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In pondering this MO question and in particularly its 1st answer, and answers to
this one recently posed, I realized there ought to be a dodecahedral K3 surface $X$. This $X$ would fiber as an elliptic surface over $CP^1$ with 12 singular fibers, each of type $I_2$. The corresponding singular points on $CP^1$ would form the vertices of the icosahedron (centers of the dodecahedron). The automorphism group of this $K3$ will then project onto the symmetry group of the dodecahedron. Do you know this $K3$? Do you have a reference?

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I could use some help getting the URLs to these questions appear as a link in the question -RM – Richard Montgomery Oct 1 '11 at 1:29
Done (also tweaked your AG tag) – Yemon Choi Oct 1 '11 at 1:32
Edited only to tweak the remaining tags. – Noam D. Elkies Oct 1 '11 at 1:56
up vote 14 down vote accepted

There's a pretty K3 with icosahedral symmetry and 12 singular fibers each of type II (so a double root of the discriminant but with additive reduction); would that do? It's $y^2 = x^3 + P(t)$ where $P$ has icosahedral symmetry. Explicitly one can take $P(t) = t^{11} - 11 t^6 - t$. This surface is isotrivial (i.e. with constant $j$-invariant, here $j=0$), and is closely related with the isotrivial surface of Chahal, Meijer, and Top, which also has $j=0$ but with $a(t) = t^{12} - 11 t^6 - 1$. Both surfaces attain the maximum of $18$ for the Mordell-Weil rank of an elliptic K3 surface over ${\bf C}(t)$. The Chahal-Meijer-Top paper is on the arXiv (9911274), and is published in Comment. Math. Univ. St. Pauli 49 (2000), 79–89. My recent student G.Zaytman studied the icosahedral surface and its Mordell-Weil lattice in the course of his thesis research.

EDIT Come to think of it, this is up to isomorphism the unique elliptic K3 surface over ${\bf C}(t)$ with icosahedral symmetries acting on the base: for the narrow Weierstrass form $y^2 = x^3 + a(t) x + b(t)$ to have icosahedral symmetry, the same must be true of the coefficients $a$ and $b$; but $a$ has degree $8$, so must vanish identically, while $b$ has degree 12 and is thus determined up to a scalar factor.

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Is it obvious that the group action on ${\mathbb P}^1$ lifts to the $K3$ surface? if it does, then there is a ${\mathbb Z}_5$ that fixes a singular fiber, so in particular it fixes the singular point. On the other hand, the action on near a fixed point has to be analytically of the form $(x,y)\mapsto (ax, a^{-1}y)$ for $a$ a $5$-th root of 1 and I'm having trouble reconciling these two facts. – rita Oct 1 '11 at 19:24
Yes, it lifts. Put the singular fiber at $t=0$ (as the formula I gave does for one of the fibers) so the surface is $y^2 = x^3 + t Q(t^5)$ for some polynomial $Q$. Then if $a^5=1$ the automorphism $t \mapsto at$ of the $t$-line lifts to $(x,y,z) \mapsto (a^2 x, a^3 y, a t)$. – Noam D. Elkies Oct 1 '11 at 21:27
Noam, thanks for the Chahal et al ref. Page 4, sec. 3, par. 2 asserts that putting $P(t)=A(t^6 -b)^2$ in your representation above gives a K3 with 6 singular fibers of type IV. The corresponding 6 zeros of P lie on a hexagon so don't have the octahedral symmetry of my MO question of last week (that you'd answered). But the zeros of $P(t) = t (t^2 +1) (t^2 -1)$ form the vertices of the octahedron. Might the elliptic surface for this P be Hirzebruch's K3 with the octahedral symmetry? (How does one compute $p_g$ from $P(t)$?) – Richard Montgomery Oct 2 '11 at 5:40
@Noam: thank you, now I see it. – rita Oct 2 '11 at 11:38
@Richard: That's a good thought, but no, it's not the same. Yes, it's an elliptic K3 surface (as long as $P$ has no zeros of order $6$ or more, $y^2=x^3+P(t)$ is K3 for $6 < \deg P \leq 12$), but it can be shown that if such a surface $X$ attains the upper bound of $20$ on the rank of the Néron-Severi group ${\rm NS}(X)$ then the intersection pairing on ${\rm NS}(X)$ has discriminant $-3n^2$ for some integer $n$, whereas the Hirzebruch surface has discriminant $-16$. There's more ways for a K3 surface to have octahedral symmetry than icosahedral; ... [continued due to 600-character limit] – Noam D. Elkies Oct 4 '11 at 3:41

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