Let $\pi:E\to X$ the universal generalized elliptic curve over the compactified modular curve, with zero section $e: X\to E$. Now consider the following two sheaves on $X$:

$e^*\Omega^1_{E/X}$ and $\pi_*\Omega^1_{E^{sm}/X}(\text{log}\tilde{C})$

Where $\tilde{C}$ is the inverse image of the cusps in $E^{sm}$. I would like to show that this two sheaves are isomorphic. The reason is that the LHS is clearly locally free of rank one and compatible with base-change while the RHS has a de Rham interpretation and Gaus-Manin connection.

If we look at the inclusion $\pi^*e^*\Omega^1_{E^{sm}/X}\cong \Omega^1_{E^{sm}/X}\to \Omega^1_{E^{sm}/X}(\text{log}\tilde{C})$, then using the adjunction we get a morphism $e^*\Omega^1_{E/X}=e^*\Omega^1_{E^{sm}/X}\to\pi_*\Omega^1_{E^{sm}/X}(\text{log}\tilde{C})$ which is indeed an isomorphism if we restrict the sheaves to the open modular curve because there the generalized elliptic curve is an actual elliptic curve (and we can use proper base-change for example).

As I have already written, the sheaf $e^*\Omega^1_{E/X}$ is locally free of rank one (we can compute easily the stalks) and commutes with base-change by construction, but I cannot prove these properties for the other sheaf as $E^{sm}\to X$ isn't proper and I cannot use proper base-change.

My guess is that the sheaf of logarithmic differential should be related to the dualing sheaf, but I would like to avoid that if possible.

Does anyone have an idea about how to prove the sheaves are isomorphic?



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