Is a locally free sheaf of modules free on some affine open basis?
Jul 13, 2022 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for …
The definition of locally Lipschitz - Mathematics Stack Exchange
Actually, a continuously differentiable function is locally Lipschitz, but since the derivative isn't assumed continuous in the theorem, one has only the weaker property that might be dubbed …
Distribution theory and locally integrable functions
May 1, 2021 · $\begingroup$ there is a natural injective linear transformation of locally integrable functions (strictly speaking equivalence classes) into the space of distributions given by …
Locally Complete Metric Space - Mathematics Stack Exchange
Nov 27, 2018 · The Locally Complete Metric Space is a metric space where each point has a neighbourhood (which is closed) is a complete metric space. It is so obvious that every …
Difference Between Holder Continuous and Locally Holder …
Oct 17, 2022 · I have seen that a function is Locally Holder Continuous on a set if it is Holder Continuous on every open subset of that set. Also I have read that Locally Holder Continuous …
the equivalency of two definitions of locally closed sets
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Are all manifolds locally flat? - Mathematics Stack Exchange
Apr 25, 2022 · As for generality, a great many physical problems (configuration spaces, vector fields, level sets) can be set naturally on manifolds. Regarding tractability, because any two …
abstract algebra - $M$ is projective iff $M$ is locally free ...
Mar 30, 2019 · A finitely presented module is projective iff it is locally free (the localization at every prime ideal is free over the localized ring). This is given right after the statement that a finitely …
Does local flatness imply a local Euclidean metric?
Aug 12, 2021 · "Locally flat" means the metric (which is an additional structure on top of a manifold's differential structure) has 0 curvature. Any manifold is "locally Euclidean" meaning …
Locally symmetric spaces have parallel Riemannian curvature
Oct 29, 2019 · I'm trying to prove the following result: If $(M,g)$ is a locally symmetric Riemannian manifold, then the Riemannian curvature tensor is parallel: $\nabla Rm \equiv 0$.