# Questions tagged [hyperkaehler]

The hyperkaehler tag has no usage guidance.

35
questions

2
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### Describing singular fibers of the lagrangian fibration $\mathcal M^s(0, [C], 1) \to |C|$

Let $S \to \mathbb P^2$ be a two-to-one cover branched over a sextic, i.e. $S$ is a K3-surface. Let $C \subset S$ be the preimage of a (smooth) quadric, so that by Hurwitz' formula, $g(C) = 5$. ...

1
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0
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75
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### Is Stenzel's Ricci-flat metric on $T^*\mathbb{CP}^n$ hyperkahler?

In a well-known paper, Stenzel constructed complete Ricci-flat Kahler metrics on the total spaces of cotangent bundles of $S^n$, $\mathbb{RP}^n,$ $\mathbb{CP}^n$, $\mathbb{HP}^n$, and $\mathbb{OP}^2$.
...

7
votes

1
answer

450
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### 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 ...

5
votes

1
answer

204
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### Can deformation equivalent Kähler manifolds always be obtained by a deformation where all the fibers are Kähler?

Given compact Kähler manifolds $X$ and $X'$ deformation equivalent over the unit disk $\Delta \subset \mathbb{C}$. More precisely, there is a proper holomorphic surjective map
\begin{align*}
\pi\...

1
vote

0
answers

55
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### 3-Sasakian manifolds and contact Fano Kähler-Einstein manifolds

Let $(M,g)$ be a Riemannian manifold. The Riemannian cone
of $M$ is $C(M) = M \times {\Bbb R}^{>0}$ with the metric $t^2 g + dt\otimes dt$.
A manifold is called Sasakian if its cone is Kähler, ...

1
vote

0
answers

57
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### Some question on defining the space of graph homology for Jacobi diagrams

I read Chapter 2.1 "The space of graph homology" from Nieper-Wißkirchen's Chern Numbers and Rozansky-Witten Invariants of Compact Hyper-Kähler Manifolds today, and have some question. I'm ...

4
votes

0
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159
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### Can Lagrangian fibrations have multiple fibres in codimension $1$?

I know that if $\pi: S \to \mathbb P^1$ is an elliptic fibration of a K3-surface $S$, then $\pi$ does not have multiple fibers. A proof of this can be found in Huybrechts' Lectures on K3 surfaces, ...

3
votes

1
answer

227
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### Does miracle flatness always fail for a non-regular base?

In [1], Huybrechts and Mauri argue that a holomorphic Lagrangian fibration $f: X \to B$ with smooth base $B$ is flat. This is an application of so called miracle flatness [2, Thm 23.1], because ...

1
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0
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67
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### Can this embedding to double dual EPW sextic happen?

Let $\widetilde{Y}_{A^{\perp}}$ denote the double dual EPW sextic defined by the Lagrange subspace $A\subset \bigwedge^3V_6$. If $A$ is very general, then $\widetilde{Y}_{A^{\perp}}$ is a smooth ...

3
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54
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### Ricci deformation of hyperkahler ALE orbifold

Let $(X^2,g)$ be a hyperkahler ALE orbifold surface. Consider its Ricci deformation equation:
$$
\Delta h+2Rm(h)=0
$$
for $\text{div}_g h=\text{Tr}_gh=0$ and $h=O(r^{-\epsilon})$ as $r \to +\infty$. ...

2
votes

0
answers

149
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### Conics on Gushel-Mukai fourfold

Let $X$ be a very general Gushel-Mukai fourfold, let $\mathcal{U}$ be the tautological sub-bundle and $\mathcal{Q}$ be the tautological quotient bundle. Let $C\subset X$ be a $\rho$-conic, then $\...

3
votes

0
answers

153
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### Deformation to a normal cone for a holomorphically symplectic manifold

Let $X$ be a subvariety in $M$.
"Deformation to the normal cone"
is a holomorphic deformation of a neighbourhood
of $X$ in $M$ over the disk such that its central fiber
is the total space ...

9
votes

2
answers

681
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### Bialynicki-Birula decomposition for real analytic varieties

Let $X$ be a smooth complex algebraic variety endowed with a $\mathbb{C}^*$ action. We assume also to have an antiholomorphic involution $\sigma$ over $X$ such that it anticommutes with the action ...

3
votes

1
answer

151
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### Stuck on a computation with quaternions and moment maps

I am trying to understand an article by Gibbons, Rychenkova and Goto, called "Hyperkähler quotient construction of BPS Monopole Moduli Spaces". I will paraphrase the relevant notions and ...

6
votes

1
answer

227
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### The state of art of the singular Levi problem -- and hyperkähler quotients

One of the versions of the classical Levi problem asks the following:
Let $X$ be a complex manifold. Is it true that $X$ is Stein iff
$X$ admits a smooth exhaustion strictly plurisubharmonic ...

7
votes

0
answers

257
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### 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 ...

2
votes

0
answers

89
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### Monodromy operators on hyperkähler varieties

Let $X$ be a hyperkähler variety.
In an article (Conjecture 2.1) from some years ago, Markman conjectured that any monodromy operator acting trivially on $H^2(X,\mathbb Z)$ is the identity operator, ...

