Does there exist a smooth embedding $\varphi\colon S^k\to S^n$ such that $\varphi(S^k)$ has non-trivial normal bundle? I looked at some of the old papers by Kervaire, Haefliger, Massey, Levine but I couldn't find an answer. There are various related results, for example it is shown by Kervaire et al that in many dimensions the normal bundle is trivial. Furthermore Kervaire showed that there exists an immersion, such that the pullback of the normal bundle is non-trivial.
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$\begingroup$ If k<2(n-k)-1 then the normal bundle is trivial by a result of Kervaire (Theorem 8.2 in his "An Interpretation of G. Whitehead's Generalization of H. Hopf's Invariant"). If n=k+1,k+2,k+3 it is trivial as well. $\endgroup$– archipelagoJan 24, 2021 at 12:16
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1$\begingroup$ Using this, one sees that all embeddings with $k\le 6$ have trivial normal bundle. I believe it is an open question whether there is an embedding with nontrivial normal bundle for k=7 and n=11. $\endgroup$– archipelagoJan 24, 2021 at 12:22
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$\begingroup$ Do you know if Kervaire's example $\nu$ is stably trivial? Maybe I'm oversimplifying, but if you have an immersion $f : S^k \to S^n$ with $k < n$, then it factorises as $i_n\circ f_0$ where $i_n : \mathbb{R}^n \to S^n$ is the standard inclusion and $f_0 : S^k \to \mathbb{R}^n$. The map $h : S^k \to S^{n+k+1}$ given by $i_{n+k+1}\circ(f_0, \iota)$ is an embedding where $\iota: S^k \to \mathbb{R}^{k+1}$ is the standard embedding. If I'm not mistaken, the normal bundle of $h$ is $\nu\oplus\varepsilon^{k+1}$ which is trivial if and only if $\nu$ is stably trivial. $\endgroup$– Michael AlbaneseJan 24, 2021 at 12:49
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2$\begingroup$ @MichaelAlbanese Any normal bundle arising from such an embedding is stably trivial, since the tangent bundle of $S^k$ is. $\endgroup$– archipelagoJan 24, 2021 at 12:51
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$\begingroup$ According to Hsiang, Lee, and Szczarba, it is an unpublished result of Haefliger that $S^{11}$ embedds into $S^{17}$ with trivial normal bundle (see 'On the normal bundle of a homotopy sphere embedded in Euclidean space'). This would show that Kervaire's bound is sharp. $\endgroup$– archipelagoJan 24, 2021 at 12:54
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