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\title{Physics Seminar\\April 1, 2005\\Tanveer Prince}
\author{Gabriel C. Drummond-Cole}
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Last time, let me remind you of Gabriel's theorem.  
\begin{enumerate}
\item A quiver $Q$ is finite type if and only if its underlying graph $\hat{Q}$ has type $A_n,\ D_n,\ E_6,\ E_7,$ or $E_8.$  This means it has finitely many indecomposable representations.

\item If $Q$ is of finite type, then there is a bijection between positive roots of the corresponding Lie algebra and indecomposable representations.
\end{enumerate}
In particular this says how many representations it has.  So for $A_n$ this is $\frac{n(n+1)}{2};$ for $D_n$ it is $n^2-n.$

But this is stronger because it gives a bijection.  For an indecomposable representation, you can just take the dimension vector.  To go the other way is a little more tricky.

So let's go on.  We'll start with a definition and some notation.  We start with a Coxeter functor.
\begin{defi}
Say $x\in Q_0,$ the vertices of the quiver $Q.$  We can construct the quiver $S_x(Q),$ which is the same quiver but with all arrows incident on $x$ reversed.

Now let $x\in Q_0$ be a sink.  Define $C^+_x,$ a functor from representations of $Q$ to representations of $S_x(Q).$ 

So say I have a sink $x,$ I want to show how to define the functor.  The vertex $x,$ instead of being enriched with the same vector space $V(x),$ is enriched by the kernel of the map $V_1\oplus\cdots\oplus V_k\to V(x)$ with the maps the projections.

If I have a source $x$ I can define $C_x^-$ similarly by replacing $V(x)$ with $coker(V(x)\to V_1\oplus\cdots\oplus V_k).$
\end{defi}
One more definition.  Consider the simple representation where you have $k$ in one place and zero elsewhere.  This is represented by the symbol $\epsilon_x,$ where $x$ is the vertex.  We denote the corresponding reflection by $S_x.$  This is defined to be $S_x(\beta)=\beta-(\epsilon_x,\beta)\epsilon_x),$ where $(,\ )$ is the symmetric Euler form so $(\epsilon_x,\beta)=\langle\epsilon_x,\beta\rangle+\langle\beta,\epsilon_x\rangle.$

\begin{thm}
Let $Q$ be a quiver of finite type and $V$ an indecomposable representation of $Q.$  Say $x$ is a sink.  Then one of the following is true:
\begin{enumerate}
\item $V=E_x,$ and $C^+_xV$ is a representation of $S_xQ$ which is $0.$
\item $C^+_xV$ is indecomposable and the dimension vector of $C^+_xV$ will be $S_x$ applied to the original dimension vector.  Alsoo $C_x^-C_x^+V\sim V.$
\end{enumerate}\end{thm}
These representations are not equivalent, but there is a nice correspondence between the indecomposables one and the other.  

Before I prove this one, let me write down some corollaries.
I can number my vertices so that arrows go in increasing order.

Then $1$ is a source and $n$ is a sink.  Then $C_n^-\circ\cdots\circ C_2^-\circ C_1^-$ is a functor from $Rep(Q)$ to $Rep(S_n\cdots S_2S_1Q)=Rep(Q).$  Similarly, $C^+_1\cdots C_{n-1}^+C_n^+$ is a functor from $Rep(Q)$ to itself.  Denote these by $C^-$ and $C^+$ with corresponding reflections $c^-$ and $c^+$ or just $c.$

\begin{cor}
$V$ is an indecomposable representation then $C^+V=0$ or is indecomposable with dimension vector $d_{C^+(V)}=c^+(d_V).$
\end{cor}
\begin{lemma}
Let $Q$ be a quiver where $\hat{Q}\in\{A_n,D_n,E_6,E_7,E_8\}$ and $\alpha$ a dimension vector.  Then $c^k\alpha$ is a negative vector.
\end{lemma}

