Proof. We will induct on \(n \equiv \dim(V)\). If \(V\) is irreducible, then we are done, so let \(V_1 < V\) be an irreducible subrepresentation. Let \(U \equiv V/V_1\), and take a filtration \[\begin{aligned} U_0 < U_1 < \cdots < U_{k-1} = U \end{aligned}\] such that \(U_i/U_{i-1}\) are irreducible. Then let \(V_i \equiv U_i \oplus V_1\). This shows \[\begin{aligned} V_i / V_{i-1} = (U_i \oplus V_1)/(U_{i-1}\oplus V_1) \cong U_i / U_{i-1}, \end{aligned}\] which shows that the successive quotients are irreducible. Thus, \(0 < V_1 < \dots < V_{k-1} < V\) is the desired filtration. ◻

Proof. Let \(V\) be a simple \(A\)-module and \(I \leq A\) a nilpotent ideal. Let \(v \in V\) be arbitrary. If \(Iv \neq 0\), then \(Iv = V\). However, this implies that \(xv = v\) for some \(x \in I\), therefore \(x^nv = v\), and so \(x^n \neq 0\). This contradicts \(I\) being nilpotent, so we must have \(Iv = 0\), and thus \(I \leq \operatorname{Rad}(A)\).

Next, consider \(A\) as an \(A\)-module over itself, and take a filtration \[\begin{aligned} 0 = A_0 < A_1 < \ldots < A_k = A. \end{aligned}\] Since \(x \in A\) acts on the irreducible quotients \(A_i/A_{i-1}\) by zero, this means that \(x: A_i \to A_{i-1}\), and therefore \(x^k = 0\). This shows that \(\operatorname{Rad}(A)\) is nilpotent. ◻

Proof. Let \(J \subseteq I\) be the maximal subset of \(I\) such that \(f|V_J\) is injective. If \(J \neq I\) (in which case we are done), let \(i \notin J\). Then the map \(f:V_i \to U/f(V_J)\) is nonzero for some \(i\) (otherwise \(I = J\)), so by Schur’s lemma, it is injective since \(V_i\) is simple. This implies that \(f|V_{J \cup \{i\}}\) is injective, contradicting the maximality of \(J\). ◻

Proof. Suppose that \(V\) is a simple \(A\)-module, and let \(v_1, \dots, v_n\) be a basis of \(V\). The image of the map \(a \to (av_1, \dots, av_n)\) is a subrepresentation \(U \leq V^{\oplus n}\). By the above lemma, \(U \cong V^{\oplus r}\) for \(r \leq n\). Let \(\phi\) be the isomorphism \(V^{\oplus r} \cong U \leq V^{\oplus n}\). Represent \(\phi\) by the matrix \(X\), such that \(\phi(w_1, \dots, w_r) = (w_1, \dots, w_r)X\). If \(n = r\) we are done, so suppose \(r < n\).

Taking \(a = 1\), we see that \((v_1, \dots, v_n) \in U\), and therefore there exist \(w_1, \dots, w_r \in V\) such that \((w_1, \dots, w_r)X = (v_1, \dots, v_n)\). But since \(r < n\), \(X\) has a left kernel, so let \(X(q_1, \dots, q_n)^t = 0\). We then have \[\begin{aligned} (w_1, \dots, w_r)X(q_1, \dots, q_n)^t = \sum_i v_iq_i = 0, \end{aligned}\] which contradicts the choice of \(\{v_n\}\) as a basis. Therefore \(n = r\).

