[proofplan] The proof uses the standard highest-weight classification for finite normal type $A$ crystals. First, finiteness guarantees that every component contains at least one source for all raising operators. Normality then forces the full $GL_n$ weight of such a source to be a partition, because adjacent weight differences are the nonnegative string lengths $\varphi_i$. Finally, the finite normal highest-weight classification identifies the component generated by a source of weight $\mu$ with the tableau crystal $\operatorname{SSYT}_n(\mu)$, and grouping the disjoint components by this source weight gives the stated decomposition. [/proofplan] [step:Show that each connected component contains a highest weight element] For each $j\in\{1,\dots,r\}$, the set $\operatorname{SSYT}_n(\lambda_j)$ is finite because its tableaux have fixed finite shape $\lambda_j$ and entries in the finite alphabet $\{1,\dots,n\}$. Hence the [tensor product](/page/Tensor%20Product) set $B$ is finite. Let $C\subset B$ be a connected component of the underlying crystal graph. Choose an element $b_0\in C$. If some raising operator $e_i$ is defined at $b_0$, replace $b_0$ by $e_i b_0$. Repeating this procedure cannot continue indefinitely, because every replacement remains in the finite set $C$ and strictly raises the type $A_{n-1}$ weight by the simple root direction corresponding to $i$. Equivalently, it strictly increases the dominance order on the full content weights. Therefore the process terminates at an element $b\in C$ such that $e_i b$ is undefined for every $i\in\{1,\dots,n-1\}$. [guided] Fix a connected component $C\subset B$. We want to prove first that $C$ has at least one element from which no raising operator can move upward. The reason this is plausible is that $B$ is finite: there is no room for an infinite ascending path. For each $j$, the crystal $\operatorname{SSYT}_n(\lambda_j)$ consists of tableaux of one fixed finite Young diagram and with entries chosen from the finite set $\{1,\dots,n\}$. Thus each tensor factor is finite. Since $B$ is the Cartesian product of these finite sets as a set, $B$ is finite, and every connected component $C$ of its crystal graph is finite. Choose $b_0\in C$. If $e_i b_0$ is defined for some $i\in\{1,\dots,n-1\}$, replace $b_0$ by $e_i b_0$. This new element is still in $C$, because crystal edges connect an element to the result of applying a defined Kashiwara operator. Each such move changes the full $GL_n$ content weight by increasing the $i$th coordinate by $1$ and decreasing the $(i+1)$st coordinate by $1$. Hence it raises the type $A_{n-1}$ weight by the simple root corresponding to $i$. This repeated raising process cannot be infinite. If it were infinite, it would give an infinite sequence of elements in the finite set $C$ with strictly increasing type $A$ weight along each step, which is impossible. Therefore after finitely many steps we arrive at an element $b\in C$ such that no operator $e_i$ is defined at $b$. This is a highest weight element, or source, of the component. [/guided] [/step] [step:Use normality to prove that the source weight is a partition of $N$] Let $b\in C$ be an element such that $e_i b$ is undefined for every $i\in\{1,\dots,n-1\}$. Let \begin{align*} \operatorname{wt}_{GL_n}(b)=\mu=(\mu_1,\dots,\mu_n)\in\mathbb Z_{\ge 0}^n \end{align*} be its full content weight. Since each tensor factor has size $|\lambda_j|$, every element of $B$ has total content $N$, so \begin{align*} \mu_1+\cdots+\mu_n=N. \end{align*} The tableau crystals $\operatorname{SSYT}_n(\lambda_j)$ are finite normal type $A_{n-1}$ crystals by [citetheorem:8475], and the tensor product crystal is formed by the type $A$ tensor product rule [citetheorem:8459]. We use the standard normality theorem for tensor products of finite normal crystals, namely that the tensor product of finite normal crystals is normal and each connected component is finite normal. For a normal type $A_{n-1}$ crystal, define \begin{align*} \varepsilon_i(b)=\max\{k\in\mathbb Z_{\ge 0}: e_i^k b\text{ is defined}\} \end{align*} and \begin{align*} \varphi_i(b)=\max\{k\in\mathbb Z_{\ge 0}: f_i^k b\text{ is defined}\}. \end{align*} Normality gives \begin{align*} \varphi_i(b)-\varepsilon_i(b)=\mu_i-\mu_{i+1} \end{align*} for every $i\in\{1,\dots,n-1\}$. Since no $e_i b$ is defined, $\varepsilon_i(b)=0$. Since $\varphi_i(b)$ is a nonnegative integer, we obtain \begin{align*} \mu_i-\mu_{i+1}=\varphi_i(b)\ge 0. \end{align*} Thus $\mu_1\ge\cdots\ge\mu_n\ge 0$, and together with $\sum_i\mu_i=N$ this proves $\mu\in\mathcal P_{n,N}$. [/step] [step:Identify the component generated by a source with the tableau crystal of the same highest weight] Let $C\subset B$ be a connected component and let $b\in C$ be a highest weight element with $GL_n$ weight $\mu\in\mathcal P_{n,N}$. Since $C$ is a finite normal connected type $A_{n-1}$ crystal, the finite normal highest-weight classification applies. In the notation of [citetheorem:8473], a connected normal highest-weight crystal is uniquely determined by its highest weight. By [citetheorem:8475], the semistandard tableau crystal $\operatorname{SSYT}_n(\mu)$ realizes the normal highest-weight crystal with polynomial highest weight $\mu$. Therefore there exists an isomorphism of type $A_{n-1}$ crystals \begin{align*} \Phi_C:C\to \operatorname{SSYT}_n(\mu) \end{align*} sending $b$ to the highest weight tableau of shape $\mu$. The classification also gives uniqueness of the highest weight source in $C$, because the model crystal $\operatorname{SSYT}_n(\mu)$ has a unique highest weight element and $\Phi_C$ preserves the operators $e_i$. The isomorphism preserves the full $GL_n$ weight grading. Indeed, it preserves the adjacent type $A$ weight differences because it is a type $A_{n-1}$ crystal isomorphism, it sends the source of full weight $\mu$ to the source of full weight $\mu$, and every application of $f_i$ changes the full content weight by subtracting $1$ from coordinate $i$ and adding $1$ to coordinate $i+1$. Hence every path from the source determines the same full content change in $C$ and in $\operatorname{SSYT}_n(\mu)$. Since all weights have total size $N$, no additional determinant-line ambiguity remains. [/step] [step:Group the connected components by their unique source weight] The connected components of the crystal graph of $B$ are pairwise disjoint and their union is all of $B$. By the previous step, each connected component $C$ has a unique highest weight element, whose $GL_n$ weight is some $\mu\in\mathcal P_{n,N}$, and $C$ is isomorphic to $\operatorname{SSYT}_n(\mu)$ with its full $GL_n$ weight grading. For each $\mu\in\mathcal P_{n,N}$, define $m_\mu$ to be the number of connected components of $B$ whose unique highest weight element has $GL_n$ weight $\mu$. Index these components as \begin{align*} B(\mu,1),\dots,B(\mu,m_\mu) \end{align*} when $m_\mu>0$, and use the empty indexing set when $m_\mu=0$. Since the connected components partition $B$, we obtain the disjoint crystal decomposition \begin{align*} B\cong\bigsqcup_{\mu\in\mathcal P_{n,N}}\bigsqcup_{a=1}^{m_\mu} B(\mu,a). \end{align*} Each component $B(\mu,a)$ is isomorphic to $\operatorname{SSYT}_n(\mu)$ by construction, so the decomposition has exactly the asserted form. This completes the proof. [/step]