So $c$ has finite order $m.$  Then consider $\beta=\sum_0^{m-1} c^i\alpha.$  Then $c\beta=\beta,$ so that $\beta=0.$  This is because $\langle \alpha,c\beta\rangle=-\langle\beta,\alpha\rangle.$  So if $c\beta=\beta$ then $(\beta,\beta)=2\langle \beta,\beta\rangle=\langle \beta,\beta\rangle+\langle\beta,c\beta\rangle=0.$

So next, keep the same assumptions and let $V$ be an indecomposable representation of the quiver.  I need to show that $d_V$ is a positive root.  From the corollary we have that $d_{(C^+)^k V}=c^k d_V.$  Then there exists a $k$ for which this is a negative vector.  So some power of it must give zero.

So consider $W=(C^+)^{k-1}V.$  We know that $C^+(W)=0.$  But this is $C_1^+\cdots C_n^+W.$  So then at some point I will get $E_l$ for some $l.$  Then $W=C_n^-\cdots C_{l+1}^- E_l.$  Then $V=(C^-)^{k-1}C_n^-\cdots C_{l+1}^-E_l$ so that $d_V=(c^-)^{k-1}s_n\cdots s_{l+1}\epsilon_l.$  This makes this root positive.

This is one direction.  Now we start with a positive root and want to show that it is a dimension vector.  I'll go over an example.

[What is going on is, we want to get all indecomposables.  We only know of the $E_i$ to start.  So we apply reflections to get more, but we can only use the reflections when these are sinks or sources.]

Now for the other direction we start with a positive root $\alpha.$  From the lemma we can choose a minimal $k$ such that $c^{k+1}\alpha$ is negative.  Consider $\beta=c^k\alpha.$  So then $s_1\cdots s_n\beta$ is negative.  Now $s_i$ permutes positive roots except for $\epsilon_i.$  So then $s_{l+1}\cdots s_n\beta=\epsilon_l,$ so $\beta=s_n\cdots s_{l+1}\epsilon_l.$  Then $C^-_n\cdots C_{l+1}^- E_l$ is an indecomposable with dimenstion vector $\beta;$ then $(C^-)^kC^-_n\cdots C_{l+1}^- E_l$ is an indeocomposable with dimension vector $\alpha.$

Consider the indecomposable
$$\xymatrix{k\ar[r]^1&k\ar[r]^1&k}$$ with corresponding dimension vector $e_1-e_4=\left(\begin{array}{c}1\\1\\1\end{array}\right).$

How do we go the other way?  We have
$$
s_1\alpha=\left(\begin{array}{c}1\\1\\1\end{array}\right)-\left[\left\langle\left(\begin{array}{c}1\\1\\1\end{array}\right),
\left(\begin{array}{c}1\\0\\0\end{array}\right)\right\rangle+
\left\langle
\left(\begin{array}{c}1\\0\\0\end{array}\right),
\left(\begin{array}{c}1\\1\\1\end{array}\right)
\right\rangle\right]
\left(\begin{array}{c}1\\0\\0\end{array}\right)
=
\left(\begin{array}{c}0\\1\\1\end{array}\right).$$  So $s_2s_1\alpha=\left(\begin{array}{c}0\\0\\1\end{array}\right)=\epsilon_3.$

Now $s_1A_3$ looks like
$$\xymatrix{0&k\ar[l]\ar[r]&k}$$ while $s_2s_1A_3$ is
$$\xymatrix{0\ar[r]&0&k\ar[l]}$$
Now if I apply $C_2^+$ I get
$$\xymatrix{0&k\ar[r]\ar[l]^1&k}$$ and then $C_1^+$ to get $$\xymatrix{k&k\ar[l]\ar[r]&k}$$

Let's look at a trickier example,

$$\xymatrix{
k\ar[r]&k^2&k\ar[l]\\&k\ar[u]}
$$ with dimension vector $\alpha=e_2+e_1=\left(\begin{array}{c}1\\1\\1\\2\end{array}\right)$ and $s_3s_2s_1\alpha=\alpha$ and $s_4s_3s_2s_1\alpha=\left(\begin{array}{c}1\\1\\1\\1\end{array}\right).$

