Questions tagged [kahler-manifolds]

A complex manifold with a Hermitian metric is called a Kähler manifold if the (1,1) form that gives its Hermitian metric is a closed differential form.

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Where the 'Kahler' condition is used in the Kodaira Embedding theorem?

In the Griffiths and Harris, Principles of Algebraic Geometry, p.181, he states Kodaira Embedding theorem as follows : Kodaira embedding theorem (Ver.1). Let $M$ be a compact complex manifold and $L \...
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If $D$ is the Riemannian covariant differentiation associated to $g$ and $J$ is an almost complex structure, then $DJ=0 \implies g$ is a Kähler metric

If $D$ is the Riemannian covariant differentiation associated to $g$ and $J$ is an almost complex structure, then $DJ=0 \implies g$ is a Kähler metric. Since $Dg = 0$ and $\omega(X,Y)=g(JX,Y)$ the ...
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Kähler structure on a surface

Fix $n\ge 1$, we work over $\Bbb C$. Let $U_k:={(u_k,v_k)\in \Bbb C^2: |u_kv_k|<1}$ for $k\in \Bbb Z$. Glue $U_k$ and $U_{k+1}$ by identifying $(u_{k+1},v_{k+1})$ with $((v_k)^{-1},u_k(v_k)^2)$. ...
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Finding a Kähler manifold from a given exponential family. Intuition?

I'm wondering how to go from: $$ \mathrm{Exponential~ families} \implies \mathrm{Kähler~manifolds} $$ I read that the tangent bundle of an exponential family naturally forms a Kähler manifold. I also ...
John Zimmerman's user avatar
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Regarding proof that Quaternion-Kähler manifold is Einstein

I am interested in this statement ( taken from Arthur L. Besse - Einstein Manifolds ): 14.39 Theorem(M. Berger [ Ber 7]). A quaternion-Kähler manifold (M,g) is Einstein (in dimension 4n $\ge$ 8 ). ...
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On Marsden's 'Introduction to Mechanics and Symmetry' Exercise 5.3-4. (fubini study form is closed)

In exercise 5.3-4. in Marsden's book I'm asked to prove that $\mathbf d \Omega^{fs} = 0$ on $\mathbb P \mathcal H$ directly, where $\mathbb P \mathcal H$ is an arbitrary projective Hilbert space (...
Alfons Winkel's user avatar
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Why Kähler 4-manifold with two compatible complex structure are Hyperkähler

Let $(M,g)$ be a 4-dimensional Riemannian manifold which is Kähler with respect to two independent complex structures $J_1, J_2$. Why if these induce the same complex orientations then $(M,g)$ is ...
Federico's user avatar
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Example of reflexive coherent sheaves that are not locally free

The question is in the title, do you have simple examples of reflexive sheaves on a complex manifold $X$ (it can be Kähler if you prefer, or even smooth algebraic) that are not locally free (i.e. ...
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bochner-kodaira formula on kahler manifold

Let $(M,\omega)$ be a compact Kahler manifold. Given a vector field $V = V^i \partial_{z_i} + V^{\overline{i}} \partial_{\overline{z_i}}$ on $M$, how do I show the Bochner-Kodaira formula: $||\nabla V|...
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Classifications of Nearly Kaehler manifolds with constant holomorfic curvature

I read that six dimmensional sphere is only nearly Kaehler, not Kaehler manifolds which has constant holomorfic sectional curvature. Can someone help me to find this paper which has a proof of this - ...
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notation question Hessian on compact Kahler manifold

On a compact Kahler manifold, what does the notation $\nabla \bar{\nabla} u$ for a smooth function $u$ stands for?
CuriousAlpaca's user avatar
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Kaehler manifolds with constant sectional curvature

I am reading "Structure on Manifolds" by Yano and Kon, and I don't understand the last step in the proof of this proposition that Kähler manifolds with constant sectional curvature is flat - ...
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Show that one can write $\omega = \frac{i}{2\pi}\partial \bar{\partial} \varphi$ for some positive function $\varphi$ and determine $\varphi$.

Let $\omega = \frac{i}{2\pi} \sum dz_i \wedge d\bar{z_i}$ be the standard fundamental form on $\mathbb{C}^n$. Show that one can write $$\omega = \frac{i}{2\pi}\partial \bar{\partial} \varphi$$ for ...
Iman's user avatar
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Structure on a complex manifold and the Kähler form on $\mathbb{C}^2$.

