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…or rather I meant to. Something went wrong. I’d still want to, but I refrain from doing it now in order not to generate a messy flood of empty threads here
I have expanded and re-organized the material on free actions of finite groups on $n$-spheres a little, merging its subsection (now here) with the previously puny subsection on spherical space forms.
In particular, and for what it’s worth, I have made explicit (here) the example of the free action of finite subgroups of $Sp(1)$ on all $S^{4n+3}$-spheres.
added also pointer to the Lefschetz fixed point theorem and the implication (here) that therefore the only free action of a finite group on an even-dimensional sphere must be by $\mathbb{Z}/2$ and must be orientation reversing.
This almost shows that it must be the antipodal action, but still needs an argument for why there can be no other orientation-reversing action of $\mathbb{Z}/2$.
For what it’s worth, the hinted proof of Remark 2.6 should also suffice as an independent proof of Proposition 2.4:
Every free continuous $G$ action on $S^{2n}$ induces a group morphism $deg \colon G \to End_{Ab}(H_{2n}(S^{2n}))^{\times} \simeq \mathbb{Z}/2$ (where $End_{Ab}(H_{2n}(S^{2n}))^{\times}$ is the group of units of the endomorphism ring of $H_{2n}(S^{2n})$), and Lefschetz says that any element of the kernel of $deg$ has a fixed point and is thus the identity, i.e., that $ker(deg) = 1$.
As for that the nontrivial map of any such $\mathbb{Z}/2$ action is homotopic to the antipodal action, is there any simple approach that doesn’t boil down to that $\pi_{2n}(S^{2n}) \simeq H_{2n}(S^{2n})$ naturally by Hurewicz?
Thanks. True, the $\mathbb{Z}/2$-action is necessarily unique up to homotopy, but when comparing continuous actions, we’d want to classify them up to homeomorphism. But yeah, I suppose that follows. Will make an edit…
Edit: Ignore the following.
Per this paper, there exist continuous involutions of $S^{4n}$ with quotient not homeomorphic to $\mathbf{P}(\mathbb{R}, 4n)$, making such a classification apparently difficult.
Thanks, interesting. That would explain why I am getting stuck here…
But does that article give a pair of non-homomorphic involutions on an actual even-dim sphere, or just on a homotopy sphere?
Ah, their other article here is very explicit at least about non smoothly-homomorphic involutions on $S^{2n}$s. Will dig into this, thanks again for the pointer.
I have now added here a list of numbers of isomorphism classes of examples of non-standard smooth free involutions on $n$-spheres for low $n$, with pointers to the literature.
My understanding is that these all refer to the standard smooth structure on the spheres, with just the involution being non-standard. But particularly for the reference Lopez de Medrano 1971 this detail remains somewhat hidden (to me) behind the notation used there.
Re comments 10 and 11, I agree and now doubt whether my claim in comment 9 is even true. I asked a few friends to no avail and subsequently submitted the question to MO. Hopefully someone will know!
Thanks again.
The reply by I. Belegradek here seems to again be concerned with double covers by homotopy-spheres instead of actual spheres, though I haven’t yet followed the references given.
On the other hand, comparing to the the book by Lopez de Medrano, which is mostly concerned with classification of $\mathbb{Z}/2$-actions on $S^{n}$ in the piecewise-linear category (with some remarks for the smooth category but apparently no comments on the continuous category), the counting turns out at least similar to what Belegradek gives (which is maybe not surprising, just saying it for the record):
By the classification result that Lopez de Medrano previews on p. 2 (11 of 114) there are always at least four involutions up to piecewise-linear homomorphisms on the $n$-sphere for $n \gt 4$, and there is a finite number of them unless $n-3$ is divisible by 4.
At least for $n \in \{5,6,7\}$ this pattern also applies to involutions up to smooth homomorphism, as LdM discusses in Sec. V.6.1. By the discussion there, the counting up to diffeomorphism agrees with that up to pl-homeo for $S^5$ and $S^6$, but where the 7-sphere has $\mathbb{Z}/4 \oplus \mathbb{Z}$ worth of involutions up to pl-homeo, it has $\mathbb{Z}/2 \oplus \mathbb{Z}/28 \oplus \mathbb{Z}$ worth up to diffeos. (I am unsure whether this refers to non-standard involutions on the standard smooth 7-sphere, or if it also allows exotic smooth structure on the 7-sphere. On first reading/scanning of the book I thought the former, now I think maybe the latter. I wish the author had made this a little more transparent. )
It’s somewhat strange that, as far as i can see, LdM does not at least comment on the situation up-to-homeomorphism.
