\section{Induction is a transfer on \texorpdfstring{$D_p$}{Dp}}
\subsection{First example: \texorpdfstring{$S_3$}{S3} and Swan's lemma}
The symmetric group $G=S_3$ has one $2$-Sylow and one $3$-Sylow, which are the cyclic groups $C_2$ and $C_3$, respectively. Of course, they're both abelian.
\subsubsection{The graded ring of \texorpdfstring{$S_3$}{S3}}
There are three irreducible representations: two of degree one, and one of degree two. In degree one: the trivial one $1$ and the signature $\varepsilon$. The representation $r$ of degree 2 is obtained as the quotient of the representation of degree 3 that permutes the basis vectors of a vector space, by the trivial representation.
\begin{Remark}
These are valid for any field $\KK$ of characteristic not 2 or 3. In particular, $R^*(G,\RR) = R^*(G,\QQ) = R^*(G,\CC)$, so I just write $R^*(G)$ in the sequel.
\end{Remark}
Some relations between representations:
\begin{align}
\varepsilon r &= r \ \ \ \text{ \small{(there's only one irreducible rep of dimension 2)}} \\
\varepsilon^2 &= 1 \\
r^2 &= 1+\varepsilon +r \ \ \ \text{ \small{(looking hard enough at a character table)}} \label{relation}
\end{align}
\begin{Proposition} \label{swanslemma_S3} Let $c_1(r) = x$ and $c_2(r) = y$, then
\[
R^*(S_3) = \frac{\ZZ\left[x,y\right]}{(2x,3y,xy)}
\]
where $\vert x \vert = 1$ and $\vert y \vert = 2$.
\end{Proposition}
\begin{proof}
\begin{itemize}
\item First we show the relations do hold. Since $\varepsilon^2 = 1$, we have indeed
\[
2x = 2c_1(\varepsilon) = c_1(\varepsilon^2) = c_1(1) = 0.
\]
Moreover
\[
c_1(r) = c_1(\mathrm{det} r) = c_1(\varepsilon).
\]
Using this, we obtain the last two relations by applying the total Chern class to \cref{relation} above:
\begin{equation} \label{totalchernclassr^2}
c_t(r^2) = c_t(1+\varepsilon+r) = c_t(\varepsilon)c_t(r).
\end{equation}
Let's develop the right-hand side first.
\begin{eqnarray*}
c_t(\varepsilon)c_t(r) &=& (1+xT)(1+xT+yT^2) \\
&=& 1+2xT+(x^2+y)T^2+xyT^3 \\
&=& 1 + (x^2 + y)T^2 + xyT^3.
\end{eqnarray*}
For the left-hand side, we use the splitting principle: in some extension $L$ of $\KK$ we can write $r = \rho_1+\rho_2$ with each $\rho_i$ having degree 1. Let $s_i = c_1(\rho_i)$ for $i=1,2$, then:
\begin{eqnarray*}
c_t(r^2) &=& c_t((\rho_1+\rho_2)^2) \\
&=& c_t(\rho_1^2)c_t(\rho_2^2)c_t(2\rho_1\rho_2) \\
&=& (1 + 2s_1T)( 1 + 2s_2T)( 1 + (s_1+s_2)T)^2 \\
&=& 1 + (4s_1 + 4s_2)T + (5s_1^2 + 14s_1s_2 +5s_2^2)T^2\\ & & + (2s_1^3+14s_1^2s_2 + 14S_1s_2^2 + 2s_2^3)T^3 + (4s_1^3s_2 + 8s_1^2s_2^2 + 4s_1s_2^3)T^4.
\end{eqnarray*}
Let $\sigma_1$ (resp. $\sigma_2$) be the first (resp. second) symmetric function in $s_1, s_2$. Then, since $r = \rho_1 + \rho_2$ we have $x = c_1(r) = \sigma_1$ and $y = c_2(r) = \sigma_2$. We can rewrite the coefficients of the polynomial above in terms of symmetric functions:
\begin{eqnarray*}
c_t(r^2) = 1 + 4\sigma_1 T + (5\sigma_1^2 + 4\sigma_2)T^2 + (2\sigma_1^3+8\sigma_1\sigma_2)T^3 + 4\sigma_1^2\sigma_2 T^4.
