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The Lumenary · Research pathways

Quiver representations

Name: ______________________

Date: ______________________

You will learn to

Define representations of quivers, their dimension vectors, Euler form and Tits form, and compute $\operatorname{Hom}$ and $\operatorname{Ext}^1$ dimensions with the Euler form
Prove that a quiver of finite representation type has a weakly positive Tits form, and hence is of Dynkin type, following Tits' orbit-dimension argument
State Gabriel's theorem and verify it for small Dynkin quivers using reflections, reflection functors and the Tits form
Compute Auslander–Reiten quivers of path algebras of Dynkin type by knitting
State Kac's theorem and describe the links with Lie theory, cluster algebras and geometry, separating established results from open questions

1.
How many indecomposable representations (up to isomorphism) does the quiver $1 \to 2 \leftarrow 3 \to 4$ have over $\mathbb{C}$ ?
2.
For $D_4$ with the centre written second, $q(a, c, b, d) = a^2 + c^2 + b^2 + d^2 - ac - bc - dc$ . Which vectors are dimension vectors of indecomposable representations? (Choose all that apply.)
- A. $(1, 2, 1, 1)$
- B. $(1, 1, 1, 1)$
- C. $(1, 2, 1, 0)$
- D. $(2, 2, 1, 1)$
- E. $(1, 0, 1, 0)$
3.
For the Kronecker quiver $1 \rightrightarrows 2$ (two arrows from $1$ to $2$ ), compute $\dim \operatorname{Ext}^1(S(1), S(2))$ using the Euler form.
4.
The extended Dynkin graph $\tilde D_4$ is a star with centre $c$ and four leaves. Compute its Tits form at $\delta = (1, 1, 1, 1, 2)$ (leaves first, centre last).
5.
Two connected quivers $Q$ and $Q'$ have the same underlying graph but different orientations. Which statement is correct?
- A.They have the same number of indecomposables, but a given dimension vector may be realised by quite different representations
- B.Their categories of representations are always equivalent
- C.One may have finite type while the other has infinite type
- D.Their Tits forms differ
6.
In Tits' proof that finite type forces $q(d) \ge 1$ , where does the “ $-1$ ” in $\dim(\text{orbit}) \le \sum_i d_i^2 - 1$ come from?
- A.The scalar matrices $\lambda \cdot (I_{d_i})_i$ act trivially on $\mathrm{Rep}(Q, d)$ , so every stabiliser has dimension at least $1$
- B. $\mathrm{Rep}(Q, d)$ is a projective variety of dimension one less than the affine space
- C.Each indecomposable has a one-dimensional endomorphism ring
- D.The zero representation is a fixed point
7.
Which statements about the Kronecker quiver $1 \rightrightarrows 2$ over $\mathbb{C}$ are true? (Choose all that apply.)
- A.For each $n \ge 0$ there is exactly one indecomposable of dimension vector $(n, n+1)$
- B.The indecomposables of dimension $(1, 1)$ are parametrised by $\mathbb{P}^1(\mathbb{C})$
- C.It has finite representation type
- D.Its Tits form is positive definite
8.
Which Lean statement correctly formalises “the positive roots of $D_4$ are exactly the twelve vectors in D4_roots”? (titsD4 takes four integers, the centre third.)
- A.theorem t (a b c d : ℕ) : titsD4 a b c d = 1 ↔ (a, b, c, d) ∈ D4_roots
- B.theorem t (a b c d : ℤ) : titsD4 a b c d = 1 ↔ (a, b, c, d) ∈ D4_roots
- C.theorem t : D4_roots.length = 12
- D.theorem t (a b c d : ℤ) : 0 < titsD4 a b c d
9.
Write positive_roots(n, edges) returning the set of positive roots (as tuples) of a Dynkin graph with vertices 0, …, n-1 and edge list edges, by closing the simple roots under simple reflections $s_i(x)_i = \sum_{j \sim i} x_j - x_i$ and keeping only non-negative, non-zero vectors.
Computational — Practice

