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An oscillating tableau is a sequence of Young diagrams, each obtained from its predecessor by adding or removing one box. These are paths in the Young branching graph with both directions allowed. A path of length $\ell$ from the empty diagram to $\lambda ⊢n$, with $\ell -n=2m\ge 0$, has count $\ell !f_{\lambda }/(2^{m}n!m!)$. This differs from a saturated upward path, whose length must equal $n$.

The Young lattice, or Young poset, has one vertex for each partition of an integer, including the empty partition, ordered by inclusion of their Young diagrams. A diagram covers another exactly when it adds one box. The Young branching graph is its graded graph: vertices at level $i$ are the partitions of $i$, and edges join diagrams differing by one box.

A descending path from $\lambda ⊢n$ to $(1)$ determines a standard Young tableau: label the successive removed boxes $n,n-1,\dots ,2$, then label the remaining box $1$. Every removed box is a Removable node of a Young diagram, so the resulting labels increase along rows and columns. Conversely, deleting boxes in decreasing label order from a standard Young tableau gives the path. Extending by the unique edge from $(1)$ to the empty diagram yields the usual path-to-tableau correspondence.

For the operator identity, compare the coefficients of each $\mu ⊢i$ in $DU(\lambda )$ and $UD(\lambda )$. If $\mu \ne \lambda$, a common upper cover exists exactly when a common lower cover exists, and each is then unique: the two diagrams differ by exchanging one box, their union is the upper cover and their intersection the lower cover. Since unions and intersections of partition diagrams are again partition diagrams, these off-diagonal coefficients agree.

The coefficient of $\lambda$ in $DU(\lambda )$ is the number of addable nodes of a Young diagram; its coefficient in $UD(\lambda )$ counts the choices of a Removable node of a Young diagram. Along the diagram's boundary the two types of corners alternate, beginning and ending with addable corners. Hence there is exactly one more addable than removable corner. For the empty partition the counts are $1$ and $0$.

Therefore the up and down operators satisfyor $D_{i+1}{U}_i-U_{i-1}{D}_i=I_i$ at level $i$. At level zero, interpret $D_0$ and the lowering contribution as zero. This makes the Young lattice a differential poset.

Use the normal ordering identity for up and down operators, placing every $U$ before every $D$. The commutator from part (a) impliesby induction on $i$. Multiplying the normal-ordered expansion on the left by $D+U$ therefore giveswhere negative indices have coefficient zero and $a_{0,0}(0)=1$.

We prove that $a_{i,j}(\ell )=0$ unless $i,j\ge 0$ and $\ell -i-j=2m$ for some integer $m\ge 0$, and otherwiseThe formula holds at $\ell =0$. For the next step with $\ell +1-i-j=2m$, the three contributions in the recurrence, after factoring out $\ell !/(2^{m}i!j!m!)$, are respectively $i$, $j$, and $2m$. Terms with a negative index or $m-1<0$ are simply absent. Their sum is $i+j+2m=\ell +1$, giving the required $(\ell +1)!$ numerator. Parity and nonnegativity exclude every remaining case. This proves the coefficient formula.

Apply the identity to the empty partition. Since $D(\varnothing )=0$, only terms with $j=0$ survive. To finish at $\lambda ⊢n$, only $U^n$ can contribute, and its coefficient of $\lambda$ is $f_{\lambda }$, by the Young branching graph correspondence with standard Young tableaux. For $\ell =n+2m$,Here $(2m-1)!!=1\cdot 3\cdots (2m-1)$ is the odd double factorial, with $(-1)!!=1$. ThusThe operator $(D+U{)}^{\ell }$ counts oscillating tableaux, allowing both upward and downward steps. A strictly upward path to level $n$ would necessarily have length $n$; the printed operator specification is what determines $N$ when $\ell >n$.

A standard Young tableau of shape $\lambda ⊢n$ is a Young tableau that uses $1,\dots ,n$ exactly once and strictly increases left to right along every row and top to bottom down every column. Formally, a bijection $T:Y(\lambda )\to \{1,\dots ,n\}$ is standard whenwhenever the indicated cells belong to the Young diagram. We write $\mathrm{SYT}(\lambda )$ for the set of such fillings and $f^{\lambda }=|\mathrm{SYT}(\lambda )|$ for its cardinality.

For $(3,2)$, the five standard Young tableaux are listed below, with each pair giving the top row followed by the bottom row:There are no others: the upper-left entry must be $1$, and choosing the two bottom-row entries fixes both increasing row orders. Of the six choices of two entries from $\{2,3,4,5\}$, only bottom row $(2,3)$ fails, because its $3$ would lie below $4$. Hence $f^{(3,2)}=5$ by direct enumeration, independently of the hook-length formula.

The cells occupied by $1,\dots ,r$ form a Young diagram: every cell above or to the left of an occupied cell has a smaller label. Consequently a standard Young tableau is equivalently a path from the empty shape to $\lambda$ in the Young branching graph, adjoining one addable node of a Young diagram at each step. For $T_2$, the successive shapes areConversely, number the cell added at step $r$ by $r$ to recover the standard Young tableau. This path interpretation also makes a standard Young tableau a linear extension of a partially ordered set defined by the row and column requirements; see Stanley's Algebraic Combinatorics, Chapter 8, on tableaux and paths in Young's lattice.

The usefulness goes beyond counting: the standard polytabloid basis of a Specht module $S^{\lambda }$ over the complex numbers is indexed by $\mathrm{SYT}(\lambda )$, so ${\dim }_{\mathbb{C}}{S}^{\lambda }=f^{\lambda }$. Each of the five pictured standard Young tableaux therefore labels one basis element of the irreducible representation $S^{(3,2)}$ of the symmetric group $S_5$, whose dimension over the complex numbers is five. The filled diagrams index the basis; the actual vectors are polytabloids. The hook-length formula computes this dimension without listing all the standard Young tableaux.

The Young diagram of a partition of an integer $\lambda$ consists of one unit square, or cell, for each pairThroughout this article we use the English convention: rows are left-aligned, row $1$ is at the top, row indices increase downwards, and column indices increase to the right. The row lengths weakly decrease from top to bottom. Thus $Y(3,2)$ has cells $(1,1),(1,2),(1,3),(2,1),(2,2)$, with three cells above two. In the French convention the longest row is drawn at the bottom; the underlying partition of an integer is unchanged, but directions in descriptions of hooks of a Young diagram and standard Young tableaux must be interpreted consistently. See Grinberg's hook-length-formula teaching slides, §§1.1–1.2, for diagrams with the same orientation.

The number of cells is $|\lambda |$, and reading the row lengths recovers the partition of an integer. Reading the column lengths instead gives its conjugate partition. This turns numerical inequalities into visible geometry: Young diagrams satisfywith missing parts treated as zero. For example, $(2,1)\subseteq (3,2)$. This inclusion is the order in Young's lattice; it differs from dominance order on partitions, which compares sums of initial parts of partitions of an integer of the same size. The partition of an integer $(3,2)$ dominates $(2,2,1)$ because their successive sums are $3\ge 2$, $5\ge 4$, and $5=5$, but neither Young diagram contains the other: one has cell $(1,3)$, the other $(3,1)$.

The beauty of a Young diagram is that its shape simultaneously records a partition of an integer, indexes a Specht module, and supports local quantities such as hook lengths that determine a global dimension. Filling its cells leads to Young tableaux; adding one cell at a time leads to the Young branching graph.