\section{Saturated rings as Tambara functors}
\subsection{Representation rings and tensor induction}
First the actual definition (from Pierre's notes).\\
Fix a group $G$ and a field $\KK$. We view a $G$-set $X$ as a category with an object for each point and arrows between objects $x,y$ for each $g \in G$ such that $g\cdot x = y$. A vector bundle is defined as a functor between $X$ and the category of $\KK$-vector spaces and linear maps. Let $K_G(X)$ be the semigroup of isomorphism classes of vector bundles over $X$, under direct sum.
\begin{Lemma}
Let $X$ be a transitive $G$-set with a distinguished point $x \in X$, and let $H = \mathrm{Stab}(x)$. Then there is an isomorphism (depending on $x$) between $K_G(X)$ and the semiring of representations $R(H)$.
\end{Lemma}
\begin{proof}
Let $W$ be a representation of $H$ and consider the induced representaiton $V = \ind_{H}^{G} = \KK[G] \otimes_\KK[H] W$. Define a vector bundle on $X$ as follows: for each $y \in X$, write $y = g\cdot x$ and let $(V)_y = g\cdot W = g\otimes W \subset V$. This depends only on $y$ and the action of $g$ takes $V_{x}$ to $V_{gx}$, so this is a vector bundle.\\
Conversely if $V$ is a vector bundle on $X$, define $W = V_{x}$. This is a well-defined $H$-module (since it is stable by $H$), so $W$ is a representation. Those two constructions are mutually inverse.
\end{proof}
\begin{Remark}
The isomorsphism above depends on $x$; choosing the basepoint $y = g\cdot x$ as a basepoint instead, one obtains the isomorphic representation of $gHg^{-1}$ which is given by precomposing the action og $H$ on $V_x$ by conjugation with $g$.
\end{Remark}
We can now define our restrictions and transfers:
\begin{Definition} Let $f : X \rightarrow Y$ be a map of $G$-sets. Then we define:
\begin{itemize}
\item The \textbf{restriction} $f^*: K_G(Y) \rightarrow K_G(X)$, $f^*(V)_{x} = V_{f^*(x)} $
\item The \textbf{induction} (or transfer) $f_*: K_G(X) \rightarrow K_G(Y)$, $f_*(V)_{y} = \oplus V_{f^{-1}(y)}$
\item The \textbf{norm} (or tensor induction) $f: K_G(X) \rightarrow K_G(Y)$, $f\sharp(V)_y = \otimes V_{f^{-1}(y)}$
\end{itemize}
\end{Definition}
\begin{Theorem}
The function $K_G(-)$ with these three maps is a Tambara functor.
\end{Theorem}
\begin{proof}
\todo{check!}
\end{proof}
In \cite{tambara}, Tambara gives a formula for the norm of a sum. This is part of the proof that semi-Tambara functors (then called semi-TNR functors) are actual Tambara functors, that is, the norm extends from semirings to rings. Let $f:X\rightarrow Y$ be a map of $G$-sets. Let
\[
V = \{ (y,C) \vert y \in Y, C\subset f^{-1}(y)\}
\]
then the pullback $X \times_Y V$ can be written as the disjoint union of two sets:
\begin{itemize}
\item $U = \{ (x,C) \vert x \in X, C\subset f^{-1}(y), x \in C\}$
\item $U' = \{ (x,C) \vert x \in X, C\subset f^{-1}(y), x \notin C\}$.
