2021
DOI: 10.31219/osf.io/3c4se
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On Peterson's open problem and representations of the general linear groups

Abstract: Fix $\mathbb Z/2$ is the prime field of two elements and write $\mathcal A_2$ for the mod $2$ Steenrod algebra. Denote by $GL_d:= GL(d, \mathbb Z/2)$ the general linear group of rank $d$ over $\mathbb Z/2$ and by $\mathscr P_d$ the polynomial algebra $\mathbb Z/2[x_1, x_2, \ldots, x_d]$ as a connected unstable left $\mathcal A_2$-module on $d$ generators of degree one. We study the Peterson "hit problem" of finding the minimal set of $\mathcal A_2$-generators for $\mathscr P_d.$ Equivalently, we need to determ… Show more

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Cited by 41 publications
(112 citation statements)
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“…The above computations confirm the dimension of (QP 5 ) ns , which is noted in [11] by using the MAGMA computer algebra system. Based upon Theorems 1.2 and 2.12, we have the following, which is our second main result.…”
supporting
confidence: 78%
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“…The above computations confirm the dimension of (QP 5 ) ns , which is noted in [11] by using the MAGMA computer algebra system. Based upon Theorems 1.2 and 2.12, we have the following, which is our second main result.…”
supporting
confidence: 78%
“…It has been surveyed by many authors. (See Kameko [2], Mothebe-Uys [4], the present author [8,9,10,11], Singer [13], Sum [14,15], Wood [17], and others. )…”
mentioning
confidence: 99%
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“…However, when s 5, it is an open problem. Recently, some authors have been studied the conjecture for s = 4, 5 (see Bruner-Hà-Hưng [4], Hưng [11], Chơn-Hà [7,8], Hà [9], Nam [22], the present author [26], [28]- [35], Sum [45,47,48,49] and others). In the present work, by using techniques of the hit problem of five variables, we investigate Conjecture 1.2 in bidegree (5, d + 5), where d is determined as in Theorem 1.1.…”
Section: Introduction and Main Resultsmentioning
confidence: 99%
“…We explicitly compute p (i;I) (S) in terms u j , 1 j 23. By a direct computation using Lemma 2.2.9,Theorem 3.1.3, and from the relations p (i;j) (S) ≡ ω (5,2) 0 with either i = 1, j = 2, 3 or i = 2, j = 3, 4, one gets Here J = {1, 2, 3,4,5,6,7,8,9,10,11,12,15,16,17,18,19,21,22,23,24,25,26,28,29,31,38,39,40,41,42,45,46,48,50,51,57 [46]). By a simple computation, we get…”
Section: We Now Prove the Setmentioning
confidence: 99%