$\require{physics}\newcommand{\cbrt}[1]{\sqrt[3]{#1}}\newcommand{\sgn}{\text{sgn}}\newcommand{\ii}[1]{\textit{#1}}\newcommand{\eps}{\varepsilon}\newcommand{\EE}{\mathbb E}\newcommand{\PP}{\mathbb P}\newcommand{\Var}{\mathrm{Var}}\newcommand{\Cov}{\mathrm{Cov}}\newcommand{\pperp}{\perp\kern-6pt\perp}\newcommand{\LL}{\mathcal{L}}\newcommand{\pa}{\partial}\newcommand{\AAA}{\mathscr{A}}\newcommand{\BBB}{\mathscr{B}}\newcommand{\CCC}{\mathscr{C}}\newcommand{\DDD}{\mathscr{D}}\newcommand{\EEE}{\mathscr{E}}\newcommand{\FFF}{\mathscr{F}}\newcommand{\WFF}{\widetilde{\FFF}}\newcommand{\GGG}{\mathscr{G}}\newcommand{\HHH}{\mathscr{H}}\newcommand{\PPP}{\mathscr{P}}\newcommand{\Ff}{\mathcal{F}}\newcommand{\Gg}{\mathcal{G}}\newcommand{\Hh}{\mathbb{H}}\DeclareMathOperator{\ess}{ess}\newcommand{\CC}{\mathbb C}\newcommand{\FF}{\mathbb F}\newcommand{\NN}{\mathbb N}\newcommand{\QQ}{\mathbb Q}\newcommand{\RR}{\mathbb R}\newcommand{\ZZ}{\mathbb Z}\newcommand{\KK}{\mathbb K}\newcommand{\SSS}{\mathbb S}\newcommand{\II}{\mathbb I}\newcommand{\conj}[1]{\overline{#1}}\DeclareMathOperator{\cis}{cis}\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}\newcommand{\norm}[1]{\left\lVert #1 \right\rVert}\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}\newcommand{\ceil}[1]{\left\lceil #1 \right\rceil}\DeclareMathOperator*{\range}{range}\DeclareMathOperator*{\nul}{null}\DeclareMathOperator*{\Tr}{Tr}\DeclareMathOperator*{\tr}{Tr}\newcommand{\id}{1\!\!1}\newcommand{\Id}{1\!\!1}\newcommand{\der}{\ \mathrm {d}}\newcommand{\Zc}[1]{\ZZ / #1 \ZZ}\newcommand{\Zm}[1]{\left(\ZZ / #1 \ZZ\right)^\times}\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\End}{End}\newcommand{\GL}{\mathbb{GL}}\newcommand{\SL}{\mathbb{SL}}\newcommand{\SO}{\mathbb{SO}}\newcommand{\OO}{\mathbb{O}}\newcommand{\SU}{\mathbb{SU}}\newcommand{\U}{\mathbb{U}}\newcommand{\Spin}{\mathrm{Spin}}\newcommand{\Cl}{\mathrm{Cl}}\newcommand{\gr}{\mathrm{gr}}\newcommand{\gl}{\mathfrak{gl}}\newcommand{\sl}{\mathfrak{sl}}\newcommand{\so}{\mathfrak{so}}\newcommand{\su}{\mathfrak{su}}\newcommand{\sp}{\mathfrak{sp}}\newcommand{\uu}{\mathfrak{u}}\newcommand{\fg}{\mathfrak{g}}\newcommand{\hh}{\mathfrak{h}}\DeclareMathOperator{\Ad}{Ad}\DeclareMathOperator{\ad}{ad}\DeclareMathOperator{\Rad}{Rad}\DeclareMathOperator{\im}{im}\renewcommand{\BB}{\mathcal{B}}\newcommand{\HH}{\mathcal{H}}\DeclareMathOperator{\Lie}{Lie}\DeclareMathOperator{\Mat}{Mat}\DeclareMathOperator{\span}{span}\DeclareMathOperator{\proj}{proj}$ Let us inspect the structure of $\GL_n(\CC)$ on $V^{\otimes N}$. The fundamental idea here is that $S_N$ and $\GL_n(\CC)$ commute and act on $V^{\otimes N}$ simultaneously, yielding rich combinatorial structure. >[!idea]- Representations are Partitions >Observe that the highest weights of polynomial representations are non-increasing sequences of nonnegative integers $(\lambda_1,\dots, \lambda_n)$ which are partitions of $\abs{\lambda} = \sum_i \lambda_i$ into $\leq n$ parts. We will say that a partition *fits* if it has $\leq n$ parts. Note that $\abs{\lambda}$ is just the eigenvalue of $\id\in \gl_n$ on $L_\lambda$, and is also the $N$ such that $L_\lambda \in V^{\otimes N}$. We know that representations of $\GL_n(\CC)$ are completely reducible. $\GL_n(\CC)$ has a natural polynomial representation over $V^{\otimes N}$, which splits naturally into polynomial irreps:$V^{\otimes N} = \bigoplus_{\lambda: \abs{\lambda} = N} L_\lambda\otimes \pi_\lambda,\qquad \pi_\lambda := \Hom_{\GL_n(\CC)} (L_\lambda, V^{\otimes N}).$Here, $L_\lambda = 0$ if $\lambda$ doesn't fit. Observe that each $\pi_{\lambda}$ is a representation of $S_n$! Oh my god this is really deep. >[!idea]- Reminder about basic representation theory >In representation theory, the $\pi_\lambda$ are called ==**multiplicity spaces**==. By Schur's lemma, any intertwiner must map isomorphically onto some copy of $L_\lambda$ in $V^{\otimes N}$. This might be some linear combination of the $L_\lambda^{\oplus \dim \pi_\lambda}$ generically, but that's why the formula works. We now let the images of $U(\gl_n)$ and $\CC S_N$ in $\End_\CC(V^{\otimes N})$ be $A$ and $B$, (which we like to call the ==**Schur algebra**== and the ==**centralizer algebra**==). Here's the deep fact: >[!theorem] Schur-Weyl Duality >1. The centralizer of $A$ is $B$, and vice-versa. >2. If $\lambda$ fits, then the representation $\pi_\lambda$ of $\CC_\lambda$ is irreducible. Such representations are pairwise non-isomorphic. >3. If $\dim V \geq N$ (such that all $\lambda$ fit), then the $\pi_\lambda$ enumerate all irreducible representations of $S_N$. # Proof of Schur-Weyl Duality >[!claim] $S^N U$ is spanned by elements $x\otimes,\dots,\otimes x$, $x\in U$. >[!proof]- >The span of these vectors forms a nonzero $\GL(U)$-subrep, but $S^N(U)$ was supposed to be irreducible. >[!claim] Basis bootstrapping $S^NR$ >For any associative algebra $R$ over $\CC$, the algebra $S^NR$ is generated by the elements >$\Delta_N(x) = x\otimes1\otimes\dots\otimes1+\dots+1\otimes\dots\otimes1\otimes x$ >for $x\in R$. >[!proof]- >There is some Newton polynomial that computes $z_1z_2\dots z_n$ as a polynomial in $\sum_{i=1}^N z_i^k$ with $k=1,\dots,N$. Then, $x\otimes\dots\otimes x= P_n(\Delta_N(x),\dots,\Delta_N(x^N))$. The centralizer of $B$ is exactly those elements of $\End_\CC(V^\otimes N)$ which commute with all permutations in $S_N$; by the averaging trick, this is exactly $S_N(\End V)$. By the preceding claim, the centralizer is generated by $\Delta_N(x)$. But this is basically the definition of $A$; each $x\in U(\gl_n)$ gets mapped right to $\Delta_N(x)$! We observe $\CC[S_n]$, as a group algebra, is semisimple, thus its image $B$ is as well, and the [[Dual Centralizer Lemma]] finishes off the first part. It actually nukes the other two parts as well: - If $\lambda$ fits, $L_\lambda$ occurs in $V^{\otimes N}$ somewhere, thus $\pi_\lambda \neq 0$. - If $\dim V \geq N$, then we can pick any independent $v_1,\dots, v_n\in V$, then acting with $\CC S_n$ on $v_1\otimes\dots\otimes v_n$ shows $\CC S_N = B$ (the map is injective).