Lets $k$ be a (perfect) field and $\Gamma=\text{Gal}(k^{s}/k)$ its absolute Galois group. Let $F$ be a field of characteristic zero (e.g. $F=\mathbb{Q}$), condsider now the category $\textrm{Rep}(\Gamma)$of continuous representations of $\Gamma$ in finite dimensional $F$-vectorspaces. I now call such a represensation $\textit{permutative}$ if the corresponding vectorspace $V$ admits a basis, such that $\Gamma$ acts on $V$ via permutations of the basis. Hence the subcategory of $\textrm{Rep}(\Gamma)$ consisting of permutative representations is equivalent to the category of $\Gamma$-sets. My question is now: If $\rho_{1}\in\text{Rep}(\Gamma)$, is there another $\rho_{2}\in\text{Rep}(\Gamma)$ such that $\rho_{1}\oplus\rho_{2}$ is a permutative representation?
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$\begingroup$ Thank you both for your answer. Not quite sure, what I was thinking ;) $\endgroup$– HenryCommented Nov 24, 2010 at 0:58
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$\begingroup$ You never said what the topology on $F$ was. From both the question and the answers, I gather it is the discrete topology? Indeed, if $F$ was a $p$-adic field with its usual topology then in general you can't hope for $\rho_2$ to exist in general. $\endgroup$– Kevin BuzzardCommented Nov 24, 2010 at 22:13
2 Answers
First, let me remark that the category of permutation representations in characteristic 0 is very far from being equivalent to the category of $\Gamma$-sets, since non-isomorphic $\Gamma$-sets can induce isomorphic representations. This is already true on the finite level: you might want to google for "Brauer relations".
The answer to your question is 'yes'. A continuous Galois representation factors through a finite quotient and any representation of a finite group in characteristic zero is projective. E.g. if it is irreducible, then it is always a direct summand of the regular representation.
Any continuous representation $V$ of $\Gamma$ will factor through some finite quotient of $\Gamma$, say $\Gamma/N=G$. We can then choose an epimorphism $F[G]^N\to V$ for some $N$, and let $V'$ be the kernel. By the usual averaging argument this will split to give $V\oplus V'\simeq F[G]^N$, and clearly $F[G]^N$ is permutative.
Incidentally, it is not actually true that the permutative subcategory of $Rep(\Gamma)$ is equivalent to that of continuous $\Gamma$-sets. The obvious reason is that the latter is not an additive category. More subtly, if $G$ is elementary abelian of order $4$, and $A$, $B$ and $C$ are the three subgroups of order $2$, then the $G$-sets $G/A\amalg G/B\amalg G/C$ and $G/1\amalg G/G\amalg G/G$ are non-isomorphic but give isomorphic representations.
