\section{Classic graded representation rings}
\subsection{Theory}
\subsubsection{Definitions, first properties}
\paragraph*{Notation.} I write $R^*_\KK(G)$ for the graded representation ring associated to $G$ with coefficients in the field $\KK$. When $a$ (lower case) is an element of the graded ring $R^*_\KK(G)$, denote by $A$ (upper case) any lift of $a$ to $R_\KK(G)$.\\
The induction of representations from a subgroup $H$ to a bigger group $G$ is denoted $\ind_H^G$, or $i_*$ for short (when $H,G$ are obvious from the context). Similarly for the restriction $\res_H^G$ or $i^*$.
\begin{Definition*}
The \textit{weight} of a monomial $t = c_{i_1}(x_1)\cdots c_{i_n}(x_n)$, is $w = \sum i_k$. The degree of $t$ is $n$.
\end{Definition*}
\begin{Definition*}(See \cite{webb}).Let $R$ be a commutative ring with a $1$ and let $R$-$\Mod$ be the category of $R$-modules. Let $G$ be a finite group. A \textit{Mackey functor} over $R$ is a function
\[
M: \{\text{subgroups of } G\} \longrightarrow R\text{-}\Mod
\]
with, for all subgroups $K\leq H\leq G$ and for all elements $g \in G$, morphisms:
\begin{align*}
I_K^H: M(K) & \longrightarrow M(H) \\
R_K^H: M(H) & \longrightarrow M(K) \\
c_g : M(H) & \longrightarrow M(^gH)
\end{align*}
where $^gH = gHg^{-1}$, such that, for all subgroups $J\leq K \leq H \leq G$:
\begin{enumerate}
\item $I_H^H, R_H^H, c_h: M(H) \rightarrow M(H)$ are the identity morphisms for all subgroups $H$ and for all $h\in H$.
\item $R_J^KR_K^H = R_J^H$
\item $I_K^HI_J^K = I_J^H$
\item $c_hc_h = c_{gh}$ for $g,h\in G$
\item $R_{^gK}^{^gH}c_g = c_gR_K^H$
\item $I_{^gK}^{^gH}c_g = c_gI_K^H$
\item $R_J^HI_K^H = \sum\limits_{x\in [J\backslash H / K]} I_{J\cap ^xK}^{J}c_xR_{J^x\cap K}^{K}$
\end{enumerate}
\end{Definition*}
\begin{Proposition} \label{R*G_G-torsion}
The graded representation ring $R^*(G)$ is of $\vert G \vert$-torsion.
\end{Proposition}
\begin{proof}
\todo{To be added.}
\end{proof}
\begin{Proposition} \ref{R*G_G-torsion}
Let $G$ and $H$ be groups with coprime order. Then
\[
R^*(G\times H) = R^*(G) \otimes_\ZZ R^*(H)
\]
\end{Proposition}
\begin{proof}
Any irreducible representation $\rho$ of $G\times H$ is of the form $\rho_G\otimes \rho_H$ for some irreducible representations $\rho_G$ of $G$ and $\rho_H$ of $H$. Thus $R(G\times H) = R(G) \otimes_\ZZ R(H)$, and the filtration on $R(G\times H)$ is
\[
\Gamma^n(G\times H) = \bigoplus_{i=0}^n \Gamma^i(G)\otimes \Gamma^{n-i}(H).
\]
Since $R^k(G)$ (resp. $R^k(H)$) is of $\vert G \vert$ (resp. $\vert H\vert$)-torsion for $k>0$, the terms $\Gamma^i(G)\otimes \Gamma^{n-i}(H)$ vanish mod $\Gamma^{n+1}(G\times H)$ whenever $i \neq 0,n$, so that:
\[
R^n(G\times H) = \frac{\left( \Gamma^0(G)\otimes \Gamma^n(H) \right) \oplus \left( \Gamma^n(G)\otimes\Gamma^0(H) \right)}{\left( \Gamma^0(G)\otimes \Gamma^{n+1}(H) \right) \oplus \left( \Gamma^{n+1}(G)\otimes\Gamma^0(H) \right) } \cong R^n(G) \oplus R^n(H).
\]
Thus:
\begin{align*}
R^*(G\times H) &= \ZZ \oplus \left( \bigoplus_{i\geq 1} R^i(G)\oplus R^i(H) \right) \\
&= R^*(G) \otimes R^*(H)
\end{align*}
\end{proof}
\subsubsection{Evaluation of characters, topology}
\paragraph{Topological considerations.} In the sequel, we view $R(G)$ as a topological ring, with the topology induced by the filtration $\lbrace \Gamma^n \rbrace$; that is, a subset $U \subseteq R(G)$ is open if for any $x\in U$ there is a $t$ such that $x + \Gamma^t \subseteq U$.
\begin{Proposition}[{\cite[Cor. 12.3]{atiyah-characters_cohomology}}] \label{gammatopology_coincides_idaictopology}
The topology induced by the $\Gamma$-filtration coincides with the $I$-adic topology, where $I$ is the augmentation ideal of $R(G)$.
\end{Proposition}
\begin{proof}
\todo{To be added.}
\end{proof}
If $G$ has exponent $m$, then each conjugacy class representative $g \in G$ gives rise to a ring morphism:
\[
\phi_g : \begin{cases} R(G) \rightarrow \ZZ[\mu_m] \\
\ \ \rho \mapsto \chi_\rho(g) \end{cases},
\]
where $\mu_m$ is a choice of primitive $m$-th root of unity. We are interested in continuity and density questions with respect to $p$-adic topologies on $\ZZ$. Note that our choice of primitive root $\mu_m$ fixes asn extension of the $p$-adic valuation to $\ZZ[\mu_m]$, for any prime number $p$.\\
Suppose we are given groups $\widetilde{\Gamma}^n$ $(n\geq 1)$ such that:
\begin{enumerate}[A.]
\item $\widetilde{\Gamma}^{n+1} \subseteq \widetilde{\Gamma}^n$ \label{admissiblefiltration1}
\item $\Gamma^n = \widetilde{\Gamma}^n + \Gamma^{n+1}$ \label{admissiblefiltration2}
\end{enumerate}
(think of $\widetilde{\Gamma}^n$ as a truncated $\Gamma^n$). Then by an immediate induction:
\begin{Lemma}
For all $k\in\mathbb{N}$,
\[
\Gamma^n = \widetilde{\Gamma}^n + \Gamma^{n+k}
\] \qed
\end{Lemma}
\begin{Definition*}
Call $(\widetilde{\Gamma}^n)_n$ an \textit{admissible approximation} for $(\Gamma^n)_n$ if it satisfies the conditions above.
