Timeline for Connection, compatible with type (1, 1) tensor field
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 20, 2016 at 17:14 | vote | accept | Andrei Konyaev | ||
| Jan 1, 2016 at 21:24 | comment | added | Andrei Konyaev | I know! But look at the formula | |
| Jan 1, 2016 at 20:48 | comment | added | Robert Bryant | Hmmm.... This is hard to believe, because the above condition that guarantees the exisence of a torsion-free connection that makes $\nabla R$ symmetric is an open condition on $R$. If your formula were correct, then this would imply that the Nijenhius tensor of a $(1,1)$-tensor vanishes generically, and this isn't true. | |
| Jan 1, 2016 at 12:08 | vote | accept | Andrei Konyaev | ||
| Jan 1, 2016 at 12:09 | |||||
| Jan 1, 2016 at 12:03 | comment | added | Andrei Konyaev | Thank you very much, for taking time to answer my question. When I was talking about the complex eigenvalues I meant the sufficient, not necessary conditions. You showed the iff-condition, which is really awesome. The thing is that in terms of torsionless connection $\nabla$ the Nijenhuis tensor of the operator field can be rewritten as $$ N_R (X, Y) = (\nabla R) (RX, Y) +(\nabla R) (X, RY) - (\nabla R) (RY, X) - (\nabla R) (Y, RX). $$ This means that if connections do exist the Nijenhuis tensor vanishes | |
| Jan 1, 2016 at 11:05 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Fixed some typos and added some interpretations of the condition for subjectivity
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| Dec 31, 2015 at 11:09 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Expanded upon the original answer and clarified what happens in the case of multiple eigenvalues
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| Dec 30, 2015 at 17:52 | history | answered | Robert Bryant | CC BY-SA 3.0 |