\end{itemize}
Then we have two commutative diagrams of $G$-sets:
\[
\xymatrix{ U \ar[r]^{r} \ar[d]_{t} & X \ar[d]^{f} \\
V \ar[r]_{s} & Y \\}
\ \ \ \ \ \ \ \
\xymatrix{ U' \ar[r]^{r'} \ar[d]_{t'} & X \ar[d]^{f} \\
V \ar[r]_{s} & Y \\}
\]
where $r, r', s$ are the projections and $t: U \rightarrow V$ sends $(x,C)$ to $(f(x), C)$ (similarly for $t'$).
\begin{Proposition}[{\cite[Prop. 4.1]{tambara}}]
Let $X$ and $Y$ be $G$-sets and $S$ a semi-TNR functor. Let $a, b \in S(X)$.
\begin{equation}
f_\sharp(a+b) = s_*\left(t_\sharp r^*(a)\cdot t'_\sharp r'^*(b) \right)
\end{equation}
\end{Proposition}
We now apply this formula to representation rings and the norm map $\nn_{H}^{G}$, for $H$ a normal subgroup of $G$ of prime index. We have $X = G/H$, $Y = G/G = \{*\}$ and $f$ is the constant map. Moreover, $V$ is just the set of subsets of $G/H$. Let $\mathcal{O}(V)$ be a complete set of representatives for the $G$-orbits in $V$, and pick a set $\mathcal{S} \subset G$ of coset representatives of $G/H$. Then, for a representation $\rho$ of $H$ and a subset $C = \{g_1, \cdots, g_i\}$ of $G/H$ with $g_j \in \mathcal{S}$, write
\begin{equation}
\rho^{\otimes C} = \rho^{g_1}\otimes \rho^{g_2} \otimes \cdots \otimes \rho^{g_i} \label{equation_addition_formula}
\end{equation}
where $\rho^g$ is the conjugate of $\rho$ by $g$. Note that this is still a representation of $H$.
\begin{Proposition}
Let $H$ be a subgroup of $G$ and let $\tau, \sigma$ be representations of $H$.
\begin{equation}
\nn_{H}^{G}(\tau + \sigma) = \sum_{C\in \mathcal{O}(V)}\ind_{\mathrm{Stab} C}^{G}\left(\tau^{\otimes C}\sigma^{\otimes (X\setminus C)}\right)
\end{equation}
This formula does not depend on the set of orbit representatives $\mathcal{O}(V)$.
\end{Proposition}
\begin{proof}
We apply Tambara's formula:
\[
\nn_{H}^{G}(\tau+\sigma) = s_*\left(t_\sharp r^*(\tau)\cdot t'_\sharp r'^*(\sigma) \right)
\]
The map $r$ sends a pair $(x,C)$ to $x$, thus $r^*(\tau)_{(x,C)} = (\tau)_x = \tau^x$. Similarly $r'^*(\sigma)_{(x,C)} = \sigma^{x}$. Now for any $C \in V = \mathcal{P}(X)$,
\[
\left(t_\sharp r^*(\tau)\right)_C = \bigotimes_{x \in C}r^*(\tau)_{(x,C)} = \bigotimes_{x \in C} \tau^x
\]
and $\left(t'_\sharp r'^*(\sigma)\right)_C = \bigotimes_{x \notin C} \sigma^x$. Let
\[
\rho_C^{x} = \begin{cases} \tau^x & \text{if} \ x\in C \\
\sigma^x & \text{if} \ x \notin C \end{cases}
\]
then $\left( t_\sharp r^*(\tau) \cdot t'_\sharp r'^*(\sigma) \right)_C = \bigotimes_{x\in X}\rho_C^x$. Let $\rho_C = \bigotimes_{x\in X}\rho_C^x$, then $\rho$ is a representation of $\mathrm{Stab} C$, and $\rho_{g\cdot C} = (\rho_C)^g$, the conjugate representation.
\begin{align*}
s_*\left( t_\sharp r^*(\tau) \cdot t'_\sharp r'^*(\sigma) \right)_y &= \bigoplus_{C \in \mathcal{P}(X)}\rho_C = \bigoplus_{C \in \mathcal{O}(V)}\left( \bigoplus_{D \in G\cdot C} \rho_{D} \right) \\
&= \bigoplus_{C \in \mathcal{O}(V)} \left( \bigoplus_{g \in G/\mathrm{Stab} C} (\rho_{g.C}) \right) = \bigoplus_{C \in \mathcal{O}(V)} \left( \bigoplus_{g \in G/\mathrm{Stab} C} (\rho_C)^g \right)\\
&= \sum_{C \in \mathcal{O}(V)} \ind_{\mathrm{Stab} C}^{G} \left(\tau^{\otimes C}\sigma^{\otimes (X\setminus C)}\right)
\end{align*}
As promised, this formula does not depend on the set of orbit representatives $\mathcal{O}(V)$: picking another representative boils down to conjugating $\tau^{\otimes C}\sigma^{\otimes(X\setminus C)}$ by some element $g \in G$, which does not change the induced representation.
