\section{\texorpdfstring{$R^*$}{R*} and \texorpdfstring{$PR^*$}{PR*} are not Mackey functors}
\subsection{The graded ring of \texorpdfstring{$A_4$}{A4}}
Let $A_4$ be the alternating group on $4$ elements, generated by the permutations $(12)(34)$ and $(123)$. There are $4$ irreducible complex representations of $A_4$:
\begin{itemize}
\item Of dimension 1: the trivial representation $\mathbb{1}$, and the representations $\rho$ (resp. $\bar{\rho}$) that send $(123)$ to $e^{2i\pi/3}$ (resp. $e^{-2i\pi/3}$) and $(12)(34)$ to $1$.
\item Of dimension 3: the representation $\theta$, which is the quotient of the representation $\bar{\theta}$ acting on $\CC^4$ by permutation of the basis vectors, by the trivial representation. The character $\chi_\theta$ of $\theta$ sends 3-cycles to 0 and $(12)(34)$ to $-1$.
\end{itemize}
There are the following relations between the representations:
\begin{align}
\rho^2 = \bar{\rho} \\
\rho\theta = \theta \\
\theta^2 = \mathbb{1}+\rho+\bar{\rho}+2\theta \label{A4_relationtheta2}
\end{align}
Additionally $\lambda^2(\theta) = \theta$ (by a direct calculation of the exterior power) and $\mathrm{det }(\theta) = \mathbb{1}$.
\begin{Conjecture} \label{A4_repringconjecture}
Let $x = c_1(\rho)$ and $y = c_2(\theta)$, then
\[
R^*_\CC(A_4) = \frac{\ZZ[x,y]}{(3x,12y,4y+x^2)}
\]
\end{Conjecture}
I do not have a complete proof of this. However, I can show:
\begin{Proposition}
Let $x = c_1(\rho)$ and $y = c_2(\theta)$, then
\begin{align*}
R^*_\CC(A_4)\otimes \FF_2 &= \FF_2[y] \\
R^*_\CC(A_4)\otimes \FF_3 &= \frac{\FF_3[x,y]}{(y+x^2)}
\end{align*}
\end{Proposition}
\begin{proof}\textbf{Generators.}
In degree $1$, we have $x = c_1(\rho) = -c_1(\bar{\rho})$ and $3x = 0$. Moreover $c_1(\theta) = c_1(\mathrm{det } \theta) = c_1(\mathbb{1}) = 0$, so $x$ is the only generator of degree 1.
In degree $2$ we let $y = c_2(\theta)$.
In degree $3$:
\[
C_3(\theta) = \gamma^3(\theta-3) = \lambda^3(\theta-1) = -1 +\theta -\lambda^2(\theta) + 1 = 0
\]
so there is no additional generator in degree 3.\\
\textbf{Relations.} Apply the total Chern class $c_t$ to both sides of (\ref{A4_relationtheta2}) to obtain:
\begin{align*}
c_t(\theta^2) &= c_t(1+\rho+\bar{\rho}+2\theta) \\
&= c_t(\rho)c_t(\bar{\rho})c_t(\theta)^2 \\
&= (1+xT)(1-xT)(1+yT^2)^2 \\
&= 1+ (2y-x^2)T^2 + (y^2-2yx^2)T^4 + zy^2T^6 \numberthis
\end{align*}
while on the left-hand side (looking only at even terms of degree $\leq 6$ and keeping in mind that $c_1(\theta) = c_3(\theta) = 0$):
\begin{align*}
c_t(\theta^2) &= c_t((\sigma_1+\sigma_2+\sigma_3)^2) \\
&= 1 + 6yT^2 + 9y^2T^4 + 4y^3T^6 \numberthis
\end{align*}
Putting the previous two equations together yields $4y = -x^2$. In particular this means that the order of $y$ is a multiple of $3$. To show it is not $3$, we restrict to $H:= (\ZZ/2)\times(\ZZ/2)$. By \cref{gradedrepring_Cpk} :
\[
R^*_\CC(\ZZ/2\times \ZZ/2) = \frac{\ZZ[t_1,t_2]}{(2t_1,2t_2,t_1^2t_2-t_1t_2^2)}
\]
Then $\res_H(y) = t_1^2 + t_1t_2 + t_2^2$, which has order $2$. So the order of $y^i$ is a multiple of $2$.
