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The following result was mentionned earlier in this thread, I searched a bit in the related threads and couldn't find a proof. I would really like to see a proof of it:

Let $G$ be a finite group and $\rho : G \rightarrow GL(\mathbb{C}, n)$ a faithful representation of $G$. Then every irreducible representation of $G$ is contained in some tensor power of $\rho$.

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I just remembered about… –  darij grinberg Mar 14 '10 at 20:57

6 Answers 6

See problem 3.26 in Etingof's "Introduction to representation theory". If you have troubles with understanding the hint, feel free to ask me. (The first sentence uses the fact that if a vector space over an infinite field is the union of finitely many subspaces, then one of these subspaces is the whole vector space. The surjectivity of the map $SV\to F\left(G,\mathbb C\right)$ is because a polynomial can take any arbitrary finite set of values at some given distinct points. In order to conclude from this, note that this map $SV\to F\left(G,\mathbb C\right)$ is a homomorphism of representations of $G$.)

This proof works over any algebraically closed field of characteristic $0$. This can't quite be said about the proof in Fulton-Harris, if I remember it right.

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Another proof is in Curtis-Reiner: Representation theory of finite groups and associative algebras, Thm 32.9 in the first edition. To my mind the proof is very cute, it considers the generating series $\sum_k a_k t^k$, where $a_k$ is the number of times the irreducible representation occurs in the $k$'th tensor power of $\rho$. Using the orthogonality formulas for characters they sum this up as a rational function which, because $\rho$ is faithful, has a simple pole at $t=1/n$ and hence many of the $a_k$ are non-zero. –  Torsten Ekedahl Mar 14 '10 at 18:59
I THINK you mean the same proof as in Fulton-Harris. Alas, I don't see how to show that the pole is indeed a pole (i. e., not cancelled by another term) without the use of complex numbers. –  darij grinberg Mar 14 '10 at 19:10
There are two additional short proofs in Proc. AMS just after Curtis-Reiner first appeared, avoiding Burnside's use of complex numbers. (These back issues are to appear on the AMS journal page after scanning but haven't yet. I can access them through the UMass library or JSTOR, but these are restricted.) The two papers appear in the collected papers of the two authors: R. Steinberg, Complete sets of representations of algebras, Proc. Amer. Math. Soc. 13 (1962), 746-747 R. Brauer, A note on theorems of Burnside and Blichfeldt, Proc. Amer. Math. Soc. 15 (1964), 31-34 –  Jim Humphreys Mar 14 '10 at 20:01
Thanks a lot, I'll look into these. –  darij grinberg Mar 14 '10 at 20:06
How does Brauer conclude this: "Since X is faithful, one of the A_j consists only of the unit element"? –  darij grinberg Mar 14 '10 at 20:59

Another proof (not really that different from Geoff's, but appealing to a somewhat different intuition): Let $W$ be the representation corresponding to $\rho$, let $\underline{1}$ be the trivial representation, and let $V$ be the representation which we want to appear in some representation of $W^{\otimes N}$. I will show instead that $V$ appears in some representation of $(W \oplus \underline{1})^{\otimes N}$; this is equivalent because $(W \oplus \underline{1})^{\otimes N} = \bigoplus_{k=0}^N \binom{N}{k} W^{\otimes k}$.

Let $\chi$ be the character of $V$ and let $\psi$ be the character of $W$. Then $$\dim \mathrm{Hom}_G(V, (W \oplus \underline{1})^{\otimes N}) = \frac{1}{|G|} \sum_{g \in G} \overline{\chi(g)} (\psi(g)+1)^N. \quad (*)$$ We want to show that this Hom space is nontrivial for large $N$.

We have $|\psi(g)| \leq \dim W$ for all $g \in G$ and, since $W$ is faithful, $\psi(g)$ is $\dim W$ if and only if $g=e$. So $|\psi(g)+1| \leq \dim W + 1$, with equality precisely for $g=e$. So the right hand side of $(*)$ is a finite sum of exponentials, and the term $(\dim V) (\dim W + 1)^N$ has a larger base than any of the others. So the right hand side is positive for large $N$, and we see that the irrep $V$ appears in $(\underline{1} \oplus W)^{\otimes N}$ for sufficiently large $N$.

I explain how to modify this for compact Lie groups in this answer.