\end{eqnarray*}
Replacing $\sigma_1$ by $x$ and $\sigma_2$ by $y$ and equating both sides of \cref{totalchernclassr^2} we get:
\[
1 + 4xT + (5x^2+4y)T^2 + (2x^3+8xy)T^3 + 4x^2yT^4 = 1 + (x^2+y)T^2 + xyT^3
\]
which yields, keeping in mind that $2x = 0$,
\begin{align*}
x^2+4y = x^2 +y \Leftrightarrow 3y = 0 \\
xy = 0
\end{align*}
\item We still need to show there are no further relations. Since $x$ is 2-torsion and $y$ is 3-torsion one cannot have a relation of the type $x^{2n}+y^n = 0$ for some $n$. So the only other possible relations are $x^n = 0$ for some $n$ and $y^m = 0$ for some $m$. To show this is impossible, we restrict to subgroups. \\
For $x = c_1(\varepsilon)$, we restrict to $C_2 \subset S_3$. The restriction of the signature $\varepsilon$ on $C_2$ is the signature $\overline{\varepsilon}$, whose Chern class $\overline{x}$ generates $R^*(C_2) = \ZZ\left[x\right]/(2x)$. In particular, $\overline{x}^n \neq 0$ for all $n$, so its preimage $x^n$ must be nonzero as well. \\
For $y = c_2(r)$ we restrict to $C_3 \subset S_3$. Let $\sigma$ generate $C_3$, then the restriction of $r$ to the subgroup $C_3$ is a two-dimensional representation with character
\[
\chi : \begin{cases} 1 \mapsto 2 \\ \sigma \mapsto -1 \\ \sigma^2 \mapsto -1 \end{cases}
\]
which means it has to be $\rho + \overline{\rho}$, where $\rho$ is the one-dimensional representation of $C_3$ such that $\rho(\sigma) = e^{2i\pi/3}$. So $y$ restricts to
\[
c_2(\rho+\overline{\rho}) = c_1(\rho)c_1(\overline{(\rho}) = -c_1(\rho)^2
\]
When $\KK = \CC$, the graded ring $R^*(C_3,\CC) = \ZZ\left[z\right]/(3z)$ is generated by $z = c_1(\rho)$. Thus the image of $c_2(r)$ is $-z^2$, whose powers are all nonzero. When $\KK = \RR$ or $\QQ$, the graded ring $\ZZ\left[t\right] / (3t)$ is generated by $t = -c_1(\rho)^2$, and so the image of $y$ is again non nilpotent.
\end{itemize}
\end{proof}
\begin{Remark} If I'm not mistaken, the signature $\varepsilon$ restricts to the trivial representation on $C_3$. I think this gives a nicer argument for the fact that there's no linear dependency between powers of $x$ and $y$.
\end{Remark}
\subsubsection{Swan's lemma for \texorpdfstring{$S_3$}{S3}}
We have already done the legwork so this is quick.
\paragraph*{}The action of the generator $\tau$ of $C_2 = S_3/C_3$ on $C_3$ (which is the only $3$-Sylow of $S_3$ and a normal subgroup) exchanges the two generators $\sigma$ and $\sigma^2$. Thus it sends the representation $\rho$ to its inverse $\overline{\rho} = \rho^{-1}$, and so
\[
\tau^*c_1(\rho) = -c_1(\rho) .
\]
Thus $\tau$ sends the generator $z$ of $R^*(C_3,\CC)$ to $-z$, and the generator $t$ of $R^*(C_3, \RR)$ and $R^*C_3, \QQ)$ to itself. The stable part of $R^*(C_3)$ is, in all cases, the even degrees, which happens to be the image of the restriction, as we showed during the proof of \cref{swanslemma_S3}. Thus Swan's lemma is true for the $3$-Sylow $C_3$ of $S_3$.
\paragraph*{}Pick a $2$-Sylow $C_2$ of $S_3$. Its normalizer is itself, so the action of the normalizer is trivial. Thus
\[
R^*(C_2)^{N_{S_3}(C_2)} = R^*(C_2)
\]
As we saw above, the restriction to $C_2$ from $S_3$ is surjective, so Swan's lemma is again (trivially) true.
\subsection{Generalization: the dihedral groups \texorpdfstring{$D_p$}{Dp}}
\subsubsection{The graded ring of \texorpdfstring{$D_p$}{Dp}}
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}
\subsubsection{Swan's lemma for \texorpdfstring{$D_p$}{Dp}}
We have $D_{p} = C_2\rtimes C_p$.
On the one hand, $C_2$ is an abelian 2-Sylow of $D_p$ which is self-normalizing. Thus its action on itself is trivial and the stable subring $R^*(C_2)^{N_{D_p}(C_2)}$ is the whole ring $R^*(C_2)$. The restriction $R^*(D_p) \rightarrow R^*(C_2)$ is surjective, so Swan's lemma is true in that case.
On the other hand, the quotient $D_p/C_p = C_2$ acts on the (normal, $p$-Sylow) subgroup $C_p$ via $\sigma \mapsto \sigma^{-1}$. Thus:
\begin{enumerate}
\item On $R^*(C_p,\CC) = \ZZ\left[c_1(\rho)\right]/(pc_1(\rho0)$, this induces $c_1(\rho) \mapsto -c_1(\rho)$ and so the stable subring is
\[
R^*(C_p, \CC)^{C_2} = R^{2*}(C_p, \CC).
