Let $A$ be a $C^*$-algebra: then one defines topological $K_1$ group as $GL_{\infty}(A^+)/\Big(GL_{\infty}(A^+)\Big)_0$ where $A^+$ denotes $A$ with the unit adjointed (even if $A$ already had a unit: in this case $A^+$ is isomorphic to $A \oplus \mathbb{C}$ but this definition turns out to be equivalent if we insert $A$ instead of $A^+$). The algebraic $K_1(A)$ is defined as $K_1(A)=GL_{\infty}(A)/[GL_{\infty}(A),GL_{\infty}(A)]$ where ${H,H}$ is a commutator subgroup and it is known that it is not the same as topological $K_1$. I read somewhere that $K_1^{top}(A)$ may be defined somehow similarly to the algebraic $K_1$ as $K_1^{top}(A)=GL_{\infty}(A)/ \overline{[GL_{\infty}(A),GL_{\infty}(A)]}$, at least for unital $C^*$-algebras. My question is the following:

How to show that these two aproaches are equivalent?


One must show that $$ \overline{[GL_\infty(A),GL_\infty(A)]}= (GL_\infty(A))_0. $$ The proof that the left side is in the right side is the easier one: Let $a$ be in the left side. Then $a=bc$, where $b\in [GL_\infty(A),GL_\infty(A)]$ and $\|c-1\|<1$. Commutators belong to $(GL_\infty(A))_0$, because $K_1(A)$ is abelian (or because of the proof of this fact). So $b\in (GL_\infty(A))_0$. Also, $c=e^h$ for some $h\in M_\infty(A)$, and so it is connected to $1$ by the path $t\mapsto e^{th}$.

Choose now $a\in (GL_\infty(A))_0$. It can be written as finite product of exponentials $e^h$, with $h\in M_\infty(A)$. So it is enough to show that these exponentials belong to the left side. We first prove this for $h\in M_\infty(A)$ of the form $xy-yx$. In this case one has $$ (e^{x/n}e^{y/n}e^{-x/n}e^{-y/n})^{n^2}\to e^{xy-yx}. $$ The elements on the left side are products of commutators so we get the desired result. Finally, notice that any $h\in M_\infty(A)$ is a limit of commutators. For example, assuming that $h\in A$ for simplicity we can express the matrix $$ \begin{pmatrix} 1 & & & \\ & -\frac 1 n & &\\ && -\frac 1 n &\\ &&&\ddots \end{pmatrix} $$ as a commutator in $M_{n+1}(\mathbb C)$ then multiply by $h\otimes 1_{n+1}$ to get diag$(h,-h/n,\ldots,-h/n)$ as a commutator.

| cite | improve this answer | |
  • $\begingroup$ 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 $B$? $\endgroup$ – truebaran May 5 '16 at 2:07
  • $\begingroup$ I meant multiply by the diagonal matrix constant $h$. $\endgroup$ – Leonel Robert May 5 '16 at 3:05
  • $\begingroup$ 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). $\endgroup$ – truebaran May 5 '16 at 15:23
  • $\begingroup$ Yes, that's right. $\endgroup$ – Leonel Robert May 5 '16 at 18:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.