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As I have said elsewhere on Mathoverflow, in dealing with related questions, I think the simplest and the best proof of this result is due to Blichfeldt. It is simpler than both the power series type argument which appears in Burnside's book and the Vandermonde determinant argument of Brauer. Since we are dealing with characteristic zero representations of finite groups, we need only deal with characters. Let $\chi$ be the character afforded by $\rho$, and let $\alpha_{1},\alpha_{2},\ldots,\alpha_{m}$ be all the distinct values taken by $\chi$ on non-identity elements of $G$. Note that the class function $\chi \prod_{i=1}^{m}(\chi - \alpha_{i}1)$ vanishes on all non-identity elements of $G$, but does not vanish at $1_{G}$. Hence this class function is a non-zero multiple of the regular character (in fact a rational algebraic integer multiple). On the other hand, it may be written in the form $\sum_{j=1}^{m+1} a_{j}\chi^{j}$ for certain rational integers $a_{i}$ (note that $\{\alpha_{1},\ldots \alpha_{m} \}$ is a set of algebraic integers closed under algebraic conjugation). Since any irreducible character $\mu$ of $G$ has non-zero inner product with the regular character, we see that $\langle \chi^{j},\mu \rangle \neq 0$ for some $j$ with $ 1 \leq j \leq m+1$. (It is not necessary to use the factor $\chi$ if any $\alpha_{i}$ is already zero and, in any case, the factor $\chi$ is only used to make sure that we use strictly positive tensor powers of $\rho$, avoiding the issue of the trivial module). Note that Blichfeldt's argument yields that $\prod_{i=1}^{m} (\chi(1) - \alpha_{i})$ is an integer multiple of $|G|$.

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This is an elegant proof, but it doesn't seem to generalize readily to compact groups. –  Qiaochu Yuan Apr 26 '11 at 20:55
That's probably correct. It does, however, generalize to representations over fields of finite characteristic in various ways, as I have explained elsewhere. –  Geoff Robinson Apr 26 '11 at 21:10
Mmm, presumably you could replace "non-identity elements" with "non-identity elements of order <m" and then let m increase, but it's true that this sort of wrecks the elegant nature of the proof. –  Ben Webster Apr 27 '11 at 0:40

Sorry to resurrect such an old thread, but we supply two proofs. The first proof is due to Sameer Kailasa.

Problem 2.37, Fulton-Harris. Show that if $V$ is a faithful representation of $G$, i.e., $\rho: G \to GL(V)$ is injective, then any irreducible representation of $G$ is contained in some tensor power $V^{\oplus n}$ of $V$.

Let $W$ be an irreducible representation of $G$, and set$$a_n = \langle \chi_W,\chi_{V^{\oplus n}}\rangle = \langle\chi_W,(\chi_V)^n\rangle.$$If we consider the generating function $f(t) = \sum_{n=1}^\infty a_nt^n$, we can evaluate it as$$f(t) = {1\over{|G|}}\sum_{n=1}^\infty \sum_{g\in G} \overline{\chi_W(g)}(\chi_V(g))^nt^n = {1\over{|G|}} \sum_{g \in G} \overline{\chi_W(g)} \sum_{n=1}^\infty (\chi_V(g)t)^n$$$$={1\over{|G|}} \sum_{g \in G}{{\overline{\chi_W(g)}\chi_V(g)t}\over{1 - \chi_V(g)t}}.$$Note that in this sum, the term where $g = e$ evaluates to $${{(\dim W \cdot \dim V)t}\over{1 - (\dim V)t}},$$which is nonzero. If no other term in the summation has denominator $1 - (\dim V)t$, then this term can not cancel, so $f(t)$ is a nontrivial rational function. We can then conclude that not all of the $a_n$ are $0$. Thus, to complete the proof, it suffices to show $\chi_V(g) = \dim V$ only for $g = e$.

Suppose $\chi_V(g) = \dim V = n$ for $g \neq e$. Also, say $G$ acts on $V$ via $\rho: G \to GL(V)$. There is $k$ such that $\rho(g)^k = I$. If $\lambda_1, \dots, \lambda_n$ are the eigenvalues of $g$ we have$$\lambda_1^{ik} + \dots + \lambda_n^{ik} = n$$for $i = 0, 1, \dots$. Since $g^{k+1} = g$, we also see$$\lambda_1^{ik+1} + \dots + \lambda_n^{ik+1} = n.$$It follows that $$\lambda_1^{ik}(\lambda_1 - 1) + \dots + \lambda_n^{ik}(\lambda_n - 1) = 0,$$which implies for all polynomials in $\mathbb{C}[x]$, we have$$P(\lambda_1^k)(\lambda_1 - 1) + \dots + P(\lambda_n^k)(\lambda_n - 1) = 0.$$Choosing appropriate polynomials with roots at all but one of the eigenvalues, we see that all the eigenvalues must be $1$. Since $\rho(g)$ is diagonalizable, it follows $\rho(g) = I$. This contradicts the faithfulness of $V$.

