7
$\begingroup$

Let $S^n \times \mathbb{T}^n$ be the trivial Torus bundle over $S^n$. Assume that we have a continuous fiber preserving map $\phi :TS^n \to S^n \times \mathbb{T}^n$ which restriction to each fiber gives us a covering space structure.

Does this imply that $n=1,3,7$?

Example: Existence of such structure gives us a covering space structure $\alpha: TS^n \to S^n \times \mathbb{T}^n$. Then a lifting process gives us a non vanishing section for the tangent bundle of sphere. Hence for $n=2k$ such structure can not exist.

More generally assume that $E$ and $X$ are $n$ dimensional vector and (not necessarily trivial) torus bundles over a topological space $Y$, respectively. What type of obstructions would appear for existence of a fiber preserving covering map $\alpha :E\to X$?

Added: According to comment of Mark Grant we ask :"Is there a nontrivial vector bundle $E$ and torus bundle $X$ of the same dimension over a space $Y$ which admit a fiber preserving map $\alpha: E \to X$ whose restriction to each fiber is a covering map?

Improve this question
$\endgroup$
10
  • 1
    $\begingroup$ Is this equivalent to reduction of the structure group from $GL_n (\mathbb{R})$ to $GL_n (\mathbb{Z}) $? If so then the answer to the sphere question is yes. $\endgroup$
    – Mark Grant
    Commented Oct 11, 2017 at 12:25
  • 5
    $\begingroup$ I was thinking about the clutching functions $S^{n-1}\to GL_n(\mathbb{R})$ of the tangent bundles, and somehow this would seem to say that they preserve a lattice. Also, torus bundles clearly have something to do with principal $GL_n(\mathbb{Z})$-bundles, since $\operatorname{haut}(T^n)=GL_n(\mathbb{Z})$. I haven't yet been able to make this argument precise. $\endgroup$
    – Mark Grant
    Commented Oct 11, 2017 at 15:36
  • 3
    $\begingroup$ Is it true that the spheres with trivial tangent bundle are the only ones with trivial tangent microbundle? $\endgroup$
    – Tom Goodwillie
    Commented Oct 13, 2017 at 21:16
  • 1
    $\begingroup$ Here are my thoughts. The tangent bundle has a zero section, so let's assume that the fibers are based. The tangent bundle as a vector bundle is classified by a map $S^n \to BGL_n(\Bbb R)$, but as a (based) bundle is classified by a map $S^n \to BHomeo_*(\Bbb R^n)$. Asking for it to be equivalent to the trivial torus bundle is asking for a lift $S^n \to \{*\} \to BHomeo_*(\Bbb T^n) \to BHomeo_*(\Bbb R^n)$ (where the second map classifies "fiberwise universal cover"). As a result I think that this is asking when the tangent bundle is a trivial $\Bbb R^n$-bundle. $\endgroup$
    – Tyler Lawson
    Commented Oct 13, 2017 at 22:17
  • 2
    $\begingroup$ $BHomeo_\ast(\mathbb R^n)$ is a classifying space for microbundles (the Kister-Mazur Theorem). $\endgroup$
    – Tom Goodwillie
    Commented Oct 14, 2017 at 0:00

0

Reset to default

You must log in to answer this question.