# $M$ is orientable $\Leftrightarrow$ determinant bundle $\det(TM)$ is trivial

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 $M$.

• Notice that you did not ask any question in what you wrote. – Mariano Suárez-Álvarez Feb 6 '17 at 17:10
• You are correct. I edited my message. – un umile appassionato Feb 6 '17 at 17:32
• 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... – Mariano Suárez-Álvarez Feb 6 '17 at 18:21
• Then this means there exists a volume form on $M$, and therefore $M$ is orientable. I guess the converse is also true. – un umile appassionato Feb 6 '17 at 18:32
• 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. – 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.
• 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. – un umile appassionato Feb 6 '17 at 17:35