Timeline for If $E\oplus_\phi E \cong E\oplus_\psi E,$ does it imply that $\phi= \psi$?
Current License: CC BY-SA 4.0
10 events
when toggle format | what | by | license | comment | |
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Jan 4, 2019 at 2:19 | vote | accept | Idonknow | ||
Aug 9, 2018 at 2:52 | history | bounty ended | Idonknow | ||
Aug 8, 2018 at 21:12 | comment | added | Duchamp Gérard H. E. | OK, I understand. | |
Aug 8, 2018 at 21:10 | comment | added | Mikhail Ostrovskii | @ Duchamp Gérard H. E. I use the notation $\ell_p^2$ for two-dimensional $\ell_p$, it also can be written as $\mathbb{R}\oplus_p\mathbb{R}$. | |
Aug 8, 2018 at 20:25 | comment | added | Duchamp Gérard H. E. | Thanks. I suppose that you note $\ell_p^2$, the space $\ell_p\oplus \ell_p$. | |
Aug 8, 2018 at 20:15 | comment | added | Mikhail Ostrovskii | @Idonknow It is a strange and impolite approach: to open a bounty and to pay no attention to an answer. | |
Aug 8, 2018 at 15:31 | history | edited | Mikhail Ostrovskii | CC BY-SA 4.0 |
added 898 characters in body
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Aug 8, 2018 at 15:31 | comment | added | Mikhail Ostrovskii | @ Duchamp Gérard H. E. I have inserted the corresponding argument. | |
Aug 8, 2018 at 7:26 | comment | added | Duchamp Gérard H. E. | +1 but I'd like to know more about your argument in [if we can find ... ($p,q$ are fixed).] or a reference. | |
Aug 7, 2018 at 21:30 | history | answered | Mikhail Ostrovskii | CC BY-SA 4.0 |