Finally you get $s_3s_2s_1s_4s_3s_2s_1\alpha=\left(\begin{array}{c}0\\0\\0\\1\end{array}\right)$
so that $\alpha=s_1s_2s_3s_4s_1s_2s_3\epsilon_4.$

So you can start with $E_4$ in the quiver
$$\xymatrix{0&k\ar[r]\ar[l]\ar[d]&0\\&0&}$$ and apply $C_3^+$ to get
$$\xymatrix{0&k\ar[l]\ar[d]&k\ar[l]^1\\&0&}$$
and then $C_2^+$ yields
$$\xymatrix{0&k\ar[l]&k\ar[l]^1\\&k\ar[u]^1&}$$
and then $C_1^+:$
$$\xymatrix{k\ar[r]^1&k&k\ar[l]^1\\&k\ar[u]^1&}$$
Here it gets a little tricky because I have a kernel.
At the end I get
$$\xymatrix{\{(\beta,\gamma)|\beta+\gamma=0\}\ar[r]&R=\{(\alpha,\beta,\gamma)|\alpha+\beta_\gamma=0\}&\{(\alpha,\beta)|\alpha+\beta=0\}\ar[l]\\&\{(\alpha,\gamma)|\alpha+\gamma=0\}\ar[u]}$$

Why does the Coxeter functor applied to an indecomposable give either $0$ or an indecomposable?

The idea is as follows.  Say $x\in Q_0$ is a sink.  Then I have $C_x^+:Rep(Q)\to Rep(S_xQ).$  I can also go in the other direction with $C_x^-.$

First we'll construct a natural transformation $i_x$ from $C_x^-C_x^+$ to the identity.  In a similar way we construct a functor $p_x$ from the identity to $C_x^+C_x^-.$  Then the proof depends on some characteristics of these natural transformations.

$C_x^-C_x^+$ at $x$ is the cokernel of the projection from the kernel of the map into $V(x).$  This is the quotient of the sum $\oplus V_i$ by the kernel, which is the original space.  This is pure linear algebra.

There are some properties of these two natural transformations.
\begin{enumerate}
\item If $i_x$ is an isomorphism then $d_{C^+xV}=s_xd_V$ and similarly for the other one.
\item If $V=C_x^-W$ then $i_x$ is an isomorphism.
\item if $x\in Q_0$ is a sink then $V=C_x^-C_x^+\oplus \tilde{V},$ the cokernel of $i_x.$ 
\end{enumerate}
The proof of all of this is not hard.  Let me skip most of it.  Let $V$ be an indecomposable representation.  If $V=E_x$ then $C_x^+V=0.$  If $V$ is not this, then we will show that $C_x^+V$ is indecomposable.  

Since $V$ is indecomposable, either $V=\tilde{V}$ or $V=C^-_x C^+_x V.$  If $V=\tilde{V}$ then $V=E_x$ because it's concentrated at $x.$  Otherwise $V=C_x^-C_x^+V.$  We want to show that $C_x^+V$ is indecomposable.  Say it can be written $W_1\oplus W_2.$  Then $C_x^-$ preserves the direct sum so this is $C_x^-W_1\oplus C_x^- W_2.$  Since $V$ is indecomposable, now one of these has to be zero.  Then apply $C_x^+$ to that piece to get $C_x^+C_x^- W_2=0.$  But we have $p_x$ an isomorphism from the identity to $C_x^+C_x^-.$  This implies $W_2=0,$ so that $C_x^+V$ is indecomposable, as desired.

[Next week Justin will be talking about ...]

I have an article about this at \~sawon/borel\_weil.ps.gz

This is the most important thing about geometric representations.  He'll be talking about projective spaces.

This allows you to calculate cohomology groups of vector bundles etc.
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