rt.representation theory - Faithful representations and tensor powers

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$$a_n = langle chi_W,chi_{V^{oplus n}}rangle = langlechi_W,(chi_V)^nrangle.$$If we consider the generating function $f(t) = sum_{n=1}^infty a_nt^n$, we can evaluate it as$$f(t) = {1over{|G|}}sum_{n=1}^infty sum_{gin G} overline{chi_W(g)}(chi_V(g))^nt^n = {1over{|G|}} sum_{g in G} overline{chi_W(g)} sum_{n=1}^infty (chi_V(g)t)^n$$$$={1over{|G|}} sum_{g in G}{{overline{chi_W(g)}chi_V(g)t}over{1 - chi_V(g)t}}.$$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$$U_g={uin V^*text{ }|text{ }(rho_{V^*}(g)-I)u= 0};$$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 $U notsubseteq W_i$ 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$$[Phi_k(h(v_1dots v_k))](g) = [Phi((hv_1) dots (h v_k))](g) = prod_{i=1}^k[(gu)(gv_i)]$$$$= prod_{i=1}^k [(h^{-1}gu)(v_i)] = [Phi_k(v_1 dots v_k)](g^{-1}g) = {h[Phi_k(v_1 dots v_k)]}(g).$$(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 $F(G, mathbb{C}) cong Gmathbb{C}$ via $Phi$, $Gmathbb{C}$ 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 $Gmathbb{C}$ 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 $i le n-1$.