34 votes
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What are the main contributions to the mathematics of general relativity by Sir Roger Penrose, winner of the 2020 Nobel prize?

It seems (as mentioned by Sam Hopkins above) that the Singularity Theorem is the official reason for the Nobel Award. But that is by no means the only (and perhaps not even the most important) ...
28 votes
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Why is $\mathbb{Z}$ not a Kähler group?

If $X$ is a compact Kähler manifold, then $h^{p,q}(X) = h^{q,p}(X)$ and $b_k(X) = \sum_{p+q=k}h^{p,q}(X)$, so in particular, $b_1(X) = h^{1,0}(X) + h^{0,1}(X) = 2h^{1,0}(X)$ is even. Now, $$b_1(X) = ...
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26 votes
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Does every group arise as the fundamental group of a complete Kähler manifold?

Any Stein manifold admits a complete Kähler metric. Start with a connected real analytic manifold with the given fundamental group. A suitable tubular neighbourhood of the complexification will be ...
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25 votes

What are the main contributions to the mathematics of general relativity by Sir Roger Penrose, winner of the 2020 Nobel prize?

I answered about the incompleteness theorem in the other thread. Let's talk about some of his other contributions here. (This list is definitely incomplete*, but just some stuff off the top of my head....
22 votes

Fundamental groups of compact Kähler manifolds

This is a bit off the top of my head. So if someone feels I omitted some important development, feel free to mention it to me in a comment, by email, or in your own answer. One can also look at Burger'...
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22 votes

What are the main contributions to the mathematics of general relativity by Sir Roger Penrose, winner of the 2020 Nobel prize?

A very interesting contribution (not directly related to relativity) is joint with Moore on the so-called Moore-Penrose inverse or generalized inverse, which is crucial in inverse problems theory and ...
20 votes
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Darboux-like theorems

While Francois' answer is fine, as far as it goes, I think that it is important to bear in mind the history of this problem. The original problem of 'flatness' or integrability (and more generally, ...
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19 votes
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Does a Kähler manifold always admit a complete Kähler metric?

Grauert proved that a relatively compact domain with real analytic boundary in C^n has a complete Kahler metric iff the boundary is pseudoconvex .For any relatively compact domain in C^n which is the ...
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19 votes

Are most Kähler manifolds non-projective?

See Claire Voisin's amazing results on the subject, or the published version: On the homotopy types of compact kaehler and complex projective manifolds, Inventiones Math. 157 2 (2004), 329 - 343. (...
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18 votes
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Is a symplectic submanifold of a Kähler manifold Kähler?

No. In $\mathbf C^2$ with standard 2-form and complex structure, the real span of $U=\left(\begin{smallmatrix}1\\0\end{smallmatrix}\right)$ and $V=\left(\begin{smallmatrix}i\\1\end{smallmatrix}\right)$...
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16 votes

What are the main contributions to the mathematics of general relativity by Sir Roger Penrose, winner of the 2020 Nobel prize?

I would say Penrose is a mathematical physicist and I don't think he can be considered (at least not primarily) to be a pure mathematician. For example, his argument for the Penrose inequality is a ...
14 votes

Why is $\mathbb{Z}$ not a Kähler group?

Every free group is not Kähler (fundamental group of compact Kähler manifold). In particular, a free group of rank 1 is not Kähler. This follows since the rank of the abelianization of a Kähler ...
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13 votes

Hard Lefschetz Theorem for the Flag Manifolds

To add to David's answer a bit: The Borel description of $H(G/P)$ is the $W_P$-invariant subring of the $W$ coinvariants, $S^{W_P}_W$, where $W$ is the associated Weyl group and $W_P\subseteq W$ is ...
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13 votes
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Equivariant Almost Complex Structures on the Full Flag Manifolds

Actually, though this may seem pedantic, there are two almost-complex structures on $\mathbb{CP}^m$ that are invariant under $\mathrm{SU}(m{+}1)$, namely the 'standard' one and its conjugate. Of ...
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13 votes

Fundamental groups of compact Kähler manifolds

It is still open whether or not all Kahler groups occur as the fundamental groups of smooth complex projective varieties. However, there has been some interesting work on which groups can occur as ...
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12 votes
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Hard Lefschetz Theorem for the Flag Manifolds

I'll spell out Hard Lefschetz as an explicit combinatorial statement about the Grassmannian $G(d,n)$. I can definitely give you a version of this for the full flag manifold if you want it and probably ...
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12 votes
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What makes a Kähler manifold projective?

If I understand the question correctly, I think that the answer is given by the main result in S. Ji: Currents, metrics and Moishezon manifolds, Pac. J. Math. 158, No. 2, 335-351 (1993). ZBL0785.32011....
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12 votes
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Does the Kähler form $\omega$ satisfy $d^*\omega=0$?

Recall that $d^* = -\ast d\ast$ and $\ast\omega = \frac{1}{(n-1)!}\omega^{n-1}$, see Example 1.2.32 of Huybrechts' Complex Geometry: An Introduction for example. Therefore $$d^*\omega = -\ast d\ast\...
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11 votes
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Algebraic vs. homological equivalence for curves on a smooth complex projective surface

Super vast generalisation: for divisors on a smooth projective variety over an algebraically closed field of any characteristic, the notions of algebraic, homological (for any Weil cohomology theory), ...
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11 votes

Darboux-like theorems

I think you have in mind the integrability (a.k.a. "flatness") problem for $G$-structures. Beyond the cases mentioned at that link (symplectic, Kähler, and complex structures, corresponding to $G=Sp(n,...
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11 votes

Are most Kähler manifolds non-projective?

I agree that in a sense it's harder to get your hands a non algebraic Kähler manifold, because you can't simply write an equation for one, but I would argue that there are plenty of them. You won't ...
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11 votes
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Is there a Kähler manifold with no anti-holomophic involution?

The moduli space $\mathcal{M}_1^\mathbb{R}$ of real algebraic curves of genus $1$ equals the real part of the moduli space $\mathcal{M}_1=\mathbb{C}$. In particular, the general elliptic curve has no ...
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10 votes

Why can we not always take a Kähler class to be in rational cohomology?

Since Artie Prendergast-Smith is not expanding his comment in an answer, let me do it. As I said in the comments, his comment is essentially THE answer to the OP question. But let me give some more ...
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10 votes

List of Applications of the $\partial\overline{\partial}$-lemma

This is a list of length one :) The $\partial \bar \partial$-Lemma allows a parameterization of the cohomology class $[\omega]$ of a compact Kaehler manifold $(M, \omega)$ by means of scalar ...
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  • 590
10 votes
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The logarithm of Kähler metric is not globally defined

Question 1: Note, $\omega + \partial\bar{\partial}\phi$ cannot be zero. Recall that $\phi$ is chosen so that $\omega + \partial\bar{\partial}\phi$ is another metric (in particular, a Kähler-Einstein ...
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10 votes
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Atiyah-Singer for Riemannian and Kaehler manifolds

I highly recommend the discussion in Shanahan's book, The Atiyah-Singer Index Theorem (An introduction), Lecture Notes in Math 638. In addition to a sketch of the proof, he gives a nice discussion of ...
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10 votes
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The period map and the Kodaira--Spencer map

Differential of period map $d P^{p+q,p}$ is composition of KS-map $T_{B,0} \to H^1(X_0,T_{X_0})$ with natural map $H^1(X_0,T_{X_0}) \to Hom(H^{p,q}(X_0),H^{p-1,q+1}(X_0))$ (given by the cup product ...
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