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For a standard Young tableau $T$, let $c_T(r)=j-i$ when its entry $r$ lies in cell $(i,j)$. Its content vector of a standard Young tableau is $C(T)=(c_T(1),\dots ,c_T(n))$. We use the following sign convention for axial distance in a Young tableau:This is the axial distance from $j$ to $i$. In particular, write $d=c_T(r+1)-c_T(r)$ for the adjacent transposition $s_r=(rr+1)$.

Here is Young orthogonal form, with one orthonormal basis vector $w_T$ for each standard Young tableau of a fixed shape:If $s_{r}T$ is not standard, omit that term. In this case $r,r+1$ occupy adjacent cells in one row or column, so $d=1$ or $d=-1$, respectively, and the scalar action is $+1$ or $-1$. On a standard pair $T,R=s_{r}T$, $|d|\ge 2$ and the matrix isThis is a real orthogonal involution.

To prove the formula, explicitly use the permitted Young seminormal form theorem: the Specht module has a tableau basis $v_T$ whose vectors are joint eigenvectors of the Young–Jucys–Murphy elements, and whose actions, on an admissible pair oriented so that tableau length increases from $T$ to $R$, areNonadmissible swaps have the scalar actions just described. The theorem supplies these as one globally consistent representation, not as independently chosen two-dimensional blocks.

Average a positive definite Hermitian inner product over $S_n$. Distinct tableau lines are orthogonal because their commuting self-adjoint Young–Jucys–Murphy elements have distinct joint weights. Also $s_r$ is self-adjoint, so the two seminormal equations implyUpon setting $w_T=v_T/\Vert v_T\Vert$, the coefficient from $w_T$ to $w_R$ becomes $\Vert v_R\Vert /\Vert v_T\Vert =\sqrt{1-d^{-2}}$; the reverse coefficient is the same. All phases are consistent because the seminormal basis was global, and the group relations remain true under this change of basis. This proves Young orthogonal form.

For $\lambda =(n)$, there is exactly one standard Young tableau, the row $1,2,\dots ,n$. Its content vector of a standard Young tableau isEvery adjacent distance is $1$, so $s_r{w}_T=w_T$ for every $r$. Thus $\boxed{S^{(n)}\cong {\mathbb{C}}_{\mathrm{triv}}}$ is the trivial representation.

A near Young tableau is a filling of a Young diagram by distinct entries from a totally ordered alphabet, increasing along rows and down columns; the entries need not be $1,\dots ,n$. For row insertion of $x$, scan the first row for its leftmost entry larger than $x$. If one exists, replace it by $x$ and insert the displaced entry into the next row by the same rule. Otherwise append $x$ to the row and stop. Continue until a new outer-corner box is appended.

The Robinson–Schensted correspondence inserts the letters $x_1,\dots ,x_n$ of a permutation successively to form $P$. Whenever the $k$th insertion creates a new cell, put $k$ in that cell of a second tableau $Q$. Row insertion preserves increasing rows and columns, so $P$ is standard; the sequence of growing diagrams makes $Q$ standard, with the same shape. To reverse the construction, remove the box carrying the largest label in $Q$, and reverse the bumping in $P$: in each row above, exchange the carried entry with the rightmost smaller entry, continuing up to the first row. The final expelled entry is the last letter of the permutation. Repeating recovers the whole permutation, establishing the bijection with pairs of standard Young tableaux of a common shape.

For the bumping inequality, each displaced entry is the first entry strictly larger than the incoming one. Thus every step replaces a larger entry with a smaller entry and carries that larger value downwards. Consequently the bumped values strictly increase:

Use the Young–Jucys–Murphy elements in the group algebra $\mathbb{C}{S}_n$. Their joint spectrum isEquivalently one can use the regular representation, which contains every irreducible representation. The commuting elements are self-adjoint in a unitary representation, so there is a simultaneous eigenbasis.

A standard Young tableau has entries increasing along each row and down each column. If entry $r$ occupies cell $(i_r,j_r)$, its Content of a Young-diagram cell is $c_T(r)=j_r-i_r$. Define the set of content vectors of standard Young tableaux byThe shape here is a partition of an integer; it is not itself the vector of eigenvalues. For $n=1$ both sets consist of $(0)$.

Necessity is a property of standard Young tableaux alone. The first entry occupies $(1,1)$, with Content of a Young-diagram cell zero. Every subsequent cell has a cell immediately above or to its left, of content one larger or one smaller, already present. If two entries occupy the same diagonal, the later cell lies strictly southeast of the earlier. The cells immediately right of and immediately below the earlier one exist and have contents $a+1,a-1$; their entries lie strictly between the two given entries. This proves all three conditions for content vectors of standard Young tableaux.

For sufficiency, build the Young diagram one cell at a time. All coordinates are integers, because each new coordinate differs by one from an earlier coordinate, starting at zero. Cells of any fixed content $a$ are ordered northwest to southeast. For $a\ge 0$ their positions are $(r,r+a)$, $r=1,2,\dots$; for $a<0$ they are $(r-a,r)$.

