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I want to know what is known about the cofibre of the $n$-fold transfer map $\mathbb{R}P^{\wedge n}_+\to S^0$, for $n>1$. I am happy to know of any specific example worked out. The case $n=1$ is known, due to James I think, to be $\Sigma\mathbb{R}P_{-1}$.

It appears that the below strategy does not work. But, I keep it for the comments below to make sense.

In particular, if it was possible to find some representation $U$ of $O(1)^{\times n}$ with a $O(1)^{\times n}$ invariant metric so that we have a homeomorphism $O(1)^{\times n}\to S(U)$ as $O(1)^{\times n}$-spaces, then I would be able to identify the cofibre (in some sense at least). Here, $S(U)$ is the sphere within $U$ with respect to the metric of $U$. Again, for the case $n=1$ we have $O(1)\to S(\mathbb{R})$. If you think such a representation $U$ cannot exist, which I think is the case, then how do you see this?

EDIT: I still, look forward other methods of identifying cofibre of the transfer maps, specifically those transfer maps associated to the diagonal embedding $1<G^{\times n}$ where we allow $G$ to be a compact Lie group. As far as I know, there is not much known about this? But, I appreciate any reference to any work or any unpublished work.

Addendum. For $n=2$, if I am not mistaken, I think playing with cofibrations, one can show that there exists a fibration of spectra as $$\mathbb{R}P_{-1}\to \mathbb{R}P_+\wedge\Sigma\mathbb{R}P_{-1}\to C_{\lambda^2}$$ where $\lambda^2:\mathbb{R}P^{\wedge 2}_+\to S^0$ is the double transfer. I think one can play around to find similar fibrations to fit in the coufibre of the other folded transfer maps.

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    $\begingroup$ $S(U)$ is a sphere, but $O(1)^{\times n}$ isn't, unless $n=1$. $\endgroup$ Sep 19, 2015 at 10:16
  • $\begingroup$ @OscarRandal-Williams Thanks. That is what I guess. But, how do you see this? That is, what is the obstruction for the existence of some representation $U$ so that $O(1)^{\times n}$ is homeomorphic to $S(U)$? $\endgroup$
    – user51223
    Sep 19, 2015 at 11:28
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    $\begingroup$ I have no idea what you mean here. $O(1)^n$ is a discrete set of $2^n$ points, so it cannot be homeomorphic to $S(U)$. $\endgroup$ Sep 19, 2015 at 13:08
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    $\begingroup$ This is more convincing, that the topology itself, and the nature of continuity is the main obstruction, that we can see by $\pi_0$. Since, I was looking at vector spaces with certain metric, I just felt there might be some `exotic' metric, invariant under the action, so that allows $O(1)^n$ to become a sphere, which now appears to be nonsense! $\endgroup$
    – user51223
    Sep 20, 2015 at 8:17

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