\]
\item The induced action on $R^*(C_p,\RR) = \ZZ\left[c_1(\hat{\rho})\right]/(pc_1(\hat{\rho}))$ is the identity. Therefore the stable subring is
\[
R^*(C_p,\RR)^{C_2} = R^*(C_p,\RR).
\]
\end{enumerate}
In both cases, we have
\[
\Ima (res_{C_p}^{D_p}) = R^*(C_p)^{N_{D_p}(C_p)}.
\]
\subsection{Induction as a transfer for \texorpdfstring{$D_p$}{Dp}} \label{Dp_is_saturated}
Fix a finite group $G$ and a subgroup $H$ of $G$. Then any (complex) representation $\rho$ of $H$ induces a complex representation $\mathrm{Ind}_{H}^{G}(\rho)$ of $G$. This induction can be extended by linearity to an additive group homomorphism $T: R(H) \rightarrow R(G)$. However, $T$ is not in general a ring homomorphim, since $T(1) = \CC\left[G/H\right]$.
\paragraph*{Question.} \textit{Is $T$ compatible with the graded ring structure on $R(H)$ and $R(G)$?}\\
More precisely, we'd like the following property to be true:
\[
T(\Gamma^N(H)) \subset \Gamma^N(G)
\]
for any $N \in \NN$.
\paragraph*{Useful properties.} \begin{enumerate}
\item Projection formula: $T(x\cdot\res(y)) = T(x)y$
\item Augmentation: $\varepsilon(T(x)) = \left[G:H\right]\varepsilon(x)$, so in particular $T(I_H) \subset I_G$ where $I_H, I_G$ are augmentation ideals.
\end{enumerate}
\begin{Example}
Let $p$ be an odd prime. Consider the dihedral group $G=D_p$ of order $2p$ and its normal subgroup $H = C_p$. Fix $\rho$, an irreducible (nontrivial) representation of $H$, and recall that
\[
R^*(G) = \frac{\ZZ\left[x,y\right]}{(2x,py,xy)}
\]
with $\vert x \vert = 1$, $\vert y \vert = 2$. Let $t = C_1(\rho) \in R(H)$, then $\res(Y) = t^2$ (where $C_1(\rho), Y $ are some lifts of $c_1(\rho), y$). Since $H$ is abelian, we have $\Gamma^N(H) = (\rho-1)^N = (t^N)$.\\
Recall that $R(H)$ is generated by powers of $t$; so we need to look at the effect of $T$ on those. We have:
\begin{itemize}
\item $T(t) \in I_G = \Gamma^1$ as we noticed above.
\item $T(t^2) = T(1\cdot \res(Y)) = T(1)\cdot Y$ by the projection formula, and $Y \in \Gamma^2$ by definition.
\item In general, $T(t^n) = T(t^{2k+i}) = T(t)^iY^k \in \Gamma^i\cdot\Gamma^{2k} \subset \Gamma^{2k+i}$ for $i=0,1$.
\end{itemize}
So in this case we have $T(\Gamma^N(H)) \subset \Gamma^N(G)$.
\begin{Remark} This also works for the other nontrivial subgroup $K=C_2$ of $D_p$, since the generator $z$ of $R^*(K)$ is the restriction of the generator $x$ of degree 1 of $R^*(G)$.
\end{Remark}
\end{Example}
\begin{Remark} Suppose that some element $x \in \Gamma^1(H)$ satisfies $x^k = \res(y)$ for some $y \in R(G)$. Then, proceeding as in the above calculation, we can show that if $T(x^i) \in \Gamma^i$ for all $i = 1,\cdots,k-1$ (and $y\in\Gamma^k$) then for all $N$,
\[
T(x^N)= T(x^{mk+i}) = T(x^i)y^m \in \Gamma^i\Gamma^{mk}
\]
\end{Remark}
Another idea is to quantify the obstruction to $T$ being a ring homomorphism.
\begin{Lemma}
Suppose $H$ is normal in $G$ and $y \in R(H)$ is stable by $G$, that is:
\[
y(\sigma\tau\sigma^{-1}) = y(\tau)
\]
for any $\tau \in H$, $\sigma \in G$. Then for any $x \in R(H)$
\[
T(x)T(y) = T(xy)T(1)
\]
\end{Lemma}
This gives some hope for calculations when, for example, $H$ is a $p$-Sylow: in that case $T(1) = \CC\left[G/H\right]$ is $G/H$-torsion in the graded ring $R^*(G/H)$ and would be "somewhat invertible" in $R^*(G)$. Note also that $y$ being stable by $G$ seems a nontrivial condition: it is not satisfied by any irreducible representation of $H=C_p$ in the example above.
\begin{proof}
If $y$ is stable by $G$ then the induced character $T(y)$ is $T(1)y'$, where the values of $y'$ are those of $y$ on $H$ and 0 on $G\setminus H$. In other words we have $\res (y') = y$. Thus:
\[
T(x)T(y) = T(x)T(1)y' = T(x \res(y'))T(1) = T(xy)T(1).
\]
\end{proof}