Currently learning about complex and Kähler manifolds and I'm reading this article about Kähler forms. Specifically I'm trying to understand how do they conclude that on $\mathbb{C}^2$ the Kähler form ...
Louie's user avatar
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On compact Kahler manifold, Lie derivative of Killing vector field commutes with $dd^c$

Let $(M,\omega)$ be a compact Kahler manifold, $X$ a Killing vector field, $f \in C^\infty(M,\mathbb{R}). $ I would like to show that: $L_X(dd^cf) = dd^cL_Xf$. If I write it out locally, $X = X^i \...
CuriousAlpaca's user avatar
3 votes
1 answer
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Relation between Ricci and Kähler forms on a cscK surface

I'm studying the paper https://arxiv.org/abs/dg-ga/9506002 from LeBrun and I'm stuck on the the last part of the proof of Theorem 2. The author claims that on a Kähler surface $(M,g,J)$ with constant ...
Federico's user avatar
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Computer algebra system for applying Cartan's test to systems of PDEs

It is my understanding that if one has a (possibly overdetermined) system of PDEs, one can check for compatibility by applying Cartan's test (see for example [1], Chapter 7). It involves first writing ...
Gateau au fromage's user avatar
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1 answer
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Volume form on Kähler manifold

Consider a Kähler manifold of complex dimension $n$ with Kähler metric $\omega$ and Riemann metric $g$ which we complex linearly extend to $h$. My ultimate goal is to prove that $\omega^n/n!$ is the ...
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A contraction of curvature form on Kahler manifold

In Huybrechts' Complex Geometry- an introduction, the author defined a contraction of curvature form: The contraction of the curvature $F_{\nabla} \in \mathcal{A}^{1,1}\left(\operatorname{End}\left(T^{...
eulershi's user avatar
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Equivalence of Hyper Kahler structure and three non isomorphic symplectic structures

Suppose that $(M,g, I, J,K)$ is a Hyper Kahler manifold, then the two forms: \begin{alignat*}{3} \omega_I(v,w)=&g(Iv,w),\qquad &&\omega_J(v,w)=g(Jv,w),\qquad &&\omega_K(v,w)=...
Chris's user avatar
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Cohomology of local systems and cohomology of vector bundles

Let $X$ be a compact Kähler manifold, $L$ be a local system of finite dimensional $\mathbb{C}$-vector spaces on $X$. Let $E=L\otimes_{\mathbb{C}}O_X$ be the induced holomorphic vector bundle on $X$. ...
Doug's user avatar
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Â-genus of a complex manifold

I am trying to understand the proof of the Riemann-Roch-Hirzebruch theorem using the index theorem (Heat Kernel and Dirac operators, [BGV]), and at the end they say that since $$TM \otimes \mathbb{C} =...
zarathustra's user avatar
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First cohomology group of Kähler-Einstein manifolds [closed]

I have read that the first cohomology group of a Kahler-Einstein manifold of dimension four, with positive scalar curvature, is zero. But I cannot find a proof. How can I prove it? Where can I find a ...
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The Fibrewise Multiplication on a Normal Vector Bundle of a subset of a Compact Kähler Surface

Let $\mathbb{A}$ be a (non-empty) subset of a compact kähler surface $\mathbb{X}$. Denote by $\mathcal{N}(\mathbb{A})$ the normal vector bundle of $\mathbb{A}$. What do they mean by saying "let $\...
Neil hawking's user avatar
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What can we say about the divergence of Hamiltonian vector fields?

Let $M$ be a smooth $n$-dimensioanl manifold. To set some notations $C^\infty(M)$ denote smooth functions $M \to \mathbb{R}$ $\Omega^k(M)$ denote $k$-forms on $M$ $\tau(M)$ denote vector fields on $M$...
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Diffeomorphism between complex hyperbolic space and unit ball

Let $h$ be the bilinear map: $$h:\mathbb{C}^{n+1}\times \mathbb{C}^{n+1}\to \mathbb{C},\ \ \ \ (z,w)\mapsto -z_0\overline{w_0}+\sum_{i>0} z_i\overline{w_i}$$ and let $\mathfrak{I}:h(z,z)=-1$. There ...
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3 votes
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Proving that Fubini metric is well-defined

Let $$\pi:S^{2n+1}\to S^{2n+1}/S^1=\mathbb{P}^n\mathbb{C}$$ be the Hopf fibration. I already know that $d_z\pi$ is a vector space isomorphism for every $z\in S^{2n+1}$ (when restricted to the ...
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Kähler Geometry: How to diagonalize two hermitian matrices simutaneously to get this result?

