Timeline for Why is the Leibniz rule a definition for derivations?
Current License: CC BY-SA 3.0
3 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
|
|
| Dec 28, 2012 at 3:51 | comment | added | user30180 | Introducing local rings and $M^2$ is too abstract. Your basic point can be said more simply: any $\mathbf{R}$-linear assignment $\partial:f \mapsto \partial(f) \in \mathbf{R}$ on smooth functions $f$ near $P$ which kills constants and depends on $f$ only "to first order at $P$" (i.e., kills any $f$ whose Taylor series at $P$ in some - hence, any - local coordinate system begins in degree $\ge 2$), as we'd expect of "generalized directional derivatives", necessarily is a directional derivative in local coordinates and so satisfies the Leibnitz Rule at $P$. But this doesn't answer the question! | |
| Dec 28, 2012 at 3:27 | history | answered | KConrad | CC BY-SA 3.0 |