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The most general version of Weitzenbock identities (with coefficients in appropriate universal enveloping algebras) is due to Uwe Semmelmann and Gregor Weingart: http://arxiv.org/abs/math/0702031 "The …

Lempert, László The Dolbeault complex in infinite dimensions. III. Sheaf cohomology in Banach spaces. Invent. Math. 142 (2000), no. 3, 579-603. Lempert, László The Dolbeault complex in infinite dimen …

Here is a list of literature which I compiled when I taught the course on Ricci flow. Basic differential geometry: Einstein Manifolds (Besse). Riemannian geometry (Gallot S., Hulin D., Lafontaine J.) …

The corresponding Kahler metric is called "Weil-Petersson metric"; it is often constructed using infinite-dimensional determinants of the corresponding Laplace operators. The standard reference is a s …

A generic deformation of a Hilbert scheme of K3 and a generic torus have no subvarieties, hence they are "simple" in the above sense. For a torus it's well known, for a Hilbert scheme of K3 it's in my …

The proof of simple connectedness of a resolution of quotient singularities is in this paper of mine http://arxiv.org/abs/math/9903175 Theorem 4.1 (published in Asian J. Math. 4, 2000, no. 3, 553-56 …

Suppose that $H^0(L^{\otimes n})|_ Y$ is non-zero for all $Y$ and $n$ sufficiently big, and has a section which vanishes somewhere on $Y$. Then it follows that the base set of $L$ is trivial: indeed, …

There are some references where it is computed D. Mall, {\em The cohomology of line bundles on Hopf manifolds}, Osaka J. Math. {\bf 28} (1991), 999--1015. D. Mall, {\em Contractions, Fredholm operator …

There are manifolds without non-constant global meromorphic functions, such as generic K3 or a torus. Among these K3 surfaces, there are ones with (-2)-curves, which give a counterexample to the quest …

Let $M$ be the vector space $V$ of quadratic forms on ${\Bbb R}^n$ (or its projectivization), and $G=GL(n)$. The open orbits correspond to non-degenerate quadratic forms, and there are precisely $n+1$ …

It seems that the answer is very simple (and should have been known to Uhlenbeck-Yau). Let $E$ be a stable bundle, equipped with an Hermitian-Einstein metric and connection, and $End(E)$ its authomorp …

For $d$ odd these two components are disconnected, because $I$ and $-I$ induce opposite orientation. For $d$ even, you have an involution, which takes a lattice to a complex conjugate lattice. This in …