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.

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


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

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.


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.

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

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