1
$\begingroup$

My questions are

  1. whether there exists a product structure for $K^{-1}(X)$? Here $K^{-1}$ is the odd topological $K$-group, and $X$ is a compact space (or a manifold), say.

  2. If such a ring structure exists then how to multiply two elements? For example, if we take $K^{-1}(X):=[X, GL(\mathbb{C})]$, where $GL(\mathbb{C})$ is the infinite general linear group, then for $f, g\in K^{-1}(X)$, I could have guessed that the product structure is given by $f\otimes g$, and I would think of it as the tensor product of two matrices of finite but not equal size.

  3. Moreover, if a ring structure for $K^{-1}$ really exists, may you provide me a reference?

I know $K^0(X)$ is a ring and $K^0(X)\oplus K^{-1}(X)$ is a ring. But I don't know whether $K^{-1}(X)$ along is a ring or not.

Thanks a lot.

$\endgroup$
8
  • 5
    $\begingroup$ The product of two elements of $K^{-1}(X)$ belongs to $K^{-2}(X)$ which is identified with $K^0(X)$ via the Bott periodicity. Thus $K^{-1}(X)$ is not a ring. $\endgroup$
    – asv
    Commented May 28, 2014 at 9:36
  • $\begingroup$ You might try to use the suspension isomorphism $\tilde K^{-1}(X)\cong\tilde K(\Sigma X)$... $\endgroup$ Commented May 28, 2014 at 9:38
  • $\begingroup$ @semyonalesker: Your comment does not prove that there is no natural ring structure on $K^{-1}$. $\endgroup$ Commented May 28, 2014 at 18:06
  • $\begingroup$ @StevenLandsburg: You are right. But I hoped that it does prove that there is no KNOWN canonical ring structure. $\endgroup$
    – asv
    Commented May 28, 2014 at 18:40
  • 1
    $\begingroup$ @მამუკაჯიბლაძე In your first comment you suggest to consider the ring structure coming from $\tilde K(\Sigma X)$. I just want to point out that any cup product on a suspension is trivial, no matter in which extraordinary multiplicative cohomology theory you work. On the other hand, suggestion 2 in the question seems to make sense and might actually produce a ring. $\endgroup$ Commented Dec 31, 2015 at 19:21

0

You must log in to answer this question.