Timeline for Isomorphism classes of rings of differential operators
Current License: CC BY-SA 3.0
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| Apr 25, 2018 at 15:59 | comment | added | Carlos | Good, thank you, that's what I was aiming at first, but wasn't so sure. So, essentially, at least in the smooth setting and in characteristic zero, the algebra of differential operators is a complete invariant of the underlying manifold. The "Milnor exercise" gives a rather explicit construction of the underlying manifold via the evaluation functional on the algebra of smooth functions. Looking at the Grabowski and related papers, it doesn't, however, seem so straightforward to obtain the same manifold from its Lie algebra of vector fields, or is it? | |
| Apr 23, 2018 at 18:44 | history | answered | Peter Michor | CC BY-SA 3.0 |