Let $R$ be a commutative unital ring and let $i$ be a non-negative integer such that $K^i_{alg}(R)$ is finitely generated abelian group. Is it possible that there does not exist weak homotopy type of finite CW complex $X$ such that $KO^i(X)\approx K^i_{alg}(R)$?
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5$\begingroup$ Topological $K$-theory satisfies Bott periodicity, algebraic $K$-theory not. $\endgroup$– user19475Commented Feb 17, 2019 at 9:27
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2$\begingroup$ For $K_0$ it is true that topological K-theory $K_0$ of a $C^*$-Algebra is the same as algebraic K-theory $K_0$ of the underlying ring. But even then not every ring is a $C^*$-Algebra. $\endgroup$– ThiKuCommented Feb 17, 2019 at 10:06
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2$\begingroup$ @ThiKu That's not true for higher degrees. You need a stability condition: algebraic and topological K-theory of a $C^*$-algebra $A$ coincide if there is an isomorphism $\mathcal{K}\hat\otimes A\cong A$ $\endgroup$– Denis NardinCommented Feb 17, 2019 at 10:50
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2$\begingroup$ For a finite complex $X$, $K_i(X)$ is a finitely generated abelian group. It is probably possible to exhibit any fg ab group. But algebraic $K$-theory is not finitely generated. Eg, $K_1(Q)=Q^\times$, and for $R=F_2[t,e]/e^2$, $K_1(R)\sim R^\times$ is an infinite rank $F_2$-vector space. I think that there is an open conjecture that a smooth complete variety over a finite field has fg $K$-theory. Maybe even if just one of smooth/complete. Example with different flavor: if $E/C$ elliptic curve, $R$ functions on $E-pt$ then $K_0(R)\sim E(C)$. $\endgroup$– Ben WielandCommented Feb 18, 2019 at 20:37
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1$\begingroup$ Compact Hausdorff spaces are more general than finite complexes. But beyond finite complexes, there are multiple definitions of the topological $K$-theory of the Cantor set or the Hawaiian earrings. $\endgroup$– Ben WielandCommented Feb 20, 2019 at 19:23
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