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

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.
• 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 $B$? – truebaran May 5 '16 at 2:07
• I meant multiply by the diagonal matrix constant $h$. – Leonel Robert May 5 '16 at 3:05
• 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). – truebaran May 5 '16 at 15:23