3
$\begingroup$

Let $(M,g)$ be a Riemannian manifold. We equip the tangent bundle $TM$ with the Sasaki metric $g_s$.

A smooth vector field $X:M \to TM$ is called a transversal vector field if $X(M)$ is transverse to the zero section. We say that a vector field $X$ is "Isometric to the zero field" if there is an isometry $\phi$ of $(TM,g_s)$ which maps $X(M)$ to the zero section.

Question:Let $(M,g)$ be a compact Riemannian manifold. Is there a uniform upper bound $N$, depending only on $(M,g)$, such that every transversal vector field $X$ isometric to the zero field has at most $N$ singularities?

We denote by $n_0$, the infimum of all $N$ with the above property. This $n_0$ is a geometric invariant associated to the Riemannian manifold $(M,g)$.

If the answer of the main question is not affirmative, what about if we merely consider the space of real analytic vector fields, rather than the wider class of all smooth vector fields?

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ Do you assume that $M$ is compact? Also, are you aware of any example of a compact manifold $M$ such that isometries of $TM$ with Sasaki metric don't preserve the structure of a fibration? $\endgroup$
    – Dmitri Panov
    Commented Dec 2, 2018 at 13:21
  • 1
    $\begingroup$ @DmitriPanov I am really sorry for my delay. Yes I assume M is compact. Regarding your second question, if I understand it correctly, I guess that such an example exist. When M is not compact, say $M=\mathbb{R}$. there are isometries of $T\mathbb{R}=\mathbb{R}^2$ which does not preserves the fiber structure, the rotations of the plane. So I think that one can construct a similar example for cylinder the tangent bundle of the circle. $\endgroup$
    – Ali Taghavi
    Commented Dec 5, 2018 at 8:11

0

Reset to default

You must log in to answer this question.