I am trying to prove the following:
Let $M$ be a smooth manifold. Then $\bigwedge^k T^*M$ is a smooth subbundle of dimension $\binom{n}{k}$ of $\bigotimes^kT^*M$.
To do this, I think the following theorem will be helpful:
Theorem: The image and kernel of a constant rank smooth map between smooth vector bundles over a manifold $M$ are both subbundles of their ambient bundles.
So a natural thing to try is to define a map $f:\bigotimes^k T^*M\to \bigotimes^k T^*M$ as $f(\omega)=\text{alt}(\omega)$ for all $\omega\in \bigotimes^k T^*M$, where $$ \text{Alt}(\omega)=\frac{1}{k!}\sum_{\sigma\in S_k}\text{sgn(}\sigma){^\sigma}\omega $$
The image of $f$ is $\bigwedge^k T^*M$.
But I am unable to see that the rank of $f$ is constant.
Can somebody help?
Thanks.
I guess I should turn my comment into an answer. As the OP noted, the image of the map $f$ is exactly $\Lambda^k T^*M$. In other words, for each $x\in M$, we have $f\big(\bigotimes^k T^*_xM\big) = \Lambda^k T^*_kM$, which is a vector space of dimension $\binom n k$. Thus $f$ has constant rank equal to $\binom n k$.