Timeline for Heisenberg group: function without vertical derivative
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| May 21, 2015 at 20:44 | comment | added | Martin Hairer | Writing $\Delta$ for the horizontal Laplacian, you could use $(1-\Delta)^{-\alpha} \xi$, where $\xi$ is a realisation of white noise on $\mathbb{H}$ and $\alpha$ is suitable. (You probably need $\alpha > 3/2$ to guarantee that you have one horizontal derivative and $\alpha < 2$ to make sure you don't create a vertical derivative, but I haven't checked the details, these exponents are just based on scaling.) | |
| May 21, 2015 at 12:57 | history | edited | Nikita Evseev | CC BY-SA 3.0 |
typo
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| May 15, 2015 at 11:29 | history | edited | Nikita Evseev | CC BY-SA 3.0 |
added 222 characters in body
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| May 4, 2015 at 8:51 | comment | added | Nikita Evseev | about preserving Lie bracket math.stackexchange.com/q/365535/23566 | |
| May 4, 2015 at 7:02 | comment | added | TaQ | @Nate Eldredge: You are obviously right. I was too sloppy in my comment above. If we write $g^{-1}$ as $(x,y,t)\mapsto(u,v,w)$ , then we get the requirements that $(u_x,v_x,w_x)=(1,0,-\frac 12\,y)$ and $(u_y,v_y,w_y)=(0,1,\frac 12\,x)$ and $(u_t,v_t,w_t)=(0,0,1)$ must hold. So in particular $w_x=-\frac 12\,y$ and $w_y=\frac 12\,x$ must hold but this gives $-\frac 12=\frac 12$ at least in the distributional sense which is impossible. So such a diffeomorphism $g$ cannot exist even locally. | |
| May 4, 2015 at 6:55 | comment | added | Nate Eldredge | @TaQ: Won't any diffeomorphism $g$ preserve the Lie bracket? So $g$ can't pull back $\partial_x, \partial_y$ to $X,Y$ since the former commute and the latter do not. I think this is true even if $g$ is only $C^1$. | |
| May 4, 2015 at 6:08 | comment | added | TaQ | Since you are not imposing any continuity condition on $f$ , you may even take $h(t)=0$ for $t\in\mathbb Q$ and $h(t)=1$ for $t\in\mathbb R\setminus\mathbb Q$ . So the only problem is the construction of a diffeomorphism $g$ . | |
| S May 4, 2015 at 6:04 | history | suggested | TaQ | CC BY-SA 3.0 |
typo corrected vertival --> vertical
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| May 4, 2015 at 6:01 | review | Suggested edits | |||
| S May 4, 2015 at 6:04 | |||||
| May 4, 2015 at 5:55 | comment | added | TaQ | If you can find a $C^1$ diffeomorphism $g:U\to V$ pulling $\partial_x,\partial_y,\partial_t$ back to $X,Y,T$, taking as $h:\mathbb R\to\mathbb R$ any continuous nowhere differentiable function, then you are done with $f=h\circ p\circ g$ where $p$ is given by $(x,y,t)\mapsto t$ . Since $X,Y,T$ are locally linearly independent, at least locally this can be done around every point. | |
| May 4, 2015 at 5:28 | history | edited | Nikita Evseev | CC BY-SA 3.0 |
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| May 4, 2015 at 5:20 | comment | added | Nikita Evseev | @MikeBenfield You are right. It is (if one can say) nonsmooth analysis on Carnot group. And we assume that functions could have only first derivatives (of derivatives in weak sense). | |
| May 3, 2015 at 22:10 | comment | added | Michael Benfield | It doesn't seem relevant that you are working in the Heisenberg group. Since $T = [X,Y]$, any function that is twice differentiable for $X$ and $Y$ will also be differentiable for $T$. If the function you're seeking does exist, it will not be very regular. | |
| May 2, 2015 at 19:06 | history | edited | Nikita Evseev | CC BY-SA 3.0 |
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| May 2, 2015 at 18:36 | history | asked | Nikita Evseev | CC BY-SA 3.0 |