I am reading the book "An Introduction to Contact Topology" by Geiges.

In the proof of Lemma 1.1 where he proves that any co-dimension 1 hyperplane field distribution is locally kernel of a 1-form, he obtains the line bundle $TM/\xi=\xi^\perp$ and states that $\xi^\perp$ is locally trivial. I can not see why this is true, why is $\xi^\perp$ locally trivial? Thank you for any help.

Here $M$ is the manifold, $TM$ its tangent bundle, $\xi$ is the hyperplane field decomposition and $\xi^\perp$ is the orthogonal complement of $\xi$ with respect to a Riemannian metric on M.


Let $x\in M$ you can construct a vector field $X$ on a neighborhood $U$ of $x$ such that $X(x)$ is not $\xi_x$ (suppose that $U$ is contractible, $TM_{\mid U}$ is trivial, take any $u\in T_xM$ which is not in $\xi_x$, $y\rightarrow(y,v)$ is $X$). This implies that there exists a neighborhood $V\subset U$ such that $X(y)$ is not an element of $\xi_y$. There exists an isomorphism $V\times\mathbb{R}\rightarrow (TM/\xi)_V$ defined by $(y,u)\rightarrow p(uX(y))$ where $p:TM\rightarrow TM/\xi$ is the quotient map.

The results follows from the fact that $TM/\xi$ is isomorphic to $\xi^{\perp}$.

  • $\begingroup$ Thanks @Tsemo Aristide . $\endgroup$ – selfmathish Jan 24 at 15:56

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.