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Results tagged with complex-geometry
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user 1648
Complex geometry is the study of complex manifolds, complex algebraic varieties, complex analytic spaces, and, by extension, of almost complex structures. It is a part of differential geometry, algebraic geometry and analytic geometry.
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Did Kahler say "a long list of miracles occur"?
I've been reading Moroianu's Kahler geometry notes and found a unattributed quote that says that if the Kahler properties hold, then
"a long list of miracles occur"
I am guessing that this quote belo …
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0
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0
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Are the Dolbeault Operators for a Quotient Space Equivariant?
Let $G$ be a Lie Group and $H$ a closed subgroup such that $G/H$ (the set of right cosets) is a complex manifold manifold. Now $\Omega^1(G/H)$, the space of complex one forms, is a $H$-equivariant bun …
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Atiyah--Singer for the Complex Projective Line
I'm trying to understand Atiyah--Singer by looking at the usual starting point of $CP^1$ and the Dirac--Dolbeault operator. If I've reduced everything down correctly, then in this case the theorem giv …
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Questions Suggested by the Parabolic Subgroup Definition
Take the following definition:
"A parabolic subgroup of a linear algebraic group defined over $\mathbb{C}$ is a subgroup, closed in the Zariski topology, for which the quotient space is a projective a …
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The Relationship between Complex and Algebraic Geomety
I have recently begun to study algebraic geometry, coming from a differential geometry background. It seems that there is a deep link between complex manifolds and complex varieties. For example, one …
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2
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Principal Bundles over Complex Projective Varieties
For various reasons, I'm interested in working with complex projective varieties that are also principal bundles. I began by looking at projective spaces themselves $\mathbb{CP}^n = SU(n+1)/U(n)$, the …
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2
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1
answer
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When does the Anti-Holomorphic Chain Complex Exist for Non-Kahler Manifolds?
Given an $N$-dimensional Riemannian manifold $M$, with associated Hodge $\ast$-mapping $\ast$, we have the chain complex
$$
\Omega^{0} {\buildrel {\text d}^\ast \over \longleftarrow} \Omega^{N} {\bui …
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answer
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Relation between the de Rham and Hodge Laplacians on the Exterior Algebra
For a Kahler manifold $M$ we have two well-known Laplacians: the de Rham Laplacian $\Delta_{\text{d}} = ($d$ + $d$^\ast)^2$, and the Dolbeault Laplacian $\Delta_{\overline{\partial}} = (\overline{\par …
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Equivariant Tangent Bundle Decomposition
Given a $G$-homogeneous space $M$, for $G$ a (Lie) group, we have a canonical $G$-action on the tangent bundle $T(M)$ of $M$. If $M$ is a complex manifold, then we have a decomposition of $T(M) \otime …
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5
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When are the Smooth Sections of a Bundle Generated as a Module (over Smooth Functions) by th...
For a holomorphic vector bundle $E$ over a complex manifold $M$, we denote its space of smooth sections by $\Gamma^{\infty}(E)$, and its space of holomorphic sections by $\Gamma^{hol}(E)$. Now I've be …
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Why can the Dolbeault Operators be Realised as Lie Algebra Actions
I've been looking at an example in the non-commutative geometry literature and I'm having trouble figuring out what the classical motivation is. I'll just describe the classical case here: Recall that …
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4
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1
answer
443
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Dolbeault Operators for $CP^1$ as $\mathfrak{su}(2)$ Actions.
This question is related to a previous question of mine. More specifically, it results from my attempts to understand the simplest incarnation of a phenomenon mentioned therein.
Put a grading on the …
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4
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2
answers
654
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Semi-Simple Kahler Groups?
We say that a Kahler manifold is a Kahler group if it is also a Lie group. I would like to know which semi-simple Lie groups are also Kahler groups?
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answer
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Kähler Structure for Projective Varieties over a Finite Field
(i) In 1960 Serre proved a famous analogue of the Weil conjectures for Kähler manifolds. This poses an obvious question: Does there exist an analogue of a Kähler structure for (non-singular) projectiv …
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Hyper-complex and quaternionic Kähler Geometry
What is the exact relationship hyper-complex and quaternionic Kahler manifolds? From Wikipedia we get that hyper-Kahler manifolds are both hyper-complex and quaternionic Kahler. Thus, the two families …
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