Timeline for Degeneration of Kaehler-Einstein metric of negative Ricci curvature
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| Dec 14, 2017 at 20:31 | comment | added | user21574 | .....So after base change the central fiber can admit KE metric with negative ricci curvature outside of Zariski open subset, but I am not sure we can extend such metric on the whole of central fiber after base change? But at least we can say that we can have some sort of twisted KE metric | |
| Dec 14, 2017 at 20:27 | comment | added | user21574 | End of comment:Let me more explain my previous comment: Let $π^0∶ (X^0, B^0) → S^0$ be a family of irreducible slc pairs$ (X,B)_t$ with ample $K_{X_t }+ B_t$. Then, possibly after a finite base change $S′ → S$, this family can be uniquely extended to a complete family over $S$ such that the central fiber $(X_0, B_0)$ is slc with ample $K_{X_0} + B_0.$ see Theorem 1.3.6 springer.com/us/book/9783034809146 and combine with the result of link.springer.com/article/10.1007/s00039-014-0301-8 . .... | |
| Dec 14, 2017 at 7:57 | comment | added | user21574 | Let me mention that the central fiber which type of canonical metric can admit. If semi-stable degenerations become smooth and we include the logarithmic differential form and if we take Kahler forms $ω_t$ on fibers $X_t$, then on central fiber $X_0$, we can think $ω_0$ is logarithmic differential form $Ω^2(\logX_0)$. Read pages 6 and 7. arxiv.org/pdf/1212.0553.pdf, So if the general fibers $\omega_t$ admits Kahler-Einstein metric, then by allowing Mumford's semi-stable degeneration the central fiber $X_0$ admits twisted Kahler-Einstein metric $Ric(\omega_0)=\lambda\omega_0+[PD(E)]$ | |
| Nov 7, 2017 at 6:10 | comment | added | user21574 | I have a conjecture that for $π:X→Δ$ if the general fibers $X_s$ admits Kahler-Einstein metric and central fiber $X_0$ be Kahler manifold, then central fiber also admits Kahler-Einstein metric (this can be considered as new view of theorem of Kodaira) | |
| Oct 24, 2017 at 13:07 | comment | added | user21574 | See Corollary 30. of the paper of Berman arxiv.org/pdf/1002.3717.pdf | |
| Oct 24, 2017 at 13:01 | comment | added | user21574 | .....continued: now by taking derivative on both sides and comparing their Kahler Ricci flow equations, we get $\frac{\partial f(s)}{\partial s}=-f(s)+I_{\omega}^{\omega'}$, so if $s\to 0$, then on central fibre we must study the behaviour of the solution of this equation. Here $I_{\omega}^{\omega'}=log\frac{\omega^n}{\omega'^n}$. In fact conditions such that $f(s)$ remain constant when $s\to 0$. This may help | |
| Oct 24, 2017 at 13:01 | comment | added | user21574 | @Dima Now assume each $X_s$ admit Kahler-Einstein metric with negative Ricci curvature then we have a variation of Kahler-Ricci flow $\frac{\partial \omega(s,t)}{\partial t}=-Ric(\omega(s,t))-\omega(s,t)$ on each fibre $X_s=\pi^{-1}(s)$, so in this case by theorem of Yau-Aubin, we have unique Kahler-Einstein metric up to scaling , but the word of scaling, in this case, is highly non-trivial, so we get $\omega=\omega'+f(s)$, on each fibre which $f(s)$ is fiberwise constant, which may not be constant in general | |
| Oct 15, 2017 at 15:23 | comment | added | user21574 | About my previous comment see the paper of S. Ivashkovich :Theorem3, Theorem4, gdz.sub.uni-goettingen.de/en/dms/loader/img/… | |
| Oct 15, 2017 at 14:45 | comment | added | user21574 | Another sad part of story is that $X_0$ admit Kähler-Einstein metric outside of an analytic subvariety $S$. i.e $X\setminus S$, hence we must extend it as current to whole of $X_0$ and this exactly means that we extend the $K_{X_0\setminus S}$ to whole of $X_0$ . But we may have luck of extension theorem. But if the curvature be $L^2$ bounded then we get the desired result. So this tells us that find KE outside of analytical subvariety is not big deal, but if this KE be extended to whole of $X$, then we have done! | |
| Sep 22, 2017 at 21:04 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Aug 23, 2017 at 20:32 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jul 27, 2017 at 8:02 | comment | added | user21574 | In fact $X_0$ can admit continuous Kahler-Einstein metric outside of an analytical subvariety $S$. But the sad part of story is that such $S$ may no longer be Zariski open subset. | |
| Jul 27, 2017 at 7:48 | comment | added | user21574 | It seems my previous comment work even $X_0$ has canonical singularites. But we must impose central fiber has mild singularities to consider KE metric | |
| Jul 24, 2017 at 19:51 | answer | added | Henri | timeline score: 1 | |
| Jul 24, 2017 at 19:03 | comment | added | user21574 | Related answer: Central fibre $X_0$ can admit Kähler-Einstein metric with negative Ricci curvature outside of a analytical subvariety if general fibers are projective. Let state Dan Popovici theorem: Let $π:X→Δ$ be a complex analytic family of compact complex manifolds such that the fibre $X_t:=π^{−1}(t)$ is projective for all $t≠0$. Then $X_0:=π^{−1}(0)$ is Moishezon. We know , Let $X$ be any compact complex variety. Then $X$ is a Moishezon space if and only if there is a proper analytic subset $S⊂X$, such that $X∖S$ admits a complete Kähler-Einstein metric with negative Ricci curvature. | |
| Jul 24, 2017 at 18:34 | answer | added | Jason Starr | timeline score: 2 | |
| Jul 24, 2017 at 18:18 | history | edited | Ben McKay | CC BY-SA 3.0 |
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| Jul 24, 2017 at 17:52 | comment | added | Jason Starr | I do not know the answer to this question, but I do have a family in mind that might be relevant. For a Lefschetz pencil of hypersurfaces $X\to \mathbb{P}^1$ in $\mathbb{P}^3$ of degrees $d\geq 5$, the general member $X_t$ has canonical bundle isomorphic to $\mathcal{O}(d-4)|_{X_t}$. There are finitely many singular fibers over $\{t_1,\dots,t_\delta\}\subset \mathbb{P}^1$. For a ramified double cover of $\mathbb{P}^1$ branched over the points $t_i$, there are small resolutions of the singular fibers. So those fibers are smooth, but the canonical divisor class is not ample on those fibers. | |
| Jul 24, 2017 at 17:12 | review | First posts | |||
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| Jul 24, 2017 at 17:12 | history | asked | Dima | CC BY-SA 3.0 |