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Reynolds Operator on functors

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Amelia Alvarez, Carlos Sancho, Pedro Sancho

Algebraic Geometry · Submitted 10 Nov 2006

abstract compiled · 2026-09-23 · no language-model review

The authors' main claim If $\mathbb M$ is a dual functor of $G$-modules and $w_G := (1,0) \in R \times B^* = A^*$, we prove that $\mathbb M^G = w_G \cdot \mathbb M$ and $\mathbb F = w_G \cdot \mathbb M \oplus (1-w_G) \cdot \mathbb M$; hence, the Reynolds operator can defined on $\mathcal M$.

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arXiv IDmath/0611311 (version 5)
Pith IDLF77WPS5, Pith's permanent record of this paper and its review
CitationAmelia Alvarez, Carlos Sancho, Pedro Sancho. "Reynolds Operator on functors." arXiv:math/0611311 (2006). https://pith.science/paper/LF77WPS5
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The paper's own sentences, arranged. Each label is drawn from a closed vocabulary; the number beside it is the probability the classifier gave that label. Numbers in the tables are the authors', checked by code. No sentence in this record was written by a model.

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Compiled by classifier-1.13.0 via TypeSafe API, question bank abstract-p1-20260923, 2026-09-23: 1 request, 3,907 input tokens.

What the abstract itself states, as probabilities: claims first 0.03 · states limitation 0.08 · human participants 0.01 · controlled comparison 0.01 · randomized assignment 0.01 · states quantitative result 0.04 · compares with prior results 0.05 · mentions code or data availability 0.01.

The abstract, sentence by sentence

arXiv listing abstract v5. Results and the main claim are underlined; .

Let $G= {\rm Spec} A$ be an affine $R$-monoid scheme. We prove that the category of dual functors (over the category of commutative $R$-algebras) of $G$-modules is equivalent to the category of dual functors of ${\mathcal A}^*$-modules. We prove that $G$ is invariant exact if and only if $A^*= R \times B^*$ as $R$-algebras and the first projection $A^* \to R$ is the unit of $A$. If $\mathbb M$ is a dual functor of $G$-modules and $w_G := (1,0) \in R \times B^* = A^*$, we prove that $\mathbb M^G = w_G \cdot \mathbb M$ and $\mathbb F = w_G \cdot \mathbb M \oplus (1-w_G) \cdot \mathbb M$; hence, the Reynolds operator can defined on $\mathcal M$.

background contribution method result interpretation main claim

What the classifier reports for each sentence. The number is the probability it gave the label.
#sentencerolethe authors' own workstates the main conclusionstrength
s0 Let $G= {\rm Spec} A$ be an affine $R$-monoid scheme. background 0.99 0.12 0.02 asserted 0.90
s1 We prove that the category of dual functors (over the category of commutative $R$-algebras) of $G$-modules is equivalent to the category of dual functors of ${\mathcal A}^*$-modules. result 0.94 0.95 0.46 asserted 1.00
s2 We prove that $G$ is invariant exact if and only if $A^*= R \times B^*$ as $R$-algebras and the first projection $A^* \to R$ is the unit of $A$. result 0.99 0.90 0.21 asserted 1.00
s3 If $\mathbb M$ is a dual functor of $G$-modules and $w_G := (1,0) \in R \times B^* = A^*$, we prove that $\mathbb M^G = w_G \cdot \mathbb M$ and $\mathbb F = w_G \cdot \mathbb M \oplus (1-w_G) \cdot \mathbb M$; hence, the Reynolds operator can defined on $\mathcal M$. result 0.97 0.93 0.36 asserted 1.00

What kind of paper the classifier reports this is: theoretical result 1.00 · evidence from proof or derivation 1.00.

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Every quoted sentence on this record is the authors'. Every label is from Pith's closed vocabulary, with the probability shown beside it. Every number in the measurements table is the authors', checked by code. Record as JSON

The paper
Original abstract

Let $G= {\rm Spec} A$ be an affine $R$-monoid scheme. We prove that the category of dual functors (over the category of commutative $R$-algebras) of $G$-modules is equivalent to the category of dual functors of ${\mathcal A}^*$-modules. We prove that $G$ is invariant exact if and only if $A^*= R \times B^*$ as $R$-algebras and the first projection $A^* \to R$ is the unit of $A$. If $\mathbb M$ is a dual functor of $G$-modules and $w_G := (1,0) \in R \times B^* = A^*$, we prove that $\mathbb M^G = w_G \cdot \mathbb M$ and $\mathbb F = w_G \cdot \mathbb M \oplus (1-w_G) \cdot \mathbb M$; hence, the Reynolds operator can defined on $\mathcal M$.

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