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Apr 13, 2017 at 12:58 history edited CommunityBot
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Feb 29, 2016 at 20:19 comment added Ali Taghavi @YonatanHarpaz Of course we can not give a counter example in matrix algebra or operator algebra or Von Neumann algebras, because all nilpotent elements are equivalent to each other. But what about other algebras? Thanks again for your very interesting comment.
Feb 29, 2016 at 20:08 comment added Ali Taghavi @YonatanHarpaz Thank you for your attention to my question. I confess that I did not pay attention to this prerequisit,. The nilpotent version of proposition 4.2.4 of page 26 of Blackadar book on K theory. The reason that i did not pay attention:I was saying myself $a\simeq..a'$ and $b\simeq b'$ imply that $\begin{pmatrix} a&0\\0&b\end{pmatrix} \simeq \begin{pmatrix}a'&0\\0&b'\end {pmatrix}$,. So according to your comment, my construction collaps if that prerequisite can not be proved.
Feb 29, 2016 at 19:32 comment added Yonatan Harpaz Let $A,B \in \cup_n M_n(R)$ be a pair of nilpotent matrices such that $AB = BA = 0$ and let $A',B' \in \cup_n M_n(R)$ be another pair of nilpotent matrices over such that $A'B' = B'A' = 0$. If $A \simeq A'$ and $B \simeq B'$, is there any reason to think that $A+B \simeq A'+B'$? This seems like a prerequisite result one would need in order to show that the desired semi-group structure on $\cup_n M_n(R)/\simeq$ is well-defined.
Feb 27, 2016 at 20:55 comment added Ali Taghavi @YemonChoi This construction for $\mathbb{C}$ gives the trivial group(As I explained here:.mathoverflow.net/questions/231328/…)
Feb 27, 2016 at 20:50 comment added Yemon Choi What does this construction give for $R={\mathbb C}$?
Feb 27, 2016 at 20:20 history edited Ali Taghavi CC BY-SA 3.0
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Feb 27, 2016 at 19:38 history edited Moritz Firsching CC BY-SA 3.0
typo for Hausdorff
Feb 27, 2016 at 19:34 history edited Ali Taghavi CC BY-SA 3.0
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Feb 27, 2016 at 19:27 history edited Ali Taghavi CC BY-SA 3.0
added 269 characters in body
Feb 27, 2016 at 19:21 history asked Ali Taghavi CC BY-SA 3.0