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Existence of Kähler Metric of Bounded Geometry on the Hermitian Vector Bundle on Projective Spaces

A Riemannian manifold $(M,g)$ is said to be of bounded geometry if the Riemannian curvature tensor and its derivatives are bounded, and it has positive injectivity radius. I am working with the ...
Jaewon Yoo's user avatar
2 votes
0 answers
142 views

Is curvature of the canonical line bundle always $(1,1)?$

Let $(M,g,\omega)$ be a symplectic manifold with $g$ and $\omega$ denoting the Riemannian metric and the symplectic form respectively. If $J$ is a compatible almost-complex structure, then is the ...
Partha's user avatar
  • 954
1 vote
0 answers
210 views

Ricci-flat metrics on complex tori of dimension $n \geq 3$

Let $\mathbb{T}^n = \mathbb{C}^n /\Lambda$ be a complex torus of (complex) dimension $n$. If $n=2$, it is a theorem of Berger that the Ricci-flat metrics on $\mathbb{T}^2$ are flat. This follows from ...
AmorFati's user avatar
  • 1,379
7 votes
1 answer
607 views

Kähler metric with two compatible complex structures

Let $(M^4,g)$ be a complete $4$-dimensional Riemannian manifold such that two almost complex structures $I$ and $J$ are compatible with $g$ and $\nabla_g I=\nabla_g J=0$. Can we prove that $(M,g)$ is ...
Zhiqiang's user avatar
  • 891
4 votes
0 answers
73 views

Representing homotopy classes of Kähler manifolds by harmonic maps

Let $(M,g_M)$ be a compact Kähler manifold with negative bisectional curvature. Let $\alpha : (S,g_S) \to M$ be a continuous map from a compact Riemannian manifold $(S,g_S)$. Is $\alpha$ homotopic to ...
AmorFati's user avatar
  • 1,379
0 votes
0 answers
321 views

Why are holomorphic $p$-forms parallel?

Let $X$ be a complex compact Kähler manifold, $\Omega_X$ its cotangent bundle and $D$ the Chern connection. It is, I believe, a standard fact that parallel $k$-forms $\alpha \in \mathcal{A}^k(X)$, i.e....
Nico Berger's user avatar
2 votes
0 answers
185 views

Norm of a $(1, 1)$ form on a Kähler manifold

Given a Kähler manifold $(M, g)$ what is the convention for defining the inner product on two $(1,1)$ forms $\alpha = \sqrt{-1}\alpha_{i \bar k} dz_{i} \wedge d \bar{z}_{k}$ and $\beta = \sqrt{-1}\...
Adam's user avatar
  • 153
5 votes
0 answers
131 views

Examples of compact Kähler manifolds whose Bochner curvature tensor has constant norm?

The Bochner curvature tensor is the Kähler analog of the Weyl curvature tensor in the curvature decomposition of a Kähler, discovered by Bochner in 1949. The article on Bochner-Kähler metrics by ...
AmorFati's user avatar
  • 1,379
2 votes
0 answers
203 views

Yau proof of $K_X>0$ using a non-smooth metric which restricts to a metric of negative holomorphic sectional curvature on all curves

In this lecture of Yau's on the Existence of complete Kähler-Einstein metrics with negative scalar curvature he mentions the following, I quote: Negative holomorphic sectional curvature is a rather ...
AmorFati's user avatar
  • 1,379
6 votes
0 answers
249 views

Why do we always need the Schwarz lemma when bounding the trace of a Kähler metric?

I posted this question on MSE, and while it has received some upvotes, it is not getting much attention. Perhaps it is more relevant here? My undergraduate thesis topic is Kähler geometry. The general ...
Geometer in the making's user avatar
3 votes
1 answer
210 views

Ricci curvature of the Weil-Petersson metric?

