Timeline for Do hyperKahler manifolds live in quaternionic-Kahler families?
Current License: CC BY-SA 2.5
8 events
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| Aug 2, 2017 at 13:25 | comment | added | Malkoun | Consider $\mathbb{C}^4$ with a complex symplectic form $\omega$ on it. This induces the null correlation bundle $E$ of rank 2 over $\mathbb{C}P^3$. I think that $E(1) \simeq E \otimes \mathcal{O}(1)$ might be an example of what you are looking for, since $\mathbb{C}P^3$ is the twistor space of the QK manifold $\mathbb{H}P^1 \simeq S^4$. Please check the details. It may be a trivial example, I don't know. Just a suggestion. | |
| Jul 29, 2010 at 20:33 | answer | added | Simon Salamon | timeline score: 11 | |
| Jun 1, 2010 at 7:22 | history | bounty ended | Marty | ||
| May 25, 2010 at 15:53 | comment | added | Marty | I understand that one can find a family of holomorphic symplectic compact manifolds, parameterized by a base space $P^1(C)$, by hyperKahler rotation. But I'm looking for a quaternionic-Kahler base manifold $X$, such that each point $x$ of this base manifold corresponds to such a $P^1(C)$-family of holomorphic symplectic manifolds (where the $P^1(C)$ is the fibre of the twistor cover $Z_x$). Can you explain further "you can do it in a family"? In a quaternionic-Kahler family? | |
| May 25, 2010 at 8:49 | comment | added | Victor Protsak | It is rather difficult to write an $\textit{explicit}$ hyper-Kahler metric (for reasons that I won't go into), whereas holomorphic symplectic structure is often manageable and by Calabi – Yau, in the compact case it guarantees the existence of the h-K metric. In fact, my recollection from 15 years ago is that the only general method was hyper-Kahler reduction (ADHM and generalizations). It is symmetric in the sense you indicated and you can do it in a family if you like. Is this in the direction that you wanted? | |
| May 25, 2010 at 6:55 | history | bounty started | Marty | ||
| May 12, 2010 at 1:30 | history | edited | j.c. |
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| Jan 27, 2010 at 21:11 | history | asked | Marty | CC BY-SA 2.5 |