Timeline for Algebraic $K_1$ group for a $C^*$-algebra

8 events

when toggle format	what		by	license	comment
May 5, 2016 at 19:59	vote	accept	truebaran
May 5, 2016 at 18:56	comment	added	Leonel Robert		Yes, that's right.
May 5, 2016 at 15:23	comment	added	truebaran		Ok, just to be sure whether I understood everything correctly: so now everything commutes, $diag(h,-\frac{h}{n},...,-\frac{h}{n})$ is a commutator and $diag(h,0,...,0)$ is a limit of commutators. Now we apply the same argument for $h \in M_n(A)$ where the matrix $diag(1,-\frac{1}{n},...,-\frac{1}{n})$ is understood as a block diagonal matrix where the size of each block is $n$, and blocks are diagonal matrices (the scalar matrix is a commutator iff its trace is $0$ so it is again a commutator).
May 5, 2016 at 3:05	comment	added	Leonel Robert		I meant multiply by the diagonal matrix constant $h$.
May 5, 2016 at 3:02	history	edited	Leonel Robert	CC BY-SA 3.0	added 65 characters in body
May 5, 2016 at 2:07	comment	added	truebaran		Thank you for the great answer! However I don't see how does the final argument (that each element in $M_{\infty}(A)$ is a limit of commutators) works. The problem is that if we express the matrix $diag(1,-\frac{1}{n},...,-\frac{1}{n})$ as a commutator $AB-BA$ and then multiply by $diag(h,0,...,0)$ then how do we know that $h(AB-BA)$ are also commutators (this would be fine if $diag(h,0,...,0)$ will commute with $A$ and $B$?
May 5, 2016 at 0:36	history	edited	Leonel Robert	CC BY-SA 3.0	added 4 characters in body
May 5, 2016 at 0:28	history	answered	Leonel Robert	CC BY-SA 3.0