Let $\omega_{\text{WP}}$ denote the Weil-Petersson metric associated to a family of Calabi-Yau manifolds. That is, let $f : X \to Y$ be a surjective holomorphic map with connected fibres such that, ...
AmorFati's user avatar
  • 1,379
8 votes
1 answer
328 views

An integration identity on $\mathbb{P}^{n-1}$

Let $\omega_{\text{FS}}$ denote the Fubini–Study metric on $\mathbb{P}^{n-1}$ with unit volume, and let $[w_1 : \cdots : w_n]$ be standard unitary homogeneous coordinates. On page 5 of Yang–Zheng's ...
GradStudent's user avatar
21 votes
1 answer
1k views

Does every group arise as the fundamental group of a complete Kähler manifold?

The fundamental group of a manifold is countable, and every countable group $G$ arises as the fundamental group of a (smooth) manifold; see this comment or this answer for a construction of an open ...
Michael Albanese's user avatar
6 votes
1 answer
338 views

Atiyah-Singer for Riemannian and Kaehler manifolds

I am trying to understand the proof of the Atiyah--Singer index theorem, and would like to see how it works for two "simple" examples. Could somebody direct me to a proof for the special ...
Dick Johnson's user avatar
3 votes
1 answer
278 views

Curvature of varieties of log general type

Let $X$ be a projective manifold and $\Delta$ a divisor with simple normal crossings. Consider $X$ as the compactification of a quasi-projective variety $X_0$ with boundary $\Delta$, i.e. $X_0 = X \...
AmorFati's user avatar
  • 1,379
1 vote
1 answer
218 views

Fixed locus of a Kahler $S^1$-action

Given a compact Kahler manifold $M$ with an $S^1$-action by Kahler isometries, we know that Its fixed loci $F=M^{S^1}$is a smooth Kahler submanifold. It splits $F=\sqcup_{\alpha \in A} F_{\alpha}$ ...
Filip's user avatar
  • 1,677
9 votes
1 answer
333 views

Complex structures on Hermitian symmetric space

Let $(M_1,g_1,J_1)$ and $(M_2,g_2,J_2)$ be two simply-connected Hermitian symmetric spaces, which are isometric as two Riemannian manifolds. Can we find an isometry $\varphi:M_1 \to M_2$ such that $$ \...
Totoro's user avatar
  • 2,535
7 votes
0 answers
295 views

Examples of non-compact, holomorphically symplectic Kähler manifolds which are not hyperkähler

Let $(M,\omega_{1},I_{1})$ be a non-compact Kähler manifold. If $M$ admits a holomorphic symplectic form $\Omega$, is it possible M not be hyperkähler? Is there any example? (*)Under the assumption ...
Eder Moraes's user avatar
1 vote
0 answers
348 views

Sectional curvature in complex manifold

Let $(X, \omega)$ be a Hermitian manifold .Say that the sectional curvature of X is negative is the same to say that the sectional curvature of the Hermitian metric $\omega$ is negative, otherwise, ...
Samir's user avatar
  • 11
3 votes
1 answer
354 views

Holomorphic structures for line bundles over projective manifolds

Let $M$ be a compact K\"ahler manifold, which is assumed to be projective, i.e. there exists an ample line bundle over $M$ giving an embedding into $\mathbb{C}P^n$. Let $\mathcal{L}$ be a smooth line ...
Max Schattman's user avatar
1 vote
0 answers
126 views

Condition for Integrability of an Almost Complex Structure

The following question concerns a remark made in the paper: Lebrun, C., Complete Ricci-flat Kähler metrics on $\mathbb{C}^n$ need not be flat, Proceedings of Symposia in Pure Mathematics, Volume 52 ...
AmorFati's user avatar
  • 1,379
8 votes
2 answers
695 views

Kronheimer's results on ALE spaces as hyperkahler quotients

Background: In his two papers from late 80s Kronheimer proved that any 4-dimensional ALE space is given by a hyperkahler quotient, say $X_{{\zeta_\mathbb{R}},{\zeta_\mathbb{C}}}(Q)$ where Q is a ...
Filip's user avatar
  • 1,677
2 votes
0 answers
147 views

