TITLE: Adjoint of norm map
QUESTION [5 upvotes]: Let $G$ be a finite group acting trivially on a spectrum $X$. Then the norm map $X_{hG} \to X^{hG}$ may be thought of as a map $\Sigma^\infty_+ BG \wedge X \to F(\Sigma^\infty_+ BG , X)$. By adjointness, this corresponds to a map $\Sigma^\infty_+BG\wedge \Sigma^\infty_+ BG \wedge X \to X$. By naturality, it really suffices to consider the case where $X=S$ is the sphere spectrum, so we have a pairing $\Sigma^\infty_+ BG \wedge \Sigma^\infty_+ BG \to S$.
Question: What is this pairing?
All I can tell is that it restricts to the multiplication-by-$|G|$ map on $S$.
I'd be happy to understand the case $G = C_p$, $p$-locally.

REPLY [4 votes]: To elaborate the information in the comments: 
Inside $G\times G$ you have the diagonal subgroup $\Delta$.  This gives a map 
$$ p\colon E(G\times G)/\Delta\to E(G\times G)/(G\times G) $$
This is a covering map, so it induces a transfer map $p^!$ of suspension spectra in the opposite direction.  We can identify $E(G\times G)/(G\times G)$ with $B(G\times G)=BG\times BG$, and $E(G\times G)/\Delta$ with $B\Delta\simeq BG$, so we have $p^!\colon BG_+\wedge BG_+\to BG_+$.  There is also a collapse map $BG_+\to S^0$, and by composing these, we get a stable map $BG_+\wedge BG_+\to S^0$.  This is the map that you are asking about.  A proof is contained in Theorem 8.3 of my paper "K(n)-local duality for finite groups and groupoids".  As you mention, that paper works in the $K(n)$-local context, but that does not really play any role in the proof of Theorem 8.3; it is easy to extract a proof of the unlocalised statement.  Nonetheless, the proof is more complicated than I feel it ought to be; it would be interesting to revisit the question using more modern foundations.