Timeline for Smooth action on cotangent space of the plane
Current License: CC BY-SA 4.0
26 events
| when toggle format | what | by | license | comment | |
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| Oct 2, 2023 at 14:27 | comment | added | Ali Taghavi | mathoverflow.net/questions/451247/… | |
| Oct 2, 2023 at 14:27 | comment | added | Ali Taghavi | Any way your question has a beautiful philosophy when certain property of f on M carries to the same property for f^* on the cotangent bundle. on can ask some questions of this type in for example Ergodic theory. Assume f is ergodic is the pull back an ergodic map too? please see the following: | |
| Oct 2, 2023 at 13:32 | comment | added | Ali Taghavi | In the other word what is a map (time one map of flow) on the plane which does not come from a circle action? | |
| Oct 2, 2023 at 12:18 | comment | added | Ali Taghavi | @kvicente the isotopy you are talking about works for both R action and circle action so I mean what is the raeson (and difference) for emphasis on circle action | |
| Oct 2, 2023 at 11:30 | comment | added | kvicente | and to get the invariance I'm pretty sure is basically, the equivalent to this setitng, of @TobiasDiez condition on the differential. | |
| Oct 2, 2023 at 11:29 | comment | added | kvicente | @AliTaghavi Diffeomorphisms coming from a circle action are isotopic (by diffeoemorphisms) to the identity, I'm not sure if in the plane every diffeoemorphism is isotopic to the identity, or said otherwise, the group $Diff(\mathbb{R}^2)$ is connected. Try for example to connect $(x,y)\to(-x,y)$ to the identity via diffeoemorphisms. The emphasis of the circle action is basically that instead of just one diffeomorphisms I have an $S^1$-family of diffeomorphisms. Now, this $K^{\circ}$ that you define in the Riemannian sense is no other that the disk cotangent bundle of K, by definition. | |
| Oct 2, 2023 at 10:27 | comment | added | Ali Taghavi | and do they form a subgroup of the general linear group? | |
| Oct 2, 2023 at 10:24 | comment | added | Ali Taghavi | regarding the first question I guess that $f$ must have a chaotic dynamics or it must be a finite order diffeomorphisms. BTW can one characterize all linear isomorphisms of the Euclidean space whose dynamics comes from a circle action? | |
| Oct 2, 2023 at 10:22 | comment | added | Ali Taghavi | Then ask under what conditions the dual is invariant under the dual (pull) back map | |
| Oct 2, 2023 at 10:19 | comment | added | Ali Taghavi | $K^\circ=\{\zeta \in T^*K\mid \zeta(v)\leq 1 \forall v\in T^1_p K, \forall p\in K$ where $T^1$ is the unit or disk tangent bundle of K. | |
| Oct 2, 2023 at 10:19 | comment | added | Ali Taghavi | @kvicente Any way the question was interesting however the differential can destroy what you desire. But your question is motivating some other question: First what is a diffeomorphism of the plane not coming from a circle action because I understood you emphasis o circle action? Another question which I am thinking to(as a generalization of materials you pointed out to) is to nonlinearize the concept of dual: Let $M$ be a Riemannian manifild and $K$ is a submanifold of $M$ then one may define | |
| Oct 2, 2023 at 7:53 | comment | added | kvicente | I think you are right @TobiasDiez, thanks for your comment it as really helpful! | |
| Oct 2, 2023 at 4:33 | comment | added | Tobias Diez | By invariance of the pairing, this is true if the derivative of the action, viewed as $\phi'_g: \mathbb{R}^2 \to \mathbb{R}^2$ preserves $K$. In particular, it works for linear actions (generalizing your example of the rotation action). If I'm not mistaken, the condition on $\phi'$ is also necessary: if there is a $z \in K$ such that $\phi'_g z$ is not again in $K$ for some $g$, then choose $y \in K^0$ with $y (\phi'_g z) = 2$. But then $\phi^*_g y$ has value $2$ at $z$, showing that it is not an element of $K^0$. | |
| Oct 1, 2023 at 9:02 | comment | added | kvicente | Let us continue this discussion in chat. | |
| Oct 1, 2023 at 9:02 | history | edited | kvicente | CC BY-SA 4.0 |
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| Oct 1, 2023 at 8:58 | comment | added | Ali Taghavi | To have a true understanding of your previous 2 comments I need to confirm that by $K^\circ$ you mean $\{y\mid y.x\leq 1 \forall x\in K$ yes? | |
| Oct 1, 2023 at 8:42 | comment | added | kvicente | Answering your first question, the role of the $S^1$-action is crucial for applying some other symplectic geometry arguments, replacing it with a $\mathbb{Z}_p$ action would render those other arguments useless. But your question is a good start, is a necessary condition for my question to be positive | |
| Oct 1, 2023 at 8:40 | comment | added | kvicente | I don't think that such a diffeomorphism with large differential at the origin will clearly disrupt the dual of $K$ as you say, the thing about symplectic maps is that, heuristically, large deformations that you perform on the basis ($x_1,x_2$ coordinates), are compensated by the an inversally proportional deformation in the fiber ($y_1,y_2$ coordinates), so as the global deformation balances itself, so if we still preserve the basis, I don't see why we could not preserve the fiber. I need more convincing beyond hand-waving arguments with that approach. | |
| Oct 1, 2023 at 8:33 | comment | added | Ali Taghavi | So I personally guess that the answer is not affirmative. In the rotation example you considered the differential is not big. But in general the invariance of D does not implies that the action has a small differentiation | |
| Oct 1, 2023 at 8:29 | comment | added | Ali Taghavi | Is not possible a diffeomorphism of $\mathbb{R}^2 $ keeps the disk invariant but has a very larg differential at the origin then disrupt the polar dual equation then not keep invariant $K^\circ$? | |
| Oct 1, 2023 at 8:26 | comment | added | Ali Taghavi | I think a standard notatio for $f^*$ is $f^*(x,y)=(f^{-1}(x), df(f^{-1}(x)).y$ slightly different from what you wrote in group action case | |
| Oct 1, 2023 at 8:24 | comment | added | Ali Taghavi | So for a given diffeomorphism of $f:\mathbb{R}^2 \to \mathbb{R}^2 $ assume that $f(D)=D$ can one say that $f^*:$ maps $D\times D^\circ$ to itself? | |
| Oct 1, 2023 at 8:22 | comment | added | Ali Taghavi | What is the roll of $S^1$ in your question? I mean that is it sensitive wrt circle? can we restate the question for $\mathbb{Z}$ action? | |
| Oct 1, 2023 at 8:09 | comment | added | kvicente | Maybe I should have been a little bit more specific, thanks for the question. In the ball, if you take the standard circle action given by rotation of angle $t$, this will lift to the cotangent bundle as a rotation too by angle $t$ (this is a linear action and to dualize it we take the inverse of the transpose). Given that the dual of the ball is the ball itself we have that the action preserves $K\times K^{\circle}$ . | |
| Oct 1, 2023 at 8:01 | comment | added | Ali Taghavi | What is a proof for the case of $K=$ ball | |
| Sep 27, 2023 at 19:00 | history | asked | kvicente | CC BY-SA 4.0 |