Finally, restricting $x$ to the subgroup generated by $(123)$ shows that $x$ is not nilpotent, as well as restricting $y$ and $yx$ to the subgroup generated by $(12)(34)$.
\end{proof}
$H = \ZZ/2\times \ZZ/2$ is a normal, abelian, $2$-Sylow of $G$. The image of $G$ under the restriction map
\[
\res_H^G: R^*_\CC(A_4) \longrightarrow R^*_\CC(\ZZ/2\times \ZZ/2)
\]
is generated by powers of $\res_H(y) = t_1^2 + t_1t_2 + t_2^2$, since $\res_H(x) = 0$. On the other hand, $G$ acts on $R^*_\CC(H)$ by cyclic permutations of the elements $t_1, t_2,t_1+t_2$. The element $z = t_1^3 + t_2^3+ t_1^2t_2$ is invariant under this action. But $z$ is not a combination of powers of $t_1^2 + t_1t_2 + t_2^2$ since it has odd degree, and thus does not belong to the image of the restriction map. Therefore:
\[
\Ima(\res_H^G) \subsetneq R^*_\CC(H)^{N_G(H)}
\]
which means that $R^*$ is not a Mackey functor.
\subsection{The permutation ring \texorpdfstring{$PR^*(A_4)$}{PR*(A4)}}
There are $5$ conjugacy classes of subgroups of $A_4$, which correspond to isomorphism classes of subgroups: $1, C_2, C_3, C_2\times C_2, A_4$. The permutation representations associated to each of these classes are:
\begin{itemize}
\item $\QQ\left[A_4/A_4\right] = \mathbb{1}$
\item $\QQ\left[A_4/C_2\times C_2\right] = \mathbb{1}+\rho+\overline{\rho}$
\item $\QQ\left[A_4/C_3\right] = \mathbb{1} +\theta$
\item $\QQ\left[A_4/C_2\right] = \mathbb{1} + \rho +\overline{\rho} + \theta$
\item $\QQ\left[A_4\right] = \mathbb{1} + \rho +\overline{\rho} + 3\theta$
\end{itemize}
So $PR(A_4)$ is generated by $\mathbb{1}, \rho + \overline{\rho}$ and $\theta$, and (with notation as above) the graded permutation ring $PR^*(A_4)$ is generated by $z = c_2(\theta)$ and $t = c_2(\rho+\overline{\rho})$. As we did above, we can apply the total Chern class $c_t$ to both sides of the equation:
\[
\theta^2 = 1 + \rho + \bar{\rho} + 2\theta
\]
and obtain that $4z = t$. The natural map $PR^*(A_4) \rightarrow R^*_\CC(A_4)$ induced by the inclusion $PR(A_4) \hookrightarrow R_\CC(A_4)$ maps $z$ to $y$, and thus the order of $z$ in $PR^*$ is a multiple of $2$ and $3$ (it is $12$ if \cref{A4_repringconjecture} is true), and $z$ is not nilpotent. We have proven:
\begin{Proposition}
Let $y = c_2(\theta)$, then
\begin{align*}
PR^*(A_4)\otimes \FF_2 &= \FF_2[y] \\
PR^*(A_4)\otimes \FF_3 &= \FF_3[y]
\end{align*} \qed
\end{Proposition}
The permutation ring $PR^*(C_2\times C_2)$ is isomorphic to $R^*_\CC(C_2\times C_2)$: there are $5$ conjugacy classes of subgroups of $C_2\times C_2$, which are the two trivial ones and three subgroups isomorphic to $C_2$; The permutation representations corresponding to these groups are $\rho_{(1,0)} + \rho_{(0,1)}$, $\rho_{(1,1)} + \rho_{(0,1)}$ and $\rho_{(1,0)} + \rho_{(1,1)}$ respectively. This means that the ring $PR^*$ is generated by $t_1+t_2$, $t_1 + (t_1+t_2) = t_2$ and $t_2 + (t_1+t_2) = t_1$. In other words,
\[
PR^*(C_2\times C_2) = \frac{\ZZ[t_1,t_2]}{(2t_1,2t_2,t_1^2t_2-t_1t_2^2)} = R^*_\CC(C_2\times C_2)
\]so Swan's lemma also fails for $PR^*$, as is shown by restriction of $y$.