According to Yau’s proof of Calabi conjecture: If we choose another coordinate system so that $g_{i\bar{j}}=\delta_{ij}$ and $\varphi_{i\bar{j}}=\delta_{ij}\varphi_{i\bar{i}}$, then we have …… Here $...
yuanqing's user avatar
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Kähler geometry: Computation of curvature tensor in normal coordinate w.r.t $g^{i\bar{j}}$

Tian’s book Canonical metrics in Kähler geometry, page 52-53 says: Assume that we have normal coordinates at the given point, so $g_{i\bar{j}}=\delta_{ij}$ and that the first order derivatives of $g$ ...
yuanqing's user avatar
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Why only projective varieties can be Kahler?

Say we have a complex variety given by: $$x^5+y^4+3=0$$ with $(x,y)\in \mathbb{C}^2$. Then apparently this is not a Kähler surface. (Or is it?) But if we do the projective completion: $$x^5+y^4z+3z^5=...
zooby's user avatar
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Kähler geometry: coefficient 1/2 of the Kähler form and the sign of curvature

I’m studying Kähler geometry. I have several quiet basic questions: The Kähler metric is usually denoted by $g_{\mathbb{C}}=\sum _{i,j}g_{i\bar{j}}dz^i\otimes d\bar{z}^j$, Kähler form is defined to ...
yuanqing's user avatar
2 votes
0 answers
86 views

Laplacian comparison with Lefschetz decomposition

Set-up: Let $X$ be a complex manifold. Let $A^k$ be the sheaf of sections of the differential $k$-forms on a differentiable manifold, and let $A_{\mathbb{C}}^k$ be the sheaf of sections of $\Omega^k_{...
asking's user avatar
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First chern class of canonical line bundle on $CP^n$

I am trying to calculate the first chern class $c_1(K)$ of the canonical bundle $K = \Lambda^n(T^*\mathbb{CP}^n)^{1,0}$, where my definition of the first chern class is $c_1(K)=\frac{i}{2\pi}[F(A)] \...
Steven Mai's user avatar
4 votes
1 answer
81 views

Products of antiharmonic forms (or functions) with harmonic forms

Let $X$ be a compact Kähler manifold, with fixed Kähler form $\Omega$. Then, the wedge product of two harmonic forms is not necessarily harmonic, as explained for instance here. This prompts the ...
Riccardo Pengo's user avatar
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1 answer
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Motivation behind $g(X,Y)=g(JX,JY)$.

In Ballmann's book p.23, there is Let $M$ be a complex manifold with corresponding complex structure $J$. We say that a Riemann metric $g$ is compatible with $J$ if $$g(X,Y)=g(JX,JY)$$ for all vector ...
Tom's user avatar
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Representation of the C*-algebra of a Kähler manifold

This question is inspired from mathematics of quantum mechanics. In quantum mechanics, we start with a Kähler manifold $\mathcal{M}$ (which is $\mathbb{CP}^n$), with the symplectic form $\omega$ and ...
johnnydines's user avatar
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Expressing "the current $T$ gives no mass to the subset $E$" in terms of differential forms on the Complex Projective Plane $\mathbb{CP}^2$

Let $(\mathbb{CP}^2, \omega)$ be the Complex Projective Plane, where $\omega$ is a Hermitian Metric (or, the Kähler Form). Let $D^{(1,1)}(\mathbb{CP}^2)$ be the space of $(1,1)$-differential forms on ...
Neil hawking's user avatar
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8 votes
1 answer
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Is a finite covering of a $\partial\bar{\partial}$-manifold still $\partial\bar{\partial}$-manifold?

