It is known that the Haefliger trefoil $S^3\hookrightarrow S^6$ is PL trivial but non-trivial smoothly. I wonder, where exactly does the problem come? Consider its tubular neighborhood $T\cong S^3\times D^3$. By the disc bundle theorem its closed complement $C$ is diffeomorphic to $S^2\times D^4$. Therefore, $S^6$ is presented as a union of $S^3\times D^3$ and $S^2\times D^4$ along some non-trivial diffeomorphism $\phi$ of $S^3\times S^2$. But it follows from the smoothing theory (Burghelea-Lashof) that for any closed 5-fold $M$, $\pi\_0 Diff(M)=\pi\_0 PL(M)$, because $\pi\_i PL\_5/O\_5 = \pi\_i PL/O=0$, $i\leq 6$. In our case, as example, since $S^3\times S^2$ is parallelizable, one has a fiber sequence
$$
Diff(S^3\times S^2) \to PL (S^3\times S^2) \to Map(S^3\times S^2,PL\_5/O\_5).
$$
Therefore, the non-trivial self-diffeomorphism of $S^2\times S^3$ is also non-trivial as PL self-homeomorphism. But why we then get a non-trivial knot in the smooth category but not in PL? I thought maybe the problem is that $\phi$ is PL concordant to the identity but is not so smoothly. But these sets of concordance classes are also the same $\pi\_0\widetilde{Diff}(S^3\times S^2)=
\pi\_0\widetilde{PL}(S^3\times S^2)$ thanks to the fiber sequence
$$
\widetilde{Diff}(S^3\times S^2) \to \widetilde{PL} (S^3\times S^2) \to Map(S^3\times S^2,PL/O).
$$
So, can the smooth non-triviality of the Haefliger trefoil be detected in this decomposition? I am puzzled because it also almost looks like a proof that the Haefliger trefoil is smoothly trivial.



Haefliger trefoil $S^3\hookrightarrow S^6$
The problem comes from the fact that $S^3\subset S^6$ admits more $PL$-framings than smooth ones: $\pi\_3 SO\_3=\mathbb{Z}$, while $\pi\_3 PL\_{6,3}=\mathbb{Z}^2$. (The group $PL\_{6,3}$ is the group of PL self homeomorphisms of $\mathbb{R}^6$ preserving $\mathbb{R}^3\subset\mathbb{R}^6$ pointwise.) Thus, when we PL isotop the Haefliger trefoil identically to the trivial knot, and we lift this isotopy to its tubular neighborhood $T=S^3\times D^3$, the obtained $PL$-framing of the trivial knot is not smoothable i.e., it does not come from an $SO\_3$-framing. In both smooth and PL categories the space of embeddings $T\hookrightarrow S^6$ that send $S^3\times \*$ identically to itself is homotopy equivalent to the space of such immersions, while the latter space by Smale-Hirsch-Haefliger-Poenaru is equivalent to $Map(S^3,SO\_3)$ and $Map(S^3,PL\_{6,3})$, respectively.


