Orientation of a Manifold with Trivial Tangential Bundle

Question

Let $M$ be a smooth (eg $C^{\infty}$) manifold. Let assume that $M$ has trivial, oriented tangent bundle $TM$, so $TM \cong M \times \mathbb{R}^n$ for appropriate $n$ and orientable.

How to conclude that in this case $M$ is also orientable?

My considerations:

I use the orientability criterion with charts: eg a manifold $N$ is orientable iff there exists an atlas $(\phi_i:U_i \to V_i)_{i \in I}$ with $U_i \subset N, V_i \subset \mathbb{R}^n$ with following property:

for all $i,j \in I$ with $U_i \cap U_j \neq \varnothing$ the differential $D_p(\phi_j \circ \phi_j^{-1}) =T_p(\phi_j \circ \phi_j^{-1})$ of restricted map $\phi_j \circ \phi_j^{-1} \vert _{U_i \cap U_j}$ at every $p \in U_i \cap U_j $ has positive determinant.

An attempt is to start with product charts for the oriented (by assumption) tangent bundle of the shape $\phi_i \times id_{\mathbb{R}^n}$ which form an oriented atlas for $TM \cong M \times \mathbb{R}^n$ and trying to restrict them to charts $\phi_i$ is a sophisticated way (modyfying them by multiplying if neccessary with $\pm 1$ in appropriate cases ) to get an induces oriented atlas on $M$. But I'm not sure how and if that could work. Does anybody have a better idea?

Remark: Since every tangent bundle of a manifold is oriented I think that the triviality of $TM$ is the main ingredient here.

Answer 1

Hint: $$ (dx_1 \wedge ... \wedge dx_n)(v_1,...,v_n)= det(V), $$ where $V$ is the matrix whose column vectors are $v_1,...,v_n$. This simply comes from the definition of the wedge product $$ u_1\wedge ... \wedge u_n $$ as the alternation of the tensor product $u_1\otimes ... \otimes u_n$: $$ u_1\wedge ... \wedge u_n= Alt(u_1\otimes ... \otimes u_n)= \sum_{\sigma} sign(\sigma) u_{\sigma(1)} \otimes ... \otimes u_{\sigma(n)} $$ where the sum is taken over all permutations $\sigma$ of $\{1,...,n\}$. You also use the definition of pairing of covariant and contravariant tensors: $$ dx_{\sigma(1)}\otimes ... \otimes dx_{\sigma(n)}(v_1,...,v_n)= (dx_{\sigma(1)}(v_1))...(dx_{\sigma(n)}(v_n))= v_{{\sigma(1)}1}\cdots v_{{\sigma(n)}n}.$$

If $A$ is a (linear) endomorphism of $R^n$, then $$ (A^*(dx_1 \wedge ... \wedge dx_n))(v_1,...,v_n)= dx_1 \wedge ... \wedge dx_n(w_1,...,w_n), $$ where $w_i=A v_i, i=1,...,n$. Hence, if $W$ is the matrix with column-vectors $w_1,...,w_n$ then $det(W)= det(A) det(V)$.

Edit. For those who did not read the exchange in the comments section, here are relevant details. Since $TM$ is a trivial bundle, so is its $n$-th exterior power, $\wedge^n TM$. In particular, $\wedge^n TM$ admits a nonvanishing section, a volume form $\eta$. This is all what is needed to construct an orientation-preserving atlas on $M$.

As a general remark, if $E\to B$ is a rank $n$ vector bundle then $\wedge^n E\to B$ is naturally isomorphic to the determinant bundle $det(E)$, i.e. the rank one bundle over $B$ whose transition maps (between the fibers) are the determinants of the transition maps of the original bundle. This is what the above hint is about. Applying this to the bundle $E=TM$, we obtain that if $\eta$ is a degree $n$ form on $M$ and $\eta_i=\nu_i(x)dx_1\wedge ... \wedge dx_n$ are the expressions of $\eta$ in local coordinate charts $\phi_i: U_i\to M$, then for any two charts $\phi_i, \phi_j$ the local expressions are related by the formula $$ \eta_i= f_{ij}^* \eta_j, f_{ij}= \phi_j^{-1}\circ \phi_i, $$ or $$ \nu_i= det(D f_{ij}) \nu_j. $$ In particular, if $\eta$ and the charts are chosen so that $\nu_i(x)>0$ for all $x$ and all $i$ (i.e. the forms $\eta_i$ are "positive") then the transition maps $f_{ij}$ are orientation-preserving, i.e. $M$ is oriented. Assuming that charts are chosen so that $U_i\cap f_{ij}^{-1}(U_j)$ is connected for each $i, j$ and the form $\eta$ is a volume form (i.e. is nowhere zero), we obtain that the charts $\phi_i$ can be "corrected" (by composing them with reflections in $R^n$ if necessary) so that each $\eta_i$ is positive. Thus, if $M$ admits a volume form then it is orientable in the sense that it admits an orientation atlas.

Answer 2

Your assumption of a trivialization $f: M\times \Bbb R^n \to TM$ yields, for each $p\in M$, a linear isomorphism $f_p:\Bbb R^n \to T_pM$ given by the composition of $f|_{\{p\}\times \Bbb R^n}$ with the identification $\{p\}\times \Bbb R^n \approx \Bbb R^n$.

Now let $\cal A$ be the maximal atlas for $M$. For any chart $(\phi_j,U_j) \in \cal A$, say that $\phi_j$ is "good" if for all $p\in U_j$, $$\det\left[(D_p \phi_j) \circ f_p\right] > 0,$$ and that $\phi_j$ is "bad" otherwise. Now remove all of the "bad" charts from $\cal A$ to get a collection $\cal{B}$. Since any good chart can be obtained from a bad chart (and vice versa) by multiplying one of the coordinate functions by $-1$, $\cal{B}$ is an (nonmaximal) atlas.

You can check that the atlas $\cal{B}$ satisfies your specified orientation-preserving criterion (use the chain rule).

Edit (more details): for any pair of "good" charts, by the chain rule it follows that $$D_{\phi_j(p)}(\phi_i\circ \phi_j)^{-1} = \left[(D_{p}\phi_i)\circ f_{p}\right] \circ \left[(D_{p}\phi_j)\circ f_{p}\right]^{-1}, $$ which has positive determinant as desired.