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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 $\phi = M_{\phi(1)}$.

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up vote 7 down vote accepted

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.

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Thanks! This is very helpful. –  Ellen May 1 '13 at 17:23
No problem, you're welcome. –  Nik Weaver May 1 '13 at 18:43
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