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Timeline for Configuration space of flags

Current License: CC BY-SA 3.0

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Oct 1, 2011 at 8:46 comment added Xin Nie Ahh, you are right, I made a mistake on crossratios.
Sep 30, 2011 at 12:53 comment added Michael Joyce $PGL(2)$ acts simply transitively on triples of distinct points, so any element of $(\mathbb{P}^1)^4$ can be brought to a unique element of the form $(0, 1, \infty, z)$, where $z$ is any element of $\mathbb{P}^1$ except $0$, $1$, or $\infty$. The book An Invitation to Quantum Cohomology by Koch and Vainsencher has a very good discussion of this moduli space.
Sep 30, 2011 at 7:15 comment added Xin Nie Thanks, that paper is enlightening, although it treats a different question from mine. However, I can't understand your first paragraphe... To me, the quotient is just $\mathbb{P}^1$, no need to remove three points. And can't image how can stalbe curves be involved
Sep 29, 2011 at 21:49 history answered Michael Joyce CC BY-SA 3.0