This shows us that \(U \cong V^n\). Therefore if \(c \in \operatorname{End}(V)\), then \((cv_1, \dots, cv_n)\) is in the image of the map \(a \mapsto (v_1, \dots, v_n)\), so \(av_i = cv_i\) for each \(i\). ◻

Proof. Consider the homomorphism \[\begin{aligned} \rho = \bigoplus_{V_i \in \hat A} \rho_{i} : A \to \bigoplus_{V_i \in \hat A} \operatorname{End}V_i \end{aligned}\] Let \(B_i\) be the image of \(\rho_i\) and \(B\) be the image of \(\rho\). By Lem. 2, we know that \(B\) is semisimple. By the density theorem, \[B_i \cong \operatorname{End}(V_i) \cong V_i^{\dim V_i}\] so \(V_i\) occurs in \(B\) with multiplicity \(\dim V_i\). However this implies \[\bigoplus_i \operatorname{End}_i(V_i) \leq B \leq \bigoplus_i \operatorname{End}_i(V_i)\] which proves that \(\rho\) is a surjection. This implies that there are only finitely many non-isomorphic \(A\)-modules, all of which are finite dimensional. The kernel of \(\rho\) is \(\operatorname{Rad}(A)\), which proves the claim. ◻

Proof. Let \(V\) be a finite-dimensional \(A\)-module, and let \(v_1, \dots, v_n\) be a basis for \(V\). Then the map (of \(A\)-modules) \(\psi:A^n \to V\) via \(\psi(a_1, \dots, a_n) = a_1v_1 + \cdots + a_nv_n\) is surjective (by choosing \(a_1, \dots, a_n\) to be scalars). Therefore \(V\) is isomorphic (as an \(A\)-module) to a subspace of \(A^n\), and since we have shown in Thm. 1 that \(A^n\) is semisimple, so is \(V\). ◻

Proof. Since \(V\) is semisimple, we have \[\begin{aligned} V \cong \bigoplus_{V_i \in \hat A}V_i^{\oplus n_i} \cong \bigoplus_{V_i \in \hat A}k^{n_i} \otimes V_i \end{aligned}\] (\(k\) is the field). By Schur’s lemma, we find \[\begin{aligned} k^{n_i} \cong \operatorname{Hom}_A(V_i, V) \end{aligned}\] This proves the two are isomorphic. Furthermore, by Schur’s lemma, \(f \in \operatorname{Hom}_A(V_i, V)\) is given by element-wise scalar multiplication, so \(f \otimes x \mapsto f(x)\) is injective. This proves the claim. ◻

Proof. By Cor. 3, \(E\) is semisimple as an \(A\)-module, and so by Lem. [lem:altshur], we have a decomposition \[\begin{aligned} E \cong \bigoplus_{V_i \in \hat A}\operatorname{Hom}_A(V_i, E) \otimes V_i \end{aligned}\] Let \(W_i \equiv \operatorname{Hom}_A(V_i, E)\). Next, since \(A \cong \bigoplus_{i}\operatorname{End}(V_i) \subseteq \operatorname{End}(E)\), this shows that \(V_i\) occurs as a subspace of \(E\) for each \(i\), and thus \(W_i\) is nonzero for each \(i\). By Schur’s lemma, we have \[\begin{aligned} B = \operatorname{End}_A(E) &\cong \operatorname{Hom}_A(\bigoplus_{V_i \in \hat A}W_i \otimes V_i, \bigoplus_{V_j \in \hat A}W_j \otimes V_j) \notag\\ &\cong \bigoplus_{V_i,V_j \in \hat A}\operatorname{Hom}(W_i, W_j)\otimes \operatorname{Hom}_A(V_i, V_j) \notag\\ &\cong \bigoplus_{V_i \in \hat A}\operatorname{End}(W_i) \end{aligned}\] From this decomposition, we can see that \(W_i\) are irreducible as \(B\)-modules (precisely because \(W_i\) is irreducible as an \(\operatorname{End}(W_i)\)-module). By the same argument, \(\operatorname{Rad}(\operatorname{End}(W_i)) = 0\), so \(B\) is semisimple. Lastly, by Prop. 2, this shows that \(W_i\) are exhaustive of the irreducible \(B\)-modules, thus showing that irreducible \(B\)-modules are in 1-to-1 correspondence with irreducible \(A\)-modules. This proves all the statements of the theorem. ◻

