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1. • CommentRowNumber1.
   • CommentAuthorDavidRoberts
   • CommentTimeAug 18th 2016
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexI've added a proof that rank 0 virtual vector bundles are nilpotent elements in K-theory to [[virtual vector bundle]], and made some small edits to [[topological K-theory]]. The latter page had the definition of K-theory mixed up with the definition of _reduced_ K-theory, so I made a small correction there, but the proof that $K(X)$ is a ring is not there now, as the existing proof was for reduced K-theory only. I will edit this in the coming days. Incidentally, for 'compact support' vector bundle K-theory of a non-compact space (so **not** representable K-theory), this implies that the $\tilde K(X)$-adic topology is discrete, as opposed to the case where for non-compact spaces representable K-theory has something interesting going on (cf work of Atiyah-Segal on equivariant K-theory and completion, where they compute $K(BG)$ in the case of representable K-theory for $G$ compact).

   I’ve added a proof that rank 0 virtual vector bundles are nilpotent elements in K-theory to virtual vector bundle, and made some small edits to topological K-theory. The latter page had the definition of K-theory mixed up with the definition of reduced K-theory, so I made a small correction there, but the proof that $K(X)$ is a ring is not there now, as the existing proof was for reduced K-theory only. I will edit this in the coming days.

   Incidentally, for ’compact support’ vector bundle K-theory of a non-compact space (so not representable K-theory), this implies that the $\tilde K(X)$-adic topology is discrete, as opposed to the case where for non-compact spaces representable K-theory has something interesting going on (cf work of Atiyah-Segal on equivariant K-theory and completion, where they compute $K(BG)$ in the case of representable K-theory for $G$ compact).

2. • CommentRowNumber2.
   • CommentAuthorDmitri Pavlov
   • CommentTimeAug 18th 2016
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   Author: Dmitri Pavlov
   Format: MarkdownItexWhy does e^{n+1} belong to K_{⋃Y_i ∪ Z}(X) in this proof?

   Why does e^{n+1} belong to K_{⋃Y_i ∪ Z}(X) in this proof?

3. • CommentRowNumber3.
   • CommentAuthorDavidRoberts
   • CommentTimeAug 18th 2016
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   Author: DavidRoberts
   Format: MarkdownItexWhoops, the fact mentioned in the first sentence of the proof had $\cap$ instead of $\cup$. It should be ok now.

   Whoops, the fact mentioned in the first sentence of the proof had $\cap$ instead of $\cup$. It should be ok now.