Timeline for Simple question regarding shape operator/minimal surface equation
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Apr 22, 2012 at 5:12 | comment | added | Nick | Originally I was calculating symmetric polynomials of the second fundamental form in higher dimensions. The complexity of the calculation grows quickly when dimension increases and when taking more complicated functions of the second fundamental form than the trace. Hand calculation becomes prohibitively time consuming. In retrospect, I should have tried the hand calculation in the case of mean curvature in dimension 2, and would have if that's what I had originally been computing. However, I (mistakenly) didn't think to do this because I had already been doing the calculation using software. | |
| Apr 21, 2012 at 21:44 | comment | added | Will Jagy | Software? For this? Why would you do that? | |
| Apr 21, 2012 at 9:29 | vote | accept | Nick | ||
| Apr 20, 2012 at 20:58 | answer | added | Nick | timeline score: 0 | |
| Apr 20, 2012 at 20:56 | comment | added | Nick | The difficulty arose from the software I was using for the matrix product, that's why I kept running into this problem. The claim in Prof. Yang's comment is correct, I had thought that this claim was true but kept obtaining a different result. Thank you for helping to clear this up. | |
| Apr 19, 2012 at 14:48 | comment | added | Deane Yang | It should be noted that Guan and Spruck use $\gamma^{-1}h\gamma^{-1}$ because they want all of the principal curvatures and not just their sum (the mean curvature). | |
| Apr 19, 2012 at 8:29 | comment | added | Deane Yang | The two different ways of doing the calculation have to lead to the same answer, since the trace of $(\gamma^{-1} h) \gamma^{-1}$ equals the trace of $\gamma^{-1}(\gamma^{-1} h)$, which equals the trace of $g^{-1}h$. | |
| Apr 19, 2012 at 8:26 | comment | added | Deane Yang | How many times have you redone your calculation? My advice is not to post a question like this unless you have already redone the calculation rather carefully at least, say, 10 times. | |
| Apr 19, 2012 at 7:30 | comment | added | Robert Haslhofer | Are you sure that you have computed $s$ correctly? If $g_{ij}=\delta_{ij}+f_if_j$ then the inverse metric is $g^{ij}=\delta_{ij}-\tfrac{f_if_j}{1+|Df|^2}$ ... | |
| Apr 19, 2012 at 6:08 | history | edited | Nick | CC BY-SA 3.0 |
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| Apr 19, 2012 at 5:46 | comment | added | Will Jagy | I give a pretty clear description of this in Michigan Mathematical Journal, volume 38 (1991), pages 255-270. I had not seen explicit formulae for what I needed so I included some. | |
| Apr 19, 2012 at 5:41 | history | edited | Nick | CC BY-SA 3.0 |
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| Apr 19, 2012 at 5:19 | comment | added | Anton Petrunin | My solution to such questions: never do calculations because you(=I) never do it right. | |
| Apr 19, 2012 at 5:12 | history | edited | Nick | CC BY-SA 3.0 |
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| Apr 19, 2012 at 5:03 | history | asked | Nick | CC BY-SA 3.0 |