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visits | member for | 4 years, 6 months |

seen | 19 hours ago | |

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Aug
23 |
comment |
Volume form on pair (X,D)
The part about wedge products of currents is exactly what I mentioned above, Demailly's results. Also, in BBEGZ's definition 3.3 that you quote there is no wedge product of singular currents. |

Aug
19 |
comment |
Removable base loci for non-projective varieties
Short answer: Yes. This is proved in Fujita's original paper. More generally he proves this result with $X$ just a compact complex analytic space, and this includes all complete algebraic varieties over $\mathbb{C}$. |

Aug
19 |
comment |
Volume form on pair (X,D)
The "non-pluripolar" product defined in the paper that you mention is not what I had in mind in my comment, and it does not have the usual "standard" properties of the complex Monge-Ampere operator, e.g. it is in general discontinuous along decreasing sequences, and furthermore its total integral could be less than the cohomological volume of the class. For the statements of Demailly (which gives a different Monge-Ampere operator from BEGZ, which satisfies the two properties that I just mentioned), see www-fourier.ujf-grenoble.fr/~demailly/manuscripts/trento.pdf, Theorem 2.5. |

Aug
19 |
comment |
Volume form on pair (X,D)
Not really, you need some assumptions in order to do that (e.g. that the locus where the current is singular has sufficiently high codimension). For example, to apply his result to get the top product $\omega^n$ you would need $\omega$ to be smooth everywhere except at finitely many points. Or, you could make some other assumption on $\omega$ (e.g. that it is quasi-isometric to some reasonable model near its singularities). But you did not specify anything at all in your question. |

Aug
18 |
comment |
Volume form on pair (X,D)
And what is the volume form of a Kahler current? In general you cannot multiply currents by other currents. |

Aug
17 |
comment |
How can we define constant scalar curvature Kahler or cscK on pair $(X,D)$
I don't see a reasonable way to define the scalar curvature over $D$ (say e.g. as a distribution), even in the model case of $(\mathbb{C},0)$ with the standard cone metric $|z|^{2(\beta-1)}idz\wedge d\overline{z}$. |

Aug
12 |
revised |
Bigness of a symplectic form on pair $(X,D)$
added 2 characters in body |

Aug
9 |
comment |
Vafa's semi-Ricci flat metric
This question has been solved in the affirmative by Y-J Choi in arxiv.org/abs/1508.00323 |

Jul
1 |
awarded | ag.algebraic-geometry |

Jun
29 |
comment |
Complex structure on $S^6$ gets published in Journ. Math. Phys
I am quite worried by the fact that the subbundle $H\subset TG_2$ that the author defines is not integrable. Just before (13) the author says that the Levi-Civita connection $\nabla_0$ preserves the splitting $TG_2=V\oplus H$, but this would certainly imply that $H$ is closed under Lie bracket. |

Jun
22 |
comment |
Lie derivative and taking trace
If $X$ has a zero then there is a complex-valued function $u$ such that $X=g^{i\bar{j}}\partial_{\bar{j}}u \partial_i$, and so $\int_M X(f) \omega^n=-\int_X u\Delta f\omega^n=-\int_X ug\omega^n$. |

Jun
17 |
comment |
Lie derivative and taking trace
Yes, if you allow non-local operators in your formula, such as the Green operator of $\omega$ (and then it is trivial to derive a formula for $X(f)$). If you insist on a purely local formula, then I am afraid the answer is no, because $X(f)$ involves only one derivative of $f$, while $g$ involves two derivatives. On the other hand, if all you care about is an integral relation between these quantities, then often you can do this, by using integration by parts (in particular you can do it if your vector field $X$ has a zero, in which case $X$ has a holomorphy potential). |

Jun
13 |
revised |
Vafa's semi-Ricci flat metric
added 2 characters in body |

May
15 |
awarded | Enlightened |

May
15 |
awarded | Nice Answer |

Apr
20 |
comment |
Acyclicity of the sheaf of real analytic differential forms
Cartan's paper can be downloaded here: archive.numdam.org/ARCHIVE/BSMF/BSMF_1957__85_/… |

Mar
2 |
comment |
Why can we not always take a Kähler class to be in rational cohomology?
"may very well have empty intersection" is not correct since the $0$ vector is always in this intersection |

Feb
22 |
awarded | Yearling |

Feb
17 |
revised |
An identity for Futaki-Donaldson invariant
edited tags |

Feb
16 |
comment |
Complex manifolds with trivial canonical bundle
See also the long list of references on page 1 of arxiv.org/pdf/1401.4797.pdf, which will provide you with many more examples of compact complex non-Kähler manifolds with trivial canonical bundle. |