13
votes
Kähler metric with two compatible complex structures
No, you cannot prove this because it is not true. For example, consider $M$ to be the product of two oriented, complete Riemannian surfaces $M=\Sigma_1\times\Sigma_2$ where $g$ is the product metric ...
- 103k
8
votes
Accepted
Do these definitions of integrable quaternionic structure agree?
These two 'definitions' do not agree. Also, you should be careful about your choice of sources. Most differential geometers use the terminology 'almost quaternionic' to mean that the structure group ...
- 103k
6
votes
Accepted
Bialynicki-Birula decomposition for real analytic varieties
No, consider the following $\mathbb{C}^*$-action on $\mathbb{CP}^2$ : $$z.[z_0:z_1:z_{2}] = [z_0:z.z_1:z^2.z_2] ,$$ along with the antiholomorphic map $\sigma ([z_0:z_1:z_2]) = [\bar{z_{0}}:\bar{z_{1}...
- 6,783
5
votes
Accepted
Are there hyperkahler manifolds with multi-isotropic hyperkahler submanifolds?
NB: New evidence has changed my conclusions.
In the first nontrivial case where this question makes sense, i.e., when $X$ has dimension $16$ and $S\subset X$ has dimension $4$, a preliminary ...
- 103k
5
votes
compact manifold as a hyperkahler quotient of an infinite-dimensional affine space
Wow, my thesis; it's been a while! Perhaps I was/am confused but I'll add a few remarks in case it helps.
I don't know of any examples of a compact hyperkahler manifold obtained as an infinite-...
- 1,404
4
votes
Bialynicki-Birula decomposition for real analytic varieties
There's some simple topology behind why this will work in complex cases, and not consistently in real ones: complex affine spaces are even dimensional as real manifolds, and real affine spaces are ...
- 42.7k
3
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 ...
- 1,444
2
votes
Are there non-projective, but algebraic, hyperkahler varieties?
With the correct definition of hyperkähler (which as Jason said requires $H^0(X,\Omega^2_X)$ to be generated by a holomorphic symplectic form), there are examples constructed by Yoshioka in ...
- 6,498
2
votes
Submanifold of a Hyperkahler manifold which is 'Lagrangian' w.r.t. all three symplectic structures
Such submanifold does not exist (in physics terminology, there are no
$(A,A,A)$-branes). A submanifold which is Lagrangian with respect to two symplectic forms, say $\omega_2$ and $\omega_3$, is ...
- 6,720
2
votes
Accepted
Small contraction for Hyperkähler Varieties
Let $f:X\to Y$ be a birational contraction where $X$ is hyperkähler, then $K_X\sim 0$ and $K_Y=f_*K_X\sim 0$, and hence $K_X=f^*K_Y$. In particular, this means that $Y$ has canonical singularities. ...
- 1,054
2
votes
Accepted
Does miracle flatness always fail for a non-regular base?
The answer lies in Theorem 23.7 from Matsumura's Commutative Ring Theory:
Theorem 23.7. Let $(A, \mathfrak m, k)$ and $(B, \mathfrak n, k')$ be local Noetherian local rings, and $A \to B$ a local ...
- 875
1
vote
Accepted
Can deformation equivalent Kähler manifolds always be obtained by a deformation where all the fibers are Kähler?
I don't think this is known. For hyperkahler manifolds, conjecturally,
all smooth complex deformations are class C and birational to hyperkahler.
If this is true, your conjecture would follow ...
- 8,828
1
vote
Accepted
Stuck on a computation with quaternions and moment maps
I was able to finally prove that
$$d(\omega.d\mathbf{r}) = - \frac{1}{2r^3}(d\mathbf{r}\,\mathbf{r} \wedge d\mathbf{r}).$$
In the process, I have learned a lot. The main issue for me was that I was ...
- 4,103
1
vote
The state of art of the singular Levi problem -- and hyperkähler quotients
Privet, Anya.
Your reference [FN80] actually seems to contain an answer to this problem! They state (in particular, see the question 1.5 in the introduction) that the class of weakly psh functions, i....
- 1,822
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