\end{Definition*}
\begin{Proposition}\label{evaluation_approximation}
Let $p$ be a prime number, and fix an extension suppose the evaluation morphisms
\[
\phi_1,\cdots \phi_k : R(G) \mapsto \ZZ[\mu_m]
\]
are continuous with respect to the topology induced by the filtration $\lbrace \Gamma^n \rbrace$ on $R(G)$, and the $p$-adic topology on $\ZZ[\mu_m]$. Then for all $x \in \Gamma^n$, there is an element $\widetilde{x} \in \widetilde{\Gamma}^n$ such that for all $i = 1,\cdots,k$:
\[
v_p(\phi_i(x)) = v_p(\phi_i(\widetilde{x}))
\]
\end{Proposition}
\begin{proof}
Let $x \in \Gamma^n$. Write $x = \widetilde{x}+r$ with $\widetilde{x}\in\widetilde{\Gamma}^n$ and $r \in \Gamma^N$, choosing $N$ large enough that $v_p(\phi_i(r)) > v_p(\phi_i(x))$ for all $i$. So
\[
v_p(\phi_i(x)) \geq \min\lbrace v_p(\phi_i(\widetilde{x})), v_p(\phi_i(r)) \rbrace
\]
with equality if $v_p(\phi_i(\widetilde{x})) \neq v_p(\phi_i(r))$.
Now, we must have
\[
v_p(\phi_i(\widetilde{x})) < v_p(\phi_i(r))
\]
since otherwise $v_p(\phi_i(\widetilde{x})+\phi_i(r)) \geq v_p(\phi_i(r)) > v_p(\phi_i(x))$. Thus
\[
v_p(\phi_i(x)) = v_p(\phi_i(\widetilde{x}))
\]
\end{proof}
\begin{Proposition}
Let $G$ be a $p$-group. Then the morphisms $\phi_g$, for $g \in G$, are all continuous with respect to the $p$-adic topology on $\ZZ[\mu_m]$.
\end{Proposition}
\begin{proof}
Fix an element $g \in G$ and let $|G| =: p^n$. By \cref{gammatopology_coincides_idaictopology}, it suffices to show that $\phi_g$ is continuous with respect to the $I$-adic topology on the left, and it is enough to show that for any irreducible character $\rho$ of $G$,
\[
v_p(\phi_g(\rho - \varepsilon(\rho))) > 0.
\]
Since $G$ is a $p$-group, every character is a sum of $p^n$-th roots of unity, so
\begin{align*}
\phi_g(\rho-\varepsilon(\rho)) &= \rho(g) - \varepsilon(\rho) \\
&= \sum_{l = 1}^{\varepsilon(\rho)}(\mu_{p^n}^{i_l} - 1),
\end{align*}
and each $(\mu_{p^n}^{i_l} - 1)$ has positive valuation.
\end{proof}
\subsection{Examples}
\subsubsection{From the definitions}
\paragraph{Cyclic groups.}
\begin{Proposition}[{\cite[Prop. 3.4]{guillot-minac}}]
Let $C_N$ be the cyclic group of order $N$. Whenever $\KK$ is an algebraically closed field of characteristic prime to $N$,
\[
R^*_\KK(C_N) = \frac{\ZZ\left[z\right]}{(Nz)}
\]
with $\vert z \vert = 1$ and $z = c_1(\rho)$ for some nontrivial, one-dimensional representation $\rho$ of $C_N$.
\end{Proposition}
\begin{Proposition}
\[
R^*(C_N,\RR) = \begin{dcases} \ \ \ \ \ \ \ \ \frac{\ZZ\left[t\right]}{(Nt)} &N = 2m+1 \\
\ \ \ \ \ \frac{\ZZ\left[t,s\right]}{(Nt,2s,s^2)} &N \equiv 0 \ (\mathrm{mod} \ 4) \\
\frac{\ZZ\left[t,s\right]}{(Nt,2s,s^2-\frac{N}{2}t)} &N \equiv 2 \ (\mathrm{mod} \ 4) \end{dcases}
\]
where $\vert t \vert = 2$ and $\vert s \vert = 1$.
\end{Proposition}
\begin{proof}\textit{Representations.} Let $g$ be a choice of generator of $C_N$. The real representations of $C_N$ are obtained by pairing complex representations, and are of the form:
\[
\rho_k(g) = \begin{pmatrix} \cos\left(\frac{2k\pi}{N}\right) & -\sin\left(\frac{2k\pi}{N}\right) \\
\sin\left(\frac{2k\pi}{N}\right) & \cos\left(\frac{2k\pi}{N}\right)
\end{pmatrix}
\]
When $N = 2m+1$ is odd, we obtain $m$ irreducible representations in this way. When $N = 2m$, we have $(m-1)$ such representations, as well as the $1$-dimensional representation $\varepsilon: g \mapsto -1$. We have the relations:
\begin{align}
\rho_k\rho_l &= \rho_{k+l}+\rho_{k-l} \label{CN_rhokrhol} \\
\varepsilon^2 &= \mathbb{1} \label{CN_epsilon2}\\
\varepsilon\rho_k &= \rho_{k+m} \label{CN_epsilonrho}
\end{align}
where the last two relations, of course, only apply when $N = 2m$, and indices are taken mod $N$.
\textit{Generators and relations.} Computing gamma operations explicitly, we have
\begin{align*}
C_1(\rho_k) &= \rho_k -2 \\
C_2(\rho_k) &= \gamma^2(\rho_k-2) = \lambda^2(\rho_k-1) = \lambda^2(\rho_k) - \rho_k +1 = \rho_k + 2 \\
C_1(\varepsilon) &= \varepsilon -1
\end{align*}
We see immediately that $C_1(\rho_k) = -C_2(\rho_k) \in \Gamma^2$ and so $c_1(\rho_k) = 0$.
