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Sep 10 at 7:31 comment added Arkadij Thank you! Perhaps let me comment that this argument works very strict Calabi-Yau n-fold. Also, you don't need to use the adjunction at the end. The statement follows from unwrapping your $H^1(X, \mathcal{O}_X) \oplus H^1(X, \mathcal{O}_X) = H^1(X \times X, \mathcal{O}_{X \times X}) \stackrel{s}\to H^1(X \times X, \Delta_*\mathcal{O}_X) = H^1(X, \mathcal{O}_X)$.
Sep 10 at 7:26 vote accept Arkadij
Sep 10 at 5:26 history answered Sasha CC BY-SA 4.0