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 $M\times N$ )



Proper action on product manifold
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](https://mathoverflow.net/questions/424165/how-to-prove-the-product-of-whitehead-manifold-and-mathbbr-is-homeomorphic)), hence, $d(X\times {\mathbb R})=4$.


The same example works in the smooth category.




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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**](http://dx.doi.org/10.1016/0040-9383(88)90005-5), Topology 27, No. 1, 27-35 (1988). [ZBL0658.57007](https://zbmath.org/?q=an:0658.57007).


(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 $c(X) 
 + c({\mathbb R})=1$.


