Timeline for The only rotation fields satisfying this PDE are constant
Current License: CC BY-SA 4.0
13 events
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| Mar 22, 2021 at 14:51 | comment | added | Willie Wong | $\nabla R$ is certainly not generic in $Hom(\mathbb{R}^2, \mathbb{R}^2)$. You cannot just forget that $R$ takes values on a one dimensional manifold: as a $2\times 2$ matrix $\nabla R$ must be rank 1. | |
| Mar 21, 2021 at 8:43 | comment | added | Asaf Shachar | by passing from $\mathrm{d} R=0$ to $\text{div}(R)=0$; part of the complication is that the $\text{div}$ operator actually acts on the larger space $\text{Hom}(\mathbb{R}^2,\mathbb{R}^2)$. Of course, writing $R=\begin{pmatrix} c & -s \\\ s & c \end{pmatrix}$ one can see that if $\text{div}(R)=0$, then $c,s$ satisfy the CR equations, hence are constant, but In general $\mathrm{d} R$ contains more information than $\text{div}(R)$. | |
| Mar 21, 2021 at 8:43 | comment | added | Asaf Shachar | Thanks, your last comment is very interesting; let's write $ R=\exp(r g)$. I agree that $(\mathrm{d} R ) R^{-1}= \mathrm{d}r\otimes g$, so if $\mathrm{d} R=0$, then $\mathrm{d}r=0$. However as you say, here we don't exactly have $\mathrm{d} R=0$ but $\text{div}(R)=\text{tr}(\nabla R)=0$, which involves some contraction of the total derivative $\nabla R$--which considers $R$ as a map $\Omega \to \text{Hom}(\mathbb{R}^2,\mathbb{R}^2)$. (while $\mathrm{d} R$ views $R$ as a map $\Omega \to \text{SO}(2)$). I wonder whether there is a nice way to see that we don't lose information | |
| Mar 18, 2021 at 17:05 | history | edited | Willie Wong | CC BY-SA 4.0 |
just to make sure that we are always using the right "type" of vectors.
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| Mar 18, 2021 at 17:04 | comment | added | Willie Wong | Re 2up: that was my mistake. $\frac{d}{dx} e^{ix} = i e^{ix}$, I dropped the $i$. // Re 1up: general fact about Lie groups. Let $\mathfrak{G}$ be a lie group and $\mathfrak{g}$ its Lie algebra. Fix $g\in \mathfrak{g}$ and given a function $r: \Omega\to \mathbb{R}$, then you have $(\mathrm{d} \exp(r g) ) \exp(-rg)= \mathrm{d}r\otimes g$. For divergence, you contract, and so $g$ acts on $dr$ by multiplication (on the left or on the right, by your choice). | |
| Mar 18, 2021 at 16:53 | history | edited | Willie Wong | CC BY-SA 4.0 |
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| Mar 18, 2021 at 16:51 | vote | accept | Asaf Shachar | ||
| Mar 18, 2021 at 16:51 | comment | added | Asaf Shachar | Thank you for the comments and clarifications. If I did not make any mistake, then assuming your convention, I got that if $R=e^{ir}$, then $(\text{div}(R))R^{-1}=-(\partial_x r) dy+(\partial_y r) dx$, which is not exactly your claimed $(\text{div}(R))R^{-1}=dr$, but gives the same conclusion. (Quite embarrassingly, I am not fluent in abstract index notation, so I really just tried to do everything here by hand; that should be entirely doable, since we are in dimension $2$ after all...). | |
| Mar 18, 2021 at 16:06 | history | edited | Willie Wong | CC BY-SA 4.0 |
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| Mar 18, 2021 at 16:00 | comment | added | Willie Wong | Basically, if you $f:\Omega\to \mathbb{R}$ and $F = exp(if)$ where $i = (0, -1; 1,0)$, you have that the tensor $\nabla_a F_{bc} = \nabla_a f (i F)_{bc} = \nabla_a f (F i)_{bc}$. And so if you want to contract $ac$ instead of $ab$, you can multiply on the left by $(F^T)_{db}$. | |
| Mar 18, 2021 at 15:56 | comment | added | Willie Wong | (I thought when you say "acting row by row" you meant taking $\partial_x$ of first row, and $\partial_y$ of second row...) | |
| Mar 18, 2021 at 15:53 | comment | added | Willie Wong | My convention is that $\mathrm{div} M = (\partial_x, \partial_y)^T M$. This corresponds to $\nabla^i M_{ij}$. If after the divergence you end up with a column vector, you are acting on the right, so you should flip all the right multiplications to left multiplication... so where I conjugated on the right with the reflection by $R$ you should conjugate on the left instead. | |
| Mar 18, 2021 at 14:34 | history | answered | Willie Wong | CC BY-SA 4.0 |