Proof. Recall the exponential map \(\mathrm{e}: \mathfrak g \to G\). In particular, \[\begin{aligned} \pdv{t}\Big|_{t=0} \mathrm{e}^{tX} = X \end{aligned}\] Then we can obtain the action of \(X\) based on the action of \(G\): \[\begin{aligned} \pdv{t}\Big|_{t=0}[\mathrm{e}^{Xt}(v_1 \otimes \cdots \otimes v_n)] &= \pdv{t}\Big|_{t=0}[\mathrm{e}^{Xt}v_1 \otimes \cdots \otimes \mathrm{e}^{Xt} v_n] \\ &= \sum_{i = 1}^n v_1 \otimes \cdots \otimes Xv_i \otimes \cdots \otimes v_n \end{aligned}\] which follows by the product rule. ◻

Proof. We will induct on \(n\). When \(n = 1\), the claim is obvious. Since \(k\) is a complete field and \(U\) is finite-dimensional, it is closed, and therefore contains \[\begin{aligned} \pdv{t}|_{t= 0} \sum_{i = 1}^n t^i v_i = v_1 \end{aligned}\] Thus, \(\sum_{i = 1}^{n-1} t^iv_{i+1} \in U\), and the claim follows from induction. ◻

Proof. By multinomial expansion, we find \[\begin{aligned} \qty(\sum_{i=1}^dt_iv_i)^{\otimes n} = \sum_{k_1 +k_2 + \cdots + k_d = n} \binom{n}{k_1, \dots, k_d}\pi_{\text{sym}}\bigotimes_{i=1}^d t_i^{k_i}v_i^{\otimes k_i} \end{aligned}\] where \(\pi_{\text{sym}}\) is a projector onto the symmetric subspace. Applying the lemma above sequentially to each \(i\), we get \(\pi_{\text{sym}}\bigotimes_{i=1}^d v_{i}^{\otimes k_i}\) in the span of \(\{v \otimes \cdots \otimes v\}\) for each \(k_1 + \dots + k_d = n\), which clearly span \(\operatorname{Sym}^n(V)\). ◻

Proof. Since \(V\) is finite-dimensional, we have \(\operatorname{End}_{\mathbb C\mathrm{S}_n}(V^{\otimes n}) \cong \operatorname{Sym}^n(\operatorname{End}(V))\). Let \(\Delta\) represent the action of \(\mathcal U(\mathfrak{gl}(V))\), i.e. \[\begin{aligned} \Delta(X)(v_1 \otimes \cdots \otimes v_n) = \sum_{i}v_1 \otimes \cdots \otimes Xv_i \otimes \cdots v_n \end{aligned}\] By the fundamental theorem on symmetric polynomials, for any \(v \in \operatorname{End}(V)\), we can write \(v^{\otimes n}\) as a polynomial in \(\Delta(v), \Delta(v^2), \Delta(v^3) \dots\). By Lem. 5, \(v^{\otimes n}\) span \(\operatorname{Sym}^n \operatorname{End}(V)\). This proves the claim. ◻

Proof. Note that \(\mathfrak{u}\otimes \mathcal C\), also known as the complexification of \(\mathfrak{u}\), is just \(\mathfrak{gl}(V)\). Passing to the complexification does not change the centralizer. ◻

Proof. This is a consequence of the more general tensor-hom adjunction, which we will state here:

From this, Frobenius reciprocity follows: observe that \[\operatorname{Hom}_{\mathbb CG}(\mathbb CG \otimes_{\mathbb CH} V, U) \cong \operatorname{Hom}_{\mathbb CH}(V, \operatorname{Hom}_{\mathbb CG}(\mathbb CG, U)) \cong \operatorname{Hom}_{\mathbb CH}(V, U)\] where we applied the isomorphism \(\operatorname{Hom}_{A}(A, V) \cong V\) for any ring \(A\) and \(A\)-module \(V\). ◻

Proof. By Frobenius reciprocity, \[\begin{aligned} \operatorname{Hom}_{S_n}(I^{-}_{\lambda}, I^+_{\lambda}) &\cong \operatorname{Hom}_{Q_\lambda}(\mathrm{sgn}, \mathbb C[S_n/P_{\lambda}]) \cong \operatorname{Hom}_{Q_{\lambda}}(\mathbb C[S_n/P_\lambda], \mathrm{sgn}) \\ &\cong \{f:S_n \to \mathbb C: f(\tau g\sigma) = \operatorname{sgn}(\tau)f(g) \ \forall \ \tau \in Q_\lambda, \sigma \in P_\lambda\} \end{aligned}\] where we have identified \(\operatorname{Ind}_H^G(\mathrm{triv}) \cong \mathbb C[G/ H]\), and also used \(\operatorname{Hom}_G(U, V) \cong \operatorname{Hom}_G(V, U)\) for completely reducible representations.