Problem 3.26, Etingof. Let $G$ be a finite group, and $V$ a complex representation of $G$ which is faithful, i.e., the corresponding map $G \to GL(V)$ is injective. Show that any irreducible representation of $G$ occurs inside $S^nV$ (and hence inside $V^{\otimes n}$) for some $n$.

Let $n = |G|$.

Step 1.

There exists $u \in V^*$ whose stabilizer is $1$.

For given $g \neq 1$, since $\rho_V:G \to GL(V)$ is injective, $\rho_V(g)^{-1} - I = \rho_V(g^{-1}) - I \neq 0$. Thus there exists $u \in V^*$ for which $(\rho_{V^*}(g) - I)u$ is not the zero transformation. (We make the observation that $((\rho_{V^*}(g) - I)u)(v) = u((\rho_V(g)^{-1} - I)v)$; just define $u$ so that it sends something in the range of $\rho_V(g)^{-1} - I$ to $1$.) Define$$U_g=\{u\in V^*\text{ }|\text{ }(\rho_{V^*}(g)-I)u= 0\};$$that is, $U_g$ is the kernel of the linear transformation $\rho_{V^*}(g) - I$ on $V^*$. Then when $g \neq 1$, $U_g$ is a proper subspace of $V^*$. Hence, the union $\bigcup_{g \in G,\,g \neq 1} U_g$ cannot be the entire space $V^*$. (See the following lemma.)

Lemma. Let $W$ be a complex vector space and $W_1, \dots, W_m$ proper subspaces of $W$. Then$$W \neq \bigcup_{i=1}^m W_i.$$

Proof. For each $i$, choose a vector $w_i \notin W_i$. Let $U = \text{span}(w_1, \dots, w_m)$. Note that $U \not\subseteq W_i$ for any $I$. Replacing $W_i$ with $W_i \cap U$ and $W$ with $U$ as necessary, we may assume that $W$ is finite-dimensional.

For each $i$, find a linear functional $f_i$ such that $\text{ker}(f_i) = W_i$. Choose a basis $e_1, \dots, e_k$ of $W$. Then$$f(x_1, \dots, x_k) := \prod_{i=1}^m f_i(x_1e_1 + \dots + x_ke_k)$$is a polynomial in the $x_1, \dots, x_k$ over an infinite field, so there exists $(x_1, \dots, x_k)$ such that $f(x_1, \dots, x_k) \neq 0$. This point is not in any of the $W_i$.$$\tag*{$\square$}$$Taking $u \in V^* - \bigcup_{g \in G} U_g$, we get that$$u \notin U_g \implies \rho_{V^*}(g)u \neq u$$for any $g \in G$, $g \neq 1$. In other words, $\rho_{V^*}u = u$ if and only if $g = 1$, and the stabilizer of $u$ is $1$.

Step 2.

Define a map $SV \to F(G, \mathbb{C})$.

Define the map $\Phi: SV \to F(G, \mathbb{C})$ by sending $f \in SV$ to $f_u$ defined by $f_u(g) = f(gu)$. In other words, we define $\Phi$ as follows.

  1. First, define $\Phi_k: S^kV \to F(G, \mathbb{C})$ as the linear map induced by the symmetric $k$-linear map $\beta_k: V^k \to F(G, \mathbb{C})$ given by$$[\beta_k(v_1, \dots, v_k)](g) = \prod_{i=1}^k[(\rho_{V^*}(g)u)(v_i)] = \prod_{i=1}^k [i(\rho_V(g)^{-1}v_i)].$$Note that $\Phi_k$ is a homomorphism of representations since$$[\Phi_k(h(v_1\dots v_k))](g) = [\Phi((hv_1) \dots (h v_k))](g) = \prod_{i=1}^k[(gu)(gv_i)]$$$$= \prod_{i=1}^k [(h^{-1}gu)(v_i)] = [\Phi_k(v_1 \dots v_k)](g^{-1}g) = \{h[\Phi_k(v_1 \dots v_k)]\}(g).$$(For $k = 0$, the map is the map $\mathbb{C} \to F(G, \mathbb{C})$ sending a number to its constant function.)
  2. Define $\Phi: SV \to F(G, \mathbb{C})$ by$$\Phi = \bigoplus_{k=0}^\infty \Phi_k.$$

Step 3.

$\Phi$ is surjective; in fact, the map restricted to $\bigoplus_{i \le n-1} S^i V$ is surjective.

