\classheader{2012-10-29}

Last time, we finished with the definition of a $G$-equivariant $A$-module.

\begin{examples}
$\text{}$

\begin{itemize}
\item Let $A=k\{X\}$, and let $G\lcirclearrowright X$. Then there is an obvious induced action $G\lcirclearrowright A$: because $g$ sends $x$ to $gx$, it should send $\mathbf{1}_x$ to $\mathbf{1}_{gx}$, and hence it sends any $f=\sum_{x\in X}\lambda_x\mathbf{1}_x\in A$ to
\[gf=\sum_{x\in X}\lambda_x\mathbf{1}_{gx}.\]
Note that $(gf)(x)=\lambda_{g^{-1}(x)}$. Given an $A$-module $M$ and a (possibly unrelated) $k$-linear $G$-action $G\lcirclearrowright M$, what does it mean for $M=\bigoplus_{x\in X}M_x$ to be a $G$-equivariant $A$-module?  

$M$ is $G$-equivariant when, for any $g\in G$, the action of $g$ on $M$ consists of maps $g_*:M_x\to M_{g(x)}$ (in general, it could be any $k$-linear map, even one that doesn't respect the decomposition of $M$ over the elements of $X$) with the property that, for any $f=\sum_{x\in X}\lambda_x\mathbf{1}_x=(\lambda_x)_{x\in X}\in A$ and $m=(m_x)_{x\in X}\in M$, \[g_*((\lambda_x)(m_x))=(\lambda_{g^{-1}(x)})g_*((m_x)).\]

Visualizing $M$ as the vector space $M_x$ attached to the corresponding point $x$, 
\begin{center}
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\node at (0,-0.2) {$x$};
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\node at (4,-0.2) {$h(g(x))$};
\node at (0,0.5) {$M_{x}$};
\node at (2.01,0.49) {$M_{g(x)}$};
\node at (4.02,0.49) {$M_{h(g(x))}$};
\node at (1,-0.3) {$g$};
\node at (1,1.06) {$g_*$};
\node at (3,-0.3) {$h$};
\node at (3,1.06) {$h_*$};
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\end{center}
\item For any $k$-algebra $A$ and any $k$-vector space $V$, we can make a free $A$-module $M=A\otimes_k V$. Then, for any action $G\lcirclearrowright A$ and any $G$-representation $G\to\GL(V)$, we can let $G$ act on $M=A\otimes_k V$ by $g(a\otimes v)={}^ga\otimes g(v)$. Let's do an example:

Let $G\lcirclearrowright E$ where $E$ is a set. Then $G\lcirclearrowright A=k[E]$, the polynomial ring in the elements of $E$. Let $M=k[E]\otimes V$. Then we get what in geometry would be called a vector bundle, and moreover one carrying a $G$-action:
\begin{center}
\begin{tikzcd}
E\times V\ar{d} & & (e,V) \ar{r}{g_*} \ar{d} & (g(e),V) \ar{d}\\
E & & e \ar{r}[swap]{g} & g(e)
\end{tikzcd}
\end{center}
Because $G$ acts on $k[E]$ by automorphisms, knowing how $G$ acts on $E$ tells us how it will act on any element of $k[E]$.
\end{itemize}
\end{examples}
\begin{question}
Can we define an algebra $A\# G$ such that the data of a $G$-equivariant $A$-module is equivalent to the data of an $A\# G$-module?
\end{question}

\begin{definition}
Given a $k$-algebra $A$, the algebra $A\# G$ is defined to be a free $A$-module with basis $G$, so a general element looks like $\sum_{g\in G}a_g g$. We define
\[(ag)(bh)=(a\cdot{}^gb)\cdot (gh)\]
where $a,b\in A$, and $g,h\in G$. This construction answers the question in the affirmative.
\end{definition}
Note the similarity to the definition of the semi-direct product.

Note that we have an inclusion $A\hookrightarrow A\# G$ defined by $a\mapsto a\cdot 1_G$. 

When $G$ is finite (so that we can speak of $kG$), we also have an inclusion $kG\hookrightarrow A\# G$ mapping $\lambda\cdot g\mapsto \lambda\cdot g$. In particular, if $A=k$, then $k\# G=kG$.  Note that $A\# G\cong A\otimes_k kG$ as $k$-vector spaces, but \textbf{not} as $k$-algebras, and certainly not as $A$-algebras.

