Correspondence between the Algebraic K1K_1 and the topological K1K_1

By Serre Swan theorem we have a nice correspondence between Topological K0K_0 and Algebraic K0K_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 K1K_1. The definition I am using for Topological K1K_1 is K1(X)=K0(SX)K_1(X)=K_0(SX) where SXSX is the suspension of space XX. The definition I am using for Algebraic K1K_1 is K1(R)=GL(R)/[GL(R),GL(R)]K_1(R)=GL(R)/[GL(R),GL(R)] where GL(R)GL(R) is the direct limit of the diagram GL(1,R)↪GL(2,R)↪GL(3,R)↪….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→GL(R)X \to GL(\mathbb{R}). So this gives an element of K1(C(X,R))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 EE over SXSX by FF such that E⊕εn=F⊕εnE \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.

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You have a comparison map K1(C(X))→K1(X)K_1(C(X))\to K^1(X) which is induced by the identity GLdisc(C(X))→GLtop(C(X))GL^{\mathrm{disc}}(C(X))\to GL^{\mathrm{top}}(C(X)). However, it is not necessarily an isomorphism, for example K1(C)=C×K_1(\mathbb C)=\mathbb C^\times, but K1(pt)={0}K^1(\mathrm{pt})=\{0\}. This is discussed in Chapter 3 of “Elements of Noncommutative Geometry”.
– MaoWao
2 days ago

  

 

@MaoWao i dont really understand your notation what is K1K^1?
– happymath
2 days ago

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K1(X)K^1(X) is what you call K1(X)K_1(X) (I prefer this notation because K1K^1 is contravariant on topological spaces). Anyway, the answer to this question should answer yours as well: math.stackexchange.com/questions/608514/…
– MaoWao
2 days ago

  

 

You also need to assume that X X is compact and Hausdorff.
– Berrick Caleb Fillmore
2 days ago

  

 

@MaoWao Thank you for the counter example but still I am interested in knowing where it goes wrong, which I am having difficulty grasping from the counter example.
– happymath
yesterday

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1 Answer
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As someone already pointed out, it would be better to write K1(X)K^1(X) for topological K-theory of the space XX. Note that this is the same as operator K-theory K1(C(X))K_1(C(X)) of the C∗C^\ast-algebra C(X)C(X).

Now, let us denote algebraic K-theory by Kalg1K_1^{\text{alg}}. You want to compare K1(C(X))K_1(C(X)) and Kalg1(C(X))K_1^{\text{alg}}(C(X)).

As you say, K1(C(X))K_1(C(X)) can be defined via K0K_0 and suspension, but for our purposes, the following equivalent formulation is more useful:
(∗)K1(A)=GL(A)GL(A)0, (\ast)\qquad\qquad K_1(A)= \frac{\mathrm{GL(A)}}{\mathrm{GL(A)^0}},
where the 00-superscript stands for “connected component of the identity”. Note that for this to make sense AA must be, say, a Banach algebra.

This already gives a feeling for comparison since as you wrote
Kalg1(A)=GL(A)[GL(A),GL(A)]. K_1^{\text{alg}}(A)= \frac{\mathrm{GL(A)}}{[\mathrm{GL(A)},\mathrm{GL(A)}]}.

One can also define SK1S\!K_1 by replacing GL\mathrm{GL} with SL\mathrm{SL} in (∗)(\ast). Then if AA is commutative, therefore A≅C(X)A\cong C(X), the following result holds:
Kalg1(A)≅A∗⊕SK1(A),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.