It suffices to show the functions $1_h$ defined by$$1_h(g) = \begin{cases} 1 & \text{if }g = h \\ 0 & \text{if }g \neq h \end{cases}$$are in the image of $\Phi$, since they span $F(G, \mathbb{C})$. Given $h$, we will find a vector $f \in SV$ such that $\Phi(f) = k1_h$ for some $k \in \mathbb{C} - \{0\}$.

Let $K$ be the kernel of $u$; since $u$ is a nontrivial linear transformation $V \to \mathbb{C}$,$$\dim(K) = \dim(V) - \dim(\mathbb{C}) = n-1.$$For each $g \in G$, let$$V_g = gK = \rho_V(g)K.$$So $V_g$ is the subspace of vectors $v$ such that $g^{-1}v \in \text{ker}(u)$, i.e. $u(g^{-1}v) = 0$. We define $v_g$ for $g \neq h$; consider two cases.

  1. If $V_g \neq V_h$, define $v_g \in SV$ to be a vector in $V_g - V_h \subseteq V$. Note each $V_g$ has dimension $n-1$ since $g$ is invertible. ($V_g$, $V_h$ both have the same dimension, so neither is contained in the other.) Then$$[\Phi(v_g)](h) = u(h^{-1}v_g) \neq 0,\text{ }[\Phi(v_g)](g) = u(g^{-1}v_g) = 0.$$
  2. If $V_g = V_h$ and $g \neq h$, then let $v_g'$ be a vector in $V - V_g$. Then $u(g^{-1}v_g') = \lambda$ for some nonzero $\lambda$. Define $v_g \in SV$ to be the vector $v_g' - \lambda$. Note that$$[\Phi(v_g)](g) = u(g^{-1}v_g') - \lambda = 0.$$If $u(h^{-1}v_g') = \lambda$, then $gu = u(g^{-1}*)$ and $hu = u(h^{-1}*)$ would be identical linear transformations (they already agree on $V_g$ as they are identically zero there; $V_g + \text{span}(v_g') = V$), contradicting the fact that $U$ has stabilizer $1$. Hence, $u(h^{-1}v_g') \neq \lambda$ and$$[\Phi(v_g)](h) \neq 0.$$Now consider$$f = \prod_{g \neq h} v_g \in \bigoplus_{i \le n-1} S^i V.$$We have $[\Phi(f)](g) = 0$ for all $g \neq h$ since $[\Phi(v_g)](g) = 0$ for $g \neq h$. On the other hand, $[\Phi(v_g)](h) \neq 0$ for all $g \neq h$, so $[\Phi(f)](h) \neq 0$. Thus, $\Phi(f)$ is a multiple of $1_h$. Since this works for all $h$, $\Phi$ is surjective.

Step 4.

$W := \bigoplus_{1 \le n-1} S^i V$ contains every irreducible representation of $V$.

Note that$$F(G, \mathbb{C}) \cong \text{Hom}_\mathbb{C}(\mathbb{C}G, \mathbb{C}) \cong (\mathbb{C}G)^* \cong \mathbb{C}G.$$The last isomorphism follows since $\chi_{\mathbb{C}G}$ is real, (as each $\rho_{\mathbb{C}G}(g)$ is real) and hence equal to its conjugate $\overline{\chi_{\mathbb{C}G}} = \chi_{(\mathbb{C}G)^*}$. Since $W$ maps surjective to $F(G, \mathbb{C}) \cong G\mathbb{C}$ via $\Phi$, $G\mathbb{C}$ must actually occur inside $W$. This is since$$\chi_W = \chi_{\text{ker}(\Phi)} + \chi_{W/\text{ker}(\Phi)} = \chi_{\text{ker}(\Phi)} + \chi_{\mathbb{C}G}.$$Since $G\mathbb{C}$ contains every irreducible representation, so does $\oplus_{i \le n-1} S^i V$. Thus, every irreducible representation occurs inside $S^i V$ for some $i \le n-1$.

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By Satz 90, the fraction field of the symmetric algebra of a faithful representation of a finite group contains all irreducible representation. It remains to get rid of the denominators, just multiplying by the product of their conjugates.

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How does this follow from Hilbert 90? –  darij grinberg Apr 19 '13 at 14:33

Any semilinear Galois representation is trivial (the skew product of a field $K$ with a finite group $G$ of its automorphisms is the endomorphism algebra of $K$ as vector space over the fixed subfield), so extension to $K$ of coefficients of any irreducible representation of $G$ is isomorphic to a direct sum of copies of $K$.

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