I am working on the 1972 paper A Note on 4-Dimensional Handlebodies by F. Laudenbach and V. Poénaru, and I had two questions. I will use their notations to simplify things, since the paper is very short (8 pages long) and is easy to read.

First, I am willing to prove that any diffeomorphism $\phi$ of $\#^k(S^1\times S^2)$ can be extended to a diffeomorphism of $\natural^k(S^1\times B^3)$. By making use of their Lemma 1, and by inspecting the proof, I have set $h=\beta^{-1}\phi^{-1}\alpha$, to use what they did at page 342 of the journal (the article's sixth page). If we assume that $\beta H_1 h\alpha^{-1}=\beta H_1\beta^{-1}\phi^{-1}$ is orientation-preserving, then we are done, because of the remark they make : “mark that no diffeomorphism of $X^p$ was needed here !”.

However, is it always possible to assume that we are in this setting ? More precisely, if I require $\phi$ to be orientation-preserving from the start, it only suffices to have $H_1$ to be orientation-preserving. This $H_1$ is the restriction of $H\in\text{Diff}(Y^p)$, which was obtained from Lemma 2 by surjectivity. Is it always possible to assure that this $H$ is orientation-preserving ? Or even better : Is it always possible to prescribe the orientation of $H$ (to force it to be the same as $\phi$) ?

If I understand well their proof for that Lemma 2, they are obtaining the Nielsen transformations that generate $\text{Aut}(\pi_1Y^p)$. I have worked out that

  • the permutation of two generators is obtained by permuting the corresponding handles in the decomposition of $Y^p$,
  • and the one that inverses one generator is obtained by a diffeomorphism of the corresponding handle. Edit : now that I think about it, such a diffeomorphism would be orientation-reversing (wouldn't it ?), which is problematic regarding what follows !

For the third one, it seems to me that it boils down to a handle sliding. Is that correct to say that all these elementary diffeomorphisms are orientation-preserving, and thus that we can always choose $H$ to be this way ? And for the case where we need $H$ to be orientation-reversing, can we simply use an additional elementary diffeo that is orientation-reversing on $Y^p$ but that acts trivially on the free group ?

Edit : I will comment my own question with the remark that I find it unlikely to be able to prescribe the orientedness of $H$, for the reason that in their proof, Laudenbach and Poénaru didn't use this. If that were possible, they'd just pick $H$ so that their map is orientation-preserving and they are in the favorable case all the time... So what is wrong with my arguments regarding the Nielsen transformations ?

Now, my next question is probably easy, yet I cannot seem to see it... I have read several times that the paper implies that if a (closed, oriented) 4-manifold has a handle decomposition with one 0- and 4-handles, $g$ 1- and 3-handles, and no 2-handles, then it has to be diffeomorphic to $\#^g(S^1\times S^3)$. I totally agree with the case $g=0$, which is just copy-pasting their Theorem A'. However, I cannot see how this implies this more general version of the statement... How to go from the genus zero case to the more general case ?

A related question that may help me in understand what I'm missing is how to justify another I read : “the remarkable paper [...] tells you that there is only one way to attach the 3-handles”. Why is it so ? (it also seems it is a consequence of my first question, as this document relates in section 3.5 page 13... And yet I cannot see the link)

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    $\begingroup$ In your penultimate paragraph, that closed manifold had best be diffeomorphic to a g-fold connected sum of $S^1 \times S^3$. $\endgroup$ Apr 21, 2021 at 18:31
  • $\begingroup$ @DannyRuberman You're 100% correct, that's a typo on my side, it's fixed now, thanks ! $\endgroup$ Apr 21, 2021 at 18:41


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