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2

$\mathfrak{gl}(1)$ isn't semisimple! For that matter, neither is $\mathfrak{gl}(n)$ for higher values of $n$.


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$\varphi$ is a mapping from the representation $G$ to the general linear group of $V$. $\varphi_g $ is probably shorthand to represent $\varphi(g) $ for $g \in G $. $\varphi(g) $ is an element of $GL(V) $, and therefore an isomorphism from $V$ to $V$.


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The idea here is that we want to find an invariant subspace, that is, a subspace $W$ of $V$ such that if $w \in W$ we have that $gw \in W$ for all $g \in S_3$. As Tobias mentioned, we want to pick a subspace where no matter how we permute the basis vectors, we get the same subspace. The subspace $W = <(1,1,1)>$ is invariant, and it's complement is a ...


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Let's go over how these representations are defined: We have a certain partition of the set $\{1,2,\ldots,n\}$; this partition is into an ordered collection of unordered sets. Such a partition might look like $[\{1,3\}, \{2,4\}]$; note this is exactly the same partition as $[\{3,1\}, \{2,4\}]$, but not the same as $[\{2,4\}, \{1,3\}]$. The problem with ...


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Here are some examples. The page number refer to Dummit and Foote's Abstract Algebra 3/e. The gray region means that no module can satisfy the condition. Focus on the suffix of the term "decomposable" and the first word of the term "completely reducible". Decomposable just means that it can be able to be decomposed. Completely reducible means that it can ...


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The horizontal maps are not $\varphi$ and $\psi$, but $\varphi_g$ and $\psi_g$ for some particular $g\in G$. That is, the top horizontal map $\varphi_g$ is the image of $g$ under the homomorphism $\varphi$ (more commonly written $\varphi(g)$), which is a linear isomorphism $V\to V$, and similarly for $\psi_g$.


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Suppose that $G$ is abelian, let $V_c$ be the eigenspace of $f(g)$ associated to $c$. For every $g'\in G$, and $x\in V_c, f(g)(f(g')(x))=f(gg')(x)=f(g'g)(c)=f(g')(f(g)(x))=f(g')(cx)=cf(g')(x)$. Suppose that $F$ is algebraically closed and $V$ is irreducible, for every $g$, $V_c$ is a submodule, thus $V_c=V$.


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Let us for the moment forget all about algebraic geometry and just consider a group $G$ and a $G$-module $V$. The definition of being rational makes sense in this context, so let us just find an example of a non-rational $G$-module which will work whenever $G$ is infinite (so it will also work for whatever algebraic group you choose as long as it is ...



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