Calabi $C^3$ estimate

I have a question regarding a computation analogous to the Calabi $C^3$ estimate which is used in the proof of the Calabi--Yau theorem. Motivation: Establishing Liouville type theorems for complex ...
AmorFati's user avatar
  • 1,379
3 votes
1 answer
269 views

Hyperkähler ALE $4$-manifolds

It is well known that Kronheimer classified all hyperkähler ALE $4$-manifolds. In particular, any such manifold must be diffeomorphic to the minimal resolution of $\mathbb C^2/\Gamma$ for a finite ...
Totoro's user avatar
  • 2,535
7 votes
1 answer
458 views

Large isometry groups of Kaehler manifolds

Let $M$ be a closed simply-connected Kaehler manifold that is not isomorphic to a product of lower-dimensional Kaehler manifolds. Pick an orientation for $\mathbb{C}$; this endows $M$ with an ...
user avatar
5 votes
1 answer
291 views

Bishop-Gromov for Kähler metrics

Let $(M, g)$ be a (complete) Kähler manifold with Ricci curvature $\geq c$. Is it true that the volume ratio of geodesic balls in $M$ with respect to balls in the corresponding (simply connected) ...
Martin de Borbon's user avatar
2 votes
1 answer
269 views

Function is $L^p$-integrable for $p >1$ [Kähler Geometry]

I am reading through a proof in W. Ding and G. Tian's 1992 paper on the generalised Futaki invariant. To provide context, we are looking for obstructions to the existence of Kähler--Einstein metrics ...
AmorFati's user avatar
  • 1,379
1 vote
0 answers
497 views

(Real) holomorphic vector fields on compact Kähler manifolds

I am trying to prove Proposition 2.1.1 of Gauduchon's note on Kähler extremal metrics (page 67). In order to show that, for compact Kähler manifolds, the complex Lie algebra of real holomorphic vector ...
Cracovia's user avatar
4 votes
1 answer
215 views

Examples of surfaces with negative Kahler curvature operator

Compact ball quotients are examples of compact Kahler surfaces with negative curvature operator. Are there any other examples ? What about nonpositive (other than the product of two Riemann surfaces ...
Vamsi's user avatar
  • 3,383
7 votes
1 answer
967 views

Riemannian holonomy of generic manifolds

It is well known, as well as absolutely intuitive, that the Riemannian holonomy of a generic Riemannian manifold is $O(n)$, the Riemannian holonomy of a generic orientable Riemannian manifold is $SO(n)...
diverietti's user avatar
  • 7,902
8 votes
1 answer
256 views

Do all symmetries of a Kähler quotient come from the original space?

For a Kähler manifold $M$, let $\operatorname{Iso}_{\mathbb{C}}(M)$ denote the group of holomorphic isometries. Suppose that $K$ is a compact subgroup of $\operatorname{Iso}_{\mathbb{C}}(M)$ and ...
user116804's user avatar
4 votes
0 answers
237 views

Terminology: Almost hyper-Hermitian vs Almost hyper-Kähler vs

After non-thorough literature search, it seems to me that there is no consensus on the usage of the terminology "(almost) hyper-Hermitian" vs "almost hyper-Kähler" vs "almost quaternion-Hermitian" etc....
seub's user avatar
  • 1,347
7 votes
1 answer
701 views

Negatively curved manifolds with many totally geodesic submanifolds

I'm curious about the following question and have not been able to find any literature on the topic: Suppose that $M$ is a closed negatively curved Riemannian manifold with a "large" quantity of ...
Clark's user avatar
  • 179
15 votes
2 answers
829 views

Darboux-like theorems

Related to Kahler version of Darboux's Theorem I know a few theorems that feel like Darboux's theorem. By that, I mean some kind of geometry based around a "pointwise" condition and the existence ...
user44191's user avatar
  • 4,991
2 votes
1 answer
590 views

Ricci form is closed?

