Timeline for Isothermal-related functions in higher dimensions
Current License: CC BY-SA 3.0
22 events
| when toggle format | what | by | license | comment | |
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| Dec 1, 2014 at 17:44 | comment | added | Ali | @Robert Bryant: no Robert, their answers are perfectly valid and correct thanks for your time | |
| Dec 1, 2014 at 14:12 | comment | added | Deane Yang | Robert, thanks for the clarification. I misremembered the difference between hypoelliptic and subelliptic and mistakenly thought hypoelliptic was a stronger condition. And I've completely forgotten (assuming I once knew) the subtleties of such operators when the coefficients are real-analytic. How can the generic solution be real-analytic and yet non-real-analytic solutions exist? | |
| Dec 1, 2014 at 13:04 | comment | added | Robert Bryant | @DeaneYang: Well, I agree that the problem is always subelliptic, but it's also clear that the linearization of the equation at a generic solution really is hypoelliptic (because the $2$-plane field spanned by $\nabla u$ and $\nabla v$ will be bracket-generating). It's curious because, while there apparently can be nonanalytic solutions (even when the metric is real-analytic), it seems (when the metric is real-analytic) that the generic solution must be real-analytic. | |
| Dec 1, 2014 at 12:58 | comment | added | Robert Bryant | @Ali: I had a look at the other question, but I don't understand what you would like me to do. The answers given there are correct: For most pairs of functions $(\phi_1,\phi_2)$ on $(M^3,g)$ the Lie bracket of their gradient vector fields will be linearly independent from those gradients, in which case, there is no nonconstant solution $u$ to the problem, just as they wrote (and for the reasons they stated). Is there something else you wanted to know? | |
| Nov 30, 2014 at 20:41 | comment | added | Ali | The reason I ask this, is that it has some level of similarity to the question at hand here. | |
| Nov 30, 2014 at 20:40 | comment | added | Ali | Thanks to both for insightful comments! @Robert Bryant Could I ask you to share an insight on this topic as well? mathoverflow.net/questions/188244/… | |
| Nov 30, 2014 at 16:34 | history | edited | Deane Yang | CC BY-SA 3.0 |
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| Nov 30, 2014 at 16:32 | comment | added | Deane Yang | The correct terminology might be "subelliptic" rather than "hypoelliptic". The operator also reminds me of the $\bar\partial_b$ operator on a generic CR-manifold. Perhaps that theory could provide some useful guidance on how to prove local existence. Checking Google, I also see that there has been a fair amount of work on nonlinear subelliptic PDE's. | |
| Nov 30, 2014 at 11:44 | comment | added | Robert Bryant | @DeaneYang: I agree with your comments above about the hypoelliptic nature of the problem. However, I think that, at least in higher dimensions, the smooth tame estimates might be hard to come by because the linearization of the problem at a solution $\Phi = u+ i\,v$ will be hypoelliptic only if the $2$-plane field spanned by $\nabla u$ and $\nabla v$ is bracket-generating. This is generic in all dimensions, as you say, but as the dimension $n$ increases, it's higher and higher order and thus, presumably, more difficult to analyze. When $n=3$, though, it should be fairly straightforward. | |
| Nov 30, 2014 at 6:00 | history | edited | Deane Yang | CC BY-SA 3.0 |
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| Nov 30, 2014 at 5:11 | history | edited | Deane Yang | CC BY-SA 3.0 |
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| Nov 30, 2014 at 3:09 | comment | added | Deane Yang | Robert, thanks! Will have to rethink this. | |
| Nov 30, 2014 at 2:49 | comment | added | Robert Bryant | Deane, I think that there is a misunderstanding. The equations that say that $\Phi= x^1+i\,x^2$ satisfies $\langle\mathrm{d}\Phi,\mathrm{d}\Phi\rangle_g=0$ are not $g_{11}-g_{22}=g_{12}=0$. They are $g^{11}-g^{22}=g^{12}=0$. The former are two equations on the entire coordinate system, while the latter are two equations on the two functions $x^1$ and $x^2$, so they are different if $n>2$. In fact, if $\Phi = u + i\,v$, then $$0=\langle\mathrm{d}\Phi,\mathrm{d}\Phi\rangle_g= g(\nabla\Phi,\nabla\Phi) = \bigl(g(\nabla u,\nabla u){-}g(\nabla v,\nabla v)\bigr) + 2i\, g(\nabla u,\nabla v).$$ | |
| Nov 30, 2014 at 2:06 | history | edited | Deane Yang | CC BY-SA 3.0 |
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| Nov 30, 2014 at 2:01 | history | edited | Deane Yang | CC BY-SA 3.0 |
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| Nov 30, 2014 at 1:45 | comment | added | Deane Yang | There's no constraint in $g$ or $g_0$. Restricting to dimension $3$, I'm starting with a single co-ordinate function $x^3$ and solving for two more co-ordinate functions $x^1$ and $x^2$ such that $\partial_1\cdot\partial_1 = \partial_2\cdot\partial_2$ and $\partial_1\cdot\partial_2 = 0$. | |
| Nov 30, 2014 at 1:04 | comment | added | Ali | What I am saying is precisely this: let $\Sigma_0$ be an embedded surface in M and suppose you are trying to find $\Phi$ subject to the constraint that $<d\Phi,d\Phi>_{g0}=0$ This is simply impossible | |
| Nov 30, 2014 at 0:49 | comment | added | Deane Yang | No, the equation $\partial_\nu\Phi = 0$ does not necessarily hold, because it implies that the co-ordinates $x^1$ and $x^2$ are invariant under the flow induced by the normal vector (in dimension 3). This however does not necessarily hold for the almost-isothermal co-ordinates. | |
| Nov 29, 2014 at 22:33 | comment | added | Ali | Well we have $ <d\Phi,d\Phi>_g = (\partial_{\nu} \Phi)^2 + <d\Phi,d\Phi>_{g^0}$ where $g^0$ is the induced metric on the surface... | |
| Nov 29, 2014 at 22:18 | comment | added | Deane Yang | Why is $\partial_\nu\Phi = 0$? | |
| Nov 29, 2014 at 21:47 | comment | added | Ali | Thanks you for your response. Suppose as per your suggestion I pick a foliation of 2-dimensional surfaces $\Sigma_t$ and I pick $\Phi|_{\Sigma_0}$ to be isothermal coordinates on $\Sigma_0$. Then it is clear to see that we would have $\partial_\nu \Phi = 0$ where $\nu$ is the unit normal to the surface and that would ofcourse be problematic. so I am thinking you do need the condition that Robert mentioned in his response, i.e that h be non vanishing | |
| Nov 29, 2014 at 21:18 | history | answered | Deane Yang | CC BY-SA 3.0 |