\end{proof}
So far the representation ring $R(-)$ is only a semi-Tambara functor, since we only defined the norm map for actual representations. However, the formula can be extended to all virtual representations, following the method outlined in \cite[Th. 6.1]{tambara}; first, define
\[
V^{(2)} = \{ C_1, C_2 \subset G/H | C_1\cap C_2 = \emptyset \}
\]
and consider the three maps $m$, $p_1$, $p_2 : V^{(2)} \rightarrow V$ which send a pair $(C_1, C_2)$ to $C_1 \sqcup C_2$, $C_1$, $C_2$, respectively. Then put, for any $a,b \in K_G(V)$
\[
a\vee b := m_*(p_1^*(a)\cdot p_2^*(b)) = \bigoplus_{C_1\sqcup C_2 = C}(a_{C_1}\otimes b_{C_2}) \in K_G(V).
\]
The operation $\vee$ is associative with unit element $i_*(1)$, where $i: \emptyset \rightarrow V$, and we have:
\begin{Proposition}[{\cite[Prop. 4.4]{tambara}}]
\[
t_\sharp r^*(a+b) = t_\sharp r^*(b) \vee t_\sharp r^*(b).
\]
\end{Proposition}
Put $\chi := t_\sharp r^*$, then $\chi$ is a morphism from the additive monoid $K_G(X)$ to the multiplicative monoid $K_G(V)$ with multiplication $\vee$. Note that if $j:Y \rightarrow V$ denotes the map sending the only point of $Y$ to $G/H$, then $j^*\chi = \nn_{H}^{G}$. Finally, write $V = \sqcup V_n$ with $V_n = \{ C\subset G/H | \sharp C = n\}$. Then $K_G(V)$ is a graded semi-ring for $\vee$ with grading $K_G(V) \cong \bigoplus K_G(V_n)$ \todo{(check! it says so in Tambara)}, and $\Ima \chi \subset 1 + \prod_{n = 1}^{|G/H|} K_G(V_n)$.
\begin{Proposition}
Let $\tau \in R^+(H)$. Putting
\[
\nn_{H}^{G}(-\tau) = (-1)^{|G:H|}\nn_{H}^{G}(\tau)
\]
turns $R(-)$ into a Tambara functor, and the addition formula \cref{equation_addition_formula} applies to virtual representations.
\end{Proposition}
\begin{proof}
We follow the proof of \cite[Th. 4.4]{tambara} : for a monoid $M$, denote by $\gamma M$ the universal abelian group on this monoid and by $k_M: M \rightarrow \gamma M$ the corresponding monoid map. If $M$ has the structure of a semi-ring, then $\gamma M$ as a unique ring structure such that $k_M$ is a semi-ring map. This map sends $1 + \prod K_G(V_n)$ to $1+ \prod \gamma K_G(V_n)$, which is a group for $\vee$ \todo{(check! Tambara says it too)}. Thus $\chi$ can be uniquely extended to a monoid map $\widetilde{\chi}: \gamma K_G(G/H) \rightarrow \gamma K_G(V)$. In other words, we put $\chi(-\tau) = \chi(\tau)^{-\vee}$, the inverse of $\chi(\tau)$ for $\vee$. Then $\nn_{H}^{G} = j^*\widetilde{\chi}$ extends $\nn_{H}^{G} = j^*\chi$. \\
Let $a \in \gamma K_G(V)$ satisfy $\chi(\tau)\vee a = \mathbb{1}$. Keeping in mind that $\chi(\tau)_C = t_\sharp r^*(\tau)_C = \bigotimes_{x \in C}\tau^x$, write:
\[
(\chi(\tau)\vee a)_C = \sum_{C_1 \sqcup C_2 = C}\left[ (\prod_{x \in C_1}\tau^x) \cdot (a_{C_2})\right]
\]
Since $K_G(V) = \oplus K_G(V_n)$, we work our way up by induction on $n$, so that on each $\gamma K_G(V_n)$ (except $K_G(V_0)$) our product is zero. More explicitly, we show by induction on $n$ that we need
\[
a_{C_n} =: \prod_{x \in C_n}(-\tau^x) = (-\tau)^{\otimes C_n}.
