Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The Barratt-Quillen-Priddy theorem says in one interpretation that there is a weak equivalence of spectra $K(FinSet) \simeq \mathbb{S}^0$. In other words K-theory groups of finite sets are the stable homotopy groups of spheres $\pi_*^s$.

If $A$ is a commutative ring, $K_0(A)$ has a simple definition as the free abelian group of projective finitely generated $A$-modules modulo exact sequences. On this group we use the exterior powers $\Lambda^k$ to get so-called Lambda-operations $\lambda^k$. These have nice properties and one can use them to alternatively construct Adams operations $\Psi^i$. This construction can be extended to all $K_n(A)$, giving $K_*(A)$ the structure of a Lambda-ring. This can found in sections II.4 and IV.5 of Weibel's book.

There is a strong analogy between finite sets and vector spaces. This tells you that an analogue of the exterior power $\Lambda^k$ is given by construction that sends a finite set $X$ to its set of $k$-element subsets ${X \choose k}$. This gives the standard Lambda-ring structure on $\mathbb{Z} = \pi_0^s$, i.e. the one on Wikipedia.

It seems that Weibel's construction of the Lambda-operations on higher K-theory groups works in this context as well. Is this correct? If so, we get $\lambda^i$ and $\Psi^i$ on the stable homotopy groups of spheres. What is known about these? Have they been used for anything?

share|improve this question
3  
I think that Stefan Schwede has studied this question, but I am not sure if he has published anything. –  Neil Strickland Nov 4 '12 at 19:24

3 Answers 3

up vote 14 down vote accepted

You can refine this. Let's take $k=2$ to give the idea. To a based set $X$ you can associate $(X\wedge X)/X$, a based set with free action of $\Sigma_2$. This leads to an operation going from stable homotopy of $S^0$ to stable homotopy of $B\Sigma_2$, such that when followed by transfer it gives the difference between the identity and squaring.

This leads to a proof of the Kahn-Priddy Theorem, a proof due to Kahn and Priddy I believe. (EDIT: No, I guess it was Segal.)

Waldhausen adapted the same idea to prove an important result about his $A(X)$, the "vanishing of the mystery homology theory". (That's why I know about it.) This involved extending the construction of these operations from the algebraic $K$-theory of sets to the algebraic $K$-theory of spaces.

share|improve this answer

The operation which sends a finite set $S$ to its set of $k$-element subsets, $\binom{S}{k}$, gives rise to the $k$-th stable Hopf invariant. There is additional structure in this: the set $\binom{S}{k}$ has a canonical $k$-fold covering so the operation is better viewed as a map $$ QS^0 \to Q(B\Sigma_k)_+ $$ rather than as a map $$ Q S^0 \to QS^0 , $$ where for a based space $X$, the space $QX$ is $\Omega^\infty\Sigma^\infty X$ is the representing space for the stable homotopy of $X$, i.e., $\pi_j(QX) = \pi_j^{\text{st}}(X)$. So the operation induces a homomorphism $$ \pi_j^{\text{st}}(S^0) \to \pi_j^{\text{st}}((B\Sigma_k)_+) . $$

These operations satisfy certain axioms (Cartan Formula, compatibility with transfers, etc.). A good place to read about these operations is:

Segal, Graeme: Operations in stable homotopy theory. New developments in topology (Proc. Sympos. Algebraic Topology, Oxford, 1972), pp. 105–110. London Math Soc. Lecture Note Ser., No. 11, Cambridge Univ. Press, London, 1974.

share|improve this answer

Dear Sander

Maybe this paper by Pierre Guillot can help you: http://arxiv.org/abs/math/0612327

share|improve this answer
4  
Let me add that, for some strange reason, when I uploaded this to the arxiv there was a bug with the latex compilation, and all the lambda's appear as L's. (On one other occasion I had the O in O_n, the orthogonal group, come out as the empty set. I think some commands such as \O and \l should not be redefined...). If you send me an email I can get you a copy with proper lambdas ! –  Pierre Nov 5 '12 at 9:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.