Now let $A$ be a $k$-algebra with an action of a finite group $G\lcirclearrowright A$. Then $kG\hookrightarrow A\# G$, and therefore $A\# G$ is a $kG$-bimodule, i.e. $A\# G$ has a $G\times G$ action,
\[(g_1\times g_2):a\cdot h\mapsto g_1\cdot(a\cdot h)\cdot g_2^{-1}.\]
Let
\[e=\frac{1}{\# G}\sum_{g\in G}g\]
be the standard averaging idempotent. Recall that $e$ is central and $e^2=e$ in $kG\subset A\# G$.

For any $ag\in A\# G$, we have that
\[(ag)e=\frac{1}{\# G}\sum_{h\in G}agh=\frac{1}{\# G}\sum_{h\in G} ah=ae.\]
Thus, for any $\sum_{g\in G}a_gg\in A\# G$, we have \[\left(\sum_{g\in G}a_gg\right)e=\left(\sum_{g\in G}a_g\right)e,\] and hence $(A\# G)e\cong Ae$. Then, $e(Ae)=eAe$ consists of the elements of $Ae$ fixed by $g$ (recall that in general if $G$ acts on $V$, then $eV=V^G$), so there is a canonical isomorphism $e(A\# G)e\cong eAe\cong A^G$.

\section*{Quivers}
\begin{definition}
A quiver is a finite oriented graph. We can write it as an ordered pair $(Q,I)$ where $I$ is the set of vertices and $Q$ is the set of arrows.
\end{definition}
\begin{center}
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\begin{scope}[every node/.style={inner sep=0pt,outer sep=2pt,minimum size=5pt,fill,circle}]
\node (a) at (0,0)  {};
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\node (d) at (4,0)  {};
\end{scope}
\path[thick,->] (a) edge[out=50,in=180] (b);
\path[thick,->] (b) edge[out=280,in=140] (c);
\path[thick,->] (b) edge[out=320,in=100] (c);
\path[thick,->] (c) edge[out=20,in=250] (d);
\path[thick,->] (c) edge[out=175,in=330] (a);
\path[thick,->] (a) edge[out=150,in=280,loop,looseness=2,distance=2cm] (a);
\end{tikzpicture}\vspace{-0.3in}
\end{center}
We often denote a quiver just by the set of arrows $Q$.

A path in $Q$ is a finite sequence of arrows which meet head to tail.
\begin{center}
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\begin{scope}[every node/.style={inner sep=0pt,outer sep=2pt,minimum size=5pt,fill,circle}]
\node (a) at (0,0)  {};
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\path[thick,->] (a) edge[out=40,in=140] (b);
\path[thick,->] (b) edge[out=340,in=200] (c);
\end{tikzpicture}
\end{center}

\begin{definition}
The path algebra $kQ$ of $Q$ over a field $k$ is a $k$-vector space with basis formed by paths in $Q$, with product given by concatenation of paths which can be concatenated (paths that can't be concatenated multiply to 0). The order in which paths are concatenated is apparently a delicate issue. There are also trivial paths $\mathbf{1}_i$ at each vertex $i\in I$; each trivial path is an idempotent, and acts as an identity when concatenated with paths that start or end at $i$.
\end{definition}
\begin{examples}
$\text{}$

