Timeline for Which group is the standard group of isometries?
Current License: CC BY-SA 4.0
7 events
when toggle format | what | by | license | comment | |
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Sep 3, 2021 at 19:58 | comment | added | Kevin Beanland | Thanks for the help everyone. Tomek: I very much like the name infinite hyperoctahedral group. | |
Sep 3, 2021 at 18:08 | comment | added | Tomasz Kania | That group is sometimes called the infinite hyperoctahedral group. | |
Sep 3, 2021 at 18:07 | comment | added | Geoff Robinson | @Wojowu and I seem to be saying the same thing, only in slightly different notation. | |
Sep 3, 2021 at 18:07 | comment | added | YCor | You mean the real-valued $\ell^p$, as pointed out in your previous post. To clarify @Wojowu's comment, and also replacing $N$ with an arbitrary set $X$ (I always find awkward, albeit widely done in some communities, to write $S_\infty$ as if uncountable sets were not existing), it's the unrestricted permutational wreath product $(Z/2Z)\wr Sym(X)=(Z/2Z)^X\rtimes Sym(X)$. In the complex case, it's also the same wreath product, replacing $Z/2Z$ with the circle group. | |
Sep 3, 2021 at 18:05 | comment | added | Wojowu | It should be the wreath product $\mathbb Z_2\wr S_\infty\cong\prod\mathbb Z_2\rtimes S_\infty$ | |
Sep 3, 2021 at 18:04 | comment | added | Geoff Robinson | Isn't it $( \mathbb{Z}/2\mathbb{Z}) \wr {\rm Sym}(\mathbb{N})?$ | |
Sep 3, 2021 at 17:58 | history | asked | Kevin Beanland | CC BY-SA 4.0 |