I'm trying to understand an assertion that appears in Luck's book on $L^2$-invariants. I believe that it's elementary functional analysis, but I can't seem to figure it out.


Here goes. Let
$$T^n = \{\text{$(z\_1,\ldots,z\_n)$ $|$ $|z\_i|=1$ for $1 \leq i \leq n$}\} \subset \mathbb{C}^n$$
be the $n$-torus and let $G$ be the set of bounded linear operators from $L^2(T^n)$ to itself. The group $\mathbb{Z}^n$ acts on $G$ in the following way. Consider $L=(\ell\_1,\ldots,\ell\_n) \in \mathbb{Z}^n$ and $\phi \in G$. Then $L \cdot \phi$ is the operator that takes $f \in L^2(T^n)$ to $z\_1^{\ell\_1} \cdots z\_n^{\ell\_n} \phi(f) \in L^2(T^n)$.


For any $h \in L^{\infty}(T^n)$, there is a multiplication operator $M\_h \in G$. Clearly $M\_h$ is invariant under $\mathbb{Z}^n$. Luck claims that in fact the set of $\mathbb{Z}^n$-invariants in $G$ is exactly $L^{\infty}(T^n)$. Can anyone help me prove this?


By the way, it is clear that if $\phi \in G$ is invariant under $\mathbb{Z}^n$, then $\phi$ should be equal to the multiplication operator $M\_{\phi(1)}$. But I can't even prove that $\phi(1)$ is in $L^{\infty}(T^n)$, much less that $\phi = M\_{\phi(1)}$.



Certain bounded linear operators on L^2 of a torus
Yes, this is standard functional analysis. You're starting out the right way.


I'll sketch the argument. First observe that $\phi$ is invariant for the ${\bf Z}^n$ action if and only if it commutes with the multiplication operators $M\_h$ for $h(z\_1, \ldots, z\_n) = z\_1^{l\_1}\cdots z\_n^{l\_n}$. By taking linear combinations (and using the fact that the $l\_i$ can be negative) this implies that it commutes with $M\_h$ for $h$ any polynomial in the $z\_i$'s and $\bar{z}\_i$'s. Now these polynomials are uniformly dense in the continuous functions on the torus, $C({\bf T}^n)$, so by taking limits we can conclude that $\phi$ commutes with $M\_h$ for any continuous function $h$ on the torus. Since $C({\bf T}^n)$ is weak\* dense in $L^\infty({\bf T}^n)$, we can now take weak\* limits and deduce that $\phi$ commutes with $M\_h$ for any $h \in L^\infty({\bf T}^n)$.


So we now know that $\phi(1\_A) = \phi(1\_A\cdot 1) = 1\_A\cdot\phi(1)$ for any measurable set $A$, where $1\_A$ is the characteristic function of $A$. If $|\phi(1)| > \|\phi\|$ on a positive measure set we can take $A$ to be this set and get a contradiction. That's how you know $\phi(1)$ is bounded. The rest is easy: we have $\phi(h) = \phi(h\cdot 1) = h\cdot \phi(1)$ for any $h \in L^\infty({\bf T}^n)$, and these functions are dense in $L^2({\bf T}^n)$, so we conclude that $\phi = M\_{\phi(1)}$ by continuity.


There's some work you have to do to check the approximation arguments if you're doing this from scratch, but that's the proof in outline.


