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May 12, 2013 at 12:31 vote accept Nathaniel Bottman
Mar 8, 2013 at 21:58 history edited Sam Lewallen CC BY-SA 3.0
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Mar 8, 2013 at 21:44 comment added Nathaniel Bottman Thanks Sam! This is so cool. Can someone with enough rep change $\mathbb{P}^{2g+2}$ to $\mathbb{P}^{2g+1}$? Here's how you get the two quadrics in $\mathbb{P}^{2g+1}$: let $\Sigma_g$ double-cover $\mathbb{P}^1$ with branch points $\lambda_0, \ldots, \lambda_{2g+1}$. Then the two quadrics are cut out by $X_0^2 + \cdots + X_{2g+1}^2$ and $\lambda_0X_0^2 + \cdots + \lambda_{2g+1}X_{2g+1}^2$.
Mar 8, 2013 at 21:38 vote accept Nathaniel Bottman
May 12, 2013 at 12:31
Mar 8, 2013 at 21:28 history answered Sam Lewallen CC BY-SA 3.0