# If $TM$ is trivial, then $\Lambda^n(M)$ is also trivial and $M$ is orientable

Suppose that $$M$$ is a smooth $$n-$$manifold. Suppose $$TM=\coprod_{p\in M}T_pM$$ is the tangent bundle of $$M$$. And let $$\Lambda^n(M)=\coprod_{p\in M}\Lambda^n(T_pM)$$ where $$\Lambda^n(T_pM)$$ is alternating covariant $$n-$$tensor product. i.e.

$$\Lambda^n(T_pM)=T^*_pM\wedge T^*_pM \wedge \cdots \wedge T^*_pM.$$

and

$$dim(\Lambda^n(T_pM))=\begin{bmatrix} dim(T_pM)\\ n \end{bmatrix}$$

If $$TM$$ is trivial, then $$\Lambda^n(M)$$ is also trivial and $$M$$ is orientable.

I am trying to understand but seems difficult. I would be very appreciate any help. Thanks in advance.

## 1 Answer

If $$M$$ is $$n$$-dimensional, $$TM$$ is trivial is equivalent to saying that there exists $$n$$-vector fields $$X_1,...,X_n$$ such that for every $$x\in M$$, $$X_i(x)\neq 0$$. Take a differentiable metric on $$M$$ and define the $$1$$-form $$f_i(x)(u)=\langle X_i(x),u\rangle$$ where $$u\in T_xM$$, $$\Lambda^n(M)_x$$ is generated by $$f_1(x)\wedge...\wedge f_n(x)$$ which is also a volume form on $$M$$. This implies that $$M$$ is orientable.

• Thank you for your answer. But could you let me know why $\Lambda^n(M)$ is also trivial? – Lev Ban Nov 30 '18 at 17:09
• because it is a dim 1 and you have the volume form – Tsemo Aristide Nov 30 '18 at 17:21