After racking my brain for a long while, I think I’ve found the missing ingredient: the Poincaré conjecture for topological manifolds of the appropriate dimension! After all, a topological manifold homotopy equivalent to $\mathbf{P}(\mathbb{R}, 2n)$ will have universal cover a topological manifold homotopy equivalent to $S^{2n}$, whence homeomorphic to $S^{2n}$. (The best citable reference that I could find in the $n\geq 5$ case is Theorem 7 of this paper.)
I see. That’s of course a good point.
Just to record this, I have made a brief note in the entry on what we have now regarding continuous involutions up to homeomorphism: here.
This is not meant to do justice to the topic, but just a note not to forget. Please feel invited to edit.
I am wondering about the following:
Is this example an instance of a general pattern?
Namely, the plain dihedral groups do not have free actions on spheres, but the binary dihedral groups do. Could it be true in general that for $G$ any group which does not act freely on any sphere, there is a $\widehat G \to G$ such that $\widehat G$ does have free actions on some spheres?
I have now forwarded this question (comment #19) to MathOverflow: here.
added pointer to:
(This was kindly pointed out to me by Will Sawin in MO:a/410691. While I don’t see yet how it helps with the question I was asking, it sure is a good reference to have.)
I admit that I am puzzled by MO. I still don’t see how MO:a/410691 is more than a re-statement of my question, but voters seem to think it’s about twice as insightful. What am I missing?
[edit: I see, I was missing that the main point of the argument is to show that there must be finite groups none of whose covers acts freely on some sphere.]
It seems to me that to make use of classification results for $\mathcal{P}$-groups in this context, we would first need a compatible classification of coverings of finite groups, but maybe I am missing something.
I still find the suggestion that I make in the question more concretely promising: Look for coverings that are non-trivial over all non-cyclic subgroups and of an order such that multiplying with it enforces the $2 p$ and $p^2$ condition. E.g. if the non-trivial extension is of order $2^2$ then all subgroups of order $r$ on which this extension remains non-trivial become subgroups of order $2^2 \cdot r$, which is guaranteed to satisfy both conditions.
The problem is then to guarantee that the extension is non-trivial on all non-cyclic subgroups. This is a problem in non-abelian group cohomology, which seems more likely to be solvable. Of course, my intuition may be wrong here.
Hi Urs, I’ve not thought about this at all, but regarding your puzzlement, you asked…
Maybe all finite groups can be covered by ones that act freely on some sphere?
….and the answer says that the only possible finite groups with this property are ones for which the following holds.
their Sylow subgroups must be cyclic, generalized quaternion, or dihedral
Such finite groups are certainly way less than all finite groups, so the answer to the question I quoted certainly seems to be ’no’, even if it remains open exactly how much more general such groups with a cover admitting the required action are than those which themselves have such an action (not very, I’d guess).
even if it remains open exactly how much more general such groups with a cover admitting the required action are than those which themselves have such an action
And that is, once again, the question I am asking. :-)
I had recalled the full characterization of finite groups admitting such free actions quite explicitly, its those groups that satisfy the “$p^2$-condition and the “$2p$-condition”, and I had added pointer to examples, to counter-examples, and to a case where a class of counter-examples does have covers that are examples – all recorded in some detail in this entry, group actions on spheres.
I still think it’s a reasonable question to ask how generic this last case is and I indicated a somewhat effective strategy to solve this. I thought somebody out there may have more developed thoughts on the matter, but if that’s not the case that’s okay, I just thought I give it a try.
I haven’t thought about your strategy, but the answer gives two pieces of information: a) the only difference in passing to covers is that dihedral groups are allowed as Sylow subgroups b) apparently there is a classification of exactly what groups that can happen for. Assuming that that classification is tractable, it should be a fairly trivial if tedious matter of going through all of the possibilities. But even if one doesn’t do that, it is clear that there is not a vast degree of extra generality by passing to covers, certainly far from all finite groups, which I’d guess is why people feel it is a decent answer. But I’ll bow out here :-).
I see, true, the argument gives that the property cannot hold for all finite groups.
added pointer to:
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