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Let $G$ be a discrete group. I am wondering if there is a recipe which can be applied to find elements in the group von Neumann algebra which are not absolutely summable, i.e. $T \in VN(G)$ while $T \notin \ell^1(G)$?

PS Once I heard that for $G=\Bbb{Z}$, $(a_n)_{n \in \Bbb{Z}}$, where $a_n=1/n$ for $n \neq 0$ and $a_0=0$, is such an element. Is this a particular example of a general result on abelian discrete groups? Or it is proved only for $\Bbb{Z}$?

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The example in your PS is probably related to the Fourier series of the "sawtooth function" $f : [-\pi, \pi) \to {\bf R}$, $f(t)=t/\pi$, when we regard $f$ as an element of $L^\infty({\bf T})\cong {\rm VN}({\bf Z})$. I don't recall the precise formula for the Fourier series, but it certainly gives something which does not belong to $\ell^1({\bf Z})$.

More generally: an integrable function $h$ on ${\bf T}$ whose Fourier series belongs to $\ell^1({\bf Z})$ must be equal a.e. to a continuous function, and so any element of $L^\infty({\bf T})\setminus C({\bf T})$ would also suffice. It is worth noting that there exist functions in the disc algebra (i.e. continuous on the closed unit disc and analytic in the interior) whose Taylor series do not converge absolutely on the unit circle, and these give elements of $C({\bf T})\setminus {\ell^1}({\bf Z})$ (where by abuse of notation I am identifying $\ell^1({\bf Z})$ with a space of functions on the circle).

Anyway, regarding your original question: if $G$ contains an element of infinite order then this gives an inclusion of groups ${\bf Z} \to G$ and hence (since we are dealing with discrete groups) an inclusion of von Neumann algebras ${\rm VN}({\bf Z}) \to {\rm VN}(G)$, so you can embed the counterexample for ${\rm VN}({\bf Z})$ to get a counterexample for ${\rm VN}(G)$.

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  • $\begingroup$ Thanks for this nice observation. So I assume that a machinery which can be applied to any (even abelian) group to construct such functions is not known to you. $\endgroup$
    – Mahmood Al
    Commented Jul 24, 2016 at 20:45
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    $\begingroup$ Hi Mahmood. Well, for ${\bf Z}^k$ I guess that one could take some kind of tensor product of the sawtooth examples? Another possibility might be some kind of random-signs argument: that is, you choose a sequence in $\ell^2(G)$ but not in $\ell^1(G)$, and then you multiply each coefficient by a $\pm 1$ random variable. I have a vague memory that if the original sequence is sparse then with high probability this random construction gives an element of ${\rm VN}(G)$ $\endgroup$
    – Yemon Choi
    Commented Jul 24, 2016 at 21:45
  • $\begingroup$ The later sounds really interesting, although I have no idea what you mean by "high probability". Which realm of Mathematics implies such a result? Where I can learn more about it? $\endgroup$
    – Mahmood Al
    Commented Jul 25, 2016 at 7:15
  • $\begingroup$ The "probabilistic method" for showing existence of certain examples (but not constructing them) is usually associated with combinatorics, e.g. en.wikipedia.org/wiki/Probabilistic_method -- but it also applies in analysis; look up "Johnson-Lindenstrauss lemma". In particular, the study of "random Fourier series" goes back at least 50 years -- there are some references in the 3rd edition of Katznelson's book $\endgroup$
    – Yemon Choi
    Commented Jul 25, 2016 at 11:25
  • $\begingroup$ On reflection, I may have misremembered the correct statement about random constructions for elements of ${\rm VN}(G)$. In the context of ${\bf Z}$ it may be worth looking at Section IV.2 of Katznelson, which gives a deterministic construction $\endgroup$
    – Yemon Choi
    Commented Jul 25, 2016 at 11:54

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