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A2 King stability distinguishes the arrow map

Example

Let $Q$ have vertices $1, 2$ and one arrow $a : 1 \to 2$ , let the potential be $W = 0$ , and let $θ (d_{1}, d_{2}) = d_{1} - d_{2}$ . The representation $E$ with $E_{1} = E_{2} = C$ and $E (a) = {id}_{C}$ is $θ$ -stable under the sign convention in King theta stability. The representation $E_{0}$ with the same vertex spaces and $E_{0} (a) = 0$ is not $θ$ -semistable. Thus the dimension vector and weight alone do not decide stability; the arrow map matters.

The diagram fixes the single arrow's orientation and its endpoint labels. The algebraic verification below does not rely on the drawing.

Example references

Quiver potential and King theta stability data

Facts & Assumptions

Given: Work over $C$ . A representation of $Q$ is a pair of finite-dimensional vector spaces $V_{1}, V_{2}$ and a linear map $V (a) : V_{1} \to V_{2}$ . A subrepresentation is a pair $U_{i} \subseteq V_{i}$ with $V (a) (U_{1}) \subseteq U_{2}$ .

[F1]

Quiver potential and King theta stability data: $θ$ -semistability requires $θ (E) = 0$ and $θ (A) \leq 0$ for every subobject; stability requires strict inequality for every proper nonzero subobject.

Verification

Given: The data above and $θ (d_{1}, d_{2}) = d_{1} - d_{2}$ .

1.1givenF1algebra

Since each $E_{i} ≅ C$ , a subspace $U_{i} \subseteq E_{i}$ is either $0$ or $C$ . The subrepresentation condition for $E (a) = id$ excludes $(U_{1}, U_{2}) = (C, 0)$ . Thus the subrepresentations of $E$ are $(0, 0), (0, C), (C, C)$ .

2.1givenF1step 1.1algebra

We have $θ (E) = 1 - 1 = 0$ , and the only proper nonzero subobject in step 1.1 has weight $θ (0, C) = - 1 < 0$ . Therefore $E$ is $θ$ -stable.

3.1givenF1algebra∎

For $E_{0} (a) = 0$ , the pair $(C, 0)$ is a subrepresentation because the zero map sends $C$ into $0$ . It is nonzero and proper, and $θ (C, 0) = 1 > 0$ ; hence $E_{0}$ is not $θ$ -semistable.

Depends on

Quiver potential and King theta stability data