Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
1  
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. –  MTS Jun 29 '11 at 15:39
    
do you have any suggestions on how to attack this problem or reference, papers linked to this question? –  Jiang Jun 29 '11 at 15:44
    
Other than the standard textbooks (Takesaki, Sakai, etc) I don't have anything, sorry. –  MTS Jun 29 '11 at 16:23

2 Answers 2

up vote 1 down vote accepted

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 $L(G)$.

share|improve this answer
    
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.) –  Yemon Choi Jun 29 '11 at 18:59
2  
@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 $C^*(G)$. –  Matthew Daws Jun 29 '11 at 20:06

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.