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Let $M$ be a manifold and $V$ be an oriented vector bundle. It's well known that if the Euler class of $V$ is non zero, then $V$ can't have a non-vanishing section. The converse is not true, see Vanishing of Euler class , or the answer by John Klein below, given to the first version of this question. So:

Question. Recall that the normal bundle of a smooth orieantable submanifold in $\mathbb R^n$ always has zero Euler class. Does such a normal bundle always has a non-vanishing section? If yes, how to prove this? If no, what is a counterexample?

PS. It turns out that this question is a subquestion of the following much more informed one: Embeddings without nonvanishing normal vector fields

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    $\begingroup$ Possible duplicate of Vanishing of Euler class $\endgroup$
    – Najib Idrissi
    Commented Dec 4, 2018 at 12:06
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    $\begingroup$ @LSpice Image of the section would be another copy of the manifold not intersecting the original. But I agree the title could be better $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Dec 4, 2018 at 17:24
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    $\begingroup$ I think the title is fine; indeed they are the same question. The question feels like it should have a counterexample but I don't know one. Of course it is true in codimension 2. $\endgroup$
    – mme
    Commented Dec 4, 2018 at 18:20
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    $\begingroup$ Are you implicitly assuming that the submanifold is orientable, so that its normal bundle is orientable? $\endgroup$
    – Mark Grant
    Commented Dec 4, 2018 at 20:27
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    $\begingroup$ For anyone wondering why the normal bundle has Euler class zero (as I was), this is Corollary $11.4$ of Characteristic Classes by Milnor & Stasheff. $\endgroup$
    – Michael Albanese
    Commented Jun 3, 2023 at 23:52

2 Answers 2

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Here's a counter example. Take any embedding of $\mathbb{C} P^2$ in $\mathbb{R}^7$ (such embeddings exist by

Steer, B., On the embedding of projective spaces in Euclidean space, Proc. Lond. Math. Soc., III. Ser. 21, 489-501 (1970). ZBL0206.25501;

the construction is summarised in this answer on MSE).

If such an embedding admits a normal vector field, then by the Compression Theorem of Rourke and Sanderson it is isotopic to an embedding with normal field parallel to the last coordinate of $\mathbb{R}^7=\mathbb{R}^6\times\mathbb{R}$, and then the projection is an immersion $\mathbb{C} P^2\looparrowright \mathbb{R}^6$. Such immersions cannot exist, by András Szűcs' answer here.

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  • $\begingroup$ Thanks a lot Mark, that's really cute. May I ask you to comment on the answer of Andras? He writes that if there exists an immersion of $\mathbb CP^2$ to $\mathbb R^6$ then "the normal Euler class has the property, that its square is the normal Pontrjagin class, and that is $−3$ times the signature". I am rusty with Pontriagin classes so can't understand this line... $\endgroup$
    – aglearner
    Commented Dec 5, 2018 at 10:09
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    $\begingroup$ @aglearner: Thanks! Andras is using 3 facts about Pontrjagin classes, all of which can be found in Chapter 15 of Milnor and Stasheff: The square of the Euler class of a $2k$ dimensional bundle equals $p_k$; the Pontragin classes satisfy a sum formula $p(E\oplus F) = p(E)p(F)$ modulo $2$-torsion; and $p_1(\mathbb{C}P^2) = 3a^2$, where $a\in H^2(\mathbb{C} P^2)$ is the generator. $\endgroup$
    – Mark Grant
    Commented Dec 5, 2018 at 11:40
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    $\begingroup$ (The sum formula implies that $p_1(\nu)=-3a^2$ where $\nu$ is the hypothetical normal bundle which has $\nu\oplus T\mathbb{C} P^2$ a trivial bundle.) $\endgroup$
    – Mark Grant
    Commented Dec 5, 2018 at 11:44
  • $\begingroup$ Thanks again, Mark, really happy to learn all this! $\endgroup$
    – aglearner
    Commented Dec 5, 2018 at 13:18
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There are lots of counter-examples.

Here's one: The fibration $\text{SO}(3) \to \text{SO}(4) \to S^3$ splits, since it is the principal bundle of the tangent bundle of $S^3$, and the latter is parallelizable. In particular, $\pi_5(\text{SO}(4)) \cong \pi_5(\text{SO}(3)) \oplus \pi_5(S^3) \cong \Bbb Z/2 \oplus \Bbb Z/2$ .

Let $a$ and $b$ denote the generators for the summands. Then $(a,b)$ determines a rank $4$-vector bundle over $S^6$ with trivial Euler class (since $H^4(S^6) = 0$). But this vector bundle does not have a section, for if it did that would lead to the contradiction that $b=0$.

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  • $\begingroup$ Thank you John. I edited the question, because I do want to know the answer to a part of the original question, which was not covered in the previous question on mathoverflow. $\endgroup$
    – aglearner
    Commented Dec 4, 2018 at 12:24
  • $\begingroup$ @aglearner, is your question about a submanifold in an arbitrary manifold or a submanifold of $\mathbb{R}^n$? $\endgroup$
    – Deane Yang
    Commented Dec 4, 2018 at 21:05
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    $\begingroup$ Dean, my original question was more general, and John immediately gave a counterexample, which illustrates that vanishing of the Euler class doesn't guarantee existence of a non-vanishing section. However, since I really want the answer to my question above (about submanifolds of $\mathbb R^n$), I decided to modify the original question. So now John's answer doesn't answer the current question, but it nicely illustrates it. $\endgroup$
    – aglearner
    Commented Dec 4, 2018 at 21:29

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