Let $(M,g,J)$ be an almost Kähler manifold and let $\rho$ denote its Ricci form $$ \rho(X,Y) = \operatorname{ric}^{\mbox{c}}(JX,Y) $$ where $\operatorname{ric}^{\mbox{c}}$ is the $J$-invariant part of ...
LafontaineMath's user avatar
19 votes
2 answers
2k views

Does a Kähler manifold always admit a complete Kähler metric?

Every smooth manifold admits a complete Riemannian metric. In fact, every Riemannian metric is conformal to a complete Riemannian metric, see this note. What about in the Kähler case? Does a Kähler ...
Michael Albanese's user avatar
1 vote
0 answers
58 views

Is Fano Kahler surface with reverse orientation also Kahler?

In particular, do Fano Kahler surfaces with reverse orientation admit Kahler-Einstein metrics?
user38600's user avatar
  • 399
2 votes
0 answers
309 views

Darboux-like coordinates on a Kähler manifold

If $(M, g, J, \Omega)$ is a Kähler manifold, do there exist local coordinates in which 2 out of the 3 geometrical structures look nice? I have Darboux coordinates in which $\Omega$ looks nice, but ...
Alex M.'s user avatar
  • 5,407
4 votes
2 answers
665 views

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 $c>0$?

On a compact Kahler manifold, let $g$ be the unique Kahler-Einstein metric with $\mathrm{Ricc}(g)=-g$, proved to exist by Yau and Aubin when the first Chern class $C_1(M)<0$. Question: Does $g$ ...
mdg's user avatar
  • 376
4 votes
2 answers
306 views

Reference for when a metric on a four-manifold is Kahler?

In a paper of Derdzinski1 (Proposition 5), he proved that if $\delta W^+=0$ and $W^+$ has at most two distinct eigenvalues, then the metric is (locally) conformally Kahler, and if in addition the ...
user38600's user avatar
  • 399
10 votes
2 answers
2k views

Infinite dimensional Riemannian geometry

My current research has brought me into an area the requires me to learn some infinite dimensional Riemannian and Kähler geometry. Can someone recommend some good books or survey articles to help me ...
Wintermute's user avatar
1 vote
1 answer
201 views

Derivative of (the length of) the Ricci tensor

I was wondering, have you ever seen a formula in the Riemannian (more specially Kahlerian but not essential) setting for the derivative $X \cdot |Ric|^2 = 2 g(\nabla_X Ric, Ric)$ for a vector field $X$...
David P's user avatar
  • 585
0 votes
1 answer
738 views

Is there any open Ricci-flat ALE 4-manifold other than Hyper-Kahler ALE 4-manifolds?

Concerning my previous question Non simply connected HyperKähler 4-manifolds without ALE metrics the following question occurred to me: Is there any open Ricci-flat ALE 4-manifold other than ...
Hamed's user avatar
  • 1,236
2 votes
0 answers
215 views

Geometric meaning of a certain form in almost-Kähler geometry

I have difficulties finding an appropriate reference for the following question: Let $(M^{2n},g,J,\omega)$ be a compact almost Kähler manifold. Let $\operatorname{ric}$ the usual Ricci tensor of $(M^{...
Susana Valdez's user avatar
1 vote
2 answers
674 views

Non simply connected HyperKähler 4-manifolds without ALE metrics

In a 1989 paper Peter Kronheimer showed that each simply connected HyperKähler 4-manifold possesses an ALE metric. What do we know about the non-simply connected cases?
Hamed's user avatar
  • 1,236
11 votes
1 answer
540 views

Minimum requirements for a Kähler manifold to be hyperkähler

In 'panoramic view of Riemmannian geometry' when introducing hyperkähler manifolds, Berger states, informally, that a hyperkähler manifold is a Riemmannian manifold which is Kähler for more than one ...
Thomas Richard's user avatar
2 votes
1 answer
1k views

recognizing Kahler manifolds of complex dimension n

Is there new classification of Kahler manifolds of complex dimension n and new results for necessary and sufficient conditions for a manifold being Kahler? I know if redactivity of Lie algebra on ...
user avatar