For the other relations, we use \cref{CN_rhokrhol} above and apply the total Chern class $c_t$ to both sides:
\begin{equation}
c_t(\rho_k\rho_l) = c_t(\rho_{k+l})c_t(\rho_{k-l}) \label{CN_totalclass}
\end{equation}
Let $y_i = c_2(\rho_i)$. Expand the right-hand side:
\begin{align*}
c_t(\rho_{k+l})c_t(\rho_{k-l}) &= (1 + y_{k+l}T^2)(1 + y_{k-l}T^2) \\
&= 1 + (y_{k+l} + y_{k-l})T^2 + y_{k+l}y_{k-l}T^4 \numberthis \label{CN_ctrhs}
\end{align*}
For the left-hand side, we use the splitting principle. In some extension of $R(D_p)$, we can write $\rho_k = \sigma_1 + \sigma_2$ and $\rho_l = \eta_1 + \eta_2$ with $\sigma_i$, $\eta_i$ of dimension 1. Then
\begin{align*}
c_t(\rho_k\rho_l) &= c_t((\sigma_1+\sigma_2)(\eta_1+\eta_2)) \\
&= c_t(\sigma_1\eta_1)c_t(\sigma_1\eta_2)c_t(\sigma_2\eta_1)c_t(\sigma_2\eta_2) \\
&= (1+(c_1(\sigma_1)+c_1(\eta_1))T)\cdot(1+(c_1(\sigma_1)+c_1(\eta_2))T)\cdot(1+(c_1(\sigma_2)+c_1(\eta_1))T)\cdot(1+(c_1(\sigma_2)+c_1(\eta_2))T)
\end{align*}
Now, let $s_1, s_2$ (resp. $t_1, t_2$) be the first and second symmetric polynomials in $(\sigma_1, \sigma_2)$ (resp. ($\eta_1,\eta_2)$). Then $c_i(\rho_k) = s_i$ and $c_i(\rho_l) = t_i$. The last equality can be rewritten:
\begin{align*}
c_t(\rho_k\rho_l) =& 1+ 2(s_1+t_1)T + (t_1^2+s_1^2+3s_1t_1+2t_2+2s_2)T^2 \\
&+(s_1^2t_1+s_1t_1^2+2s_1s_2+2t_1t_2+2s_1t_2 + 2s_2t_1)T^3 \\
&+ (t_2^2 + s_2^2 + s_1s_2t_1 + s_1t_1t_2 + s_1^2t_2 + s_2t_1^2 -2s_2t_2)T^4
\end{align*}
We eliminate all occurrences of $s_1 = t_1 = 0$ to obtain:
\begin{equation}
c_t(\rho_k\rho_l) = 1 + (2y_k+2y_l)T^2 + (y_k-y_l)^2T^4 \label{CN_ctlhs}
\end{equation}
Comparing coefficients in \cref{CN_ctrhs} and \cref{CN_ctlhs}, we obtain:
\begin{align}
y_{k+l}+y_{k-l} &= 2(y_k + y_l) \label{CN_relationsum} \\
y_{k+l}y_{k-l} &= (y_k-y_l)^2
\end{align}
Now plugging $l=k$ and $l=k+1$ in relation (\ref{CN_relationsum}), we get
\begin{align*}
y_{2k} &= 4y_k \\
y_{2k+1} &= 2y_k+2y_{k+1} - y_1
\end{align*}
which by induction gives $y_k = k^2y_1$. Let $t =y_1$. Since the order $N$ of the group $C_N$ annihilates $R^*(C_N)$, we have $Nt = 0$. To show that the order of $t$ is $N$ we look at the complexification morphism $\psi: R^*(C_N, \RR) \rightarrow R^*(C_N,\CC)$ which sends $t$ to $-x^2$, where $x$ is the degree 1 generator of $R^*(C_N,\CC)$. This incidentally shows that $t$ is not nilpotent. Thus if $N$ is odd, the ring is generated by $t$ alone, and we have indeed:
\[
R^*(C_N,\RR) = \frac{\ZZ\left[t\right]}{(Nt)}
\]
Now if $N = 2m$ is even, we take into account the other generator $s = c_1(\varepsilon)$. The fact that $\varepsilon^2 = 1$ implies immediately that $2s = 0$, and we only need to apply $c_t$ to relation \ref{CN_epsilonrho} to derive:
\[
y_{k+m} = y_k + s^2
\]
in which we can plug $y_k = k^2t$ to obtain
\[
s^2 = m(2k+m)t
\]
for all $k = 1, \cdots, m-1$.
\begin{itemize}
\item If $m$ is itself even, then this can be rewritten as $s^2 = 2m(k+\frac{m}{2})t = 0$, as $n = 2m$ kills all elements in $R^*(C_N,\RR)$. Thus we have the relation $s^2 = 0$. Moreover, since the complexification morphism $\psi$ sends $s = c_1(\epsilon)$ to $mx \in R^*(C_N,\CC)$, we have $\psi(st^i) = -mx^{2i=1} \neq 0$, thus
\[
R^*(C_N,\RR) = \frac{\ZZ\left[t,s\right]}{(Nt,2s,s^2)}
\]
\item If $m = 2l+1$ for some $l$ then $s^2 = 2m(k+s)t + mt = mt$, and yet another application of $\psi$ yields:
\[
R^*(C_N,\RR) = \frac{\ZZ\left[t,s\right]}{(Nt,2s,s^2-\frac{N}{2}t)}
\]
\end{itemize}
\end{proof}
\paragraph{Elementary abelian groups.} Let $p$ be a prime number, and let $C_p$ be the cyclic group of order $p$, generated by $g$.
