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?


  • 1
    $\begingroup$ I honestly don't think this is the way to go. Indeed, the rank of $f$ certainly is not constant even in the case $k=\dim M=2$. Just understand that $\bigwedge^{k} (\Bbb R^n)^*$ gives an $\binom nk$-dimensional subspace of $\otimes^k(\Bbb R^n)^*$, and then work with local trivializations of $T^*M$. $\endgroup$ – Ted Shifrin Jul 2 '15 at 20:53
  • $\begingroup$ @Ted: I don't understand your comment that the rank of $f$ is not constant. The image of the alternating projection is $\Lambda^k T_x^*M$ for each $x\in M$, so its rank is equal to $\binom n k$, right? $\endgroup$ – Jack Lee Jul 3 '15 at 16:32
  • $\begingroup$ @JackLee: That's what I thought, but if I consider the map $\det\colon M_{2\times 2} \to \Bbb R$, it drops rank at the zero matrix. But this is effectively the map $\otimes^2\Bbb R^2\to \Lambda^2\Bbb R^2\subset\otimes^2\Bbb R^2$. How am I being silly this time? :) $\endgroup$ – Ted Shifrin Jul 3 '15 at 19:09
  • $\begingroup$ @Ted: The determinant map is not the same as the alternating projection. For one thing, Alt is linear while det is not. If $\omega = a\,dx\otimes dx + b\, dx\otimes dy + c\, dy \otimes dx + d\, dy\otimes dy$, then $\det(\omega)=ad-bc$, while $\text{Alt}(\omega) = \tfrac12(b-c)dx\wedge dy$. $\endgroup$ – Jack Lee Jul 3 '15 at 19:17
  • $\begingroup$ Ah, duh, I was looking at the map starting with $\Lambda^1\otimes\Lambda^1$, taking $\text{Alt}(v\otimes w)$. Good thing I've retired :) Regardless, my original comment stands: Why eschew local trivializations? $\endgroup$ – Ted Shifrin Jul 3 '15 at 19:40

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 $\binom n k$.

  • $\begingroup$ Thank you professor Lee. I need to think about it for some time. $\endgroup$ – caffeinemachine Jul 3 '15 at 19:42
  • $\begingroup$ So a bundle homomorphism which is constant rank on fibres is a constant rank map. Nice. $\endgroup$ – caffeinemachine Jul 3 '15 at 22:36
  • $\begingroup$ Yes -- this is proved, for example, in Theorem 10.34 in my Introduction to Smooth Manifolds (2nd ed.). $\endgroup$ – Jack Lee Jul 3 '15 at 22:40

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