Let $G$ be a finite group and let $G$ act on a vector space $V$. Let $S(V)$ be the representation sphere. In the monoidal category of $RO(G)$-graded spectra $S(V)$ is invertible, so it's dualizable too.


Is $S(V)$ dualizable in the monoidal category of naive $G$-spectra?



Is a representation sphere dualizable inside naive G-spectra?
Let $G$ be of order $2$, and let $V$ be the nontrivial one-dimensional representation. I'll write $S(V)$ for the unit sphere in $V$ (which is just $G$); the one-point compactification of $V$ is then the unreduced suspension of $G$, which I'll call $S^V$. We can only form spectra from $G$-spaces with fixed basepoint, so we need to consider $S(V)\_+=G\_+$ rather than $S(V)$ itself. There is a cofibration $G\_+\to S^0\to S^V$, and $S^0$ is its own dual, so the question for $S^V$ is equivalent to the question for $G\_+$.


In the category of genuine $G$-spectra, $G\_+$ is its own dual, which means that there are unit and counit maps $S^0\xrightarrow{\eta}G\_+\wedge G\_+\xrightarrow{\epsilon}S^0$ making certain diagrams commute. In the naive category, $\{S^0,G\_+\wedge G\_+\}^G$ is the colimit of the unstable mapping sets $[S^n,S^n\wedge G\_+\wedge G\_+]^G$, which are trivial because $S^n\wedge G\_+\wedge G\_+$ is free away from the basepoint. Thus, we cannot construct the required map $\eta$, and $G\_+$ is not self-dual. The same applies if we replace one of the $G\_+$'s by something else, so $G\_+$ is not dualisable.


