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Complex, contact, Riemannian, pseudo-Riemannian and Finsler geometry, relativity, gauge theory, global analysis.

5 votes

Structure of Kähler cone

Explicit description of a Kahler cone for all hyperkahler manifolds is here: https://arxiv.org/abs/1401.0479 (Rational curves on hyperkahler manifolds, Ekaterina Amerik, Misha Verbitsky)
Martin Sleziak's user avatar
7 votes

Complex manifolds whose tangent and cotangent bundles are isomorphic as complex vector bundles

There are many such examples, for instance, all complex nilmanifolds (and most complex solvmanifolds) have tangent bundle which is topologically trivial. The Hopf manifolds also have topologically tri …
Misha Verbitsky's user avatar
4 votes
Accepted

Cohomology of invariant differential forms

There is no such thing: any $\phi$-invariant exact 1-form is a differential of $\phi$-invariant function. Indeed, let $\alpha$ be an exact $\phi$-invariant form, $\alpha=df$, where $f$ is not $\phi$-i …
Misha Verbitsky's user avatar
3 votes
Accepted

Bott-Chern cohomology for singular complex spaces

closed (1,1)-forms and currents on X are not necessary locally $dd^c$-exact in general What makes it different when X is singular? The obstruction to local $dd^c$-lemma is $R^1\pi_*(O_{X'})$, where …
Misha Verbitsky's user avatar
8 votes

Coincide between Chern-connection and Levi-Civita connection

It is easier to prove this result for 1-forms, instead of vector fields. On (1,0)-forms, $\nabla^{0,1}=\bar\partial$ because the Levi-Civita connection is torsion-free, hence $\bigwedge(\nabla(\eta))= …
Misha Verbitsky's user avatar
2 votes
0 answers
82 views

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, with t …
1 vote

Curvature forms of holomorphic line bundles

Sure, any closed 2-form $\eta$ with integer cohomology can serve as the curvature of a connection on a line bundle. This can be seen if you take a line bundle with the same Chern class and connection …
Misha Verbitsky's user avatar
6 votes
1 answer
441 views

Holonomy bounded in terms of area and the curvature

I suppose the following result follows from Ambrose-Singer theorem, but I cannot find a reference, and the arguments I found in the literature are usually weaker. The idea is that holonomy over a null …
2 votes

Is a linear vector field a geodesible vector field?

This observation seems to be very easy, but it takes care of many examples. Suppose that $A$ corresponds to a contraction (that is, all eigenvalues are $< 1$ in absolute value). Decompose ${\Bbb R}^n …
Misha Verbitsky's user avatar
8 votes

Hodge dual of de Rham cohomology and singular cohomology

The Hodge * operator action on cohomology is generally speaking metric-dependent, hence * is not well-defined without fixing the metric. There are some caveats. On complex curves, for example, the Hod …
Misha Verbitsky's user avatar
1 vote

Holomorphic retraction $\implies$ holomorphic tubular neighbourhood?

Question. Is there a holomorphic tubular neighbourhood of S in M? I have a counterexample. Let $\phi:\; Z^2 \to Aut(C^n)$ be a group homomorphism, with $Aut(B)$ denote the group of holomorphic autom …
Misha Verbitsky's user avatar
3 votes

Holomorphic version of Darboux's theorem

To put it differently, is it true that X locally looks like a cotangent bundle? This is false. Indeed, take an elliptic curve $C$ inside an elliptic K3 surface. If it had a neighbourhood $U$ which is …
Misha Verbitsky's user avatar
1 vote

HKT manifolds with non trivial canonical bundle

For $SU(3)$ and Hopf it is the anticanonical bundle which has many sections. In other examples of Joyce, it is also the anticanonical bundle or its powers. For nilmanifolds, the canonical bundle is ho …
Misha Verbitsky's user avatar
8 votes
1 answer
332 views

reference to a theorem about a product of harmonic and parallel forms

Let $\alpha$ be an exterior product of a harmonic and a parallel form on a Riemannian manifold. Then $\alpha$ is known to be harmonic. I have heard that this is an old result due to R. Bott, but I cou …
9 votes
2 answers
724 views

Bieberbach theorem for compact, flat Riemannian orbifolds

In his thesis, Bieberbach solved Hilbert 18 problem and proved that any compact, flat Riemannian manifold is a quotient of a torus. I need a reference to an orbifold version of this result: any comp …

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