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Let $M$ be a smooth manifold of dimension $n$. Let $U \subset M$ be an open subset such that the restriction $TM|_U$ is trivial. Let $N$ be a closed submanifold of $M$ contained inside $U$. Does it follow then that $TM|_N$ is trivial as well?

I thought this was a trivial result (simply take $n$ linearly independent sections over $U$ and restrict to $N$), until I played around with Stiefel-Whitney classes and got a contradictory answer.

In particular take $N$ to have codimension $1$, and $N$ and $M$ to be orientable. Let $N$ be not parallelizable. Then $TM|_N \cong TN \oplus \nu$ where $\nu$ is the normal bundle of $N$ inside $M$, and $\oplus$ denotes the Whitney sum. It can be shown that since $\nu$ is a line bundle and $N$ and $M$ are orientable, $\nu$ is trivial. Hence $w_i (TM|_N) = w_i (TN \oplus \nu) = w_i(TN)$.

But $TN$ is not trivial, so not all $w_i$ are zero for $i>0$ so $TM|_N$ cannot be trivial, a contradiction.

Clearly I made a mistake somewhere, either at the beginning with the $n$ sections or when working with the Stiefel-Whitney classes, but I can't spot it. Any help in pointing out my mistake would be appreciated.

EDIT: Correction from the discussion in the comments, in the second part, $N$ should have some nonzero Stiefel-Whitney class instead of being just not parallelizable. In this case, there cannot be such an open set $U$ containing it.

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  • $\begingroup$ This is a trivial result exactly as you say. The second half of your question is not a contradiction yet! You haven't actually provided a counterexample showing that all of the conditions you just wrote down are actually satisfiable (in particular I see no mention of the $U$ on which the restriction is supposed to be trivialized). I don't understand what argument you have in mind for showing that the normal bundle is trivial either. $\endgroup$ Jan 7 '21 at 8:46
  • $\begingroup$ Ah, then that means the $N$ I set up in the second part can never be a subset of such an open set $U$? $\endgroup$ Jan 7 '21 at 8:48
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    $\begingroup$ Well, there's the additional fact that it doesn't follow that if $N$ isn't parallelizable then some Stiefel-Whitney class is nonzero; for example $S^2$ is not parallelizable but all of its Stiefel-Whitney classes vanish. But yes, if the Stiefel-Whitney classes of $TN$ are nontrivial and the normal bundle is trivial then there can be no such $U$. $\endgroup$ Jan 7 '21 at 8:49
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    $\begingroup$ Okay I see, my proof of the second part doesn't work out. Thanks. As for the fact about triviality of the line bundle, I got it from here: math.stackexchange.com/questions/1495819/… $\endgroup$ Jan 7 '21 at 8:50
  • $\begingroup$ Ah, of course, it follows from computing $w_1$. Okay, that's fine then. $\endgroup$ Jan 7 '21 at 8:52
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The first argument is correct. In addition to the discussion in the comments, a nice example to think about explicitly here is the usual codimension-$1$ embedding of $N = S^2$ into $M = \mathbb{R}^3$. They're both orientable, $N$ is not parallelizable, the normal bundle is trivial, $TM$ is trivial (so we can take $U$ to be, say, a sufficiently large open ball containing $N$, or just all of $M$), and so the pullback of $TM$ to $N$ is also trivial.

This embedding shows that the tangent bundle $TS^2$ is stably trivial (it is trivializable after adding a trivial line; the same is true for every sphere, using the embedding $S^n \hookrightarrow \mathbb{R}^{n+1}$), and in particular all of its Stiefel-Whitney classes vanish. But $TS^2$ is not trivial (by the hairy ball theorem; in terms of characteristic classes, its Euler class does not vanish, and the same is true for every even-dimensional sphere).

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  • $\begingroup$ Stably trivial tangent bundles might just be the objects I was looking for. Perhaps related, is there a general term for a codimension 1 submanifold that can be embedded in a parallelizable open subset of $M$, if this makes sense? $\endgroup$ Jan 7 '21 at 9:25
  • $\begingroup$ @Paul: I’m already not aware of a term for either of those conditions separately. $\endgroup$ Jan 7 '21 at 9:28

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