Timeline for Norm of two operators on $l^2(\mathbb{Z}_{2}*\mathbb{Z}_{2})$ different
Current License: CC BY-SA 4.0
14 events
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| May 27, 2019 at 0:03 | history | bounty ended | CommunityBot | ||
| May 21, 2019 at 21:05 | comment | added | J. E. Pascoe | 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. | |
| May 21, 2019 at 11:08 | comment | added | worldreporter | 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 | |
| May 21, 2019 at 0:23 | comment | added | J. E. Pascoe | 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.) | |
| May 20, 2019 at 22:57 | comment | added | J. E. Pascoe | 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.) | |
| May 20, 2019 at 19:54 | vote | accept | worldreporter | ||
| May 20, 2019 at 19:54 | comment | added | worldreporter | 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 non-trivial case of those | |
| May 19, 2019 at 1:45 | history | edited | J. E. Pascoe | CC BY-SA 4.0 |
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| May 19, 2019 at 1:06 | history | undeleted | J. E. Pascoe | ||
| May 19, 2019 at 1:06 | history | edited | J. E. Pascoe | CC BY-SA 4.0 |
fixed various parts. fixed sign error from previous version which changed the answer.
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| May 19, 2019 at 0:54 | history | deleted | J. E. Pascoe | via Vote | |
| May 19, 2019 at 0:31 | history | edited | J. E. Pascoe | CC BY-SA 4.0 |
[Edit removed during grace period]; added 289 characters in body
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| May 19, 2019 at 0:06 | history | edited | J. E. Pascoe | CC BY-SA 4.0 |
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| May 19, 2019 at 0:01 | history | answered | J. E. Pascoe | CC BY-SA 4.0 |