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

Question

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 $M$.

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 06 '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 06 '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 06 '17 at 18:40

Tsemo Aristide · Answer 1 · 2017-02-06T18:18:17.483

0

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.

edited Feb 06 '17 at 18:18

answered Feb 06 '17 at 17:14

Tsemo Aristide

87,475

1

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 06 '17 at 17:35

score 0 · Answer 2 · answered Jun 30 '23 at 06:59

A $p$-form $\alpha$ can be seen as a section of the vector bundle $(\Lambda^pTM)^*$ because each $\alpha_x$ is an alternate $p$-linear map $T_xM \rightarrow \mathbb{R}$, or equivalently a linear map $\Lambda^pT_xM \rightarrow \mathbb{R}$ i.e. an element of $(\Lambda^pT_xM)$. Therefore, a volume form is a positive section of $(\Lambda^nTM)^* = \det(TM)^*$ where $n = \dim(M)$.

If $\det(TM)$ is trivial, so is $\det(TM)^*$ and since the trivial bundle admits a non vanishing smooth section, so does $\det(TM)^*$, which means that $M$ is orientable.

Reciprocally, if $M$ is orientable, then $\det(TM)^*$ admits a non vanishing section $\omega$, which is a trivialisation because each $(\omega_x)$ is a basis of $\det(TM)^*$ because it is a line bundle. Therefore, $\det(TM)$ is trivial too.

score 0 · Answer 3 · answered Jun 30 '23 at 07:53

The determinant line bundle is trivial means than there exist a nowhere vanishing $n$-linear alternate form, says $\omega$. Choose a riemanian metrics. If $M$ is orientable, we can speak of direct orthonormal base at each point. The determinant in a direct o. base is then a nowhere vanishing form, and $\omega$ is constructed. Conversely if the determinant line bundle is trivial and $\omega$ a nowhere vanishing section . Then we have an orientation by saying that a frame is direct if the value of the $\omega$ on this frame is positive.

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

3 Answers3