\]
First, we need $a_\emptyset = \mathbb{1}$. Second, looking at $1$-element sets $C_1 \in V_1$, we get:
\begin{align*}
(\chi(\tau)\vee a)_{C_1} = \chi(\tau)_{\{x\}}\cdot 1 + 1\cdot a_{\{x\}}
\end{align*}
so that $a_{\{x\}} = -\tau^x$. Now suppose the induction hypothesis is true on $V_n$ and let $C_{n+1} \in V_{n+1}$:
\begin{align*}
0 = (\chi(\tau)\vee a)_{C_{n+1}} &= \sum_{D \subset C_{n+1}}\tau^{\otimes D}\cdot a_{C_{n+1}\setminus D} \\
&= a_{C_{n+1}} + \sum_{i=0}^{n}(-1)^i\binom{n+1}{i}\tau^{\otimes C_{n+1}} \\
&= a_{C_{n+1}} + (-1)^{n}\tau^{\otimes C_{n+1}}
\end{align*}
where the middle step comes from the induction hypothesis. Thus we have:
\[
a_{C_{n+1}} = (-1)^{n+1}\tau^{\otimes C},
\]
that is, $\widetilde{\chi}(\tau) = \chi(\tau)$ and $\widetilde{\chi}(- \tau)_C = (-1)^{\sharp C}\chi(\tau)_C$. In particular, composing with $j^*$, we get
\begin{align*}
\nn_{H}^{G}(-\tau) &= (-1)^{|G:H|}\nn_{H}^{G}(\tau) \\
\nn_{H}^{G}(\tau - \sigma) &= \sum_{C \subset \mathcal{P}(X)}\tau^{\otimes C}\cdot (-\sigma)^{\otimes (X \setminus C)} \\
&= \nn_{H}^{G}(\tau) + (-1)^{|G:H|}\nn_{H}^{G}(\sigma) + \sum_{C\in \mathcal{O}(V)}(-1)^{\sharp C}\ind_{\mathrm{Stab}\ C}^{G}(\tau^{\otimes C}\cdot (-\sigma)^{\otimes(X\setminus C)})
\end{align*}
\end{proof}
\subsection{Steenrod operations}
\begin{Theorem}\label{steenrod_for_reprings}
Let $p$ be a prime, and let $\rho \in R(G)^+$ be an actual representation of $G$. Then
\[
\nn_{G}^{G\times C_p}(\rho) = \left( \psi_1(\rho)^p \right) \otimes \mathbb{1} + \sum_{i=1}^{p-1}\binom{p}{i+1}\left(\frac{1}{p}\psi_1(\rho)^p - \frac{1}{p}\psi_p(\rho)\right)\otimes (\sigma-1)^i
\]
where $\sigma$ is a choice of generator for $R(C_p)$ and the $\psi_k$ are the Adams operations.
\end{Theorem}
\begin{Lemma}
With notation as above
\[
\nn_{G}^{G\times C_p}(\rho) = \left(\frac{1}{p}\psi_1(\rho)^p + \frac{p-1}{p}\psi_p(\rho)\right) \otimes \mathbb{1} + \sum_{i=1}^{p-1}\left(\frac{1}{p}\psi_1(\rho)^p - \frac{1}{p}\psi_p(\rho)\right)\otimes \sigma^i.
