Finite-field descent and restriction of Schur quiver representations

Let $L\mathrm{/}K$ be a finite cyclic extension with generator $\sigma$ of order $r$, and let $T$ be a $\sigma$-semilinear automorphism of a finite $L$-space with $T^r=1$. Fixed vectors span it over $L$. Otherwise quotient by their $L$-span and choose $\bar{v}\mathrm{\ne }0$ there. Every average ${\sum }_{j=0}^{r-1}{T}^j(av)$ is fixed, so in the quotient ${\sum }_j{\sigma }^j(a)T^j\bar{v}=0$ for all $a\in L$. A coordinate functional nonzero on $\bar{v}$ contradicts [F2]. Choose an $L$-basis among the fixed vectors. A linear combination is fixed exactly when all its coefficients are fixed by $\sigma$, so these same vectors are a $K$-basis of the fixed space. This proves semilinear vector-space descent, and invariant arrow maps descend in these bases.

Suppose a geometric Schur class is fixed by ${\mathrm{Fr}}^m$. Choose a model and an intertwiner to its transform, all over a finite extension $L$ of $K=k_m$; enlarge $L$ if necessary so it is cyclic over $K$. The intertwiner defines a semilinear operator $T$ on the sum of the vertex spaces, preserving the vertex decomposition and commuting with all arrows. Its $r$th power is an endomorphism of the Schur representation, so is a common scalar $\lambda$. Commuting $T$ with $T^r$ gives $\sigma (\lambda )=\lambda$, hence $\lambda \in K^{\times }$. The norm $L^{\times }\to K^{\times }$ is onto: by [F1] it sends a cyclic generator to an element of order $\mathrm{\mid }{K}^{\times }\mathrm{\mid }$. Choose $a$ with norm ${\lambda }^{-1}$; then $(aT{)}^r=1$. Step 1.1 supplies fixed bases and a model over $K$. If two models over $K$ are geometrically isomorphic, their intertwiner space is the kernel of a matrix of linear equations over $K$, whose dimension stays $1$ after scalar extension. A nonzero rational intertwiner is therefore a scalar multiple over $\bar{k}$ of an isomorphism and is invertible. This proves uniqueness of the rational model.

A class fixed by ${\mathrm{Fr}}^m$ with orbit length less than $m$ is fixed by ${\mathrm{Fr}}^d$ for some proper divisor $d$ of $m$. For each $d$, step 2.1 places these fixed classes in bijection with a subset of the isomorphism classes of dimension-$\delta$ representations over $k_d$. There are finitely many such representations up to isomorphism: after fixing the finite bases, there are finitely many arrow-matrix tuples over the finite field $k_d$. Thus each fixed set for a small $d$ is finite, and we may increase $C$ to cover the finitely many small extension degrees. The number of classes with orbit length less than $m$ is at most ${\sum }_{d\mid m,d<m}C{q}^d\le Cm{q}^{m\mathrm{/}2}$. Thus at least $c{q}^m-Cm{q}^{m\mathrm{/}2}\ge (c\mathrm{/}2)q^m$ classes have exact orbit length $m$ for large $m$, yielding at least $(c\mathrm{/}2)q^m\mathrm{/}m$ distinct Frobenius orbits.

For a Schur representative over $k_m$ with exact orbit length $m$, restrict every vertex space and arrow to $k$. Such a representative exists by step 2.1. After scalar extension to $\bar{k}$, the resulting representation is the direct sum of its $m$ Frobenius conjugates: the algebra $k_m{\otimes }_k\bar{k}$ is the product of $m$ copies of $\bar{k}$ under its distinct embeddings, since the finite extension is separable. Each conjugate is indecomposable, because its endomorphism ring is a field, and they are pairwise nonisomorphic by exact orbit length. If the restricted representation split over $k$, its two nonzero summands would, by [F3], partition this multiset into two nonempty Frobenius-stable subsets. Frobenius acts transitively, so no such partition exists. Hence restriction is indecomposable. Two restrictions from distinct orbits cannot be isomorphic, again by their scalar-extended decompositions and [F3]. Nilpotence of all sufficiently long paths is preserved by restriction. Combining with step 3.1 gives the stated bound.