Let $\mathfrak{S}_\mathbb{N}$ be the symmetric group of all positive integers. Let $\ell^\infty(\mathbb{N})^*$ be the dual space of $\ell^\infty(\mathbb{N})$ equipped with weak*-topology. There is a natural group action of $\mathfrak{S}_\mathbb{N}$ on $\ell^\infty(\mathbb{N})^*$, that is to define $$\sigma(u)\big(x_i\big):=u\big((x_{\sigma(i)})\big)$$ for any $\sigma\in\mathfrak{S}_\mathbb{N}$, $u\in\ell^\infty(\mathbb{N})^*$ and $(x_i)\in \ell^\infty(\mathbb{N})$.

Now let's equip $\mathfrak{S}_\mathbb{N}$ with the permutation topology, i.e. the topology generated by the basis of neighborhood at identity element in form of $V(F)=\left\{\sigma\in\mathfrak{S}_\mathbb{N}|\sigma(n)=n\text{ for all }n\in F\right\}$, where $F$ is a finite subset of $\mathbb{N}$.

My question: is there a chance that the group action defined as above will be continuous under the setting of permutation topology?

My bad idea is to start with $\ell^1(\mathbb{N})\subset \ell^\infty(\mathbb{N})^*$ and the permutation $\sigma$ acts on $\ell^1(\mathbb{N})$ as usual. If a permutation $\sigma$ sends a closer position to a very far one (as modelling $\sigma$ not in a neighborhood of identity), then $\sigma(f)-f$ differs a lot in a sufficiently far position, and this will yield a great difference after applying an element in $\ell^\infty(\mathbb{N})=\ell^1(\mathbb{N})^*$. But this does not provide a rigorous proof, besides one knows that $\ell^\infty(\mathbb{N})^*=\ell^1(\mathbb{N})\oplus c^\perp_0(\mathbb{N})$.

So I wonder if anyone can give a proof or disproof for it?


1 Answer 1


No, it's not continuous.

Indeed, fix a non-principal ultrafilter $U$ supported by the set of even numbers, and define $m\in\ell^\infty(\mathbf{N})^*$ by $m(f)=\lim_{n\to U}f(n)$.

Now define $\tau_n$ as the transposition $(2n,2n+1)$ and $s_n=\prod_{k\ge n}\tau_k$. Then $s_n$ tends to the identity map for the permutation topology.

I claim that $s_n\cdot m$ does not tend to $m$. Indeed, let $f$ be the indicator function of the set of even numbers. It is enough to see that $(s_n\cdot m)(f)$ does not tend to $m(f)$. Indeed, $m(f)=1$, while $ (s_n\cdot m)(f)=m(s_n^{-1}\cdot f)=0$ since $s_n\cdot f$ is eventually supported by odd numbers.


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