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Let us view topological K-theory as a functor $K$ from the cateory of compact pairs (that is, a compact Hausdorff spaces with a distinguished closed subset) to the category of $\mathbb Z/2$-graded Abelian groups. We could also restrict to second countable spaces and thus countable groups.

An additive operation on topological K-theory is just a natural transformation from $K$ to $K$. These natural transformations form an Abelian group under addition.

Question: What is the isomorphism class of this group?

Examples of operations are discussed in Efton Park's book and in Max Karoubi's book, but I cannot find discussion of the collection of all (additive) operations.

Edit: After looking at Karoubi's book again, I have to state that it actually contains a very satisfactory treatment of operations in K-theory.

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I thought the answer is that this is a ring and that as a ring it is isomorphic to symmetric functions. – Bruce Westbury Feb 21 '11 at 12:11
@Bruce Westbury: A ring with respect to composition? – Rasmus Bentmann Feb 21 '11 at 12:26
Aren't symmetric functions related to unstable operations? – Loop Space Feb 21 '11 at 13:30
up vote 5 down vote accepted

The most explicit answers are in work of Sarah Whitehouse and her collaborators. You could start with this paper and its references:

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Surprisingly, the group is uncountable. See

Adams, J. F.; Clarke, F. W.

Stable operations on complex $K$-theory.

Illinois J. Math. 21 (1977), no. 4, 826–829.

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I don't recall, do they cover additive operations explicitly in this paper? – Loop Space Feb 21 '11 at 13:29
I have to admit to not having actually read the paper...but doesn't uncountably many degree $0$ stable operations imply uncountably many additive ones (since stable operations are always additive)? – Mark Grant Feb 21 '11 at 13:40
Mark, yes it does. I just couldn't remember if that paper had any more detail on the specifics of additive operations. From what I understand, the request "Please write down a stable operation (other than generated by the two obvious ones)" would still generate a shrug, whereas we know lots about additive operations, so it's worth a paper to say, "There are uncountably many stable operations" but not to say, "There are uncountably many unstable operations". – Loop Space Feb 21 '11 at 14:03
(I meant "additive operations" for the last two words) – Loop Space Feb 21 '11 at 14:04
Andrew, no I don't think they consider additive operations explicitly. – Mark Grant Feb 21 '11 at 14:39

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