Answer key — Quiver representations

1.
The underlying graph is $A_4$ , so by Gabriel's theorem the indecomposables correspond to the positive roots of $A_4$ , the $\binom{5}{2} = 10$ intervals $e_i + \dots + e_j$ . The zigzag orientation changes which representation has a given dimension vector, but not the count; the lesson's experiment confirms $10$ for both orientations.
2.
(a) and (b). Gabriel's theorem identifies dimension vectors of indecomposables with the vectors $x \ge 0$ , $x \ne 0$ , with $q(x) = 1$ ; the formal section proves that there are exactly twelve, including $(1, 2, 1, 1)$ and $(1, 1, 1, 1)$ . The others have $q = 2$ .
3.
$\langle e_1, e_2 \rangle = 0 - 2 = -2$ , because both arrows go from $1$ to $2$ . As $\operatorname{Hom}(S(1), S(2)) = 0$ , Theorem 1 gives $\dim \operatorname{Ext}^1(S(1), S(2)) = 2$ : the extensions $0 \to S(2) \to E \to S(1) \to 0$ are the representations $k \rightrightarrows k$ given by a pair $(\lambda, \mu)$ , and their classes form $k^2$ .
4.
$q(\delta) = (1 + 1 + 1 + 1 + 4) - 4 \cdot (1 \cdot 2) = 8 - 8 = 0$ . So $\delta$ is an imaginary (null) root: by Kac's theorem there are infinitely many indecomposables of dimension $\delta$ , in a $1 - q(\delta) = 1$ parameter family — the cross-ratio of four lines in a plane. This null root is what makes any graph containing $\tilde D_4$ non-Dynkin (Corollary 3).
5.
(a). The Tits form and the set of positive roots depend only on the graph, so the number of indecomposables does too. But $\mathrm{rep}(Q)$ and $\mathrm{rep}(Q')$ need not be equivalent, and the representation with a given dimension vector changes its nature: for $1 \to 2 \to 3$ the indecomposable of dimension $(1, 1, 1)$ is $P(1) = I(3)$ , both projective and injective, while for $1 \to 2 \leftarrow 3$ it is the injective $I(2)$ , which is not projective. The derived categories, however, are equivalent (Happel).
6.
(a). The orbit of $V$ has dimension $\dim GL(d) - \dim \operatorname{Stab}(V)$ , and $\operatorname{Stab}(V) = \operatorname{Aut}(V)$ always contains the scalars. Hence $\dim(\text{orbit}) \le \sum_i d_i^2 - 1$ . With finitely many orbits, one must be dense in the irreducible space $\mathrm{Rep}(Q, d)$ of dimension $\sum_a d_{s(a)} d_{t(a)}$ , giving $\sum_a d_{s(a)} d_{t(a)} \le \sum_i d_i^2 - 1$ , i.e. $q(d) \ge 1$ .
7.
(a) and (b). $q(x, y) = (x - y)^2$ , so $(n, n+1)$ and $(n+1, n)$ are real roots and $(n, n)$ are imaginary. Kac's theorem gives one indecomposable for each real root (the preprojectives $(n, n+1)$ and preinjectives $(n+1, n)$ ) and a one-parameter family ( $1 - q = 1$ ) for each $(n, n)$ — for $n = 1$ , the points of $\mathbb{P}^1$ . The Kronecker quiver is tame, not of finite type.
8.
(a). The statement must quantify over non-negative vectors (ℕ) and give an equivalence. Option (b) fails because $q(-x) = q(x)$ ; (c) is only a count; (d) forgets the non-zero hypothesis.

def positive_roots(n, edges):
    simple = [tuple(int(k == i) for k in range(n)) for i in range(n)]
    found, todo = set(simple), list(simple)
    while todo:
        x = todo.pop()
        for i in range(n):
            nbr = sum(x[b] for a, b in edges if a == i) + sum(
                x[a] for a, b in edges if b == i)
            y = list(x)
            y[i] = nbr - x[i]
            y = tuple(y)
            if min(y) >= 0 and any(y) and y not in found:
                found.add(y)
                todo.append(y)
    return found

This finds $6$ , $12$ , $20$ , $36$ roots for $A_3$ , $D_4$ , $D_5$ , $E_6$ . It terminates because the Tits form is positive definite, so only finitely many vectors have $q = 1$ . That every positive root is reached from a simple root by reflections that stay positive is a theorem about root systems; the program is evidence for these cases.