Timeline for Non-commutative algebraic geometry
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 1, 2023 at 19:12 | comment | added | D.R. | Is there some other place at which I can read more about the claims in the 3rd paragraph? The ideas and overarching philosophy sound very exciting and insightful, but it's all a bit too vague for me to really understand, especially the sentence “This reflects the fact that $M$ and $N$ no longer have well-defined supports on some concrete spectrum of $A$”. | |
| Jun 18, 2010 at 0:50 | history | edited | Emerton | CC BY-SA 2.5 |
deleted 1 characters in body
|
| Feb 13, 2010 at 18:49 | comment | added | Harry Gindi | @VA, isn't QM one of the motivations for NC geometry? | |
| Feb 13, 2010 at 18:23 | history | edited | Emerton | CC BY-SA 2.5 |
added 135 characters in body
|
| Feb 13, 2010 at 18:08 | comment | added | VA. | This analogy with quantum mechanics is very enlightening. | |
| Feb 13, 2010 at 17:56 | comment | added | Harry Gindi | Yup, that's exactly how I worked with it (by taking the commutative subalgebra just like you said). That explains it. Thank you! | |
| Feb 13, 2010 at 17:52 | comment | added | Emerton | If $A$ is a $k$-algebra and $a \in A$, then certainly $k[a]$ is a commutative subalgebra of $A$, and so has a spectrum. But Spec $A$ is the simultaneous spectrum. And of course non-commuting operators cannot be simultaneously diagonalized. | |
| Feb 13, 2010 at 17:46 | comment | added | Harry Gindi | I thought that you can define the spectrum of an element of a noncommutative complex algebra. Although I don't really see how to define the spectrum of the whole noncommutative ring. The spectrum here being "morally" the set of eigenvalues. I wouldn't swear by it, since the only time I've ever seen it before was in a bonus exercise on a homework that I did. | |
| Feb 13, 2010 at 17:36 | history | answered | Emerton | CC BY-SA 2.5 |