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Let $\pi\colon E\to M$ be a smooth $S^n$-bundle with structure group $\text{Diff}(S^n)$.

Assume there is a smooth embedding $f:E\to M \times \mathbb{R}^{n+1}$ such that $\text{pr}_1 \circ f = \pi$, where $\text{pr}_1:M\times \mathbb{R}^{n+1}\to M$ is projection to the first factor.

Question. Are there examples of this setup with $\pi:E\to M$ a nontrivial bundle? In what range of dimensions? Or does this setup somehow imply that $\pi:E\to M$ is trivial?

(If specificity helps, what about $M=S^m$ ?)

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A variation of your question has a positive answer. If you take any compact manifold that is a smooth bundle over another compact manifold $\pi : M \to N$, there is a smooth embedding

$$f : M \to N \times \mathbb R^k$$

giving a commutative diagram i.e. such that $pr_1 \circ f = \pi$, for some $k$ usually fairly large.

There's a lot of ways of proving this, but if you know about the theory of classifying spaces it follows from the observation that if $F$ is the fibre of $f$, then $BDiff(F)$$ can be thought of as the space

$$Emb(F, \mathbb R^\infty) / Diff(F)$$

The space is the limit $Emb(F, \mathbb R^\infty) = \lim_k Emb(F, \mathbb R^k)$ induced by the inclusions $\mathbb R^k \to \mathbb R^{k+1}$.

The proof is an old argument of Whitney's, basically a variation of the weak Whitney embedding theorem, that $Emb(F, \mathbb R^k)$ is highly connected provided $k$ is large. Off the top of my head I don't remember the connectivity, but it's about $\lceil \frac{k}{2} -dim(F) - 1\rceil$, i.e. the greatest integer less than or equal to the above. Making $k$ small is analogous to the question of asking the question of what is the smallest dimensional Euclidean space a manifold embeds into. You are basically asking that question "with parameters".

In your specific case the vector space is one dimension larger than your fiber sphere $S^n$. So from this point of view your question is asking about the pseudoisotopy diffeomorphism group of the sphere. You are interested in if the map from the pseudoisotopy diffeomorphism group to the vanilla diffeomorphism group can be non-trivial $PDiff(S^n) \to Diff(S^n)$. Some people would also phrase the question of if you can construct (families) of diffeomorphisms of $S^n$ that are pseudoisotopically trivial. I believe the diffeomorphisms of $S^4$ constructed by Watanabe are pseudoisotopically trivial. In his original paper he has a non-constructive argument, but he may very well have a constructive argument in follow-on papers.

https://www.math.kyoto-u.ac.jp/~tadayuki.watanabe/Diff_S4.pdf

I have a "soon to appear" paper where we write down the null pseudo-isotopies constructively, i.e. you can see them at a point-set level.

So Watanabe's construction would answer your question when $N = S^2$ and the fiber is $S^4$. But there are several papers published in the past few years that likely have analogous results for $N$ a broad range of spheres and the fiber being high-dimensional spheres. I would start maybe looking at the papers of Alexander Kupers.

edit: Regarding the reduction, given an embedding $S^n \to \mathbb R^{n+1}$, this embedding is known to bound a smoothly-embedded $D^{n+1}$ provided $n \neq 3$. This is the Schoenflies problem. When $n=3$ you still have the Jordan separation theorem, so a $3$-sphere bundle bounds a bundle whose fibers are homotopy $D^4$'s. Anyhow, so this largely reduces the question to when a diffeomorphism of $S^n$ extends to a diffeomorphism of $D^{n+1}$. For this there is the observation that $Diff(S^n)$ has the homotopy-type of $O_{n+1} \times Diff(D^n)$. Similarly $Diff(D^{n+1})$ has the homotopy-type of $O_{n+1} \times PDiff(D^n)$. It's basically the same argument. This makes the fibration $Diff(D^{n+1}) \to Diff(S^n)$ equivalent to the product map $O_{n+1} \times PDiff(D^n) \to O_{n+1} \times Diff(D^n)$ where the map $O_{n+1} \to O_{n+1}$ is the identity, and $PDiff(D^n) \to Diff(D^n)$ is the pseudoisotopy fibration (restriction to the free portion of the boundary). This argument appears in my "A family of embedding spaces" paper, but I believe this general type of argument goes back to Cerf's first papers on pseudoisotopy.

Regarding proving $BDiff(M) \simeq Emb(M, \mathbb R^{\infty}) / Diff(M)$, this is just the weak Whitney embedding theorem with parameters. We want to argue $Emb(M, \mathbb R^\infty)$ is contractible, i.e. a model for $EDiff(M)$. So take a map $S^k \to Emb(M, \mathbb R^\infty)$ and try to bound it by a map $D^{k+1} \to Emb(M, \mathbb R^\infty)$. The idea is to use the straight-line homotopy to construct such an extension, and then perturb the homotopy to be a family of embeddings.

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  • $\begingroup$ +1 Thank you so much. Could you please say a bit more about why my question is asking about the pseudoisotopy group of the sphere? I would also be grateful for any pointers to where I could learn why $BDiff(F) = Emb(F, \mathbb R^\infty) / Diff(F)$.(Btw, I think a low-tech way to prove your post's first claim, in your notation, is to start with any smooth embedding $g:M\to \mathbb{R}^k$ and then take $f=(\pi,g)$.) I will look into Kupers' papers, and I will also be interested to see yours when it appears. Your answer is very helpful, but I'll wait to accept in case a definitive one comes later. $\endgroup$ Commented Sep 6 at 21:35
  • $\begingroup$ Thanks for your follow-up edits. I think I roughly follow but need more time to digest (but e.g. the explanation re $BDiff(M)$ is clear to me, thanks!). I think my major remaining point of confusion is whether your reduction via Schoenflies essentially reduces the question to whether any nontrivial $D^{n+1}$ bundles over $M$ with structure group $Diff(D^{n+1})$ can embed in $M\times \mathbb{R}^{n+1}$? If so, I guess I'm unsure whether this new question is easy or still hard. $\endgroup$ Commented Sep 7 at 15:40
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    $\begingroup$ Apologies about that "$M$ a circle" case. I meant to talk about disc bundles over circles. So I should have said $N$ a circle, but the example doesn't lend much to the narrative so I just erased it. Regarding 2) yes, the map being non-trivial in homotopy. 3) In the homotopy-equivalence $Diff(S^n) \simeq O_{n+1} \times Diff(D^n)$, these are the diffeomorphisms of $D^n$ that are the identity on the boundary. $\endgroup$ Commented Sep 7 at 20:38
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    $\begingroup$ For every $m>0$ there are examples in which the base $M$ of the bundle is a sphere of dimension $4m$; the map $P(S^n)\to Diff(S^n)$ can be shown to be nontrivial on $\pi_{4m-1}$ for large even $n$ (or maybe for large odd $n$, I forget). The point is that by Waldhausen's work and Hatcher-Igusa stability the group $\pi_j P(S^n)$ is rationally the same as $K_{j+2}(\mathbb Z)$ for large $n$, therefore of rank $1$ when $j=4m-1$. For one parity or the other, the elements of infinite order in $\pi_{4m-1}P(S^n)$ survive in $Diff(^n)$. $\endgroup$ Commented Sep 7 at 21:47
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    $\begingroup$ This can also be explained using higher Reidemeister torsion instead of algebraic $K$-theory. $\endgroup$ Commented Sep 7 at 21:47

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