A compact complex manifold is called a $\partial\bar{\partial}$-manifold if for every pair $p,q\in \mathbb{N}$, every smooth $d$-closed $(p,q)$-form $\eta$ on $X$, $\eta$ is $d$-exact iff $\partial$-...
Doug's user avatar
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Decompose a finite (not necessarily positive) measure $\nu$ into two mutually singular measures $\lambda_1$ and $\lambda_2$

Let $\nu$ be a non-zero finite (not necessarily positive) measure on a compact Kähler surface $M$. Is it always possible to decompose the measure $\nu$ as follows $\nu=\lambda_1-\lambda_2$? Where $\...
Neil hawking's user avatar
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Aubin-Yau functional as a “distance”

Let $\omega$ be a fixed Kähler form on $M$, and $$ \mathscr{H}_{\omega}=\{v\in C^{\infty}(M)\mid \omega_v:=\omega+\sqrt{-1}\partial\overline{\partial}v>0\} $$ Now Aubin-Yau functional on $\mathscr{...
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why is pushforward of a kähler current kähler?

A Kähler current $T$ on a compact complex manifold is a positive closed $(1,1)$-current such that $T > \psi$ for a positive (not necessarily closed) $(1,1)$-form $\psi$. Let $f: X \to Y$ be a ...
Dima Sustretov's user avatar
7 votes
1 answer
71 views

is pushforward of a kähler class under a finite map kähler?

If $f: X \to Y$ is a finite map of compact complex manifolds, then we can push a cohomology class in $H^2(X)$ by considering the dual homology class in $H_{n-2}(X)$, pushing it forward, and then ...
Dima Sustretov's user avatar
1 vote
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Do deformation equivalent Kähler manifolds have the same Hodge numbers?

In this mathoverflow question, it is asserted that diffeomorphic Kähler manifolds with different Hodge numbers cannot be deformation equivalent. Why is that true? I know that if $\mathcal X \to B$ is ...
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local coordinate and fame in calculation (Example in Calabi conjecture)

I am a beginner in complex geometry. I am reading the proof of the Calabi conjecture for the first Chern class is negative. In the beginning, we will let the metric $\tilde{g}_{ij}=g_{ij}+\partial_i\...
James Chiu's user avatar
8 votes
2 answers
262 views

Non-standard complex structures on $\Bbb H\times \Bbb H$ so that multiplication is holomorphic

Let $$\mu:\Bbb H\times \Bbb H\to \Bbb H, \qquad (x,y)\mapsto x\cdot_{\Bbb H} y$$ denote the product of two quaternions. With the standard identification $$\Bbb H\cong\Bbb C^2\cong \Bbb R^4, \qquad x_0+...
s.harp's user avatar
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Regarding the Defintion of Hirzebruch Surfaces $\mathbb{F}_n$

We have $\mathbb{F}_0= \mathbb{C}\mathbb{P}^1 \times \mathbb{C}\mathbb{P}^1$ and, for $n \geqslant 1$, the $n-$th Hirzebruch Surfaces $\mathbb{F}_n$ is defined as a $\mathbb{C}\mathbb{P}^1$-bundle ...
Neil hawking's user avatar
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3 votes
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Zero Chern class and trivial line bundle

Let $X$ be a compact complex Kähler manifold and $L$ an holomorphic line bundle over $X$. Let suppose $c_{1}(L) = 0$ in $H^{1, 1}(X, \mathbb{C})$. I would like to show that if $L$ is not the trivial ...
Analyse300's user avatar
1 vote
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$G$-invariance and parallelism in homogeneous spaces

Consider a homogeneous space given by the quotient of two Lie groups $G/H=M$. Suppose that $M$ is endowed with a $G$-invariant metric $g$ ($M$ has a natural left action of $G$). Then this metric ...
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Every Hermitian scalar product in $V $ defines on $M=V /\Lambda$ a translation-invariant Kähler metric

I was reading the chapter on complex tori of Griffith Harris book "principles of Algebraic Geometry" and i read that every Hermitian scalar product on $V $ defines on the torus $M=\mathbb{C^...
Leonida Zillion's user avatar
4 votes
0 answers
101 views

Homogeneous symplectic spaces

Assume that we have a homogeneous space $M\simeq G/H$, where $G$ acts on $M$ transitively, and $H$ corresponds to the stabilizer of a generic point $x_0$, identified with the trivial coset $eH$. ...
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