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In "On elliptic modular surfaces", Shioda proves some interesting theorems on smooth elliptic surfaces (admitting a section); he then focuses on "modular elliptic surfaces" and proves some more theorems, e.g. he shows that there is a one to one correspondence between modular forms vanishing at cusps of a modular curve, and holomorphic two forms on the corresponding modular surfaces.

After this short introduction, my question is: Can you give me an explicit example of a smooth elliptic surface (with section) with at least 3 singular fibers, which is not modular? What if the base curve has genus zero?

More explanation: Let $S\to B$ be an elliptic surface with section and at least 3 singular fibers. Let $\Sigma\subset B$ be the complement of singular values. Then $\Sigma=\mathbb{H}/G$, is a quotient of upper half plane. The $j$ function of this elliptic fibration $j:\Sigma \to \mathbb{H}/PSL(2,\mathbb{Z})$ (is it always holomorphic?) gives a map $$ \iota: G \to SL(2,\mathbb{Z}).$$ I then feel that $S$ should be the modular surface corresponding to $\iota(G)\subset SL(2,\mathbb{Z})$ and I can't see what might go wrong?!

Regarding the answer of Remke below: Assuming that the map $\iota$ above has finite dimensional kernel, it seems that $S$ should be the pull-back of the modular surface corresponding to $\iota(G)$ via the ramified covering map $$ \mathbb{H}/G \to \mathbb{H}/\iota(G);$$ is this true?

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The modular elliptic surfaces are quite rare. E.g., the Mordell-Weil group is finite and the Picard number of the surface equals $h^{1,1}$ (see Shioda's paper). Such elliptic surfaces are called extremal and for a fixed Euler number of the surface and fixed genus of the base curve there are only finitely many extremal elliptic surfaces with non-constant $j$-map.

The $j$-map of an extremal elliptic surfaces is unramified away from the preimages of $0$, $1728$ and $\infty$. So any elliptic surface with $j$-map which has ramification over a point different from $0,1728$ and $\infty$ is not a modular elliptic surface.

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  • $\begingroup$ I have added a comment to the end of my question regarding your answer. What do you think of it? $\endgroup$
    – Mohammad Farajzadeh-Tehrani
    Commented Oct 17, 2013 at 23:39
  • $\begingroup$ To be honest, I do not really understand the second part of your question. For example, why is the map H/G to H/PSL(2,Z) induced by a self map of H? You need this to have the map i:G->PSL(2,Z). $\endgroup$
    – Remke Kloosterman
    Commented Oct 18, 2013 at 17:34
  • $\begingroup$ We can lift the j-map to $j:\mathbb{H}\to \mathbb{H}/PSL(2,\mathbb{Z})$; since $\mathbb{H}$ is simply connected, we can then lift this to a map $j:\mathbb{H}\to\mathbb{H}$. Then for $\gamma \in G$, we have to have $j(\gamma\cdot z)=\iota(\gamma)j(z)$, for some representation $\iota:G \to PSL(2,\mathbb{Z})$ (I am assuming this lifts to $SL(2,\mathbb{Z})$ itself, which should not be a big deal for now) $\endgroup$
    – Mohammad Farajzadeh-Tehrani
    Commented Oct 18, 2013 at 19:49
  • $\begingroup$ @MohammadF.Tehrani How does the j-map lift to a self-map of $\mathbb H$? The action of PSL$_2(\mathbb Z)$ is not free, and moreover, the complex upper half-plane $\mathbb H$ is not the universal cover of the affine line PSL$_2(\mathbb Z) \backslash \mathbb H$. $\endgroup$
    – Ariyan Javanpeykar
    Commented Mar 9, 2014 at 19:52

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