1
$\begingroup$

Assuming that $S^1$ acts on $\mathbb{R}^2$ by smooth maps (which are diffeomorphisms), the induced action on the cotangent bundle given by $$g\cdot(x,\xi)=(g\cdot x,\varphi^∗_{g^{−1}}\xi)$$ acts via symplectomorphisms (where $x\in \mathbb{R}^2$, $\xi\in T^∗_𝑥\mathbb{R}^2$, and $\varphi_g$ is the map which acts as $x↦g\cdot x$). If the action on the base preserves a convex region $K\subset \mathbb{R}^2$, containing the origin and centrally symmetric ($x\in K$ iff $-x\in K$), is it true that the action on the cotangent space preserves $K\times K^{\circ}$ where $K^{\circ}$ is the dual convex body of $K$, i.e. $$K^{\circ}:=\{y\in\mathbb{R}^2\mid \langle x,y\rangle\leq 1, \forall x \in K\}.$$

This is certainly true for $K$ being the ball, but I have not been able to draw either a positive or a negative answer in general. I would very much appreciate any help or suggestion.

Improve this question
$\endgroup$
24
  • 1
    $\begingroup$ What is a proof for the case of $K=$ ball $\endgroup$
    – Ali Taghavi
    Commented Oct 1, 2023 at 8:01
  • 1
    $\begingroup$ 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 $K\times K^{\circle}$ . $\endgroup$
    – kvicente
    Commented Oct 1, 2023 at 8:09
  • 1
    $\begingroup$ 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. $\endgroup$
    – kvicente
    Commented Oct 1, 2023 at 8:40
  • 1
    $\begingroup$ 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 $\endgroup$
    – kvicente
    Commented Oct 1, 2023 at 8:42
  • 2
    $\begingroup$ 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 $K^0$. $\endgroup$
    – Tobias Diez
    Commented Oct 2, 2023 at 4:33

0

Reset to default

You must log in to answer this question.