Timeline for What is the symbol of a differential operator?
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
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| Sep 7, 2011 at 22:38 | comment | added | Pedro Lauridsen Ribeiro | It must also be remembered that the total symbol of (say, for concreteness) a scalar linear partial differential operator doesn't live in general on the cotangent bundle, but on the bundle of jets of scalar-valued maps of the same order as the order of the operator. This can also be seen from the extension of the chain rule to higher-order derivatives. There the notion of a total symbol becomes coordinate-invariant. | |
| Nov 2, 2009 at 5:51 | comment | added | Greg Muller | The phrases 'principal symbol' (the highest degree part) and 'total symbol' (for every part) are pretty useful for distinguishing between the two. | |
| Oct 31, 2009 at 19:27 | vote | accept | Theo Johnson-Freyd | ||
| Nov 1, 2009 at 3:19 | |||||
| Oct 31, 2009 at 18:00 | comment | added | Ben Webster♦ | it doesn't really fail, because there's no proof there. Baez just says there's a map and calls it "symbol." But there is no one symbol map. However, there is a unique G-invariant isomorphism of SL-> UL which sends a homogenous function to a differential operator with that element as principal symbol. You can't blame Baez for calling that map "symbol." | |
| Oct 31, 2009 at 3:56 | comment | added | Theo Johnson-Freyd | That's what I thought. But then the "proof" of PBW (which I got from TWF) fails, unless there's something special going on for left-invariant things. | |
| Oct 31, 2009 at 1:25 | history | edited | Ben Webster♦ | CC BY-SA 2.5 |
added 156 characters in body
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| Oct 30, 2009 at 22:35 | history | answered | Ben Webster♦ | CC BY-SA 2.5 |