Timeline for Gluing local holomorphic connections
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Mar 16 at 21:10 | comment | added | Richard Lärkäng | Indeed, if one wants to be really precise, the map is not $\psi_i^{-1}$, but $Id \otimes \psi_i^{-1}$. | |
| Mar 15 at 20:55 | comment | added | Johannes | One more question, in $\xi \mapsto\psi_i^{-1} ((d + A_i) (\psi_{i} \xi))$ do you identify $\psi^{-1}_i : V \times \Bbb C^r \to E|_{V}$ with a map $$\psi^{-1}_i : \Omega^1 \otimes (V \times \Bbb C^r) \to \Omega^1 \otimes E|_{V}$$ that sends $\omega \otimes s \mapsto \omega \otimes \psi^{-1}_i(s)$, since $(d + A_i) (\psi_{i} \xi)$ is a section of $\Omega^1 \otimes (V \times \Bbb C^r)$? @richard-lärkäng | |
| Mar 13 at 21:44 | vote | accept | Johannes | ||
| Mar 13 at 21:44 | comment | added | Johannes | Thanks a lot for taking the time to answer my rather elementary questions. I'm quite new to the topic. @richard-lärkäng | |
| Mar 13 at 21:43 | vote | accept | Johannes | ||
| Mar 13 at 21:44 | |||||
| Mar 13 at 21:12 | comment | added | Richard Lärkäng | I would not read anything like that into the missing composition symbol. When you compose linear maps, there is no ambiguity, so you may include or skip the composition symbol as you wish. For the parts involving $d$ (or $\partial$), it makes sense to include the composition symbol to give it the meaning that I wrote in my answer, while without the composition symbol, it would seem natural to me to interpret it as $d$ acting $\psi_i$, but as mentioned above, this has no reasonable meaning if one does not identify $\psi_i$ with a matrix, which you cannot do in an intrinsic way. | |
| Mar 13 at 18:30 | comment | added | Johannes | Thanks for updating this. It clears things up for me. About regarding $\psi_i$'s as matrices, equation 4.4 states $$\psi^{-1}_i \circ \partial \circ \psi_i -\psi^{-1}_j \circ \partial \circ \psi_j = \psi^{-1}_jA_j\psi_j -\psi^{-1}_iA_i\psi_i,$$ where the compositions on the left-hand side are compositions of linear maps, but I thought that these products on the rhs are matrix products which would indeed mean that $\psi_i$'s are identified as matrices. Is this just being lazy and dropping the composition symbol? @richard-lärkäng | |
| Mar 13 at 13:24 | history | edited | Richard Lärkäng | CC BY-SA 4.0 |
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| Mar 13 at 13:07 | comment | added | Richard Lärkäng | Sorry, I realized that there is a reasonable interpretation of what Huybrechts wrote. I have now updated my answer. Anyhow, there is no identification of the $\psi_i$ as matrices, just as linear maps. | |
| Mar 13 at 13:06 | history | edited | Richard Lärkäng | CC BY-SA 4.0 |
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| Mar 13 at 12:39 | comment | added | Johannes | I meant changing it to $A(E)=\{U_{ij}, \psi^{-1}_j(\psi^{-1}_{ij}d\psi_{ij})\psi_j\},$ since if I understood your answer, we are now identifying the trivializations $\psi_i: E|_{U_i}\to U_i \times \Bbb C^r$ with the matrices of the linear maps $f_{\psi_i}:E_p \to \Bbb C^r$ obtained from $$\psi_i(p,v)=(p, f_{\psi_i}(v)).$$ Sorry for being overly pedantic, I just really want to understand the identifications going on here. @richard-lärkäng | |
| Mar 13 at 12:33 | comment | added | Richard Lärkäng | No, I think the definition of the Atiyah class is fine. (The issue is that $d\psi_i$ and $d\psi_j^{-1}$ do not have any intrinsic meaning, but $d\psi_{ij}$ does.) | |
| Mar 13 at 11:16 | comment | added | Johannes | Right, okay. In view of this I think I should also adjust the definition of the Atiyah class? @richard-lärkäng | |
| Mar 13 at 11:08 | comment | added | Richard Lärkäng | No, I don't think it has to do with what viewpoint you have of connections. I think it is a minor mistake of Huybrechts, but which as I described may be easily corrected, and which actually makes the proof shorter than what he has written. | |
| Mar 13 at 10:57 | comment | added | Johannes | I think I might have some idea on what is happening here. Huybrecths is describing holomorphic connections as maps of sheaves $\nabla : E \to \Omega_X \otimes E$, where $\Omega_X$ is the holomorphic cotangent bundle. Could this be the reason for why the formula would make sense? @richard-lärkäng | |
| Mar 13 at 10:48 | history | answered | Richard Lärkäng | CC BY-SA 4.0 |