Let $M$ be a differentiable manifold and $TM$ be its tangent bundle. I need to prove the following:

$M$ is orientable if and only if $\det(TM)$ is trivial.

The definition of determinant bundle I'm using is the following:

Given a vector bundle $E$ over $M$ with transition functions $g_{\alpha\beta}$, then the determinant vector bundle $\det(TM)$ over $M$ is the line vector bundle whose transition functions are $\det(g_{\alpha\beta})$.

I get the orientability and $\det(TM)$ are closely related since the transition functions of $\det(TM)$ are just the determinant of the Jacobian matrix of the change of chart for an atlas of $M$.

  • 1
    $\begingroup$ Notice that you did not ask any question in what you wrote. $\endgroup$ – Mariano Suárez-Álvarez Feb 6 '17 at 17:10
  • $\begingroup$ You are correct. I edited my message. $\endgroup$ – un umile appassionato Feb 6 '17 at 17:32
  • 1
    $\begingroup$ If $n=dim M$, a section of the dual of the determinant bundle of TM is precisely the same thing as an $n$-form. If that determinant bundle is trivial, then it has a nonzero section and... $\endgroup$ – Mariano Suárez-Álvarez Feb 6 '17 at 18:21
  • $\begingroup$ Then this means there exists a volume form on $M$, and therefore $M$ is orientable. I guess the converse is also true. $\endgroup$ – un umile appassionato Feb 6 '17 at 18:32
  • 1
    $\begingroup$ Of course, you have to make very precise the claim I made that a section of the dual of det(TM) is an n-form. $\endgroup$ – Mariano Suárez-Álvarez Feb 6 '17 at 18:40

the manifold is orientable if and only if you can suppose that $det(g_{\alpha\beta})>0$. In this case $det(TM)$ has a $R^+$-reduction, since the maximal compact subgroup of $R^+$ is trivial, you deduce that $det(TM)$ is trivial.

  • $\begingroup$ I don't seem to be able to find the definition of $R^+$-deduction. Could you please expand a little? I only have been introduced basic concepts about vector bundles. $\endgroup$ – un umile appassionato Feb 6 '17 at 17:35

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.