Sub-representations of the affine group - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T14:03:30Z http://mathoverflow.net/feeds/question/30025 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/30025/sub-representations-of-the-affine-group Sub-representations of the affine group Klim Efremenko 2010-06-30T09:08:23Z 2010-06-30T11:49:04Z <p>Let $F=\mathrm{GF}\left(p^k\right)$ be any finite field. Let $G$ be the group of all affine permutations on $F$ (i.e. permutations of form $x\mapsto ax+b$). Then the set of all functions from $F$ to $\bar{F}$ is a linear representation of $G$, where $g(f)(x)=f(gx)$.</p> <p>What are all sub-representations of this representation? Is it possible to characterize them?</p> <p>Note: that in this case $\mathrm{gcd}\left(\left|G\right|,F\right)$ not equal to $1$.</p> http://mathoverflow.net/questions/30025/sub-representations-of-the-affine-group/30036#30036 Answer by Bugs Bunny for Sub-representations of the affine group Bugs Bunny 2010-06-30T11:44:00Z 2010-06-30T11:44:00Z <p>As Victor explained consider the functions $X^m$ where $X^m(\alpha)=\alpha^m$. As $m$ runs between $0$ and $p^k-1$, these functions form a basis of your space of functions. This is a nice wavy basis, i.e., its elements span one-dimensional subrepresentations under the multiplicative group.</p> <p>Now you have to take the additive group into account. All you need to do is to use binomial formula on $(X+\alpha)^m$ and observe which non-zero $X^t$-s, you can get out. This depends on the $p$-th power in $m$.</p> <p>In particular, as Victor pointed out, polynomials of degree less than $m$ will span a submodule. But there are more, for instance, polynomials of degree $p$ and zero. In general, you will be getting <strong><em>spans of $X^t$ with $t\leq m$ and $t$ is divisible by the $p$-th power present in $m$</em></strong> as well as the sums of these gadgets.</p> <p>Hint: $(X+\alpha)^{p^sn}=(X^{p^s}+\alpha^{p^s})^n$</p>