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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.

A Kähler manifold is a manifold with $3$ compatible structures;

  1. A complex structure

  2. A Riemannian structure

  3. A symplectic structure

A Kähler manifold has a Kähler potential (a smooth, plurisubharmonic function that coincides with a Kähler form) and the Levi-Civita connection.

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Intuition for Kähler manifolds?

Define a Kähler manifold to be a complex manifold whose associated (1,1) form is closed. One can show this condition leads to many interesting properties. For example, the Hodge and Lefschetz decompositions force symmetries of the Hodge numbers.…
Potato
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Compatibility of a Kähler form

On a complex manifold $M$ a Kähler form is a symplectic form $\omega$ which is compatible with the canonical almost complex structure $J$ in the following sense $$\omega({}\cdot{},J{}\cdot{})$$ is a Riemannian metric tensor, i.e. symmetric and…
Display Name
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Second derivative of Kähler potential.

Does the following second covariant (in terms of Kähler geometry) derivative of Kähler potential vanish? \begin{equation} K_{ij}\equiv\nabla_i\nabla_j K=0, \end{equation} \begin{equation} K_{i^*j^*}\equiv\nabla_{i^*}\nabla_{j^*}…
Kosm
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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: $$x^5+y^4z+3z^5=0$$ With $(x,y,z)\in \mathbb{PC}^2$ then this is…
zooby
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Constructing Kähler sub manifolds from general sub manifolds

Given a Kähler manifold $\mathcal{N}$ and an arbitrary (i.e. non-Kähler) differentiable sub manifold $\mathcal{M}\subset\mathcal{N}$, I would like to foliate $\mathcal{M}$ with maximal Kähler sub manifolds. Intuitively, I can restrict the complex…
LFH
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