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Mar 10, 2022 at 15:54 history edited Will Sawin CC BY-SA 4.0
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Mar 10, 2022 at 15:53 comment added Will Sawin @gigi There's a number of ways to see it. One is Riemann-Roch, working fiber by fiber (so assume the base $S$ is a point). We know $H^1 ( \mathbb P, j_* \mathcal O_C) = H^1 ( C, \mathcal O_C) $ is a two-dimensional vector space and $H^1(\mathbb P, \mathcal O_{\mathbb P})=0$ so it all has to come from $\mathcal L$, which forces $\deg \mathcal L=-3$.
Mar 10, 2022 at 15:37 comment added gigi Thank you very much! Now it's very clear. Do you know why the type of splitting is exactly the trivial line bundle sum with a line bundle of degree -3 on the fibers?
Mar 10, 2022 at 15:06 history answered Will Sawin CC BY-SA 4.0