Physics Seminar April 1, 2005 Tanveer Prince

Gabriel C. Drummond-Cole

April 15, 2005

Last time, let me remind you of Gabriel’s theorem.

1. A quiver Q is finite type if and only if its underlying graph Q̂ has type A_{n}, D_{n}, E_{6}, E_{7}, or E_{8}. This means it has finitely many indecomposable representations.

2. If Q is of finite type, then there is a bijection between positive roots of the corresponding Lie algebra and indecomposable representations.

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.

Definition 1 Say x ∈ 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 ∈ 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} ⊕ ⋯ ⊕ V_{k} → 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) → V_{1} ⊕ ⋯ ⊕ V_{k}).

One more definition. Consider the simple representation where you have k in one place and zero elsewhere. This is represented by the symbol ϵ_{x}, where x is the vertex. We denote the corresponding reflection by S_{x}. This is defined to be S_{x}(β) = β − (ϵ_{x}, β)ϵ_{x}), where (, ) is the symmetric Euler form so (ϵ_{x}, β) = ⟨ϵ_{x}, β⟩ + ⟨β, ϵ_{x}⟩.

Theorem 1 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:

1. V = E_{x}, and C_{x}^{+}V is a representation of S_{x}Q which is 0.

2. C_{x}^{+}V is indecomposable and the dimension vector of C_{x}^{+}V will be S_{x} applied to the original dimension vector. Alsoo C_{x}^{−}C_{x}^{+}V ∼ V.

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}^{−} ∘ ⋯ ∘ C_{2}^{−} ∘ C_{1}^{−} is a functor from Rep(Q) to Rep(S_{n} ⋯ S_{2}S_{1}Q) = Rep(Q). Similarly, C_{1}^{+} ⋯ 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.

Corollary 1 V is an indecomposable representation then C^{+}V = 0 or is indecomposable with dimension vector d_{C^{+}(V)} = c^{+}(d_{V}).

Lemma 1 Let Q be a quiver where Q̂ ∈ {A_{n}, D_{n}, E_{6}, E_{7}, E_{8}} and α a dimension vector. Then c^{k}α is a negative vector.

So c has finite order m. Then consider β = ∑_{0}^{m−1} c^{i}α. Then cβ = β, so that β = 0. This is because ⟨α, cβ⟩ = −⟨β, α⟩. So if cβ = β then (β, β) = 2⟨β, β⟩ = ⟨β, β⟩ + ⟨β, cβ⟩ = 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}^{+} ⋯ C_{n}^{+}W. So then at some point I will get E_{l} for some l. Then W = C_{n}^{−} ⋯ C_{l+1}^{−}E_{l}. Then V = (C^{−})^{k−1}C_{n}^{−} ⋯ C_{l+1}^{−}E_{l} so that d_{V} = (c^{−})^{k−1}s_{n} ⋯ s_{l+1}ϵ_{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 α. From the lemma we can choose a minimal k such that c^{k+1}α is negative. Consider β = c^{k}α. So then s_{1} ⋯ s_{n}β is negative. Now s_{i} permutes positive roots except for ϵ_{i}. So then s_{l+1} ⋯ s_{n}β = ϵ_{l}, so β = s_{n} ⋯ s_{l+1}ϵ_{l}. Then C_{n}^{−} ⋯ C_{l+1}^{−}E_{l} is an indecomposable with dimenstion vector β; then (C^{−})^{k}C_{n}^{−} ⋯ C_{l+1}^{−}E_{l} is an indeocomposable with dimension vector α.

Consider the indecomposable

\xymatrix{k \ar[r]^{1} & k \ar[r]^{1} & k}

with corresponding dimension vector e_{1} − e_{4} = ( \begin{matrix}1 \\ 1 \\ 1 \end{matrix} ) .

How do we go the other way? We have

s_{1}α = ( \begin{matrix}1 \\ 1 \\ 1 \end{matrix} ) − [⟨( \begin{matrix}1 \\ 1 \\ 1 \end{matrix} ) , ( \begin{matrix}1 \\ 0 \\ 0 \end{matrix} )⟩ + ⟨( \begin{matrix}1 \\ 0 \\ 0 \end{matrix} ) , ( \begin{matrix}1 \\ 1 \\ 1 \end{matrix} )⟩] ( \begin{matrix}1 \\ 0 \\ 0 \end{matrix} ) = ( \begin{matrix}0 \\ 1 \\ 1 \end{matrix} ) .

So s_{2}s_{1}α = ( \begin{matrix}0 \\ 0 \\ 1 \end{matrix} ) = ϵ_{3}.

Now s_{1}A_{3} looks like

\xymatrix{0 & k \ar[l] \ar[r] & k}

while s_{2}s_{1}A_{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 α = e_{2} + e_{1} = ( \begin{matrix}1 \\ 1 \\ 1 \\ 2 \end{matrix} ) and s_{3}s_{2}s_{1}α = α and s_{4}s_{3}s_{2}s_{1}α = ( \begin{matrix}1 \\ 1 \\ 1 \\ 1 \end{matrix} ) .

Finally you get s_{3}s_{2}s_{1}s_{4}s_{3}s_{2}s_{1}α = ( \begin{matrix}0 \\ 0 \\ 0 \\ 1 \end{matrix} ) so that α = s_{1}s_{2}s_{3}s_{4}s_{1}s_{2}s_{3}ϵ_{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{\{(β, γ)|β + γ = 0\} \ar[r] & R = \{(α, β, γ)|α + β_{γ} = 0\} & \{(α, β)|α + β = 0\} \ar[l] \\  & \{(α, γ)|α + γ = 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 ∈ Q_{0} is a sink. Then I have C_{x}^{+} : Rep(Q) → Rep(S_{x}Q). 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 ⊕V_{i} by the kernel, which is the original space. This is pure linear algebra.

There are some properties of these two natural transformations.

1. If i_{x} is an isomorphism then d_{C^{+}xV} = s_{x}d_{V} and similarly for the other one.

2. If V = C_{x}^{−}W then i_{x} is an isomorphism.

3. if x ∈ Q_{0} is a sink then V = C_{x}^{−}C_{x}^{+} ⊕ Ṽ , the cokernel of i_{x}.

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 = Ṽ or V = C_{x}^{−}C_{x}^{+}V. If V = Ṽ 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} ⊕ W_{2}. Then C_{x}^{−} preserves the direct sum so this is C_{x}^{−}W_{1} ⊕ 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 s̃awon/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.