Now we notice that if \(\tau = g\sigma g^{-1}\) for any transpositions \(\tau \in Q_\mu, \sigma \in P_{\lambda}\), then \[f(g) = f(g\sigma) = f(\tau g) = -f(g) \implies f(g) = 0\]

This brings us to the following lemma:

The above lemma implies the first part of our claim, because if \(\lambda > \mu\), then for any \(g\), \(f(g) = 0\), so \(\operatorname{Hom}_{S_n}(I^-_\mu, I^+_\lambda) = 0\).

Next, we will make a similar argument for \(\lambda = \mu\). We notice that \(f(g)\) is determined entirely by the double-coset \(Q_\lambda g P_\lambda\). On the identity coset, i.e. any \(g = qp\) with \(q \in Q_\lambda, p\in P_\lambda\), we have \(f(g) = f(qp) = \operatorname{sgn}(q)f(1)\). We will now argue that \(f\) must vanish on all non-identity double-cosets, which will follow from a lemma:

Now to complete the proof, if \(g \notin Q_\lambda P_\lambda\) then \(f(g) = 0\) because by the lemma above, we can find such a \(\sigma, \tau\). If \(g = qp \in Q_\lambda P_\lambda\), then \(f(g) = \operatorname{sgn}(q)f(1)\). This shows that \(\operatorname{Hom}_{S_n}(I_\lambda^-, I_\lambda^+)\) is 1-dimensional, which completes the claim. ◻

Proof. Since \(\dim \operatorname{Hom}(I^-_\lambda, I^+_\lambda) = 1\), there is exact one irreducible \(V_\lambda\) which occurs in both \(I_\lambda^-, I_\lambda^+\) by Schur’s lemma, which we call \(V_\lambda\). If \(\lambda \neq \mu\), then either \(\mu > \lambda\) or \(\lambda > \mu\). We assume the latter without loss of generality. Then \(\dim \operatorname{Hom}(I^-_\mu, I^+_\lambda) = 0\), so \(V_\lambda\) must not occur in \(I^-_\mu\), and therefore \(V_\mu\) is not isomorphic to \(V_{\lambda}\). Since there are exactly as many nonisomorphic irreps as conjugacy classes in \(S_n\) which are labeled by partitions, this completes the proof. ◻

Proof. Let \(\{g_1, \dots, g_k\}\) be a set of coset representatives for \(G/ H\). \(\operatorname{Ind}_H^G \chi\) is spanned by elements of the form \(g_i \otimes \mathbb C_\chi\), where \(\mathbb C_{\chi}\) is a one-dimensional vector space. Then for any \(g \in G\) we have \(gg_i = g_jh\) for a unique \(g_j, h\). By definition, \[\begin{aligned} g(g_i \otimes \mathbb C_{\chi}) = \chi(h)(g_j \otimes \mathbb C_{\chi}) \end{aligned}\] On the other hand, we observe that \(\mathbb CG \epsilon\) is spanned by elements \(g_i \epsilon\), and \[\begin{aligned} g g_i \epsilon = g_j\frac{1}{|H|}\sum_{k \in H}hk\overline \chi(k) = \chi(h)g_j\epsilon \end{aligned}\] Therefore the map \(g_i \otimes \mathbb C_\chi \mapsto g_i\epsilon\) is the desired isomorphism, completing the proof. ◻