\begin{Proposition} \label{gradedrepring_Cpk}
Let $G = C_p^k$. Then
\[
R^*_\CC(G) = \frac{\ZZ[x_1,\cdots,x_k]}{(px_i,x_i^px_j = x_ix_j^p)}
\]
\end{Proposition}
\begin{proof}
Denote by $g_i$ the element $(1,\cdots,1,g,1,\cdots,1)$ with $g$ in $i$-th position. Thus $G$ is generated by $g_1,\cdots,g_k$. Let $\omega = \exp^{2i\pi/p}$. There are $p^k$ irreducible complex representations of $G$, of the form $\rho_v$, with $v = (v_1,\cdots,v_k) \in \ZZ^k_{\geq 0}$, and $\rho_v(g_i) = \omega^{v_i}$. These satisfy the two relations:
\begin{align}
\rho_v\rho_{v'} &= \rho_{v+v'} \label{Cpk_rhovrhov'}\\
\rho_v^p &= 1. \label{Cpk_rhovp}
\end{align}
Let $x_v = c_1(\rho_v)$, and let $\rho_i$ be the representation associated to $w = (0,\cdots,0,1,0\cdots,0)$ with a $1$ in position $i$. Then by (\ref{Cpk_rhovrhov'}), we have
\[
x_v = \sum_{i=1}^kv_i\cdot x_i
\]
thus the $x_i$ generate $R^*_\CC(G)$. The relation (\ref{Cpk_rhovp}) yields $px_i = 0$ for any $i$.\\
The relation $x_i^px_j = x_ix_j^p$ is obtained as follows: let $X_i$ be the standard lift $\rho_i - 1$ of $x_i$ to $R_\CC(G)$. Then $(X_i+1)^p = 1$. Thus
\begin{align*}
X_i^p &= -\sum_{l=1}^{p-1}\binom{p}{l}X_i^l = X_i\left(-\sum_{l=0}^{p-2}\binom{p}{l+1}X_i^l\right) \\
&= pX_i(-1+\phi(X_i)),
\end{align*}
where $\phi(T) \in \ZZ[X]$ has no constant term. Thus for any $i,j$, we have
\begin{align*}
X_i^pX_j\left( -1 + \phi(X_j) \right) &= pX_iX_j\left(-1+\phi(X_i)\right)\left(-1+\phi(X_j)\right) \\
&= X_iX_j^p\left(-1+\phi(X_i)\right)
\end{align*}
That is, quotienting by $\Gamma^{p+2}$ on each side:
\[
x_i^px_j = x_ix_j^p.
\]
This shows that the generators of $R^*_\CC(G)$ satisfy all the required relations.
\paragraph*{}We now show that these are the only relations: the graded piece of rank $l$ is generated by monomials of the form
\[
x_1^{s_1}\cdots x_k^{s_k}, \ \ \ \ \sum_{i=1}^{k} s_i = l
\]
Now let $S_l \subset \ZZ^k_{\geq 0}$ be the set of multi-indices $(s_1,\cdots,s_k)$ such that $\sum s_i = l$ and only the first nonzero coordinate of each $s\in S_l$ is (possibly) greater than $p-1$. To show that the relations above are the only one, we must show that the monomials $x^s = x_1^{s_1}\cdots x_k^{s_k}$ are linearly independent. So let us consider a zero linear combination of those:
\begin{align}
\sum_{s\in S_l}a_sx^s = 0 \label{Cpk_linearcombination}
\end{align}
Consider the restriction $\psi: R^*_\CC(C_p^k) \rightarrow R^*_\CC(C_p)$ to the group generated by $g_1^{i_1}\cdots g_k^{i_k}$ for some $0\leq i_j \leq p-1$. Then $\psi(x_j) = i_jz$, where $z$ is the standard one-degree generator of $R^*_\CC(C_p)$. Then \cref{Cpk_linearcombination} becomes:
\begin{equation*}
\sum_{s\in S_l}a_si_1^{s_1}\cdots i_k^{s_k}z^l = 0,
\end{equation*}
that is,
\begin{equation}
\sum_{s\in S_l}a_si_1^{s_1}\cdots i_k^{s_k} = 0 \in \Fp, \label{Cpk_linearcombinationFp}
\end{equation}
for all possible strings $(i_1,\cdots,i_k)$ with $0\leq i_j \leq p-1$. In particular, grouping terms by powers of $i_k$ in \cref{Cpk_linearcombinationFp}, we get:
\begin{align}
\begin{dcases} \ \ \ a_{(0,\cdots,0,l)}i_k^{l} = 0 &\text{ when } i_1 = \cdots = i_{k-1} = 0 \\
\ \ \ \sum_{t=0}^{p-1}\left(\sum_{s\in S_{l-t}} b_si_1^{s_1}\cdots i_{k-1}^{s_{k-1}} \right) i_k^t = 0 &\text{otherwise.}
\end{dcases}
\end{align}
This implies that the coefficient of $x_k^l$ in \cref{Cpk_linearcombination} is zero; more generally, the second equation must be true for all values of $i_k$, from $0$ to $p-1$. In other words, the $\left( \sum b_si_1^{s_1}\cdots i_{k-1}^{s_{k-1}} \right)$ are the entries of a vector in the kernel of the Vandermonde matrix $(i_k^t)_{i_k = 1,\cdots,p-1}^{t = 1,\cdots,p-1}$, which is invertible in $\Fp$. Therefore
\[
\sum_{s\in S_{l-t}} b_si_1^{s_1}\cdots i_{k-1}^{s_{k-1}} = 0
\]
for all combinations $(i_1,\cdots,i_{k-1})$. An immediate induction shows that we must have each $a_s = 0$, so the monomials $\lbrace x^s\rbrace_{s\in S_l}$ are linearly independent.
\end{proof}
\paragraph{(Some) dihedral groups.} Let $p$ be an odd prime and consider the dihedral group
\[
D_p = \left\langle \tau, \sigma \vert \tau^2, \sigma^p, \tau\sigma\tau = \sigma^{-1} \right\rangle.
\]
There are $(p+1)/2$ irreducible representations of $D_p$:
\begin{itemize}
\item Two representations of degree 1, the trivial representation 1 and the "signature" $\varepsilon$ which sends elements of the form $\sigma^j$ to -1, and elements of the form $\tau\sigma^j$ to 1.
\item And $(p-1)/2$ representations $\chi_1, \cdots \chi_{(p-1)/2}$ of degree 2. These are defined over the reals as:
\begin{align*}
\chi_k(\sigma^j) &= \begin{pmatrix} \cos(\frac{2kj\pi}{p}) & -\sin(\frac{2kj\pi}{p}) \\
\sin(\frac{2kj\pi}{p}) & \cos(\frac{2kj\pi}{p}) \end{pmatrix} \\
\chi_k(\tau) &= \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}
\end{align*}
And over the complex numbers:
\begin{align*}
\chi_k(\sigma^j) &= \begin{pmatrix} e^{\frac{2ikj\pi}{p}} & 0 \\
0 & e^{-\frac{2ikj\pi}{p}}\end{pmatrix} \\
\chi_k(\tau) &= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}
\end{align*}
\end{itemize}
These generate the ring $R(D_p)$. For convenience, define $\chi_0 = 1+\varepsilon$, then we have the following relations:
\begin{align}
\varepsilon^2 &= 1 \label{Dp_order_epsilon} \\
\varepsilon\cdot\chi_k &= \chi_k \\
\chi_k\cdot\chi_l &= \chi_{k+l} + \chi_{k-l} \label{Dp_chikchil}
\end{align}
\begin{Proposition}
Let $c_1(\chi_1) = x$ and $c_2(\chi_1) = y$, then
\[
R^*(D_p) = \frac{\ZZ\left[x,y\right]}{(2x,py,xy)}
\]
where $\vert x \vert = 1$ and $\vert y \vert = 2$.
