2
$\begingroup$

Let $S^{2n+1}$ be the $m$-dimensional sphere in $\mathbb{C}^{n+1}$. Endow $S^{2n+1}$ with the standard metric. Let $S^1$ act by multiplication on $S^{2n+1}$. Then $S^1$ and the canonical action of $SU(n+1)$ are by isometries. Since the actions of $S^1$ and $S^{2n+1}$ commute, $G = SU(n+1) \times S^1$ acts by isometries on $S^{2n+1}$. Lifting the action to $TS^{2n+1}$ we get, that this action is hamiltonian.

In their paper "New examples of manifolds with completely integrable geodesic flows", Paternain and Spatzier say, that $G$ acts on $TS^{2n+1}$ multiplicity-free/coisotropic.

That means:

1) Their exists an $\operatorname{Ad}^*_G$-equivariant momentum map $$\Phi \colon TS^{2n+1} \to \mathfrak{g}^*$$ 2) For $\alpha \in \mathfrak{g}^*$ the isotropygroup $G_\alpha$ acts transitively on the connected components of $\Phi^{-1}(\alpha)$.

I'm not sure how to show now, that this action is really multiplicity-free/coisotropic, because I don't know how to calculate the momentum map and the preimage effectively.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

1) Since $S^{2n+1}$ is a Riemannian manifold one can identify $TS^{2n+1}$ with the cotangent bundle $T^*S^{2n+1}$. The latter is well-known to carry a canonical symplectic structure and a momentum map.

2) This condition can be rephrased as $\Phi/G:T^*S^{2n+1}/G\to\mathfrak{g}^*/G$ being finite to one. Let $G_0\cong SU(n)\times S^1$ be the isotropy group of a point $x\in S^{2n+1}$. Then $T^*S^{2n+1}/G=T^*_xS^{2n+1}/G_0=V/G_0$ where $V=\mathbb{C}^n\oplus \mathbb{R}$. Hence $V/G^0=\mathbb{R}_{\ge0}\times \mathbb{R}$. Now the first factor is mapped to the quotient of $SU(n+1)$ while the second to $(Lie S^1)^*$. This shows that $\Phi/G$ is finite to one.

Remark: More generally, the cotangent bundle $T^*X$ of a homogeneous space $X=G/H$ is the fiber product $G\times^H\mathfrak{h}^\perp$ where $\mathfrak{h}^\perp\subseteq\mathfrak{g}^*$ is the annihilator and the moment map is $[g,\xi]\mapsto Ad^*(g)\xi$. So multiplicity-freeness means that $\mathfrak{h}^\perp/H\to\mathfrak{g}^*/G$ is finite to one.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Thank you very much for your help. What do you mean by $\Phi/G$ is a finite to one map. Furthermore I'm not quite sure, what you mean by "the first factor is mapped to the quotient of $SU(n+1)$". $\endgroup$
    – Olorin
    Apr 12, 2016 at 13:41
  • $\begingroup$ So we have also, that if $G$ acts multiplicity-free on $T^*X$ then it acts multiplicity-free on $TX$? $\endgroup$
    – Olorin
    Apr 12, 2016 at 15:21
  • $\begingroup$ With "finite to one" I mean a map with finite fibers. $\endgroup$
    – Friedrich Knop
    Apr 12, 2016 at 16:03
  • $\begingroup$ If $L: TX \to T^*X$ is the identification given by the riemannian metric, I don't see why the canonical lifted action of $G$ on $T^*X$ should be the same as the induced action given by the identification of the lifted action of $G$ on $TX$. SInce a priori this actions are different, multiplicity-freeness of one of these shouldn't imply multiplicity-freeness of the other one. Or am I mistaken? $\endgroup$
    – Olorin
    Apr 12, 2016 at 18:16
  • 1
    $\begingroup$ First, the actions are the same since $G$ preserves the Riemannian metric. Second, there is no genuine momentum map on $TX$ since a tangent bundle does not have a natural symplectic structure. It is only via $L$ that there is a momentum map on $TX$. $\endgroup$
    – Friedrich Knop
    Apr 12, 2016 at 21:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.