Timeline for Projective-invariant differential operator
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Mar 31, 2019 at 16:09 | comment | added | Jean Marie Becker | take a look at the book of Ovsienko and Tabachnikov : math.psu.edu/tabachni/Books/BookPro.pdf | |
| Jan 12, 2019 at 20:11 | vote | accept | user76284 | ||
| Jan 12, 2019 at 9:40 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Added a remark about the S-tensor in the n=2 case.
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| Jan 12, 2019 at 9:24 | comment | added | Robert Bryant | @user76284: Some such formulae would be right, but I'm not familiar enough with your notation to pronounce that the constants or symmetrization operators that you write down are exactly right. | |
| Jan 12, 2019 at 1:14 | comment | added | user76284 | Or alternatively $(\nabla f)^{-1} \cdot \nabla \nabla f - \frac{1}{n+1} \text{sym}(I \otimes (\nabla f)^{-1} : \nabla \nabla f)$ where $:$ is a double contraction. | |
| Jan 12, 2019 at 0:47 | comment | added | user76284 | Does this mean that, in the projective case on $\mathbb{R}^n$ with standard coordinates, $T(f) = (\nabla f)^{-1} \cdot \nabla \nabla f - \frac{1}{n+1} \text{sym}(I \otimes \nabla \log \det \nabla f)$ where $\text{sym} : V \otimes V^* \otimes V^* \rightarrow V \otimes V^* \otimes V^*$ is the symmetrization map defined by $\text{sym}(A)^k_{ij} = A^k_{ij}+A^k_{ji}$? | |
| Jan 11, 2019 at 7:40 | comment | added | Robert Bryant | @user76284: Yes to your question about the case of isometries. In the case of the flat affine structure on the space $M=\mathbb{R}^n$, the coefficients of $\alpha_0$ all vanish in the standard coordinates, so, in those coordinates, $T(f)$ only has the term $\mathsf{A}(f)\circ\alpha_0$ which, in your somewhat nonstandard notation, is the expression you give. | |
| Jan 11, 2019 at 7:30 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Corrected some typos and added a little more information here and there
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| Jan 11, 2019 at 3:21 | comment | added | user76284 | Thank you for your comprehensive answer. I'm a novice in this area but I'll try to parse it as best I can. In the case of isometries, does this correspond to $T(f) = (\nabla f)^\top \cdot (\nabla f) - I$? How does one obtain $T(f) = (\nabla f)^{-1} \cdot \nabla \nabla f$ from $T(f) = A(f) \circ \alpha - \alpha$? | |
| Jan 10, 2019 at 22:09 | history | answered | Robert Bryant | CC BY-SA 4.0 |