\end{Proposition}
\begin{proof}
As above, note that $0 = c_1(1) = c_1(\varepsilon^2) = 2c_1(\varepsilon)$ while $c_1(\chi_k) = c_1(\mathrm{det} \ \chi_k) = c_1(\varepsilon)$ for any $k$.\\
So let $x = c_1(\chi_k) = c_1(\varepsilon)$.\\
For the other relations, we use \cref{Dp_chikchil} above and apply the total Chern class $c_t$ to both sides:
\begin{equation}
c_t(\chi_k\chi_l) = c_t(\chi_{k+l})c_t(\chi_{k-l}) \label{Dp_totalclass}
\end{equation}
Let $y_i = c_2(\chi_i)$. Expand the right-hand side:
\begin{align*}
c_t(\chi_{k+l})c_t(\chi_{k-l}) &= (1+xT+y_{k+l}T^2)(1+xT+y_{k-l}T^2) \\
&= 1 + 2xT + (x^2+ y_{k+l} + y_{k-l})T^2 + (xy_{k+l} + xy_{k-l})T^3 + y_{k+l}y_{k-l}T^4 \numberthis \label{Dp_ctrhs}
\end{align*}
For the left-hand side, we use the splitting principle. In some extension of $R(D_p)$, we can write $\chi_k = \rho_1 + \rho_2$ and $\chi_l = \eta_1 + \eta_2$ with $\rho_i$, $\eta_i$ of dimension 1. Then
\begin{align*}
c_t(\chi_k\chi_l) &= c_t(\rho_1+\rho_2)(\eta_1)+\eta_2)) \\
&= c_t(\rho_1\eta_1)c_t(\rho_1\eta_2)c_t(\rho_2\eta_1)c_t(\rho_2\eta_2) \\
&= (1+(c_1(\rho_1)+c_1(\eta_1))T)\cdot(1+(c_1(\rho_1)+c_1(\eta_2))T)\cdot(1+(c_1(\rho_2)+c_1(\eta_1))T)\cdot(1+(c_1(\rho_2)+c_1(\eta_2))T)
\end{align*}
Now, let $s_1, s_2$ (resp. $t_1, t_2$) be the first and second symmetric polynomials in $(\rho_1, \rho_2)$ (resp. ($\eta_1,\eta_2)$). Then $c_i(\chi_k) = s_i$ and $c_i(\chi_l) = t_i$. The last equality can be rewritten:
\begin{align*}
c_t(\chi_k\chi_l) =& 1+ 2(s_1+t_1)T + (t_1^2+s_1^2+3s_1t_1+2t_2+2s_2)T^2 \\
&+(s_1^2t_1+s_1t_1^2+2s_1s_2+2t_1t_2+2s_1t_2 + 2s_2t_1)T^3 \\
&+ (t_2^2 + s_2^2 + s_1s_2t_1 + s_1t_1t_2 + s_1^2t_2 + s_2t_1^2 -2s_2t_2)T^4
\end{align*}
We replace $s_1 = t_1 = x$ and eliminate all occurrences of $2x$ to obtain
\begin{equation}
c_t(\chi_k\chi_l) = 1 + (x^2+2y_k+2y_l)T^2 + (y_k-y_l)^2T^4 \label{Dp_ctlhs}
\end{equation}
Comparing coefficients in \cref{Dp_ctrhs} and \cref{Dp_ctlhs}, we obtain:
\begin{align}
y_{k+l}+y_{k-l} &= 2(y_k + y_l) \label{Dp_relationsum} \\
xy_{k+l} + xy_{k-l} &= 0 \label{Dp_relationproduct} \\
y_{k+l}y_{k-l} &= (y_k-y_l)^2
\end{align}
First look at \cref{Dp_relationproduct} with $k=l$. Note that $y_0 = c_2(1+\varepsilon) = 0$, and thus \cref{Dp_relationproduct} yields $x\cdot y_{2k} = 0$ for all $k$, which is equivalent to $x\cdot y_k = 0$ for all $k$ since indices are understood modulo the odd prime $p$.\\
We then show is $py_k = 0$ for all $k$. For this we use that for any group $G$, the graded ring $R^*(G)$ is $\vert G \vert$-torsion \todo{(this is also in Yohann's notes, I'll read the proof an rewrite it here)}. Here $\vert D_p \vert = 2p$, so we multiply the equation $y_{2k} = 4y_k$ by $p$ to obtain that $py_{2k} = 0$ for all $k$. Again, this imply that $y_k = 0$ for all $k$. \\
Finally, consider \cref{Dp_relationsum}, both with $l = k$ and $l = k+1$. This gives:
\begin{align*}
y_{2k} &= 4y_k \\
y_{2k+1} &= 2(y_k+y_{k+1}) - y_1
\end{align*}
together, these two relations imply that all $y_k$'s are multiples of $y_1$. We define $y = y_1$.
