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Apr 21, 2018 at 14:17 comment added Andrey Feldman The manifold is not closed. It is an $S^2 \times S^3$ conifold, which is also an $\cal{O}(-1) \oplus \cal{O}(-1)$ bundle over $\mathbb{P}^1$, so the orbits of the $\mathbb{C}^{*}$ action are infinite helices, and just two closed 1-cycles.
Apr 21, 2018 at 13:56 comment added Will Sawin Unless $b^2$ is rational, the set $(e^{ ibn}, e^{-in/b})$ is dense in $S^1 \times S^1$. So if your manifold is closed in the coordinates you're using it should extend to a $S^1 \times S^1$ action.
Apr 21, 2018 at 13:44 history edited Andrey Feldman CC BY-SA 3.0
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Apr 21, 2018 at 13:52
Apr 21, 2018 at 13:16 history asked Andrey Feldman CC BY-SA 3.0