2
$\begingroup$

I have some aside computations to make in order to understand a proof in some online Morse Theory notes.

Here, let $M$ be a $n$-dimensional manifold. $TM$ and $NM$ are respectively the tangent and normal bundles of $M$.

I really do not get how the author can say that the bundle

$\Gamma:=\bigwedge^n TM\otimes_{\mathbb{R}}\bigwedge^{k}NM,$

is a trivial one. I tried to use the following identities : if $V$ and $W$ are two vector spaces over $\mathbb{R}$ then

$\wedge(V\oplus_{\mathbb{R}} W)\cong \wedge V\oplus_{\mathbb{R}}\wedge W,$

and

$\bigoplus_{k=0}^n\{\bigwedge^k V\otimes_{\mathbb{R}}\bigwedge^{n-k}W\}\cong \bigwedge^n(V\oplus_{\mathbb{R}}W).$

I do not want to pass too much time on this computation since it's just a tiny part of the proof, but I would really appreciate any help!

Thanks in advance! :-)

$\endgroup$
  • 1
    $\begingroup$ You're missing some information here. Is $k$ the codimension of $M$ inside the ambient manifold $X$? And is $X$ known to be, say, orientable? $\endgroup$ – Ted Shifrin Jul 6 '18 at 19:05
  • $\begingroup$ Sorry! Yes! $k$ is the codimension of $M$ and I assume $X$ is an oriented manifold. $\endgroup$ – DaveWasHere Jul 6 '18 at 19:10
1
$\begingroup$

Well, then you have it with the information you already knew and the information you provided. $TX|_M \cong TM\oplus NM$, and $\Lambda^{n+k}(TX|_M) \cong \Lambda^n M\otimes\Lambda^k NM$. But $\Lambda^{n+k}(TX)$ is a trivial line bundle, since $X$ is orientable, so, in particular, its restriction to $M$ is likewise trivial.

$\endgroup$
  • $\begingroup$ Oh! Yes, that was more simple than I expected. Thank you so much. :-) $\endgroup$ – DaveWasHere Jul 6 '18 at 19:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.