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

37 votes
1 answer
3k views

Is S^2 x S^4 a complex manifold?

As observed by Calabi a long time ago, the manifold $S^2\times S^4$ admits an almost-complex structure (obtained by embedding it in $\mathbb{R}^7$ and using the octonionic product), which however is n …
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  • 6,871
32 votes
2 answers
2k views

Example of a compact Kähler manifold with non-finitely generated canonical ring?

A celebrated recent theorem of Birkar-Cascini-Hacon-McKernan and Siu says that the canonical ring $R(X)=\oplus_{m\geq 0}H^0(X,mK_X)$ of any smooth algebraic variety $X$ over $\mathbb{C}$ is a finitely …
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  • 6,871
28 votes
Accepted

How to tackle the smooth Poincaré conjecture

In principle the Ricci flow (with surgery) could also be used to prove the smooth Poincaré conjecture in dimension $4$. There are some major problems to be overcome in this approach (problems which di …
YangMills's user avatar
  • 6,871
24 votes
Accepted

Relationship between sectional curvature, bisectional curvature and conjugate points

Since you seem to have some basic misunderstandings about these concepts, I will try to clarify the definitions. Let $(M,J)$ be a complex manifold, with a Riemannian metric $g$ which is Hermitian, i …
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  • 6,871
18 votes
3 answers
1k views

Moishezon manifolds with vanishing first Chern class

Suppose $M$ is a Moishezon manifold with $c_1(M)=0$ in $H^2(M,\mathbb{R})$. Does it follow that $K_M$ is torsion in $\mathrm{Pic}(M)$? This is true whenever $M$ is Kähler (and therefore projective) an …
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  • 6,871
14 votes

Hsiung on the Complex Structure of $S^6$

I just found this paper by B. Datta (later published in J. Indian Math. Soc. 60 (1994), no. 1-4, 171–190) that explains in details why one key equation in Hsiung's paper is wrong. See the whole discus …
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  • 6,871
14 votes
Accepted

Isoperimetry and Poincaré Inequality

Everything here is for closed Riemannian manifolds. If you have a lower bound on Cheeger's isoperimetric constant $h(M)$, then Cheeger's inequality $\lambda_1(M)\geq \frac{h(M)^2}{4}$ gives you a lowe …
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  • 6,871
14 votes
Accepted

Two approaches to compute the signature of a Kaehler manifold

About your last question, a recent theorem of Kotschick-Schreieder (see http://arxiv.org/abs/1202.2676 page 2) says that a linear combination of Hodge numbers equals a linear combination of Chern numb …
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12 votes
Accepted

Almost Complex Structures: 'Tame' versus 'Compatible'

If you assume that $M$ is compact and of dimension $4$ then Donaldson has conjectured that if $J$ is $\omega$-tame then $J$ is $\omega'$-compatible for some symplectic form $\omega'$, see Question 2 h …
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  • 6,871
12 votes

Two definitions of Calabi-Yau manifolds

It seems worthwile to point out that it is not true that the vanishing of the integral first Chern class of a compact Kahler manifold implies that the canonical bundle is holomorphically trivial. It o …
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  • 6,871
11 votes
Accepted

Calculating a curvature tensor by polarization

The explicit polarization formula is the following, taken from this paper of Bishop and Goldberg. Working with real tangent vectors (instead of $(1,0)$ vectors, but it's easy to switch from one poin …
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  • 6,871
10 votes
Accepted

Projective curves of constant curvature

A general result of Hulin shows that if $M$ is a compact complex manifold with a holomorphic embedding $f:M\to\mathbb{CP}^n$ such that $f^*g_{FS}$ is Einstein, then the Einstein constant is positive. …
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  • 6,871
9 votes

Are Chern connections on flat bundles flat?

The answer is yes if $M$ is compact Kähler (EDIT: and you allow a conformal change in the metric) and $L$ is a line bundle and no in general, already in the case of line bundles. Take for example $M …
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  • 6,871
9 votes
Accepted

Is S^2 x S^4 a complex manifold?

This is still an open problem. See this paper for some progress, which was prompted by this MO question.
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8 votes

Dolbeault cohomology

You seem to have some basic misunderstandings here. For non-Kähler manifolds, like the Hopf manifold that you consider, the Dolbeault cohomology is not in general topological, and it depends a lot on …
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