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Jan
25
answered Example of a non-Kähler manifold with varying plurigenera
Nov
21
revised Example of a compact Kähler manifold with non-finitely generated canonical ring?
added 6 characters in body
Oct
24
comment What is the obstruction to the existence of a global Kahler potential?
@Hassan Jolany 海桑乔朗丽 Never trust blindly any paper, unless you have checked the details yourself!
Oct
24
comment What is the obstruction to the existence of a global Kahler potential?
The statement of the result of Gauduchon is wrong, it should say $b^1=2h^{0,1}$.
Oct
24
comment What is the obstruction to the existence of a global Kahler potential?
The "standard textbook" proof shows that if $[\omega]=0$ in $H^2(X,\mathbb{R})$ and if $H^{0,1}_{\bar\partial}(X)$ vanishes, then you can write $\omega=i\partial\bar\partial K$. You can easily reconstruct the proof by yourself.
Oct
22
comment Unique Kahler-Einstein metric $g$ with $\mathrm{Ricc}(g)=-g$ when first Chern class $C_1(M)<0$: $\mathrm{Ricc}(h)=-g\,\Rightarrow\,h=cg$ for $c>0$?
If the dimension of $H^{1,1}$ is strictly larger than $1$, then as Henri explained there are other Kahler metrics (not of the form $cg$ with $c$ a constant) with the same Ricci curvature as $g$.
Oct
21
comment Are holomorphic quasi-positive line bundles on a Kähler manifold positive?
That's right, in the case that we described (the blowup of a point), if a curve is homologous to another curve which is contained in the exceptional divisor E then its intersection number with L must be zero, while if a curve passes through a point outside E then its intersection number with L is strictly positive (and the same holds for submanifolds instead of curves).
Oct
18
comment Are holomorphic quasi-positive line bundles on a Kähler manifold positive?
@Vesselin Dimitrov: sorry, for some reason I did not notice that you already gave that example in your comment, and I repeated it in my answer...
Oct
17
comment Are holomorphic quasi-positive line bundles on a Kähler manifold positive?
@Vesselin Dimitrov: the statement that $L$ is nef if and only if it admits a semipositive metric is false, see example 1.7 here www-fourier.ujf-grenoble.fr/~demailly/manuscripts/dps1.pdf
Oct
17
answered Are holomorphic quasi-positive line bundles on a Kähler manifold positive?
Oct
11
comment ricci flow on surfaces
then the answer to your question is obvious. For all $t$ large, $4C\exp(rt)$ is smaller than $-r/2$.
Oct
11
comment ricci flow on surfaces
Is $r$ a constant?
Sep
2
comment Vafa's semi-Ricci flat metric
Theorem 1.1 (iii)
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
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
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).