Let $f : X \rightarrow Y$ be a finite covering space. In Quillen's paper on the Adams conjecture in Topology, 1970, he gives the following argument for why

$$f_* \psi^p = \psi^p f_* \in KO[p^{-1}]$$

Quillen's argument is that the map $f_*$ can be interpreted as a Gysin homomorphism, and "since the normal bundle of $f$ has a canonical trivialization, this Gysin homomorphism is essentially the composite of one suspension isomorphism and the inverse of another".

I'd like to flesh out this last part. How does one make it rigorous?

  • $\begingroup$ I am confused with the notation: does $f_*$ refer to a map induced in $KO$-homology? I think a more geometric argument is that the operations $\psi^i$ can be seen as self maps of $BO$ or $BSO$. So, if $f_*$ denotes $KO$-homology of $f$ then it is natural. If you instead mean, $f_!$ then it is again some naturality property of transfer maps. $\endgroup$ – user51223 Feb 20 '17 at 6:45
  • $\begingroup$ It should be the transfer, which I guess is what you call $f_!$. (It goes from KO theory of X to Y.) I don't expect the equality to hold in total generality (probably there are twisting terms coming from the normal bundle), I would think it isn't completely formal from "naturality". $\endgroup$ – user84144 Feb 21 '17 at 0:02

Before I answer the question as stated, let me point out that you can also prove the claim as follows: the transfer is induced by a stable map $Y_+\rightarrow X_+$, and the pth Adams operation is a stable cohomology operation after inverting p, so the result is just naturality of cohomology operations.

Now for Quillen's argument. Embed $X$ into $S^n$ and so factor the map f as $X\rightarrow Y\times S^n \rightarrow Y$. If the normal bundle of the first map is, say, Spin oriented, then the Gysin map is defined as the composite: $KO(X) \rightarrow KO(Th(\nu)) \rightarrow KO(S^n \times Y) \rightarrow KO(Y)$. Here we use the Thom isomorphism, then the collapse map, then suspension isomorphism- I'm writing on a phone so I didn't put in the proper indices on the KO's or that some of those should be reduced K-theory.

Anyway, in this case the normal bundle in question is always trivial, and the Thom isomorphism for a trivial bundle is just the suspension isomorphism. This is what Quillen means by doing a suspension isomorphism and then undoing it.

The two arguments aren't unrelated of course, especially if you recall how the stable map yielding the transfer is constructed.

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    $\begingroup$ I've heard about this stable map but never seen the construction; can you give a reference? $\endgroup$ – user84144 Feb 20 '17 at 2:13
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    $\begingroup$ @user84144 It can be found in many different places, with many generalizations, but the relevant reference here is, I think, "Infinite loop spaces" by Adams, page 96-...; he refers to Eckmann's 1953 PNAS paper "On complexes with operators" $\endgroup$ – მამუკა ჯიბლაძე Feb 20 '17 at 6:39

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