Timeline for What is the group of automorphisms of $l^{\infty}$?
Current License: CC BY-SA 3.0
4 events
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| May 6, 2016 at 6:08 | comment | added | Uri Bader | @CalebEckhardt, you're argument seems to suggest that every isometry of $\ell^\infty$ would come from a permutation, which is incorrect: take the multiplication map with a function $\mathbb{Z}\to\{-1,1\}$ (or to $S^1$ over $\mathbb{C}$). | |
| May 6, 2016 at 1:35 | comment | added | Nik Weaver | @CalebEckhardt: that is nicer. On the other hand, identifying with $C(\beta\mathbb{Z})$ allows us to invoke the Banach-Stone theorem and say that every linear isometry of $l^\infty(\mathbb{Z})$ with itself has this form. | |
| May 3, 2016 at 1:20 | comment | added | Caleb Eckhardt | Equivalently, but maybe slightly easier, (in the sense that one doesn't have to use the correspondence $\ell^\infty(\mathbb{Z})\cong C(\beta\mathbb{Z})$) one notices that every automorphism must permute the minimal projections of $\ell^\infty(\mathbb{Z})$ which are in one-to-one correspondence with $\mathbb{Z}$ | |
| May 2, 2016 at 18:26 | history | answered | Uri Bader | CC BY-SA 3.0 |