[proofplan] We compute the character of the restricted module on the diagonal Cartan subalgebra of $\mathfrak{gl}_{n-1}(\mathbb C)$. The character of $L^{\mathfrak{gl}_n}(\lambda)$ is the Schur polynomial $s_\lambda(x_1,\dots,x_n)$, and restriction along the upper-left embedding is obtained by setting $x_n=1$. The semistandard tableau model gives the branching identity $s_\lambda(x_1,\dots,x_{n-1},1)=\sum_{\mu\prec\lambda}s_\mu(x_1,\dots,x_{n-1})$. Complete reducibility and [linear independence](/page/Linear%20Independence) of Schur characters then identify the irreducible summands and their multiplicities. [/proofplan]

[step:Compute the restricted character by setting the last torus variable equal to $1$] Let $\mathfrak h_n\subset\mathfrak{gl}_n(\mathbb C)$ be the diagonal Cartan subalgebra, and let $\varepsilon_i\in\mathfrak h_n^*$ denote the coordinate functional extracting the $i$th diagonal entry. Let $\mathfrak h_{n-1}\subset\mathfrak{gl}_{n-1}(\mathbb C)$ be the diagonal Cartan subalgebra for the upper-left copy of $\mathfrak{gl}_{n-1}(\mathbb C)$. Since $\lambda$ is a partition with at most $n$ parts, [citetheorem:9394] realizes the corresponding irreducible polynomial $\mathfrak{gl}_n(\mathbb C)$-module by the Schur functor $S_\lambda(\mathbb C^n)$. Its standard weight basis is indexed by semistandard Young tableaux of shape $\lambda$ with entries in $\{1,\dots,n\}$, so the $\mathfrak h_n$-character of $L^{\mathfrak{gl}_n}(\lambda)$ is \begin{align*} \operatorname{ch}_{\mathfrak h_n} L^{\mathfrak{gl}_n}(\lambda)=s_\lambda(x_1,\dots,x_n). \end{align*} Under the upper-left embedding $\mathfrak{gl}_{n-1}(\mathbb C)\subset\mathfrak{gl}_n(\mathbb C)$, an element $\operatorname{diag}(t_1,\dots,t_{n-1})\in\mathfrak h_{n-1}$ acts as $\operatorname{diag}(t_1,\dots,t_{n-1},0)\in\mathfrak h_n$. Therefore the weight variable corresponding to $\varepsilon_n$ contributes the scalar factor $1$, and the restricted character is \begin{align*} \operatorname{ch}_{\mathfrak h_{n-1}}\operatorname{Res}^{\mathfrak{gl}_n}_{\mathfrak{gl}_{n-1}} L^{\mathfrak{gl}_n}(\lambda)=s_\lambda(x_1,\dots,x_{n-1},1). \end{align*} [/step]

[step:Prove the Schur branching identity by separating the boxes labelled $n$]We use the semistandard tableau formula for Schur polynomials. For a partition $\nu$ with at most $m$ parts, let $\operatorname{SSYT}_m(\nu)$ denote the set of semistandard Young tableaux of shape $\nu$ with entries in $\{1,\dots,m\}$, using weakly increasing rows and strictly increasing columns. For $T\in\operatorname{SSYT}_m(\nu)$, let $a_i(T)$ be the number of entries equal to $i$. The Schur polynomial is \begin{align*} s_\nu(x_1,\dots,x_m)=\sum_{T\in\operatorname{SSYT}_m(\nu)}x_1^{a_1(T)}\cdots x_m^{a_m(T)}. \end{align*} Fix $T\in\operatorname{SSYT}_n(\lambda)$. Let $\mu(T)$ be the shape obtained from $\lambda$ after deleting all boxes whose entry is $n$. Because rows are weakly increasing, the boxes labelled $n$ in each row occur at the right end of that row. Because columns are strictly increasing, no column contains two boxes labelled $n$. Hence $\lambda/\mu(T)$ is a horizontal strip, equivalently \begin{align*} \lambda_i\ge \mu_i(T)\ge \lambda_{i+1}\quad\text{for every }1\le i\le n-1. \end{align*} These inequalities also show that $\mu(T)$ is a dominant polynomial weight for $\mathfrak{gl}_{n-1}(\mathbb C)$: for $1\le i\le n-2$ we have $\mu_i(T)\ge \lambda_{i+1}\ge \mu_{i+1}(T)$, and $\mu_{n-1}(T)\ge \lambda_n\ge 0$. Thus $\mu(T)\prec\lambda$. Conversely, suppose $\mu\prec\lambda$. Given a tableau $S\in\operatorname{SSYT}_{n-1}(\mu)$, fill every box of the skew diagram $\lambda/\mu$ with $n$. Since $\lambda/\mu$ is a horizontal strip, no column receives two new boxes. Since the new entries are placed at row ends and $n$ is the largest allowed entry, the resulting filling is a semistandard tableau of shape $\lambda$ with entries in $\{1,\dots,n\}$. This construction is inverse to deleting the boxes labelled $n$. Therefore the tableaux in $\operatorname{SSYT}_n(\lambda)$ are partitioned by the interlacing shapes $\mu\prec\lambda$, and after setting $x_n=1$, the boxes labelled $n$ contribute no monomial factor. Hence \begin{align*} s_\lambda(x_1,\dots,x_{n-1},1)=\sum_{\mu\prec\lambda}s_\mu(x_1,\dots,x_{n-1}). \end{align*}[/step]