\begin{itemize}
\item Consider the quiver $Q$ with one vertex and one edge (the trivial path is implied),\begin{center}
\tikz[>=stealth]{
\begin{scope}[every node/.style={inner sep=0pt,outer sep=2pt,minimum size=5pt,fill,circle}]
\node (a) at (0,0)  {};
\end{scope}
%\node at (0.22,0.22) {$\mathbf{1}$};
\path[thick,->] (a) edge[out=180,in=250,loop,distance=1cm] node[below left] {$x$} (a);}\vspace{-0.2in}
\end{center}
Then $kQ=k[x]$. Note that, without the trivial path at the vertex, we'd only get polynomials in $x$ with no constant term.
\item Consider the quiver $Q$ with $n$ loops around one vertex (the trivial path is implied),
\begin{center}
\tikz[>=stealth]{
\begin{scope}[every node/.style={inner sep=0pt,outer sep=2pt,minimum size=5pt,fill,circle}]
\node (a) at (0,0)  {};
\end{scope}
\path[thick,->] (a) edge[out=0,in=50,loop,distance=1cm] node[above right] {$x_1$} (a);%\vspace{-0.3in}
\path[thick,->] (a) edge[out=80,in=130,loop,distance=1cm] node[above left] {$x_2$} (a);
\path[thick,->] (a) edge[out=280,in=330,loop,distance=1cm] node[below right] {$x_n$} (a);%\vspace{-0.3in}
\node[rotate=340] (b) at (210:0.5)  {$\ddots$};
%\node at (210:0.25)  {$\mathbf{1}$};}
}
\end{center}
Then $kQ=k\langle x_1,\ldots,x_n\rangle$, the free associative algebra on $n$ generators.
\item Consider the quiver $Q$ with $n$ vertices in a row with edges between them,
\begin{center}
\begin{tikzpicture}[>=stealth,xscale=2]
\begin{scope}[every node/.style={inner sep=0pt,outer sep=2pt,minimum size=5pt,fill,circle}]
\node (a) at (0,0) {};
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\node (c1) at (3,0) {};
\node[gray] (c3) at (3.25,0) {$\cdots$};
\node (c2) at (3.5,0) {};
\node at (0,-0.35) {$\mathbf{1}_1$};
\node at (1,-0.35) {$\mathbf{1}_2$};
\node at (2,-0.35) {$\mathbf{1}_3$};
\node at (4.5,-0.35) {$\mathbf{1}_{n-1}$};
\node at (5.5,-0.35) {$\mathbf{1}_n$};
\path[thick,->] (a) edge[out=0,in=180] node[above] {$x_1$} (b);
\path[thick,->] (b) edge[out=0,in=180] node[above] {$x_2$} (c);
\path[thick,->,gray] (c) edge[out=0,in=180] node[above] {} (c1);
\path[thick,->,gray] (c2) edge[out=0,in=180] node[above] {} (d);
\path[thick,->] (d) edge[out=0,in=180] node[above] {$x_{n-1}$} (e);
\end{tikzpicture}
\end{center}
Then we get $kQ=$ the algebra of upper triangular matrices over $k$. The trivial path $\mathbf{1}_i$ at vertex $i$ corresponds to a matrix with 1 on the $i$th diagonal entry and 0 elsewhere.
\end{itemize}
\end{examples}


\begin{definition}
Given a finite subgroup $G\subset\GL(V)$, we define the McKay quiver $Q_G$ associated to it as follows. The vertex set of $Q_G$ is in bijection with $\widehat{G}$, say with $i\leftrightarrow L_i$. Then we set the number of edges from $i$ to $j$ to be
\[\#\{i\to j\}=\dim(\Hom_G(L_i,V\otimes L_j))=[V\otimes L_j:L_i].\]
\end{definition}
\begin{example}
Let $G=\left\{\left(\begin{smallmatrix}
\zeta & 0\\ 0 & \zeta^{-1}
\end{smallmatrix}\right)\,\middle\vert\,\zeta^n=1\right\}\subset\GL(\C^2)$. Then $G\cong\Z/n\Z$. Letting $z$ be a primitive $n$th root of unity, then the irreducible representations of $G$ are the maps $L_i:G\to\C^\times$ defined by $\left(\begin{smallmatrix}
z & 0\\ 0 & z^{-1}
\end{smallmatrix}\right)\mapsto z^i$ for $i=0,1,\ldots,n-1$. Noting that $\C^2=L_1\oplus L_{-1}$, we get
%\begin{center}
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%\begin{center}
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%\tikzstyle{every node}=[outer sep=0pt,shape=circle];
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%\path[mystyle,shorten >=13pt,shorten <=13pt]([xshift=-0.2ex] v2.center) edge ([xshift=0.3ex] w1.center);
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\begin{center}
\begin{tikzpicture}[>=stealth,xscale=3,yscale=2,mystyle/.style={thick,->},myotherstyle/.style={yshift=0.4ex},otherstyle/.style={yshift=-0.4ex}]
\tikzstyle{every node}=[outer sep=0pt,shape=circle];
\path (45:1cm) node (v2) {$i+1$};
\path (90:1cm) node (v3) {$i$};
\path (135:1cm) node (v4) {$i-1$};
\path (0:1cm) node[rotate=285] (w1) {$\cdots$};
\path (180:1cm) node[rotate=255] (w2) {$\cdots$};
\path[mystyle,shorten >=9pt,shorten <=16pt]([yshift=0.3ex] v4.center)  edge ([yshift=0.3ex] v3.center);
\path[mystyle,shorten >=16pt,shorten <=9pt] ([yshift=-0.3ex] v3.center) edge ([yshift=-0.3ex] v4.center);
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\path[mystyle,shorten >=9pt,shorten <=13pt]  ([xshift=0ex] w2.center) edge  ([xshift=-0.5ex] v4.center);
\end{tikzpicture}
\end{center}

\end{example}