Proof. Notice that \(I^-_\lambda \cong \mathbb C[S_n]b_\lambda\) and \(I^+_\lambda \cong \mathbb C[S_n]a_\lambda\) by the lemma above. Then we have \[\operatorname{Hom}_{S_n}(I^-_\lambda, I^+_\lambda) \cong \operatorname{Hom}_{S_n}(\mathbb C[S_n], b_\lambda \mathbb C[S_n] a_{\lambda}) \cong b_\lambda \mathbb C[S_n] a_{\lambda}\] so we conclude \(b_{\lambda} \mathbb C[S_n]a_\lambda\) is one-dimensional. Also, this implies \((c_\lambda^*)^2 = b_{\lambda}a_\lambda b_{\lambda}a_{\lambda} \propto c_\lambda^*\), so \(c_{\lambda}\) is proportional to an idempotent. Now we can compute \[\operatorname{Hom}_{S_n}(I^-_\lambda, \mathbb C[S_n]a_\lambda b_\lambda) \cong \operatorname{Hom}_{S_n}(\mathbb C[S_n]b_\lambda, \mathbb C[S_n]a_\lambda b_{\lambda}) \cong (b_{\lambda} \mathbb C[S_n] a_\lambda) b_{\lambda}\] Similarly, \[\begin{aligned} \operatorname{Hom}_{S_n}(\mathbb C[S_n]a_\lambda b_{\lambda}, I^+_{\lambda}) \cong \operatorname{Hom}_{S_n}(\mathbb C[S_n], a_{\lambda}b_{\lambda} \mathbb C[S_n] a_{\lambda}) \cong a_{\lambda} (b_{\lambda} \mathbb C[S_n]a_\lambda) \end{aligned}\] so both are one-dimensional. This shows that \(\mathbb C[S_n]a_{\lambda}b_{\lambda}\) occurs once in \(I^+_\lambda\) and \(I^-_\lambda\), so it must be isomorphic to \(V_\lambda\). ◻

Proof. Consider the map \[\begin{aligned} V_\lambda \cong \mathbb C[S_n]c_\lambda \ni \sigma c_{\lambda} \mapsto \mathrm{e}_{\sigma T_{\lambda}} \end{aligned}\] and we will prove this is an isomorphism of representations.

First, we say that \(T_1 \sim T_2\) iff \(T_2 = \sigma T_1\) for some \(\sigma \in P_T\). This implies that \(\pi T_2 = (\pi \sigma \pi^{-1})\pi T_1\) iff \(T_1 \sim T_2\). Since \(P_{\pi T} = \pi P_T\pi^{-1}\), this shows that \(\{\pi T\} = \pi\{T\}\). From this, we find \[\begin{aligned} \sigma \mathrm{e}_{\{T\}} = \sum_{\pi \in Q_T}\operatorname{sgn}(\pi)\{ \sigma\pi T\} &= \sum_{\pi \in Q_T}\operatorname{sgn}(\pi)\{(\sigma \pi\sigma^{-1}) \sigma T\} \notag \\ &= \sum_{\pi \in Q_T}\operatorname{sgn}(\sigma\pi\sigma^{-1})\{(\sigma \pi\sigma^{-1}) \sigma T\} \notag \\ &= \sum_{\pi' \in \sigma Q_T\sigma^{-1}}\operatorname{sgn}(\pi')\{\pi' \sigma T\} = \mathrm{e}_{\sigma T} \end{aligned}\] Therefore our map is a morphism of representations. Using this, we make the following observation: \[\mathrm{e}_{T_\lambda} = c_\lambda T\] We then see that \(\mathbb C[S_n]T\) is a faithful representation of \(S_n\) for any tableau \(T\) (just the fundamental representation!). As an element of \(\mathbb C[S_n]T\), we have \[\begin{aligned} v = \sum_{\sigma} a_\sigma \sigma \mathrm{e}_{T_\lambda} = \sum_{\sigma}a_{\sigma}\sigma c_{\lambda}T \end{aligned}\] so if \(v\) vanishes then \(\sum_{\sigma}a_{\sigma}\sigma c_{\lambda}\) must vanish as well. This shows that our map is injective. Surjectivity follows by definition. ◻