Timeline for Comparing real topological K-theory and algebraic K-theory
Current License: CC BY-SA 4.0
24 events
| when toggle format | what | by | license | comment | |
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| Feb 21, 2019 at 15:43 | comment | added | Robert Furber | @rori $(S^1)^\mathbb{N}$ is a compact Hausdorff space by Tychonoff's theorem. The fundamental group of this is not finitely generated, as the fundamental group of a finite complex must be. | |
| Feb 21, 2019 at 14:37 | comment | added | user74900 | @DavidWhite but is there a complete proof/reference to an example of a finitely generated algebraic K-theory group that is not real K-theory of finite CW complex? Comments at this moment of time do not appear to contain it. | |
| Feb 21, 2019 at 14:20 | review | Close votes | |||
| Feb 22, 2019 at 20:28 | |||||
| Feb 21, 2019 at 14:04 | comment | added | David White | I'm voting to close this question as off-topic because it is answered in the comments. | |
| S Feb 21, 2019 at 13:04 | history | suggested | heygo | CC BY-SA 4.0 |
edited in response to ben wieland
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| Feb 21, 2019 at 10:39 | review | Suggested edits | |||
| S Feb 21, 2019 at 13:04 | |||||
| Feb 20, 2019 at 19:23 | comment | added | Ben Wieland | 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. | |
| Feb 20, 2019 at 7:55 | comment | added | rori | @BenWieland is every compact Hausdorff space weakly homotopy equivalent to a finite complex? | |
| Feb 18, 2019 at 20:37 | comment | added | Ben Wieland | 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 $K_0(R)\sim E(C)$. | |
| Feb 17, 2019 at 15:49 | history | edited | Arun Debray |
+at.algebraic-topology +algebraic-k-theory
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| Feb 17, 2019 at 12:17 | history | edited | rori | CC BY-SA 4.0 |
edited body; edited title
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| Feb 17, 2019 at 12:15 | comment | added | rori | @DenisNardin on second thought, I am not so sure passing to connective K-theory was necessary. It is true that in algebraic K-theory there is not going to be such $u$, but if we have two space $X_1$, $X_2$ such that $K^i(A)=KO^i(X_1)$, $K^{i+8}(A)=KO^{i+8}(X_2)$ there is not any sort of product between the topological K-theory groups. Basically the question becomes: do there exist obstructions on the abelian group level, rather than ring level? | |
| Feb 17, 2019 at 11:55 | history | edited | rori | CC BY-SA 4.0 |
added 134 characters in body
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| Feb 17, 2019 at 11:13 | comment | added | Denis Nardin | @rori The element still exists, but it's unclear to me why multiplication by it should give an isomorphism. Let me take complex K-theory for simplicity, the fiber sequence $Σ^2ku→ku→H\mathbb{Z}$ shows that there are potential obstructions in integral cohomology to it being an isomorphism | |
| Feb 17, 2019 at 11:12 | history | edited | rori | CC BY-SA 4.0 |
added 8 characters in body; edited title
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| Feb 17, 2019 at 11:05 | comment | added | rori | @DenisNardin maybe I am wrong, but won't the element still exist for $i>8$ even if we take connective cover? | |
| Feb 17, 2019 at 10:53 | comment | added | Denis Nardin | Regarding the main question: Bott periodicity is still a powerful obstruction. I don't know if you mean complex or real topological K-theory, but still in general there's no element $u\in K_8(A)$ such that $u:K_i(A)→K_{i+8}(A)$ is an isomorphism for all positive $i$'s. You could potentially try to fix this using connective topological K-theory though | |
| Feb 17, 2019 at 10:50 | comment | added | Denis Nardin | @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 $\mathcal{K}\hat\otimes A\cong A$ | |
| Feb 17, 2019 at 10:16 | comment | added | rori | @ThiKu I am only starting to learn this stuff. Does your remark immediately lead to an explicit counterexample/obstruction? | |
| Feb 17, 2019 at 10:06 | comment | added | ThiKu | 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. | |
| Feb 17, 2019 at 9:38 | history | edited | rori | CC BY-SA 4.0 |
added 12 characters in body
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| Feb 17, 2019 at 9:35 | comment | added | rori | @TKe understood, will formulate a weaker question | |
| Feb 17, 2019 at 9:27 | comment | added | user19475 | Topological $K$-theory satisfies Bott periodicity, algebraic $K$-theory not. | |
| Feb 17, 2019 at 9:21 | history | asked | rori | CC BY-SA 4.0 |