Timeline for Comparing Dirichlet energy and area of a Surface-immersion
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Aug 27, 2015 at 6:26 | comment | added | H1ghfiv3 | Oh, so you claim that both the area and energy are not only independent of a choice of coordinates (which I was well aware of), their respective integrands at a point p can be compared using ANY basis at $T_pF $ regardless of coordinates? I wasnt aware of that actually. However, i had trouble finding basic literature on that topic | |
| Aug 26, 2015 at 22:05 | comment | added | Robert Bryant | You don't really use normal coordinates per se, just the fact that there's an orthonormal basis of vectors in each tangent space. This is a linear algebra fact, whereas the (local) existence of normal coordinates is a deeper fact that depends on the derivatives of the metric. | |
| Aug 26, 2015 at 21:46 | comment | added | H1ghfiv3 | I actually meant riemannian normal coordinates, not to mix up any terms here. They are used so the metric tensor $g_{ij}$ (and hence, $g^{ij}$) becomes trivial at the chosen point $p$. This seems to be necessary since the energy integrand uses the $g^{ij}$. Am i mistaken here ? | |
| Aug 26, 2015 at 19:02 | comment | added | Robert Bryant | You're welcome, but I should point out that it's not actually a use of geodesic normal coordinates. The derivatives of the metrics $g$ and $h$ play no role in the formulae for the integrands and hence are irrelevant for the inequality. It really is just linear algebra at a point. | |
| Aug 26, 2015 at 17:59 | comment | added | H1ghfiv3 | Of course, I forgot about geodesic normal coordinates. This solves the problem. Thank you, Sir. | |
| Aug 26, 2015 at 17:01 | comment | added | Robert Bryant | Calculate at a point: Let $e_1,e_2$ be a $g$-orthonormal basis of $T_xF$ and let $f_i = df(e_i)\in T_{f(x)}M$. Then you just need to show that $$\tfrac12\bigl(|f_1|^2+|f_2|^2\bigr) \ge \sqrt{\ |f_1|^2|f_2|^2-(f_1\cdot f_2)^2\ }$$ where the inner products come from $h$ on $T_{f(x)}M$. But this is obvious (and doesn't depend on $M$ being $3$-dimensional, by the way). Note, too, that you get equality if and only if $f_1\cdot f_2 = |f_1|^2-|f_2|^2 = 0$. | |
| Aug 26, 2015 at 16:51 | comment | added | H1ghfiv3 | I've tried this, but i can't quite deal with the arbitrariness of the metric $g$ on $F$. In appropriate charts, the area integrand should be $\sqrt{|\partial f/ \partial x_1|^2_h|\partial f/ \partial x_2|^2_h - <\partial f/\partial x_1,\partial f,\partial x_2>_h^2}$, while the energy integrand is $\sum_{i,j=1}^2 g^{ij}<\partial f/\partial x_i,\partial f/ \partial x_j>_h$. But since I have no control about how the $g^{ij}$ might look, i am stuck at a certain point in a chain of inequailites. | |
| Aug 26, 2015 at 15:50 | comment | added | Robert Bryant | This is just the integration of a pointwise inequality, and it only involves the first derivative of the map $f$ and the values of the metrics $g$ and $h$ at a fixed point in $F$. Thus, checking it in the 'flat' case that you have already done proves it in the general case. By the same calculation, you also get that equality holds if and only if $f$ is a conformal immersion. | |
| Aug 26, 2015 at 15:19 | review | First posts | |||
| Aug 26, 2015 at 16:35 | |||||
| Aug 26, 2015 at 15:19 | history | asked | H1ghfiv3 | CC BY-SA 3.0 |