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. PascoeMay 16, 2019 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? $\endgroup$– worldreporterMay 18, 2019 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. PascoeMay 18, 2019 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 ChoiMay 19, 2019 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. PascoeMay 20, 2019 at 15:12
2 Answers
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 $\endgroup$ May 20, 2019 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.) $\endgroup$ May 20, 2019 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.) $\endgroup$ May 21, 2019 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 $\endgroup$ May 21, 2019 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. $\endgroup$ May 21, 2019 at 21:05
One can use noncommutative probability theory to compute the spectral distributions of $X_\theta$ and $X_\theta  2 \tan(\theta) P$ with respect to $\delta_e$. Since $G = \mathbb{Z}_2 * \mathbb{Z}_2$, the operators $T_s$ and $T_t$ are freely independent with respect to the vector state given by $\delta_e$. Each of them has the spectral distribution given by the Bernoulli measure $(1/2)(\delta_{1} + \delta_1)$. Thus, the spectral distribution of $T_s + T_t$ is the free convolution of two Bernoulli distributions. Using the $R$transform (see Chapter 4 of https://arxiv.org/pdf/1908.08125.pdf), we can compute that the free convolution is the measure $\mu$ satisfying $$ F_\mu(z) = \sqrt{z^2  4}, $$ where $F_\mu(z) = 1 / \int (z  t)^{1}\,d\mu(t)$, also known as the $F$transform. Here we use the square root which is defined on the upper halfplane and close to $z$ when $z \to \infty$. This means that $\mu$ is an arcsine law $$ d\mu(x) = \frac{1}{\pi \sqrt{4  x^2}} \chi_{(2,2)}(x)\,dx $$ This computation is known; see the last full paragraph at the bottom of page 2 here: https://arxiv.org/pdf/1008.5205.pdf.
Next, let $\nu$ be the spectral measure of $T_s + T_t  8 \tan(\theta)$. This just shifts the measure $\mu$ by $8 \tan(\theta)$, and we have $F_\nu(z) = F_\mu(z + 8 \tan(\theta))$. Since the support of $\mu$ is $[2,2]$, the support of $\nu$ is $[2  8 \tan(\theta), 2  8 \tan(\theta)]$ and thus the norm of $X_\theta$ is $2 + 8 \tan \theta$, at least on the cyclic subspace generated by $\delta_e$. However, because $\delta_e$ is cyclic for the right shift operators by $s$ and $t$, it is cyclic for the commutant of $X_\theta$ and hence the norm of $X_\theta$ on the entire space agrees with the norm on the cyclic subspace generated by $\delta_e$.
Finally, let's compute the spectral measure of $X_\theta  2 \tan(\theta) P$, which we will call $\rho$. This can be done using rankone perturbation formulas. Alternatively, $X_\theta$ and $P$ are Boolean independent (see https://mast.queensu.ca/~speicher/papers/boolean.ps). The spectral measure of $2 \tan(\theta) P$ with respect to $\delta_e$ is $\delta_{2 \tan(\theta)}$. The theory of Boolean independence tells us that $$ F_\rho(z)  z = (F_\nu(z)  z) + (F_{\delta_{2\tan(\theta)}}(z)  z) = F_\nu(z)  z + 2 \tan(\theta). $$ Hence, $$ F_\rho(z) = \sqrt{(z + 8 \tan \theta)^2  4} + 2 \tan(\theta). $$ Thus, for $z$ in the upper halfplane, $$ G_\rho(z) = \frac{1}{\sqrt{(z + 8 \tan \theta)^2  4} + 2 \tan(\theta)}, $$ where $G_\rho(z) = \int (z  t)^{1} \,d\rho(t)$. Using the fact that this is rankone perturbation that the norm of the new operator $X_\theta  2 \tan(\theta) P$ is the maximum of the norm of $X_\theta  2 \tan(\theta) P$ on the cyclic subspace generated by $\delta_e$ and the norm on the orthogonal complement, which will be bounded by $\lVert X_\theta \rVert$. Thus, to show that the two operators have different norms, it suffices to show that the support radius of $\rho$ is strictly larger than that of $\nu$.
Looking at $G_\rho$, the support of the measure $\rho$ agrees with the support of $\nu$ except with the addition of atoms at the point where $(2 \tan \theta)^2 = (z + 8 \tan \theta)^2  4$ according to the version of the square root described above. Assume $\tan \theta > 0$ since the other case is symmetric (and recall $\theta$ was assumed to be in $(\pi/2,\pi/2)$). Then we are looking at a point on the negative side where $$(z + 8 \tan \theta)^2 = 4 + 4 \tan^2 \theta = 4 \sec^2 \theta.$$ Hence, $z = 2 \sec \theta  8 \tan \theta$ is the location of the atom. Since the support of $\nu$ was already farther on the negative side, we have $$ \lVert X_\theta \rVert = 2 + 8 \tan \theta < 2 \sec \theta + 8 \tan \theta = \lVert X_\theta  (2 \tan \theta) P \rVert. $$
Similar computations can be done for $\mathbb{Z}_2^{*n}$. Let $T_1$, \dots, $T_n$ be the operators corresponding to the generators. Then each $T_1 + \dots + T_n$ as a spectral distribution $\mu_n$ is the free convolution of $n$ Bernoulli laws. This can be done directly with the $R$transform, or using Proposition 3.1 of this paper: https://arxiv.org/pdf/math/0703295.pdf. This results in $$ F_{\mu_n}(z) = \frac{(n2)z + n \sqrt{z^2  4(n1)}}{2(n1)}. $$ Thus, similar computations can be done as in the $n = 2$ case.
Unfortunately, for general groups, it is not as easy to compute because the operators might not be selfadjoint. However, the tools of free probability could still be used in theory to compute the spectral distribution of $X^*X$.