3
votes

0
answers

97
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### Contact 3-manifolds with hyperkahler Stein fillings?

Is there any classification result on (homeomorphism type) of contact 3-manifolds $\Sigma$ that have Stein filling $W$ that is
1. Hyperkahler (s.t. Stein structure is the Kahler part of it)
2. not ...

3
votes

0
answers

76
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### Hypertoric varieties in dimension 4?

Are the only smooth hypertoric varieties in real dimension 4 obtained
as minimal resolutions of type A simple singularities $\mathbb{C}^2/\mathbb{Z}_{/n}$?

5
votes

0
answers

74
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### Hypothetical uniqueness of an embedding of a Riemannian manifold to a compact Kähler one

Inspired by this question (Isometric embedding of a real-analytic Riemannian manifold in a compact Kähler manifold) I ask the following:
Suppose $X$ is a real analytic Riemannian manifold with a ...

4
votes

0
answers

115
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### Coordinate-free B.Feix's construction of a hyperkähler metric

In the 2001's paper 'Hyperkähler metrics on cotangent bundles' B.Feix gives a construction of a hyperkähler metric on a neighbourhood of zero section in $T^*X$ where $X$ is a real analytic Kähler ...

10
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0
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353
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### Hyperkähler structure on the moduli space of tetrahedra?

Consider a moduli space of geodesic tetrahedra in the hyperbolic space $\mathbb{H}^3.$ In the Klein's model the hyperbolic space can be presented as the interior of a unit ball
$$
\mathbb{H}^3=\{(x_1,...

4
votes

1
answer

184
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### Lagrangian cores of quiver variety in different GIT chambers

Let $\pi: \mathfrak{M}_{(\theta,0)}(Q,\text{v},\text{w}) \rightarrow \mathfrak{M}_{(0,0)}(Q,\text{v},\text{w})$ be the projective morphism between Nakajima quiver varieties when complex moment ...

1
vote

1
answer

183
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### Small contraction for Hyperkähler Varieties

I have the following basic question. Everything is over $\mathbb{C}$.
Let $X$ be a hyperkähler (irreducible holomorphic symplectic) variety and we consider a small contraction $f\colon X \rightarrow ...

1
vote

1
answer

254
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### compact manifold as a hyperkahler quotient of an infinite-dimensional affine space

Is it possible to obtain K3 (or any other compact
hyperkahler manifold) with its hyperkahler structure
as a hyperkahler quotient of an infinite-dimensional affine
quaternionic vector space with an ...

6
votes

1
answer

661
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### Are there non-projective, but algebraic, hyperkahler varieties?

Let $k$ be an algebraically closed field of characteristic zero. I am not sure what the right definition of a hyperkahler variety over $k$ is, but I think the following might be close enough.
...

0
votes

0
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97
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### Extending Beauville-Bogomolov orthogonal decomposition from variety to scheme

I'm seeking to understand the de Rham cohomology of a Hilbert scheme $K3^{[4]}$ of the K3 surface. By Beauville, this 8-dimensional compact manifold is Kaehler, irreducible, holomorphically symplectic ...

1
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0
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100
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### Integral cohomology on compact irreducible holomorphic symplectic manifolds

In particular, I'm attempting to understand the integral cohomology on K3^[4], a Hilbert scheme on the projective K3 surface, and whether or not this compact Kaehler manifold admits a Kaehler form ...

4
votes

0
answers

123
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### Different components of real sections for twistor spaces of hyper-Kähler manifolds

A hyper-Kähler manifold is a Riemannian manifold $(M,g)$ equipped with 3 complex structures $I,J,K$ obeying the quaternionic relations and such that $g$ is a Kähler metric for each complex structure. ...

2
votes

0
answers

100
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### How exactly are holomorphic maps with hyperkahler targets identified as triholomorphic maps?

In Appendix A of Hyperinstantons, the Beltrami Equation, and Triholomorphic Maps by Fre et al., it is described how holomorphic maps from a Riemann surface to a hyperkahler manifold $\mathcal{N}$, ...

3
votes

1
answer

163
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### Are there hyperkahler manifolds with multi-isotropic hyperkahler submanifolds?

This is a continuation of this question. As explained in the comments, for a hyperkaehler manifold $X$, a multi-isotropic submanifold $S$ has the property that the three distinguished Kahler forms $\...

3
votes

1
answer

220
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### Submanifold of a Hyperkahler manifold which is 'Lagrangian' w.r.t. all three symplectic structures

Let $X$ be a hyperkaehler manifold. Being hyperkaehler, it has three distinguished Kaehler forms, $\omega^u$ ($u$=1,2,3), corresponding to its three distinguished almost complex structures.
How does ...

1
vote

1
answer

196
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### Do these definitions of integrable quaternionic structure agree?

I have found two different definitions for integrable quaternionic structure in the literature, and I need to know if they agree with one another.
One definition that I have found (from Differential ...

6
votes

0
answers

775
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### Isometries of hyper-Kähler manifolds

For the purposes of this question, a hyper-Kähler manifold will be a complete connected Riemannian manifold $(\mathcal{M},g)$ whose holonomy representation is isomorphic to the natural representation ...

8
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2
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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 ...