16
votes
Is this entire function a square?
$f$ is an entire function of exponential type (that is, $|f(z)|\lesssim e^{C|z|}$), and this implies that there are other zeros, in addition to $z=0$: if not, then its Hadmard factorization would be ...
10
votes
Accepted
Kahler version of Darboux's Theorem
Let $(M,g,J)$ be a Kahler manifold. This means that $J$ is a (integrable) complex structure, so that $(M,J)$ is a complex manifold, $g$ is a smooth metric such that $J$ is $g$-orthogonal, i.e.
$g(JX,...
- 5,011
8
votes
Accepted
When does a holomorphic symplectic manifold compactify to a Poisson manifold?
No, not even under the nicest possible algebraicity assumptions like quasiprojective.
Let $Z$ be the product of two curves of genus $\geq 2$. Choose a nonzero $2$-form $\omega$ on $Z$, the wedge of a ...
- 137k
5
votes
Finding the octonionic analog of the K3 surface, via (almost) hyperkahler geometry?
The following shows that such a manifold exists, but unfortunately it uses $bP_8 \cong \mathbb Z_{28}$.
According to Corollary 5.5 of this article a highly connected $8$-manifold $M$ with $p_1(M)=0$ ...
- 2,820
5
votes
Accepted
The singularties of the dicriminant loci of the Lagrangian fibration
I am afraid not. As it is usual for discriminants, the singularities are rather complicated. Take for instance a K3 surface $S\subset \mathbb{P}^g$, with Picard group generated by $\mathcal{O}_S(1)$. ...
- 37.3k
5
votes
Accepted
Equivariant quantum cohomology of conical symplectic resolutions
First of all, I don't understand why you say that there are no holomorphic curves - a typical example of such a space is $T^*{\mathbb C}{\mathbb P}^1$ and it perfectly has non-trivial holomoprhic ...
- 6,957
4
votes
Accepted
Lagrangian cores of quiver variety in different GIT chambers
Any quiver variety where all the v_i are 1's is a hypertoric variety. They are determined combinatorially by an arrangement of affine hyperplanes and one can compute the core by looking at the compact ...
- 1,544
4
votes
Learning Quantum (Co)Homology and Landau Ginzburg Superpotential
You can find some very accessible discussion on quantum cohomology of toric (Fano) manifolds in On the quantum homology algebra of toric Fano manifolds by Ostrover & Tyomkin (and references ...
- 431
4
votes
Accepted
Bubbling off of a pseudo holomorphic sphere on surface with cylindrical ends
It turns out that the Removable Singularities Theorem and the Monotonicity Lemma do not require compactness, but that the target manifold should have bounded geometry, such as in our case of inserting ...
- 17.2k
4
votes
Accepted
Deligne Mumford Compactification of Moduli Space Of Annuli
The study of the Deligne-Mumford compactification of Riemann surfaces with boundary can be reduced to the study of the Deligne-Mumford compactification of closed Riemann surfaces together with the ...
- 722
4
votes
Irreducibility of holomorphic symplectic quotients
Work of Gan and Ginzburg (https://arxiv.org/pdf/math/0409262v7.pdf) shows that one of our favorite examples, $M=T^*(\mathfrak{gl}_n\oplus \mathbb C^n)$ and $G=GL(n)$, has a 0 level of the moment map ...
- 43.9k
3
votes
Why symplectic geometry gives Poisson geometry
Look up sub-Riemannian geometry. This is a quite developed theory now, but here the Riemannian metric is defined only along a subbundle of the tangent bundle which is assumed to be completely non-...
- 25k
3
votes
Accepted
Two Lagrangian submanifolds with clean intersections
There is a spectral sequence from the cohomology of the intersection to the Floer cohomology. This is explained (and used) in Seidel's paper "Lagrangian 2-spheres can be symplectically knotted"
https:...
- 6,935
3
votes
Symplectic resolutions amongst cotangent bundles
I think @Joel is saying the following (this was too long for a comment, so I put it as an answer):
Given a conical symplectic resolution $X\mathrel{:=}T^*M \rightarrow X^\text{aff}$ whose $\mathbb{C}^*...
- 1,627
3
votes
Is this entire function a square?
If $f$ has a double zero at a point $z$ then not only $z=\sin z$ there
but also the derivatives must be the same, so that $1=\cos z$ there.
Then $\sin^2z = 1-\cos^2 z =0$ and therefore $z=\sin z=0$, ...
- 15.1k
2
votes
Core components of quiver varieties as fiber bundles of flag varieties
There is an example of an irreducible component, which is a blowup of $\mathbb P^2$ at a point. See Example 18 in https://arxiv.org/pdf/1611.10000.pdf.
- 2,831
2
votes
Local deformations of Lagrangian submanifolds in holomorphic symplectic manifold and their intersections
No for the first question. A counter-example is given by the Fano variety $X$ of lines in a cubic fourfold $V\subset \mathbb{P}^5$: for each hyperplane $H$ of $\mathbb{P}^5$, the lines contained in $H\...
- 37.3k
2
votes
What is the relation between holomorphic blow-up and symplectic blow-up?
In the case when you are blowing up a smooth submanifold this actually coincides with the symplectic blow-up (with the symplectic form forgotten). For simplicity, let's take the submanifold to be a ...
- 3,157
2
votes
Symplectic resolutions amongst cotangent bundles
In A characterization of nilpotent orbit closures among symplectic singularities, Namikawa proved that a weight 1 conical symplectic singularity must be a nilpotent orbit closure.
So I guess that this ...
- 4,467
1
vote
Accepted
Visualising the moduli space of stable disks with interior marked point and 4 marked point on the boundary
A generalization of this moduli space was used by Costello, in the paper https://arxiv.org/abs/math/0601130. In his terminology, your space is $D_{0,1,4,1} = \overline{\mathcal N}_{0,1,4,1}$.
The ...
- 3,673
1
vote
Rank 3 Lagrangian vector bundles on an elliptic curve
I think your guess is correct and one can proceed as follows (some details are missing though).
Let $V$ be a six dimensional symplectic vector space and $F$ be a rank three-vector bundle on $E$ wiht ...
- 7,200
1
vote
Locally affine varieties and du Val singularities
A chapter of the unpublished PhD thesis of Rebecca Leng, a student of Miles Reid from about 2002, presents a careful study of a natural affine cover of the minimal resolution $Y={\rm GHilb}({\mathbb A^...
- 3,074
1
vote
Disconnecting the Lagrangian Grassmannian
It looks as though $M_{K} := \Lambda(V) \setminus S_{K}$ is connected. Here is the way I can see it: First of all, we know that when $K$ is Lagrangian $M_{K}$ can be identified with $Sym^{2}(L^{\ast})$...
- 511
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