[guided]The key point is that the last variable is being specialized to $1$, so the entries labelled $n$ should be treated as invisible weight-wise but still visible shape-wise. We use the tableau definition of Schur polynomials: for a partition $\nu$ with at most $m$ parts, $\operatorname{SSYT}_m(\nu)$ is the set of semistandard tableaux of shape $\nu$ with entries in $\{1,\dots,m\}$, and \begin{align*} s_\nu(x_1,\dots,x_m)=\sum_{T\in\operatorname{SSYT}_m(\nu)}x_1^{a_1(T)}\cdots x_m^{a_m(T)}. \end{align*} Here $a_i(T)$ denotes the number of entries of $T$ equal to $i$. Now take $T\in\operatorname{SSYT}_n(\lambda)$ and remove all boxes labelled $n$. Call the remaining shape $\mu(T)$. Why is this shape interlacing with $\lambda$? In each row, weak increase forces every $n$ to appear at the far right of that row, since no entry can be larger than $n$. In each column, strict increase prevents two entries equal to $n$ from appearing in the same column. Thus the removed skew diagram $\lambda/\mu(T)$ has at most one box in each column; this is exactly the horizontal-strip condition. For partitions with row lengths $\lambda_1,\dots,\lambda_n$ and $\mu_1,\dots,\mu_{n-1}$, the horizontal-strip condition is equivalent to \begin{align*} \lambda_i\ge \mu_i(T)\ge \lambda_{i+1}\quad\text{for every }1\le i\le n-1. \end{align*} The inequalities also show that $\mu(T)$ is a dominant polynomial weight for $\mathfrak{gl}_{n-1}(\mathbb C)$: if $1\le i\le n-2$, then $\mu_i(T)\ge \lambda_{i+1}\ge \mu_{i+1}(T)$, and $\mu_{n-1}(T)\ge \lambda_n\ge 0$. So every tableau of shape $\lambda$ determines an interlacing dominant polynomial weight $\mu\prec\lambda$. Conversely, fix an interlacing shape $\mu\prec\lambda$ and a tableau $S\in\operatorname{SSYT}_{n-1}(\mu)$. Fill every box of $\lambda/\mu$ with $n$. The inequalities $\lambda_i\ge \mu_i\ge \lambda_{i+1}$ say precisely that $\lambda/\mu$ is a horizontal strip, so no column receives two added boxes. The row condition remains valid because the added boxes sit at the right ends of rows and carry the largest possible entry $n$. For the column condition, consider an added box in row $i$ and column $c$. Since it is added, $c>\mu_i$; since $\mu_i\ge \lambda_{i+1}$, row $i+1$ has no box in column $c$, so there is no existing box below the added $n$. Any existing box above it has entry at most $n-1$, because $S$ uses entries in $\{1,\dots,n-1\}$. Hence the new filling is a semistandard tableau of shape $\lambda$ with entries in $\{1,\dots,n\}$. The two constructions are inverse: deleting the $n$-boxes recovers $S$, and adding $n$-boxes to $\lambda/\mu$ recovers $T$. Therefore, after specializing $x_n=1$, the [generating function](/page/Generating%20Function) splits as \begin{align*} s_\lambda(x_1,\dots,x_{n-1},1)=\sum_{\mu\prec\lambda}s_\mu(x_1,\dots,x_{n-1}). \end{align*}[/guided]

[step:Identify the summands from character equality] The restricted module is finite-dimensional because $L^{\mathfrak{gl}_n}(\lambda)$ is finite-dimensional, and it is a polynomial $\mathfrak{gl}_{n-1}(\mathbb C)$-module because the upper-left block inclusion is a polynomial representation on the same [vector space](/page/Vector%20Space). The [Lie algebra](/page/Lie%20Algebra) $\mathfrak{gl}_{n-1}(\mathbb C)$ is reductive, since it is the [direct sum](/page/Direct%20Sum) of $\mathfrak{sl}_{n-1}(\mathbb C)$ and its one-dimensional center. [Weyl's complete reducibility theorem](/theorems/3755) for finite-dimensional modules over complex reductive Lie algebras therefore applies. Hence there exist nonnegative integers $m_\nu$, indexed by dominant polynomial weights $\nu=(\nu_1,\dots,\nu_{n-1})$, such that \begin{align*} \operatorname{Res}^{\mathfrak{gl}_n}_{\mathfrak{gl}_{n-1}} L^{\mathfrak{gl}_n}(\lambda)\cong \bigoplus_\nu \left(L^{\mathfrak{gl}_{n-1}}(\nu)\right)^{\oplus m_\nu}. \end{align*} Taking $\mathfrak h_{n-1}$-characters and applying the same Schur-functor character computation to each partition $\nu$ with at most $n-1$ parts gives \begin{align*} s_\lambda(x_1,\dots,x_{n-1},1)=\sum_\nu m_\nu s_\nu(x_1,\dots,x_{n-1}). \end{align*} The Schur polynomials $s_\nu(x_1,\dots,x_{n-1})$ for dominant polynomial weights $\nu$ are linearly independent: ordered by dominance, each $s_\nu$ has leading monomial $x_1^{\nu_1}\cdots x_{n-1}^{\nu_{n-1}}$ with coefficient $1$, and no different dominant weight has the same leading monomial. Comparing this expansion with the branching identity from the previous step gives $m_\nu=1$ if $\nu\prec\lambda$, and $m_\nu=0$ otherwise. Therefore \begin{align*} \operatorname{Res}^{\mathfrak{gl}_n}_{\mathfrak{gl}_{n-1}} L^{\mathfrak{gl}_n}(\lambda)\cong \bigoplus_{\mu\prec\lambda} L^{\mathfrak{gl}_{n-1}}(\mu), \end{align*} and each summand occurs with multiplicity $1$. [/step]