Timeline for Explicit Riemann Hilbert correspondence
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 26, 2019 at 19:42 | comment | added | Longma | Ok. I need more information to determine the induced map between isomorphic classes. Thanks anyway. | |
| Feb 26, 2019 at 14:21 | comment | added | Avi Steiner | @Longma Oh! Shoot! You’re right! Yeah, there’s no particular reason you need commuting matrices. My answer still holds , though. | |
| Feb 26, 2019 at 11:04 | comment | added | Longma | In higher dimensional cases, I totally agree with you. However, in my example, we are working on the one dimensional case, namely, every two components of the NCD divisor has no intersection, so we don't need to require that the monodromy commutates. (Since the commutativity condition is due to monodromy agrees on the intersection of each "singularity". Does it make any sense for you? | |
| Feb 25, 2019 at 16:03 | comment | added | Avi Steiner | @Longma if you don’t have commuting matrices, you don’t get an integrable connection, and in particular can’t use the Riemann-Hilbert Correspondence. In fact, you won’t even get a D-module | |
| Feb 25, 2019 at 9:03 | comment | added | Longma | Thanks a lot. Actually the reference you gave is exactly the book I am reading. In this book, they gave a way to obtain the monodromy representation, however, their method requires that all monodromy are commuting invertible matrices, which I do not get the point, do you have any idea? | |
| Feb 24, 2019 at 17:41 | history | answered | Avi Steiner | CC BY-SA 4.0 |