\paragraph*{}It remains to show that these are the only relations in $R^*(D_p)$. For this we use the standard restriction argument: the representation $\epsilon$ restricts to the generator of $R^*(C_2)$, whose powers are all nonzero. \\
The restriction of $\chi_1$ to $R(C_p,\CC)$ is $\rho+\rho^{-1}$, where $\rho$ sends the a generator $\sigma$ of $C_p$ to $e^{2i\pi/p}$. Thus the induced restriction $R^*(D_p,\CC) \rightarrow R^*(C_p,\CC)$ sends $y$ to $-c_1(\rho)^2$, which is not nilpotent.\\
Finally, the restriction of $\chi_1$ to $R(C_p,\RR)$ is the representation $\hat{\rho}$ obtained by pairing the complex representations $\rho$ discussed above with its complex conjugate $\overline{\rho} = \rho^{-1}$. This induced a map $R^*(D_p,\RR) \rightarrow R^*(C_p,\RR)$ sending $y$ to the (non-nilpotent) generator $t$ of $R^*(C_p,\RR)$. So the powers of $y$ are also nonzero in this case. \\
Note that in all cases, the restriction of $\varepsilon$ to $C_p$ is the trivial representation, so there cannot be any linear dependency relation between powers of $x$ and powers of $y$. \\
\end{proof}
\paragraph*{}Now take the dihedral group $D_4$ of order 8. It has four nontrivial irreducible representations:
\begin{itemize}
\item In degree 1, the representations $\rho: r \mapsto -1, \ s\mapsto 1$ and $\eta: r \mapsto 1, s\mapsto -1$ and their product $\rho\eta$.
\item In degree 2, the representation $\Delta$, which sends $s$ to $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ and $r$ to $\begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}$.
\end{itemize}
With the relations:
\begin{align}
\rho^2 &= \eta^2 = 1 \\
\rho\Delta &= \eta\Delta = \Delta \\
\Delta^2 &= 1+rho+\eta+\rho\eta \label{D4_relationdelta}
\end{align}
Then
\begin{Proposition}
Let $c_1(\rho) = x$; $c_1(\eta) = y$ and $c_2(\Delta) = b$. Then
\[
R^*(D_p) = \frac{\ZZ\left[x,y,b\right]}{(2x,2y,4b,xy,xb-yb)}
\]
where $\vert x \vert = |y| = 1$ and $\vert b \vert = 2$.
\end{Proposition}
\begin{proof}
Note that $c_1(\rho\eta) = x+y$ and $c_1(\Delta) = c_1(\det\Delta) = c_1(\rho\eta)$. So the graded ring is indeed generated by $x,y,b$. We have $2x = 2y = 0$ from the relations above, and $XB = YB = XY$, so $XY \in \Gamma^3$ and $xy = 0$, and $xb = yb$. Finally, applying $c_t$ to \cref{D4_relationdelta}, we get that $5c_1(\Delta)^2 + 4b = x^2+3xy+y^2 = x^2+ y^2$, and since $c_1(\Delta) = x+y$, we obtain $4b = 0$. \\
To see these are the only relations, we use the computation of $R^*(D_4)\otimes\FF_2 = \frac{\ZZ[x,y,b]}{(xy,xb-yb)}$ from \cite{guillot-minac}: tensoring with $\FF_2$ shows that none of $x,y,b$ is nilpotent. Finally, restriction to $C_4 = <r>$ show that $2b$ is non-nilpotent, so that any power of $b$ is 4-torsion.
\end{proof}
\subsubsection{Topological method}
\paragraph*{The quaternion group of order 8.} Let $G = Q_8 = \langle \ i,j,k \ \vert \ i^2 = j^2 = k^2 = ijk \ \rangle$. We prove:
\begin{Theorem} \label{Q8_gradedring}
\[
R_\CC^*(Q_8) = \frac{\ZZ[x,y,u]}{(2x,2y,8u,x^2,y^2, xy-4u)}
\]
where $\vert x \vert = \vert y \vert = 1$ and $\vert u \vert = 2$
\end{Theorem}
The group $Q_8$ has 5 conjugacy classes: $\lbrace 1\rbrace$, $\lbrace -1\rbrace$, $\lbrace \pm i\rbrace$, $\lbrace \pm j\rbrace$, $\lbrace \pm k\rbrace$ so 5 irreducible representations on $\CC$:
\begin{itemize}
\item In dimension 1, the trivial representation $\mathbb{1}$, $\rho_1 \begin{cases} i \mapsto 1 \\ j\mapsto -1 \end{cases}$, $\rho_2 = -\rho_1$ and $\rho_3 = \rho_1\rho_2$.
\item In dimension 2, the representation $\Delta$, which is in matrix form:
\begin{equation*}
\Delta (-1) = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}, \ \ \ \Delta(i) = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}, \ \ \ \Delta(j) = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \ \ \ \Delta(k) = \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix}
\end{equation*}
\end{itemize}
These representations satisfy the relations
\begin{align}
\rho_i^2 &= 1 \\
\rho_3 &= \rho_1\rho_2 \\
\Delta\rho_i &= \Delta \\
\Delta^2 &= \mathbb{1} + \rho_1 + \rho_2 + \rho_3
\end{align}
Let us first take a look at the generators and relations of $Q_8$:
\begin{Lemma}
The graded ring $R_\CC^*(Q_8)$ is generated by the elements
\[
x := c_1(\rho_1), \ \ \ \ y := c_1(\rho_2), \ \ \ \ u := c_2(\Delta)
\]
and those satisfy the relations in \cref{Q8_gradedring}, that is
\[
2x = 2y = 8u = 0 , \ \ \ \ x^2 = y^2 = 0, \ \ \ \ xy = 4u.
\]
\end{Lemma}
\begin{proof}
First we eliminate the redundant generators: $c_1(\rho_3) = c_1(\rho_1\rho_2) = c_1(\rho_1) + c_1(\rho_2)$, and $c_1(\Delta) = c_1(\det (\Delta) ) = c_1(\mathbb{1}) = 0$. So $R^*_\CC(Q_8)$ is indeed generated by $x,y$ and $u$. \\
Now since $\rho_1^2 = \rho_2^2 = \mathbb{1}$, we have $2x = 2y = 0$, and the order of $Q_8$ kills $R^*_\CC(Q_8)$ so $8u = 0$. For the relations in degree 2, we apply the total class $c_t$ to the relation $\Delta\rho_i = \Delta$, splitting the 2-dimensional representation $\Delta$ into $\sigma_1+\sigma_2$ where $\sigma_i$ has dimension 1. On the left-hand side we have:
\begin{align*}
c_t(\Delta\rho_i) &= c_t(\sigma_1\rho_i)c_t(\sigma_2\rho_i) \\
&= 1 + \left[ c_1(\sigma_1) + c_1(\sigma_2) + 2c_1(\rho_i) \right]T + \left[ c_1(\sigma_1)c_1(\sigma_2) + c_1(\sigma_1)c_1(\rho_i) + c_1(\sigma_2)c_1(\rho_i) + c_1(\rho_i)^2 \right]T^2
\end{align*}
While on the right-hand side:
\begin{align*}
c_t(\Delta) = 1 + \left[c_1(\sigma_1)+c_1(\sigma_2)\right]T + \left[c_1(\sigma_1)c_1(\sigma_2)\right]T^2.