\]
\end{Lemma}
\begin{proof} Note that
\[
R(G\times C_p) \cong R(G) \otimes R(C_p) \cong \bigoplus_{i = 0}^{p-1} R(G)\otimes \sigma^i.
\]
Recall that $\nn_{H}^{G}(\rho)(g\times \gamma)$ acts diagonally as $\rho(g)$, with a cyclic permutation corresponding to $\gamma$. Fix $g \in G$ and let $\mathcal{B} = \{ e_i \}_{i\in I}$ be a basis of eigenvectors for $\rho(g)$, with $\rho(g)\cdot e_i = \lambda_i e_i$. Let $\mathcal{B}^{\otimes p} = \left\{ e_{i_1}\otimes\cdots\otimes e_{i_p} \vert e_j \in \mathcal{B} \right\}$, a basis for $V^{\otimes p}$. Pick a generator $\gamma$ of $C_p$, then $\gamma$ acts on $V^{\otimes p}$ by cyclic permutation of the factors; as such $V^{\otimes p}$ can be decomposed into isotypical components, on which $\gamma$ acts as multiplication by $\omega^k$ for $k = 0, \cdots, p-1$, with $\omega$ a primitive $p$-th root of unity. Let $E_k$ be the subspace on which $\gamma$ is multiplication by $\omega^k$. Then $\nn_{G}^{G\times C_p}(\rho) = \bigoplus (\rho\otimes\cdots\otimes\rho)\vert_{E_k}\otimes\sigma^k$, and
\begin{align*}
E_0 &= \left\langle \left. \sum_{i = 0}^{p-1}\sigma(\gamma)^{i}\cdot u \ \right\vert \ u \in \mathcal{B}^{\otimes p} \right\rangle \\
E_k &= \left\langle \left. \sum_{i = 0}^{p-1}\omega^{-ik}\sigma(\gamma)^{i}\cdot u \ \right\vert \ u \in \mathcal{B}^{\otimes p} \right\rangle
\end{align*}
Let $u = e_{i_1} \otimes\cdots\otimes e_{i_p} \in \mathcal{B}^{\otimes p}$ and $w = \sum \sigma(\gamma)^i\cdot u \in E_0$. Then $\rho(g)\cdot w = \lambda_{i_1}\cdots\lambda_{i_p}w$, thus
\[
\Tr \rho(g)\vert_{E_0} = \sum_{(i_1,\cdots,i_p) \in I^p}\lambda_{i_1}\cdots\lambda_{i_p} = \frac{1}{p}\left( \sum_{i \in I}\lambda_i \right)^p + \frac{p-1}{p}\sum_{i \in I}\lambda_i^{p}.
\]
So $(\rho\otimes\cdots\otimes\rho)\vert_{E_0} = \frac{1}{p}\rho^{p} + \frac{p-1}{p}\psi_p(\rho)$. Now, since all the $E_k$ are isomorphic, we can specialize to $E_1$:
\[
\Tr \rho(g)\vert_{E_1} = \sum_{(i_1,\cdots,i_p) \in I\setminus\Delta} \lambda_{i_1}\cdots\lambda_{i_p} = \frac{1}{p}\left(\sum_{i \in I}\lambda_i\right)^p - \frac{1}{p}\sum_{i \in I}\lambda_i^{p}
\]
where $\Delta$ is the diagonal of $I^p$. Thus we have $(\rho\otimes\cdots\otimes\rho)\vert_{E_0} = \frac{1}{p}\rho^{p} - \frac{1}{p}\psi_p(\rho)$. This proves the lemma.
\end{proof}
\begin{proof}[proof of \cref{steenrod_for_reprings}] This follows from the Lemma by a change of basis, keeping in mind that
\[
\sum_{i = k}^{n}\binom{i}{k} = \binom{n+1}{k+1}.
\]
\end{proof}
