Timeline for Why is the Leibniz rule a definition for derivations?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Jan 6, 2013 at 21:24 | comment | added | Stephan Sturm | @ Uwe Franz. I believe you forgot the biscuit monad in your discussion, see en.wikipedia.org/wiki/Leibniz-Keks | |
| Jan 6, 2013 at 20:11 | history | edited | R S | CC BY-SA 3.0 |
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| Jan 5, 2013 at 2:51 | comment | added | Uwe Franz | There is no "t" in the last name of the philosopher, mathematician and physicist Gottfried Wilhelm Leibniz, see de.wikipedia.org/wiki/Leibniz. Leibnitz is a small town in Austria, see de.wikipedia.org/wiki/Leibnitz. | |
| Jan 1, 2013 at 5:20 | comment | added | R S | I think that @ayanta's answer is the one that is closest to what I was aiming to (it is possible that other answers are also good but I lack the knowledge to understand them). I only wonder if the Leibnitz rule is the only one forcing linear operators to depend only on first order, or if there are other equivalents. | |
| Dec 29, 2012 at 0:59 | comment | added | user30180 | If you view point-derivations as operators on global smooth functions rather than on germs of smooth functions near $P$ (the latter is what I used in my preceding comment), it's still OK: the Leibnitz Rule forces $\partial(f)$ for global smooth $f$ to only depend on the germ of $f$ around $P$ (and so via bump functions, the distinction between global smooth functions and germs at $P$ doesn't matter). Indeed, if a global $f$ vanishes near $P$ then $f = (1-\phi) f$ for a smooth global bump function $\phi$ that is supported near $P$ and equals 1 near $P$, so $\partial(f)=0$ by Leibnitz. | |
| Dec 28, 2012 at 12:28 | answer | added | Peter Michor | timeline score: 7 | |
| Dec 28, 2012 at 4:31 | comment | added | user21349 | You pose the question as a question about calculus on a manifold, but is anything really lost by focusing on calculus on the reals? On the reals, either the product rule or the chain rule can be taken as the fundamental defining property of the derivative. The chain rule implies the product rule in this sense: mathoverflow.net/questions/108773/… The product rule also implies the chain rule: mathoverflow.net/questions/44774/… | |
| Dec 28, 2012 at 4:26 | answer | added | Michael Murray | timeline score: 5 | |
| Dec 28, 2012 at 4:25 | answer | added | innerproduct | timeline score: 1 | |
| Dec 28, 2012 at 4:02 | comment | added | user30180 | The intuition is that the Leibnitz Rule at $P$ forces the linear operator $\partial:f \mapsto \partial(f) \in \mathbf{R}$ not just to kill constants but also to depend on $f$ only to first order at $P$ (and hence to be a directional derivative in local coordinates: $\partial(f) = D_v(f)$ where $v = \sum \partial(x_i) e_i$), as a directional derivative should. Indeed, if $f$ vanishes to first order at $P$ then for local coordinates $(x_i)$ with $x_i(P) = 0$, Taylor's theorem (the real content!) implies $f = \sum x_i x_j h_{ij}$ for smooth $h_{ij}$ near $P$, so $\partial(f) = 0$ by Leibnitz. | |
| Dec 28, 2012 at 3:39 | answer | added | Tom Goodwillie | timeline score: 17 | |
| Dec 28, 2012 at 3:27 | answer | added | KConrad | timeline score: 10 | |
| Dec 28, 2012 at 1:39 | answer | added | David Corwin | timeline score: 18 | |
| Dec 28, 2012 at 1:31 | answer | added | Qiaochu Yuan | timeline score: 17 | |
| Dec 28, 2012 at 1:11 | history | asked | R S | CC BY-SA 3.0 |