4
$\begingroup$

Define the map $$P:TS^{n}\to S^{n} \;\;\;\text{by}\;\; P((x,v))=\frac{x+v}{\parallel x+v \parallel}$$ where $$TS^{n}=\{(x,v)\in S^{n} \ \times \mathbb{R}^{n+1}\mid v \perp x \}$$

This map is used in the book of Alain Hatcher, Algebraic topology, to give a proof for the fact that every vector field on even spheres must vanish at some point of the sphere.

Question:

  1. Does $P$ define a (nontrivial) fiber bundle?

  2. Define the Hamiltonian function $H:TS^{n} \to \mathbb{R}$ with $H(x,v)={\parallel P(x,v)-x \parallel}^{2} $ where the latter norm is the standard Euclidean norm on $\mathbb{R}^{n+1}$. What can be said about the dynamical behavior of the corresponding Hamiltonian vector field $X_{H}$? Are there any periodic orbit?

  3. Assume that $V$ is a vector field on the sphere. To $V$, we associate a self map $f(x)=P(x,V(x))$ on the sphere. Are there some relations between the continuous dynamics of $V$ and the discrete dynamics of $f$. Note that $V$ and $f$ have the same fixed points.
$\endgroup$
  • 2
    $\begingroup$ $P$ is (up to a mild re-parametrization) the restriction of the exponential map for the tangent bundle of $S^n$. So 1, yes. $H$ is basically a re-scaled length of $v$, that answers (2). (3) can similarly be answered. $\endgroup$ – Ryan Budney Oct 31 '16 at 23:17
  • 1
    $\begingroup$ @RyanBudney $H$ is not a re-scaled length of $v$, since the lenght of $v$ is an unbounded function but $H$ is a bounded function! Could you please elaborate your comment. I think your comments is not clear. $\endgroup$ – Ali Taghavi Nov 1 '16 at 10:51
  • 1
    $\begingroup$ @RyanBudney Do you mean that $X_{H}$ equal geodesic flow vector field up to a constant multiplier? Do you mean that $H$ is globally equal to $\parallel v \parallel ^2$ up to a constant?How can (3) be answered immediately? $\endgroup$ – Ali Taghavi Nov 1 '16 at 15:34
  • 2
    $\begingroup$ $P$ is pretty much precisely the exponential map if you re-scale the input vector by a non-linear function of its length. i.e. re-scale the input vector's length by $L \longmapsto 1/\sqrt{1+L^2}$. Exponentiate that vector, this gives you $P$. So your function $H$ is a similar non-linear rescaling of the length squared of the input vector. $\endgroup$ – Ryan Budney Nov 1 '16 at 16:37
  • 1
    $\begingroup$ @RyanBudney But why does this imply that $X_{H}$ has the same dynamic as the geodesic flow?Moreover, what about the third part of my question? $\endgroup$ – Ali Taghavi Nov 1 '16 at 19:22
3
$\begingroup$

There is a wonderful trick - I think promulgated by Moser - for viewing Hamiltonian flows on the cotangent bundle of the sphere as the reduction of a Hamiltonian flow on the ambient phase space of (x,p)'s , i.e on ${\mathbb R}^{n+1} \times{\mathbb R}^{n+1}$ and which shows that your flow is in fact a geodesic flow running at twice the speed. View the sphere as the space of rays through the origin, so orbits of the ${\mathbb R}^+$ action $x \mapsto \lambda x$. The cotangent lift of this action to the ambient phase space is $(x,p) \mapsto (\lambda x, p/\lambda)$. The momentum map for this scaling action is $J(x,p) = \langle x, p \rangle$. The reduced space at $J =0$ is canonically the cotangent bundle of the sphere. A Hamiltonian H for the cotangent bundle of the sphere becomes then a function on the ambient phase space which is homogeneous of degree 0 -- i.e invariant -- with respect to this scaling. Example: the Hamiltonian for geodesic flow on the sphere is $H_{geod} = (1/2)|x|^2 |p|^2$. Your Hamiltonian, made homogeneous of degree 0 , is $H = |(x/|x|- |x|p)|^2 = (1 - 2<x,p> + |x|^2|p|^2)$. But we must fix $J = 0$. So $H = 1 + 2 H_{geod}$ -- your flow is geodesic flow, running at twice the speed.

$\endgroup$
  • $\begingroup$ Prof. Montgomery Thank you very much for your answer and your attention to my question. I am sorry for my delay. May be I am missing some thing but the Hamiltonian $H$ which I defined is a bounded function but the geodesic Hamiltonian is not. So, how they differ by a constant? $\endgroup$ – Ali Taghavi Dec 13 '16 at 12:57
  • $\begingroup$ Ali , what symplectic form did you want to use on the tangent bundlewould you write down, in some coordinates, what your symplectic form is, or what your vector field is? The reason I ask -- $TS^n$ does not have an intrinsic symplectic form, but $T^* S^n$ does. I was assuming you were using the standard Riemannian metric to identify the tangent and cotangent bundles in your question. $\endgroup$ – Richard Montgomery Dec 17 '16 at 1:49
  • $\begingroup$ sorry, edit system cut me off; 1) you have a point, about bounded/unboundednss. (2) what is your symplectic form, or your Ham vector field on the tangent bundle of the sphere? $\endgroup$ – Richard Montgomery Dec 17 '16 at 2:09
  • $\begingroup$ As you said, the sympletcic structure is coming from the cotangent bundle via isomorphism between the tangent and cotangent bundle via inner product inheriting from $\mathbb{R}^{n+1}$. But before going to the Hamiltonian vector field, please consider your equation $H=1+2H_{\text{geod}}$. The left part is a globally bounded function but the right part is not. How can they be equal? $\endgroup$ – Ali Taghavi Dec 17 '16 at 7:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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