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.

$\begingroup$ 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. $\endgroup$ – J. E. Pascoe May 16 '19 at 15:31

$\begingroup$ 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? $\endgroup$ – worldreporter14 May 18 '19 at 18:30

$\begingroup$ 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.) $\endgroup$ – J. E. Pascoe May 18 '19 at 22:11

$\begingroup$ 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? $\endgroup$ – Yemon Choi May 19 '19 at 1:07

$\begingroup$ @worldreporter14 I'm sort of interested as to the origin of the problem, as the answer seemed to depend of the combinatorics/geometry of the group. $\endgroup$ – J. E. Pascoe May 20 '19 at 15:12
The claim is true.
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 selfadjoint. Moreover the spectrum of $A$ contains $2$ and $2$ as the limits $\(2+A)^ne_0\^{1/n}$ and $\(2A)^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 $2+A, 2A$ are equal to $4.$)
Consider the function $$F_A(z) = \langle (T_s+T_tz)^{1}e_0,e_0 \rangle.$$ 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 AronszajnKrein 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.

$\begingroup$ First of all thanks for your answer! I had to think about it before I respond. You asked about the origin: The problem is related to the question whether or not certain deformations of the group $C^*$algebra of $\mathbb{Z}_2^{*L}$ are simple or not. This question (and I find this quite remarkable) leads to operators of the type above. The one I mentioned is the easiest nontrivial case of those $\endgroup$ – worldreporter14 May 20 '19 at 19:54

$\begingroup$ The sequence of values you get out for $\langle (T_s+T_u+T_v)^n e_0, e_0\rangle$ (presumably the $\mathbb{Z}_2^{*3}$ case should be interesting, since that group has a much faster growth rate.) $\endgroup$ – J. E. Pascoe May 20 '19 at 22:57

$\begingroup$ You may find OEIS sequence A089022 useful. It gives a related generating function for any degree. (Although the terms are for the $3$ case. The growth rate of the terms in the corresponding power series is like $8^{n/2}$ rather than $9^{n/2}$ which is what I would have expected. What is nice is that the generating function they give after a composition with $1/z^2$ and multiplying by $1/z$ is literally the correspond $F_A,$ I think.) $\endgroup$ – J. E. Pascoe May 21 '19 at 0:23

$\begingroup$ Thanks for the OEIS reference. It's interesting: For the next higher order case $\mathbb{Z}_2^{*3}$ the operator $X_\theta$ is $X_{\theta}:=12\cdot\tan\left(\theta\right)+T_{r}^{\left(1\right)}+T_{s}^{\left(1\right)}+T_{t}^{\left(1\right)}$. Here I would have also expected that $\left\Vert X_{\theta}\right\Vert \neq\left\Vert X_{\theta}2\tan\left(\theta\right)P_{\left\{ e\right\} }\right\Vert$ for $\theta \neq 0$. But (using the same methods as above) it seems like that in this case my claim is not true $\endgroup$ – worldreporter14 May 21 '19 at 11:08

$\begingroup$ I would guess it has something to do with the geometry of the group. $\mathbb{Z}^{*3}$ has exponential growth, but $\mathbb{Z}^{*2}$ has linear growth. $\endgroup$ – J. E. Pascoe May 21 '19 at 21:05