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when toggle format what by license comment
Aug 19, 2024 at 22:07 history edited YCor CC BY-SA 4.0
removed capitals from title
Aug 19, 2024 at 16:49 comment added James Tan Hi Sasha, I have edited my post. It is actually $H^0(\mathcal{O}(2C_{\infty}))$. It has the same dimension as $H^0(\mathbb{P}^2,\mathcal{O}(2))$.
Aug 19, 2024 at 16:47 history edited James Tan CC BY-SA 4.0
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Aug 19, 2024 at 16:40 comment added Sasha In fact $H^0(\mathcal{O}(-2C_\infty)) = 0$.
S Aug 19, 2024 at 16:07 review First questions
Aug 19, 2024 at 17:08
S Aug 19, 2024 at 16:07 history asked James Tan CC BY-SA 4.0