5
$\begingroup$

Let $\pi:E\to M$ be a rank $k$ vector bundle over the compact manifold $M$ and let $i:M\hookrightarrow E$ denote the zero-section. Then we have a splitting of the restriction of $TE$ to the zero-section (see here): $$TE|_M\cong TM\oplus E.$$

Now suppose that we are given metrics $g_{TM}$ on the bundle $TM$ and $g_E$ on the bundle $E$. By the above splitting this yields a metric $g^i$ on $TE|_M$.

Question: Can we extend the metric $g^i$ to a global metric on $TE$?
That is, is there a metric $g$ on the bundle $TE$ such that its restriction $g|_M$ to the zero section coincides with $g^i$?

Thanks for any insights.

$\endgroup$
6
  • $\begingroup$ Take a tubular neighborhood $U$ of $M$ in $E$. Extend the metric to $TE|_U$ by ignoring the extra parameters in $U$. Put whatever metric you like on $E-M$, and use a partition of unity refining $\{U, E - M\}$ to glue them up. $\endgroup$
    – user27126
    Apr 12 '13 at 6:55
  • $\begingroup$ @Sanchez: Thanks for your comment. I don't really get how you extend the metric to $TE|_U$, could you please elaborate on that point? I might be wrong but it seems to me that you are implicitly assuming that $TE|_U$ still splits as $TM\oplus E$ in order to construct this extension to $TE|_U$. $\endgroup$
    – Dave
    Apr 12 '13 at 7:14
  • $\begingroup$ Ah, good point - for some reason I keep thinking the normal bundle is trivial. Anyway, suppose that we can find some open sets $U_i$ of $E$, covering $M$, such that $U_i$ is really diffeomorphic to the product of $U_i \cap M$ and $\mathbb{R}^n$. Then we win since we can first trivially extend the metric on $M \cap U_i$ to a metric on $U_i$, and then use a partition of unity subordinate to $\{U_i, E - M\}$ to glue the metric together... $\endgroup$
    – user27126
    Apr 13 '13 at 2:46
  • $\begingroup$ The existence of such open sets is not difficult, since $M$ is a submanifold of $E$ so locally there are coordinate patches $U$ of $E$ that is diffeomorphic to $R^n$ and $U \cap M$ is diffeomorphic to $R^k$ under the same map. Adjust the coordinate patch to make them contractible, and vector bundle over contractible space is trivial. $\endgroup$
    – user27126
    Apr 13 '13 at 2:47
  • $\begingroup$ @Sanchez: Ah, very nice, thanks a lot! One could even simplify your argument a bit by just taking a (finite) trivializing (for $E$) open cover $\{V_i\}_i$ of $M$, i.e. $\pi^{-1}(V_i)\cong V_i\times \mathbb{R}^k$. Then extend the metric on $V_i\times \{0\}$ trivially to $V_i\times \mathbb{R}^k$ and glue it together via a partition of unity subordinate to the open cover $\{\pi^{-1}(V_i)\}_i$ of $E$. $\endgroup$
    – Dave
    Apr 13 '13 at 14:14

Your Answer

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

Browse other questions tagged or ask your own question.