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Ali Taghavi's user avatar
Ali Taghavi's user avatar
Ali Taghavi's user avatar
Ali Taghavi
  • Member for 11 years, 5 months
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Smooth action on cotangent space of the plane
@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
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Smooth action on cotangent space of the plane
and do they form a subgroup of the general linear group?
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Smooth action on cotangent space of the plane
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?
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Smooth action on cotangent space of the plane
Then ask under what conditions the dual is invariant under the dual (pull) back map
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Smooth action on cotangent space of the plane
$K^\circ=\{\zeta \in T^*K\mid \zeta(v)\leq 1 \forall v\in T^1_p K, \forall p\in K$ where $T^1$ is the unit or disk tangent bundle of K.
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Smooth action on cotangent space of the plane
@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
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Smooth action on cotangent space of the plane
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?
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Smooth action on cotangent space of the plane
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
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Smooth action on cotangent space of the plane
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 $K^\circ$?
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Smooth action on cotangent space of the plane
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
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Smooth action on cotangent space of the plane
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?
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Smooth action on cotangent space of the plane
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?
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Smooth action on cotangent space of the plane
What is a proof for the case of $K=$ ball
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orbits in locally compact group
Thank you very much for this very interesting answer.
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