sheaf of relative differentials on an elliptic curve

583 Views Asked by At

Let $f : E\rightarrow S$ be an elliptic curve over a scheme $S$ with identity section $e : S\rightarrow E$.

Why is it true that $e^*\Omega_{E/S}\cong f_*\Omega_{E/S}$? (I believe these should be naturally/canonically isomorphic, but I don't know why).

Why should this be true intuitively? The first ($e^*\Omega_{E/S}$) is the "restriction" of $\Omega_{E/S}$ to the identity section. Here, I think of the sections of $e^*\Omega_{E/S}$ as sections of the cotangent bundle on $E/S$, "restricted to fibral tangent spaces".

What about $f_*\Omega_{E/S}$? I have a difficult time understanding $f_*\Omega_{E/S}$ geometrically. What is its associated vector bundle? (What is the pushforward of a vector bundle anyway?) I suppose its sections over $U\subset S$ are just holomorphic relative differentials on $f^{-1}(U)$, but this doesn't obviously imply that $e^*\Omega_{E/S}\cong f_*\Omega_{E/S}$.

2

There are 2 best solutions below

0
On

You have an exact sequence $0\to \Omega_{E/S}(-e(S))\to \Omega_{E/S}\to {\Omega_{E/S}}_{|e(S)}=e^*\Omega_{E/S}\to 0$. Applying $f_*$, you get a long exact sequence, $0\to f_*(\Omega_{E/S})\to e^*(\Omega_{E/S})\to R^1f_*(\Omega_{E/S}(-e(S)))\to R^1f_*(\Omega_{E/S})\to 0$. Both the last two terms are line bundles on $S$ by semicontinuity theorems (and Riemann-Roch) and thus they must be isomorphic. So, we get that the first two terms are isomorphic too.

1
On

Here's another proof:

since $\Omega_{E/S}$ is trivial on the fibers, by usual arguments (base change/semicontinuity) the map $f^*f_*\Omega_{E/S}\to \Omega_{E/S}$ is an isomorphism. Apply $e^*$ to this, and use $e^*\circ f^*=(f\circ e)^*=id$.

EDIT: details.

Most of it is in these notes http://math.stanford.edu/~vakil/0506-216/216class4546.ps (look in particular at Proposition 2.4). The only detail is that there are some hypotheses there that you don't have, most notably integrality of the base $S$.

To fix this you can prove the statement universally: over the moduli stack $\mathcal{M}_{1,1}$ there is a universal family of elliptic curves $c\colon \mathcal{C}\to \mathcal{M}_{1,1}$, and your $E\to S$ is pulled back from $c$ via the classifying map $S\to \mathcal{M}_{1,1}$. Now $\mathcal{M}_{1,1}$ is integral, so you can apply the above (pass to an atlas and use descent for coherent sheaves, if you don't like doing stuff on a stack), and you get the conclusion for your family $E\to S$.