# Norm of two operators on $l^2(\mathbb{Z}_{2}*\mathbb{Z}_{2})$ different

In my research I encounered the following (very concrete) question: Consider the (discrete) group $$G:=\mathbb{Z}_{2}*\mathbb{Z}_{2}$$. Let $$s\text{, }t\in G$$ be the generating elements and define for $$\theta\in\left(-\frac{\pi}{2}\text{, }\frac{\pi}{2}\right)$$ the bounded operator $$\begin{eqnarray} X_{\theta}:=-8\tan\left(\theta\right)\cdot\text{id}+T_{s}+T_{t}\in{\cal B}\left(l^{2}\left(G\right)\right) \end{eqnarray}$$ on the Hilbert space $$l^{2}\left(G\right)$$ where $$T_{s}\delta_{g}:=\delta_{sg}$$ for every $$g\in G$$ (and $$T_{t}$$ is defined analogously). Let $$P\in{\cal B}\left(l^{2}\left(G\right)\right)$$ be the projection onto $$\mathbb{C}\delta_{e}$$ where $$e\in G$$ is the neutral element. I claim that $$\begin{eqnarray} \left\Vert X_{\theta}\right\Vert \neq\left\Vert X_{\theta}-2\tan\left(\theta\right)P\right\Vert \text{,} \end{eqnarray}$$ unless $$\theta=0$$. At first glance this looks obvious but I could not show it so far.

## This question has an open bounty worth +100 reputation from worldreporter14 ending in 6 days.

The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns.

• I believe you can at least see that $\|X_\theta\|\leq \|X_\theta- tP\|$ for any $t\in \mathbb{R}$ by looking at the spectral measure for $X_\theta^*X_\theta$ and using the fact that there are no minimal projections in the group von Neumann algebra to get a vector perpendicular to $\delta_e$ almost attaining the norm. – J. E. Pascoe May 16 at 15:31
• Thanks for your response! Under assuming equality of both norms and by using your suggestion I can show that $t \mapsto \left\Vert X_\theta -tP \right\Vert$ is constant on the interval from $[16\text{tan}\left(\theta\right), -2\text{tan}\left(\theta\right)]$ (assuming $\theta \leq 0$). Do you think that could help deducing a contradiction? – worldreporter14 yesterday
• That's unclear, but if the claim is true, that's probably the way to go. I don't think that you will get a contradiction without some amount of understanding of the group von Neumann algebra. (That is, this is not likely to be some property of all von Neumann algebras.) – J. E. Pascoe yesterday
• The group in question is isomorphic to ${\bf Z}\rtimes {\bf Z}_2$ (the latter group can be viewed as acting on ${\bf Z}$ by translations and by the flip $n\leftrightarrow -n)$. There is a noncommutative version of the Fourier transform that yields an isomorphism $\ell^2(G) \leftrightarrow L^2([0,1]; {\bf C}^2)$ with corresponding isomorphism of von Neumann algebras ${\rm VN}(G) \cong L^\infty\otimes {\bf M}_2$. Perhaps the explicit matricial picture can be used to carry out some of these calculations? – Yemon Choi yesterday

Any difference in norm must be picked up on the span of $$(T_s+T_t)^ne_0.$$ So we will apply perturbation theory on that subspace. The value of $$\langle(T_s+T_t)^ne_0,e_0\rangle$$ should be $${n}\choose{n/2}$$ when $$n$$ is even and zero otherwise. Note that $$A=T_s + T_t$$ is self-adjoint. Moreover the spectrum of $$A$$ contains $$2$$ and $$-2$$ as the limits $$\|(2+A)^ne_0\|^{1/n}$$ and $$\|(2-A)^ne_0\|^{1/n}$$ are both $$4$$ by Stirlings type estimates. (In fact, for each $$n$$ the quantities are equal. This says that the spectral radius of the operators $$2+A, 2-A$$ are equal to $$4.$$)
Consider the function $$F_A(z) = \langle (T_s+T_t-z)^{-1}e_0,e_0 \rangle.$$ The places where $$F_A$$ analytically continues through $$\mathbb{R}$$ is exactly the complement of the spectrum. Expanding $$F_A$$ at infinity gives: $$F_A(z) = -\frac{1}{z}\sum {{2n}\choose{n}} \frac{1}{z^{2n}}$$ Now consider $$\lim_{z\rightarrow 2^+} F_A(z)$$ and $$\lim_{z\rightarrow -2^-} F_A(z).$$ Apparently, using Stirling's formula type estimates, $$\lim_{z\rightarrow 2^+} F_A(z)= -\infty.$$ Also, as the function is odd, $$\lim_{z\rightarrow -2^-} F_A(z) =\infty.$$ By the Aronszajn-Krein formula, the spectrum of $$A + \alpha P$$ is governed by $$F_{A+\alpha P}=\frac{F}{1+\alpha F}.$$ Note the spectrum will only change if $$F(z) = -\frac{1}{\alpha}$$ has a real solution in the complement of the spectrum of $$A.$$ (Moreover, it will only change by one eigenvalue.)
So, now we consider the spectrum of $$4\alpha +A$$ and compare it to $$4\alpha+A + \alpha P.$$ If $$\alpha >0,$$ the extra eigenvalue of $$A+\alpha P$$ appears when $$F_A(z) = -1/\alpha$$ which happens to the right of the spectrum, and therefore the norm increases. Similarly, the norm increases in the other case.
Note that it is not true for a general $$\alpha + A + \beta P,$$ and has a somewhat subtle dependence on your choice of problem.