\end{align*}
In degree 2, this yields the relation:
\begin{align*}
c_1(\rho_i)(c_1(\sigma_1)+c_1(\sigma_2)) + c_1(\rho_i)^2 = 0.
\end{align*}
Bearing in mind that $c_1(\sigma_1)+c_1(\sigma_2) = c_1(\Delta) = 0$, we obtain $x^2 = y^2 = 0$.
The relation $xy = 4u$ is obtained by applying $c_t$ to the relation $\Delta^2 = \mathbb{1} + \rho_1 + \rho_2 + \rho_1\rho_2$ and identifying the terms in degree 2, which yields
\begin{align*}
5c_1(\Delta)^2+4u = x^2 + y^2 + 3xy,
\end{align*}
that is, $4u = xy$.
\end{proof}
What remains to be proven is that these are the only relations satisfied by the generators; thus we want to check that we have no extra nilpotency or torsion conditions on $u$, and that the products $xu^i$ and $yu^i$ are nonzero for any $i$. For this, we look at the $2$-valuation of the characters of $Q_8$: we first define an admissible approxiamtion $(\widetilde{\Gamma}^n)$ that only takes into accounts lifts of our generators. This allows us to restrict ourselves when we compute the $2$-valuations of our evaluation morphisms, which we use to obtain information about torsion in $R^*$.
Let $X, Y, U$ be standard lifts of $x,y,u$. We consider the approximation:
\begin{equation}
\widetilde{\Gamma}^n = \langle X^{\epsilon_1}Y^{\epsilon_2}U^k \rangle, \ \ \ \ 2k+ \epsilon_1 + \epsilon_2 \geq n, \ \ \ 0 \leq \epsilon_1+\epsilon_2 \leq 1 \ \ \
\end{equation}
\begin{Lemma}
The approximation $(\widetilde{\Gamma}^n)$ is admissible.
\end{Lemma}
\begin{proof}
Obviously we have $\widetilde{\Gamma}^{n+1} \subset \widetilde{\Gamma}^n$, so $(\widetilde{\Gamma}^n)$ satisfies \cref{admissiblefiltration1} in the definition of an admissible filtration.
To check \ref{admissiblefiltration2}, let $Z = \rho_3-1$ and $T = \Delta -2$ be lifts of $c_1(\rho_3)$, $c_1(\Delta)$, respectively. Then let
\[
\alpha = X^{\epsilon_1}Y^{\epsilon_2}Z^{\epsilon_3}U^kT^l \in \Gamma^n, \ \ \ \ 2k+l+\epsilon_1+\epsilon_2+\epsilon_3 \geq n
\]
If $\epsilon_{1} \geq 2$, then $\alpha$ contains a factor $X^2$, but the relation $x^2 = 0$ implies that $X^2 \in \Gamma^3$, so in that case $\alpha \in \Gamma^{n+1}$. The same goes for $\epsilon_2$, so we can restrict ourselves to monomials such that $\epsilon_1+\epsilon_2 \leq 1$. Similarly, since $c_1(\Delta) = 0$ we have $T \in \Gamma^2$, which means that $l \neq 0$ forces $\alpha \in \Gamma^{n+1}$. We proceed similarly for all factors and obtain
\[
\Gamma^n = \widetilde{\Gamma}^n + \Gamma^{n+1}
\]
\end{proof}
Now, there are 4 nontrivial evaluation morphisms on $R_\CC(Q_8)$:
\begin{itemize}
\item $\phi_{-1}: \rho_i \mapsto 1, \Delta \mapsto -2$.
\item $\phi_i: \rho_1 \mapsto 1, \ \ \ \rho_2,\rho_3 \mapsto -1, \ \ \ \Delta \mapsto 0$
\item $\phi_j: \rho_2 \mapsto 1, \ \ \ \rho_1,\rho_3 \mapsto -1, \ \ \ \Delta \mapsto 0$
\item $\phi_k: \rho_3 \mapsto 1, \ \ \ \rho_1, \rho_2 \mapsto -1, \ \ \ \Delta \mapsto 0$
\end{itemize}
Now we apply the morphisms to the standard lifts of the Chern classes and obtain
\begin{itemize}
\item $\phi_{-1}: X,Y \mapsto 0, \ \ \ T \mapsto -4, \ \ \ U \mapsto 4$
\item $\phi_i: X \mapsto 0, \ \ \ Y,T \mapsto -2, \ \ \ U \mapsto 2$
\item $\phi_j: X, T \mapsto -2, \ \ \ Y \mapsto 0, \ \ \ T \mapsto 2$
\item $\phi_k: X,Y,T \mapsto -2, \ \ \ T \mapsto 2$
\end{itemize}
\begin{Lemma}
The morphisms $\phi_g$ are continuous with respect to the $2$-valuations on $\ZZ$, for all $g\in Q_8$.
\end{Lemma}
\begin{proof}
Consider the value of $\phi_{-1}$ on a generic monomial $\alpha \in \Gamma^n$. Write $\alpha = X^{\epsilon_1}Y^{\epsilon_2}Z^{\epsilon_3}U^kT^l$ with $\epsilon_1+\epsilon_2+\epsilon_3+k+l \geq n$. Reading the values of $\phi_{-1}$, we see that $\phi_{-1}(\alpha) \geq 2^{n/2}$, which means that $\phi_{-1}$ is continuous. Similarly for the other $\phi_g$'s.
\end{proof}
We can now wrap up:
\begin{proof}[Proof of \cref{Q8_gradedring}.]
