1
$\begingroup$

Let $\mathbb{F}_2$ denotes the free group generated by a,b, denote this group by $G$. Then consider the von Neumann algebra $L(G)$ generated by the family $\{L_{x_g} : g \in G\}$, here, with $g \in G$, we denote by $x_g$ the function on $G$ that takes the value 1 at g and 0 at other elements of $G$. Then, note that we have the following relations:

$(L_{x_g})^*=L_{(x_g)^{-1}} , L_{x_g}L_{x_h}=L_{x_g*x_h}=L_{x_{gh}}$, then, for any $ A \in L(G),$we can set $ A=\sum_{g \in g}\mu_g L_{x_g},$ with $\mu_g \in \mathbb{C}.$

When we calculate $ ||Ax_{h}||^2 $, we find that $ \sum_{g \in G}|\mu_g|^2 < \infty.$

Then, is this condition sufficient for $ A \in L(G) $? Or some stronger condition is necessary? Like $ \sum_{g \in G}|\mu_g| < \infty,$ or something else?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ This is a very difficult question to answer. It is analogous to asking which sequences arise as the Fourier coefficients of a continuous function. I am pessimistic that there is any meaningful characterization here. $\endgroup$
    – MTS
    Commented Jun 29, 2011 at 15:39
  • $\begingroup$ do you have any suggestions on how to attack this problem or reference, papers linked to this question? $\endgroup$
    – Jiang
    Commented Jun 29, 2011 at 15:44
  • $\begingroup$ Other than the standard textbooks (Takesaki, Sakai, etc) I don't have anything, sorry. $\endgroup$
    – MTS
    Commented Jun 29, 2011 at 16:23

2 Answers 2

Reset to default
1
$\begingroup$

It is more common to just write $L_g$ for $L_{x_g}$. As $L(G)$ admits a finite trace, there is a natural injective map $L(G)$ into $\ell^2(G)$-- this is your map $A \mapsto (\mu_g)$. It is absolutely not true that this map surjects (Open Mapping Theorem). It is obviously sufficient that $(\mu_g)\in\ell^1(G)$ for there to be some $A$ giving rise to $(\mu_g)$.

With $G=\mathbb F_2$, one can say a bit more. For example, Haagerup showed in:
Haagerup, Uffe
An example of a nonnuclear C∗-algebra, which has the metric approximation property.
Invent. Math. 50 (1978/79), no. 3, 279–293.
See Lemma 1.4 that if $f$ is a function of finite support, then denoting $f_n$ the function which agrees with $f$ on the collection of words of reduced length $n$, and is zero elsewhere, we have that there is $A\in L(G)$ inducing $f$, with $\|A\| \leq \sum_{n\geq 0} (n+1) \|f_n\|_2$. From this, it's easy to construct functions not in $\ell^1(G)$, but which are nonetheless induced by members of $L(G)$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Matt, do you know of any example of something in the reduced $C^*$ algebra where replacing each "Fourier coefficient" with its absolute value results in an unbounded operator? (Classical Fourier analysis gives examples of this for the group ${\mathbb Z}$, i.e. you can't always tell if a function on the circle is continuous just by knowing the modulus of each Fourier coefficient.) $\endgroup$
    – Yemon Choi
    Commented Jun 29, 2011 at 18:59
  • 2
    $\begingroup$ @Yemon: Nice Question! In the general case, I don't know. But if $G$ is amenable, then as the trivial rep is contained in the left regular rep, if $f\in C^*(G)$ (with abuse of notation, $f$ is the "Fourier coefficient") is pointwise positive, then $\|f\|_{C^*} = \|f\|_1$. As $\ell^1(G) \not= C^*(G)$ unless $G$ is finite, it cannot be that $f\in C^*(G)$ implies that the absolute value of $f$ is also in $C^*(G)$. $\endgroup$
    – Matthew Daws
    Commented Jun 29, 2011 at 20:06
2
$\begingroup$

For convenience, let's identify $L(G)$ with its image in $\ell^2(G)$ as per @Matthew Daws' answer. For $f=\sum_{g\in G} \mu_g L_g\in\ell^2(G)$, we have $f\in L(G)$ if and only if $f* \xi\in \ell^2(G)$ for all $\xi\in \ell^2(G)$, where $*$ is convolution. Another way of saying this is that $L(G)$ is all $\ell^2$-sums which define bounded operators on $\ell^2(G)$ by convolution.

A good reference for this is Vaughan Jones' course notes/book on von Neumann algebras.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .