Let \(G\) be a finite group and $$\rho: G\to GL_r(V)$$ a faithful representation, with \(V\) a finite-dimensional complex vector space. The following is well-known:
Theorem 1. Let \(\gamma\) be any finite-dimensional irreducible representation of \(G\). Then \(\gamma\) appears as a direct summand of \(V^{\otimes n}\) for some \(n\gg 0\).
The usual proof of this uses analysis; today I want to record a short argument using a bit of algebraic geometry. This came up recently in a little discussion on Twitter with Noah Snyder. In fact, this argument will give the following, over arbitrary infinite fields:
Theorem 2. Let \(\rho: G\to GL(V)\) be a faithful representation, where \(V\) is a finite-dimensional vector space over an infinite field \(k\). Let \(\gamma\) be an irreducible representation of \(G\). Then \(\gamma\) appears as a quotient representation (resp. sub-representation) of \(V^{\otimes n}\) for some \(n\gg 0\).
Note that \(\gamma\) might not be a direct summand.
We’ll need the following representation-theoretic fact:
Lemma. Any irreducible representation of \(G\) is a quotient of the regular representation \(k[G]\). Any irreducible representation is also a sub of \(k[G]\).
Proof of Lemma. There is a natural isomorphism \(\operatorname{Hom}_{k[G]}(k[G], V)=V\) for any representaion \(V\). Thus any non-zero representation admits a non-zero map from the regular representation. If \(V\) is irreducible, such a map is necessarily surjective (as the image is a subrepresentation), giving the first claim. The second claim follows by applying the same argument to the dual representation, and then dualizing (using the autoduality of the regular representation).
Proof of Theorem 2. As the action of \(G\) on \(V\) is faithful, the set \(V^g\) of vectors fixed by a given non-identity \(g\in G\) is a proper Zariski-closed subset of \(V\) for each \(g\in G\). Hence for a general element \(v\in V\), \(G\) acts faithfully on \(v\). Fix such an element of \(V\), and let \(X\) be its orbit under \(G\). As a \(G\)-set, \(X\) is isomorphic to \(G\) with the left translation action.
Now set \(R=\text{Sym}^*(V^\vee)\). Viewing \(X\) as a closed subvariety of $$V=\text{Spec}(R),$$ we get a surjective map $$R\to k[X].$$ But \(k[X]\) is the regular representation! So every irreducible representation of \(G\) appears as a quotient of \(k[X]\) and hence of \(\text{Sym}^n(V^\vee)\) for some \(n\). Dualizing, every irreducible representation appears as a subrepresentation of \(\text{Sym}^n(V^\vee)^\vee\), which is a subrepresentation of \(V^{\otimes n}\), for some \(n\). Applying the same argument with \(V^\vee\) yields every representation as a quotient. \(\square\)