element of $|F_1|$ is the disjoint union of $4$ smooth elliptic curves. The singular fibers correspond to the lines $P_1P_i$, $i=2,3,4$. It seems to me that the preimage of each of these is the disjoint union of 4 double fibers of type $2I_2$.

In Ch. III, section 11 of Barth-Peters-Van de Ven (I only have the old edition), Proposition 11.4, there is a formula that relates the topological Euler characteristic of a fibered surface with the characteristic of the fiberse and of the base. In the case of a K3 with an elliptic fibration it gives: $24=\sum_{s\in T} e(F_s)$, where $T\subset {\mathbb P}^1$ is the set of critical values of the fibration, and $e(F_s)$ is the Euler characteristic of the fiber $F_s$ over $s\in T$. So if all the singular fibers are of type $I_2$, you have 12 singular fibers.

The Hirzebruch K3 (call it $S$) that you describe has at least 5 elliptic fibrations. If $P_1,P_2,P_3,P_4$ are the 4 vertices of the quadrilateral, these pencils are obtained by pulling back the pencil of lines through any of the $P_i$ and the pencil of conics through the points $P_1,\dots P_4$. I suspect that they can all be transformed one into the other: besides the $S_4$ symmetry that you point out, one can perfom a standard Cremona transformation centered, say, at $P_1,P_2,P_3$ and fixing $P_4$. This exchanges the pencil of lines through $P_4$ with the pencil conics through the 4 points.

Consider the pencil $|F_1|$ of lines through $P_1$: one can check that the preimage in $S$ of a general element of $|F_1|$ is the disjoint union of $4$ smooth elliptic curves. The singular fibers correspond to the lines $P_1P_i$, $i=2,3,4$. It seems to me that the preimage of each of these is the disjoint union of 4 double fibers of type $2I_2$. As I remarked above, the configuration of singular fibers should be the same for all the 5 fibrations by symmetry.

In Ch. III, section 11 of Barth-Peters-Van de Ven (I only have the old edition), Proposition 11.4, there is a formula that relates the topological Euler characteristic of a fibered surface with the characteristic of the fiberse and of the base. In the case of a K3 with an elliptic fibration it gives: $24=\sum_{s\in T} e(F_s)$, where $T\subset {\mathbb P}^1$ is the set of critical values of the fibration, and $e(F_s)$ is the Euler characteristic of the fiber $F_s$ over $s\in T$. So if all the singular fibers are of type $\tilde{A_1}$, you have 12 singular fibers.

The Hirzebruch K3 (call it $S$) that you describe has at least 5 elliptic fibrations. If $P_1,P_2,P_3,P_4$ are the 4 vertices of the quadrilateral, these pencils are obtained by pulling back the pencil of lines through any of the $P_i$ and the pencil of conics through the $P_1,\dots P_4$. I suspect that they can all be transformed one into the other: besides the $S_4$ symmetry that you point out, one can perfom a standard Cremona transformation centered, say, at $P_1,P_2,P_3$ and fixing $P_4$. This exchanges the pencil of lines through $P_4$ with the pencil conics through the 4 points.

Consider the pencil $|F_1|$ of lines through $P_1$: one can check that the preimage in $S$ of a general element of $|F_1|$ is the disjoint union of $4$ smooth elliptic curves. The singular fibers correspond to the lines $P_1P_i$, $i=2,3,4$. It seems to me that the preimage of each of these is the disjoint union of 4 double fibers of type $2\tilde{A_1}$.

In Ch. III, section 11 of Barth-Peters-Van de Ven (I only have the old edition), Proposition 11.4, there is a formula that relates the topological Euler characteristic of a fibered surface with the characteristic of the fiberse and of the base. In the case of a K3 with an elliptic fibration it gives: $24=\sum_{s\in T} e(F_s)$, where $T\subset {\mathbb P}^1$ is the set of critical values of the fibration, and $e(F_s)$ is the Euler characteristic of the fiber $F_s$ over $s\in T$. So if all the singular fibers are of type $I_2$, you have 12 singular fibers.

The Hirzebruch K3 (call it $S$) that you describe has at least 5 elliptic fibrations. If $P_1,P_2,P_3,P_4$ are the 4 vertices of the quadrilateral, these pencils are obtained by pulling back the pencil of lines through any of the $P_i$ and the pencil of conics through the $P_1,\dots P_4$. I suspect that they can all be transformed one into the other: besides the $S_4$ symmetry that you point out, one can perfom a standard Cremona transformation centered, say, at $P_1,P_2,P_3$ and fixing $P_4$. This exchanges the pencil of lines through $P_4$ with the pencil conics through the 4 points.

Consider the pencil $|F_1|$ of lines through $P_1$: one can check that the preimage in $S$ of a general element of $|F_1|$ is the disjoint union of $4$ smooth elliptic curves. The singular fibers correspond to the lines $P_1P_i$, $i=2,3,4$. It seems to me that the preimage of each of these is the disjoint union of 4 double fibers of type $2I_2$.