Correspondence between the Algebraic $K_1$ and the topological $K_1$

Question

By Serre Swan theorem we have a nice correspondence between Topological $K_0$ and Algebraic $K_0$ when we consider the ring to be continuous functions on a topological space. I am wondering if there is a correspondence like that even for $K_1$. The definition I am using for Topological $K_1$ is $K_1(X)=K_0(SX)$ where $SX$ is the suspension of space $X$. The definition I am using for Algebraic $K_1$ is $K_1(R)=GL(R)/[GL(R),GL(R)]$ where $GL(R)$ is the direct limit of the diagram $$GL(1,R) \hookrightarrow GL(2,R) \hookrightarrow GL(3,R) \hookrightarrow ....$$ Now the reason I suspect there is a correspondence here is because we know that an isomorphism class of a vector bundle corresponds to homotopy classes of maps from $X \to GL(\mathbb{R})$. So this gives an element of $K_1(C(X,\mathbb{R}))$. Now we have to check that this is independent of representative of the isomorphism class. I am unable to actually prove this. Even if I can somehow show this to get a proper correspondence I have to actually show that if I replace a bundle $E$ over $SX$ by $F$ such that $E \oplus \varepsilon ^n=F \oplus \varepsilon ^n$ then also I should get the same element in the image. Could someone point out how one does this or point me to a suitable reference. Thanks.

Answer 1

As someone already pointed out, it would be better to write $K^1(X)$ for topological K-theory of the space $X$. Note that this is the same as operator K-theory $K_1(C(X))$ of the $C^\ast$-algebra $C(X)$.

Now, let us denote algebraic K-theory by $K_1^{\text{alg}}$. You want to compare $K_1(C(X))$ and $K_1^{\text{alg}}(C(X))$.

As you say, $K_1(C(X))$ can be defined via $K_0$ and suspension, but for our purposes, the following equivalent formulation is more useful: $$ (\ast)\qquad\qquad K_1(A)= \frac{\mathrm{GL(A)}}{\mathrm{GL(A)^0}},$$ where the $0$-superscript stands for "connected component of the identity". Note that for this to make sense $A$ must be, say, a Banach algebra.

This already gives a feeling for comparison since as you wrote $$ K_1^{\text{alg}}(A)= \frac{\mathrm{GL(A)}}{[\mathrm{GL(A)},\mathrm{GL(A)}]}.$$

One can also define $S\!K_1$ by replacing $\mathrm{GL}$ with $\mathrm{SL}$ in $(\ast)$. Then if $A$ is commutative, therefore $A\cong C(X)$, the following result holds: $$K_1^{\text{alg}}(A)\cong A^\ast\oplus S\!K_1(A),$$ where the first summand stands for invertibles. You can find a bit more on this in Karoubi's book "K-theory - An introduction", Chapter II, exercise 6.13.