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Suppose that $\mathbb{R}^n$ is the maximal group that can act properly on a manifold $N$ and $\mathbb{R}^m$ is the maximal group that can act properly on a manifold $M$ ( i.e, $\mathbb{R}^{m+1}$ can't act properly on $M$ ).

Question: is $\mathbb{R}^{n+m}$ the maximal group that can act properly on the manifold $M\times N$ ? ( i.e $\mathbb{R}^{n+m+1}$ can't acy properly on $M\times N$ )

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    $\begingroup$ Good question, but I'm not sure about the formulation...is there a "maximal group that can act properly" on a manifold? Do you mean to say the maximal $n$ for which $\mathbb{R}^n$ can act properly? $\endgroup$
    – Mark Grant
    Jan 10, 2023 at 13:09
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    $\begingroup$ To add to Mark's comment: for example if $M=N$, then $\mathbb{R}^n\wr (\mathbb{Z}/2\mathbb{Z})=(\mathbb{R}^n\times \mathbb{R}^n)\rtimes \mathbb{Z}/2\mathbb{Z}$ acts properly on $N\times N$. $\endgroup$
    – Alex B.
    Jan 10, 2023 at 14:50
  • $\begingroup$ @MarkGrant yes, I mean the maximal $n$ for which $\mathbb{R}^n$ act properly on $N$. $\endgroup$ Jan 10, 2023 at 17:33
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    $\begingroup$ Just for completeness, in such questions you should specify if you are working in topological or smooth category. But for this particular question, it does not matter since the answer is negative in both. $\endgroup$ Jan 14, 2023 at 13:58

1 Answer 1

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First, let's formulate the question properly:

Given a topological space $X$, define be $$ d(X):=\sup \{ n: X~ \hbox{is homeomorphic to} ~Y\times {\mathbb R}^n\}. $$

Lemma. The following quantities are equal when $X$ is a manifold:

(1) $d(X)$

(2) P(X):=$\max\{n: \exists ~ \hbox{a principal}~ {\mathbb R}^n\hbox{-bundle with the total space}~X\}$

(3) p(X):=$\max\{n: \exists ~ \hbox{a proper}~ {\mathbb R}^n\hbox{-action on }~X\}$, where ${\mathbb R}^n$ is equipped with the standard topology.

Proof. A principal ${\mathbb R^n}$-bundle with the total space $X$ is the same thing as a proper ${\mathbb R^n}$-action on $X$. At the same time, each principal ${\mathbb R^n}$-bundle with the total space $X$ is trivial. qed.

Remark. One can avoid using this lemma in order to justify the example below, just I find the definition $d(X)$ cleaner.

Thus, working in the topological category, you are asking if there are manifolds $X, Y$ such that $d(X\times Y)> d(X)+ d(Y)$.

Now, here is an example: Let $X$ be the Whitehead manifold (or any other contractible 3-manifold not homeomorphic to ${\mathbb R}^3$). Then $d(X)=0$. (This is not entirely trivial, but for the purpose of a counter-example we just need to know that $p(X)<3$ which is obvious since $X$ is not homeomorphic to ${\mathbb R}^3$.) On the other hand, $X\times {\mathbb R}$ is homeomorphic to ${\mathbb R}^4$ (see for instance here), hence, $d(X\times {\mathbb R})=4$.

The same example works in the smooth category.


Update. It appears that you are now interested in proper ${\mathbb Z}^n$-actions. The answer in this setting is the same. You similarly define the invariant $c(X)$, detecting the highest rank of a discrete free abelian group acting properly on $X$. Then, let $X$ again be the Whitehead manifold. It turns out that $c(X)=0$. This is a nontrivial result of Bob Myers:

Myers, Robert, Contractible open 3-manifolds which are not covering spaces, Topology 27, No. 1, 27-35 (1988). ZBL0658.57007.

(One can also appeal to an older theorem of Waldhausen, but it only proves that $c(X)\le 2$, which suffices for our purposes but is suboptimal.)

At the same time, $c(X\times {\mathbb R})=4$, but $c(X) + c({\mathbb R})=1$.

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  • $\begingroup$ So by using you notation, is $d(X)$ equal to the maximum integer such that $\mathbb{R}^{d(X)}$ act properly on X? $\endgroup$ Jan 14, 2023 at 16:36
  • $\begingroup$ @Youshitz: My notation is correct, your's is sloppy. $\endgroup$ Jan 14, 2023 at 17:27
  • $\begingroup$ @Youshitz: Sorry, cannot do this. $\endgroup$ Jan 15, 2023 at 16:13
  • $\begingroup$ I think the Lemma is not true, (2) and (3) are not necessarily equal, could you please tell me how did prove that? $\endgroup$ Jan 16, 2023 at 16:46
  • $\begingroup$ @Youshitz see my answer here: math.stackexchange.com/questions/560371/… $\endgroup$ Jan 16, 2023 at 17:19

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