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ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6.1-sol)
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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One-color quiver full flags need not be semismall

Example

Assume the Axiom of Choice as inherited from the cited geometry suppliers. Take the loopless quiver with a single vertex and no arrows, dim⁡V=n≥2, and its unique word. Then EV=0 and Xν=Flag⁡(V). The proper incidence map to the point need not be semismall: in this example its defect 2dim⁡(fiber)+dim⁡(image)−dim⁡Xν is positive.

Verification

Given: the loopless one-vertex quiver with no arrows, the space V of dimension n≥2, the unique word ν of content n, and the choice assumption of the Example.

1.1givenalgebra

There are no arrows, so EV=0, and the stability conditions of the incidence lemma Quiver full-flag incidence has a smooth projective compactification are vacuous; the lemma gives Xν=Flag⁡(V) as the total space of the rank-zero subbundle, of dimension n(n−1)/2, and the map πν is the constant map to the point EV=0, which is proper since it is projective.

2.1step 1.1algebra∎

The semismall inequality on a target stratum T with fibre dimension d is 2d+dim⁡T≤dim⁡Xν; here there is just the point stratum. The image of πν is the single point, and its single fiber is all of Xν, so the defect 2dim⁡(fiber)+dim⁡(image)−dim⁡Xν equals 2⋅n(n−1)/2+0−n(n−1)/2=n(n−1)/2, which is positive for n≥2; hence this proper incidence map is not semismall. For n=0 or n=1 the flag space is a point and the defect is zero, so the non-semismall boundary is exactly n≥2, and a general quiver-flag decomposition proof must allow this case.

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