We want to show that $R^{2n}_\CC = \langle u^n \rangle = \ZZ/8\ZZ$, and $R^{2n+1}_\CC = \langle xu^n, yu^n \rangle = (\ZZ/2\ZZ)^2$. \\
We first look at $R^{2n}_\CC$, where we need to show that $4u^n \neq 0$, that is, $4U^n \notin \Gamma^{2n+1}$. We have $\phi_g(4U^n) = 2^{n+2}$, for any $g = i,j,k$. Suppose that $4U^n \in \Gamma^{2n+1}$; then by \cref{evaluation_approximation} there is an element $\widetilde{x} \in \widetilde{\Gamma}^{2n+2}$ satisfying
\[
n+2 = v_2(\phi_g(4U^n)) = v_2(\phi_g(\widetilde{x}))
\]
so that $v_2(\phi_g(\widetilde{x})) = n+2$. Write
\[
\widetilde{x} = a_1U^{n+1}+a_2XU^{n}+a_3YU^{n} + U^{n+2}P(X,Y,U) + XU^{n+1}Q(X,Y,U) + YU^{n+1}R(X,Y,U)
\]
where $P,Q,R$ are polynomials with integer coefficients. We then apply $\phi_g$ for $g = i,j,k$ to this equation, divide the 3 relations thus obtained by $2^{n+1}$ and look at the result mod $2$. The three $\phi_g$ give us a system of three equations:
\begin{align*}
0 &= a_1 + a_3 \\
0 &= a_1 + a_2 \\
0 &= a_1 + a_2 + a_3
\end{align*}
which has no nontrivial solution. But if $a_m = 0 (\mathrm{mod} \ 2)$ for $m=1,2,3$ then $v_2(\phi_g(\widetilde{x})) > n+2$, which is impossible. Thus $4U^n$ cannot be in $\Gamma^{2n+1}$, and the additive order of $u^m$ is indeed 8 for all $m$. \\
The same process shows that $xu^n, yu^n$ and $xu^n+yu^n$ are nonzero in $R^{2n+1}_\CC$ for all $n$.
\end{proof}
\paragraph*{Products of cyclic $2$-groups}.
\begin{Theorem}
\[
R^*(\ZZ/4 \otimes \ZZ/2) = \frac{\ZZ[x,y]}{(4x,2y,xy^3+x^2y^2)}
\]
with $|x| = |y| = 1$
\end{Theorem}
\begin{proof}
Let $\rho$ be a generator of $R(\ZZ/4)$ and $\sigma$ the nontrivial representation of $\ZZ/2$. Then $R^*(G)$ is generated by $c_1(\rho) =: x$ and $c_1(\sigma) =: y$. By restriction to cyclic subgroups, we see that $x$ has additive order $4$ and $y$, additive order $2$. Now consider $X = \rho - 1$ and $Y = \sigma - 1$. Then by expanding $(X+1)^4 = 1$ we get
\[
4X = -(6X^2 + 4x^3 + X^4)
\]
On the other hand, $Y^n = (-2)^{n-1}Y$, so
\[
XY^3 = 4XY = -X^4Y - 4X^3Y - 6X^2Y = -X^4Y - X^3Y^3 + 3X^2Y^2.
\]
Quotienting by $\Gamma^5$ we get that $xy^3 = x^2y^2$. We now show these are the only relations; the only other possible extra relations (that cannot be ruled out by restrictions to various subgroups) are: $x^{n-1}y = 0$, $x^{n-2}y^2 = 0$ or $x^{n-1}y = x^{n-2}y$ for some $n$. We use the continuity method to disprove all of these. Let $\widetilde{\Gamma}^n = \langle X^n, Y^n, X^{n-1}Y, X^{n-2}Y^2 \rangle$, then $\{ \widetilde{\Gamma}^n \}_n$ is an acceptable approximation for $\{\Gamma^n\}_n$.\\
First suppose that $x^{n-2}y = 0$ for some $n$. Then for $N$ arbitrarily large, there exists $\widetilde{Z} \in \widetilde{\Gamma}^{n+1}$ such that $X^{n-2}Y = \widetilde{Z}+R$ with $R \in \Gamma^N$; in particular, for any element $(i,j) \in \ZZ/4\times\ZZ/2$ we have
\[
v_2\left(X^{n-2}Y^2\vert_{(i,j)}\right) = v_2\left(\widetilde{Z}\vert_{(i,j)}\right).
\]
Write
\[
\widetilde{Z} = a\cdot X^{n+1} + b\cdot X^{n}Y + c\cdot X^{n-1}Y^2 + d\cdot Y^{n+1} + P\cdot X^{n+2} + Q\cdot X^{n+1}Y + S\cdot X^{n}Y^2 + T\cdot Y^{n+2}
\]
where $a,b,c,d \in \ZZ$ $P,Q,S,T \in \ZZ[X,Y]$, then evaluate at $(2,1)$; so that $X\vert_{(2,1)} = Y\vert_{(2,1)} = -2$. Then the $2$-valuation of $X^{n-1}Y$ is $(-2)^n$ while the $2$-valuation of $\widetilde{Z}$ is at least $n+1$. This shows that $x^{n-1}y$ cannot be zero. The same proof works for $x^{n-2}y^2$.\\
The only possible remaining relation is $x^{n-1}y = x^{n-2}y^2$. Let $Z = X^{n-1}Y + X^{n-2}Y^2$. With notations as above, we have
\[
\widetilde{Z}\vert_{(2,0)} = a\cdot(-2)^{n+1} + P\cdot (-2)^{n+2}
\]
while $Z\vert_{(2,0)} = 0$. Thus $a = 0 \ (\Mod 2)$, so write $a = 2a'$. We now evaluate at $(1,1)$:
\[
Z\vert_{(1,1)} = (i-1)^{n-2}\cdot 4 + (i-1)^{n-1}\cdot (-2).
\]
thus $v_2(Z\vert_{(1,1)}) = (n+1)/2$. On the other hand:
\[
\widetilde{Z}\vert_{(1,1)} = a'\cdot 2\cdot(i-1)^{n+1} + b\cdot(i-1)^{n}(-2) + c\cdot(i-1)^{n-1}4 + d\cdot(-2)^{n+1} + \widetilde{R}
\]
where $v_2(\widetilde{R}) \geq \frac{n+2}{2}$. We see that $v_2(\widetilde{Z}\vert_{(1,1)}) \geq (n+2)/2$, which completes the proof.
\end{proof}
\begin{Conjecture}
Let $G = \ZZ/2^{n}\times \ZZ/2$. Then
\[
R^*(G) = \frac{\ZZ[x,y]}{(2^nx, 2y, xy^{n+1} + x^2y^n)}
\]
\end{Conjecture}
