# Orientation of a Manifold with Trivial Tangential Bundle

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.

• Use the criterion in terms of a volume form. – Moishe Kohan Jan 30 at 3:44
• @MoisheCohen: Hi, do you mean the criterion that if $\Lambda^n TM$ (= volume form) is trivial then $M$ orientable? Or do you mean the other one? By the way: o you maybe see a more "elementary" argument (just using the positive determant criterion for transition maps) to verify that $M$ is orientable if $TM$ is trivial. Maybe considering the local structure of charts of $TM$? – KarlPeter Jan 30 at 18:01
• Yes, this is what I mean. The point is that if $E\to B$ is a trivial vector bundle, so are its exterior powers (and other tensor powers of course), hence, there is really nothing to be proven. – Moishe Kohan Jan 30 at 18:09
• @MoisheCohen: hmmm I'm just a bit curious if there is a more "elementary" argument using only the characterisation for orientability with transition charts as above. Maybe in some kind by analysing the local stucture of transition maps between charts and their differentials (triangle structure) for $TM$ using the triviality assumption. Do you know if it could work? – KarlPeter Jan 30 at 18:17
• @MoisheCohen: This argument with vanishing volume form is essentially the same as the overkill with Stiefel Whitney classes (if we accept that De Rham and simplicial cohomology provide the same theory then the argument essentially the same and taking into account the concrete construction for De Rham cohomology). Do you see some not too "high tech" argument? – KarlPeter Jan 30 at 18:22

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.

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.

• Ah ok so the point is that for every chart $\phi_j: U_j \to V_j$ on the whole $U_j$ we have only positive or negative $\det\left[(D_p \phi_j) \circ f_p\right]$ and modyfying all $\phi_j \circ f$ with $\det\left[(D_p \phi_j) \circ f_p\right] <0$ by reflection map provides the desired result? – KarlPeter Feb 2 at 15:01
• @KarlPeter: Yes, exactly. And by the chain rule, $D_{\phi_j(p)}(\phi_i\circ \phi_j)^{-1} = \left[(D_{p}\phi_i)\circ f_{p}\right] \circ \left[(D_{p}\phi_j)\circ f_{p}\right]^{-1}$, which has positive determinant. This directly shows that the resulting atlas satisfies your specifically requested criterion. – Matthew Kvalheim Feb 2 at 22:33