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K3 surface is Einstein, by Calabi-Yau theorem, but it is not conformally flat; this can be seen e.g. from topology, or from the reference that José Figueroa-O'Farrill has given above.

$SU(2)$ is a group of unitary quaternions $U(1,H)$, which are of form $a + bI + cJ + dK$, with $a^2+b^2+c^2+d^2=1$. This is clearly $S^3$. The action of unitary quaternions from the right preserves c …

Here is the solution (with thanks to Misha Kapovich who pointed this out). Gerhard Hiss, Andrzej Szczepanski, "On torsion free crystallographic groups", Journal of Pure and Applied Algebra Volume 7 …

If I understood your question correctly, you define a holomorphic connection as one which satisfies $\bar\partial (\nabla_x y)=0$ for any holomorphic vector fields $x, y$. Then the holomorphicity is …

Since Grassmannian $Gr(n,m)=SO(n+m)/SO(n)\times SO(m)$ is a homogeneous manifold, you can take any Riemannian metric, and average with $SO(n+m)$-action. Then you show that an $SO(n+m)$-invariant metri …

Ngaiming Mok, The uniformization theorem for compact Kähler manifolds of nonnegative holomorphic bisectional curvature, J. Differential Geom. Volume 27, Number 2 (1988), 179-214. Mok improves Mori's …

The most clear proof (through differential operators, graded Jacobi identities and superalgebra) actually works in both twisted and untwisted cases, no need to derive one from another. See http://arxi …

A connected compact complex Lie group is a torus, hence the question is apparently about principal torus bundles. Of course, product of a torus and a compact Kahler manifold is Kahler, giving a trivia …

I have looked for a while for a proof which does not use the Calabi-Yau theorem and nobody seems to know it. Also, there are plenty of non-Kaehler manifolds with canonical bundle trivial topologicall …

the construction of the canonical $J$ for a complex manifold is what I'm interested in Given a complex manifold, you have a bundle of (1,0)-forms within complexified 1-forms which is generated (o …

What is the easiest (and what is the most elementary) way of proving Newlander-Nirenberg theorem for Riemannian surfaces? I was able to reduce it to existence of non-trivial harmonic functions (locall …

Let $M$ be a compact Riemannian manifold with vanishing curvature of Levi-Civita connection. Such manifolds were classified by Bieberbach; sometimes they are called Bieberbach manifolds. According to …

A paper by Gabor Etesi was published that purports to solve a major outstanding problem: Complex structure on the six dimensional sphere from a spontaneous symmetry breaking Journ. Math. Phys. 56, 04 …

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 …

If the Chern class vanishes over integers, it's Calabi-Yau manifold, and it has a holomorphic (n,0)-form by Bogomolov's theorem (see there: Two definitions of Calabi-Yau manifolds). If you relax your …