Search Results

I was wondering whether there is some notion of "vector bundle de Rham cohomology". To be more precise: the k-th de Rham cohomology group of a manifold $H_{dR}^{k}(M)$ is defined as the set of closed …

You might also find the work of Joyce useful, which builds upon the work of Dubuc (E. J. Dubuc. C∞-schemes. Amer. J. Math., 103(4):683–690, 1981) and Moerdijk and Reyes (I. Moerdijk and G. E. Reyes. M …

WLOG the fan $\Sigma$ of our toric variety $X_{\Sigma}$ is simplicial. (So $X_{\Sigma}$ has at worst orbifold singularities and all cones $\sigma \in \Sigma$ are simplicial). The classical toric reso …

Let $M$ be a manifold (variety, scheme, your favorite object) and let $N_1,N_2$ be two submanifolds (subvarieties, closed subschemes, ideal sheafes, etc.) such that $N_1 \cap N_2 \neq \emptyset$. Deno …

Let $V$ be a smooth manifold, $E \rightarrow V$ a vector bundle over $V$ and $\Gamma$ be a finite group acting nontrivially on $V$ and $E$. Let $s \in C^\infty(E)$ be a generic $\Gamma$-equivariant se …

Just to add some things to Igor Belegradek's post: "1.That the isometry group of a Riemannian manifold is always a lie group." This is the famous Myers-Steenrod theorem, proven in 1939 (Myers, S.B. a …

Maybe this is a silly question (or not even a question), but I was wondering whether the cotangent bundle of a submanifold is somehow canonically related to the cotangent bundle of the ambient space. …

I'm not sure whether the following is too advanced, but I found "Introduction to Topological Manifolds (Graduate Texts in Mathematics) (Paperback) by John Lee" quite readable. (Edit: As Ho Chung Si …

First of all: this is from the "differential-geometric point of view" If you want to "classify" vector bundles chern classes are a very helpful tool. Well, it can happen that two bundles are "differe …

Since a $\bar{\partial}$-Operator is an elliptic Operator, you can use elliptic theory in order to prove the Serre duality. In fact the Serre duality is a kind of corollary of the"fundamental theorem" …

The Newlander-Nirenberg theorem states that an almost complex structure is integrable if and only if the Nijenhuis tensor vanishes. I heard that this statement is not true in infinite dimensions, sin …