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

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


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

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! :-)

  • 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

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.

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

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