At the first occurrence of a positive $a$, the candidate is $(1,a+1)$. The occurrence of $a+1$ earlier would already force a cell of content $a$ in its row, so the neighbor condition must instead supply the content $a-1$. This supplies the left predecessor. The negative case is symmetric, supplying the upper predecessor, and the first zero gives $(1,1)$.

At any later occurrence of $a$, take the next position on its diagonal. The preceding occurrence already supplied all but the next necessary predecessor on each neighboring diagonal. The repeated-entry condition supplies both $a-1$ and $a+1$ after that preceding occurrence, so those next predecessors are present. More explicitly, for $a>0$ and the $r$th occurrence, the left predecessor is the $r$th cell of content $a-1$, and the upper predecessor is the $(r-1)$st cell of content $a+1$. For $a=0$ both are the $(r-1)$st cells on their respective diagonals; for $a<0$ the roles are reversed. These are exactly the predecessors supplied by the two new neighboring occurrences.

The new cell is therefore an addable node of a Young diagram. Induction produces a partition of an integer at every stage, and filling the added cell by its insertion time gives a standard Young tableau. No choice was possible because addable nodes of a Young diagram have distinct contents. Thusand the construction also proves uniqueness of the tableau with a specified vector.

The local spectral interchanges preserve an irreducible representation, and standard Young tableaux of a fixed shape are connected by admissible adjacent interchanges. Different shapes have different multisets of Young-diagram cell contents. Their elementary symmetric functions in the Young–Jucys–Murphy elements are central: the identityproves this coefficient by coefficient. By the Schur lemma, one irreducible representation cannot therefore contain two different shape classes. The multiset determines a diagram because the counts on its positive and negative diagonals determine its arm and leg lengths at the diagonal cells.

Different irreducible representations cannot share a joint eigenvalue vector: the Gelfand–Tsetlin algebra is generated by the Young–Jucys–Murphy elements, so all its elements would act identically on that vector in both representations, whereas a central primitive idempotent distinguishes their two blocks. Thus distinct irreducibles give distinct shape classes.

The number of irreducible representations is the number $p(n)$ of conjugacy classes, each indexed by a partition of an integer. Every spectral vector gives a tableau by the preceding construction. Closure under admissible interchanges makes every shape class that occurs occur in full. The two counts $p(n)$ then force every shape to occur, with exactly one irreducible representation per shape. This supplies $\mathrm{Spec}(n)=\mathrm{Cont}(n)$.

On restricting to $S_{n-1}$, delete the last coordinate. For tableaux of shape $\lambda$, the entry $n$ occupies one Removable node of a Young diagram. Grouping by that node gives one copy of the corresponding module; the path decomposition has simple multiplicities. Hence the restriction branching rule for a symmetric group isThe different removable nodes produce distinct partitions. The multiplicity-free structural input defining the Gelfand–Tsetlin basis is distinct from identifying this graph with the graph of Young diagrams.

Use a Gelfand–Tsetlin basis in Young seminormal form.

For a standard Young tableau $T$, put $d=c_T(i+1)-c_T(i)$, using the Content of a Young-diagram cell. Choose the row-reading tableau $T_0$, and let $\ell (T)$ be the Coxeter length of the unique permutation sending $T_0$ to $T$. A Gelfand–Tsetlin basis can be chosen so that, when $R=s_{i}T$ is standard and $\ell (R)=\ell (T)+1$,If $R$ is not standard, the action is $+v_T$ for two consecutive entries in one row, and $-v_T$ for two in one column. This is one usual normalization of the Young seminormal form.

Here is a construction and proof of the normalization. Fix $v_{T_0}\ne 0$, let $P_T$ be the projection onto the tableau line, and defineThe permutation ${\pi }_T$ has a reduced expression consisting entirely of admissible swaps, by the reduced adjacent-swap path between linear extensions. At each swap the off-diagonal coefficient is nonzero. In its expansion, the only term that can reach a tableau at distance $\ell (T)$ uses all $\ell (T)$ swaps; omitting a swap gives a shorter path. Thus $v_T\ne 0$. This also makes its definition independent of a chosen reduced expression, because ${\pi }_T$ itself is fixed.

The relation $s_i{X}_i{s}_i+s_i=X_{i+1}$ forces the coefficient of $v_T$ in $s_i{v}_T$ to be $1/d$, and forces every other component to lie on the swapped line. If length increases, project the identity ${\pi }_R=s_i{\pi }_T$ onto the line for $R$. The shorter terms of ${\pi }_T{v}_{T_0}$ cannot reach $R$ in one step, so the coefficient of $v_R$ is exactly one. Applying $s_i^2=1$ then gives the reverse coefficient $1-d^{-2}$ and diagonal coefficient $-1/d$. In the nonstandard cases the same relation gives $d=\pm 1$ and the asserted scalar action. This proves the theorem rather than merely specifying pairwise scalings.

Assume $n\ge 2$. The place-permutation action on ${\mathbb{C}}^n$ preserves the coefficient-sum kernelIt is the augmentation subrepresentation of a permutation representation. The point action of $S_n$ is two-transitive, so the irreducible augmentation criterion for a transitive group action makes $W$ irreducible. The two-row Young permutation module decomposition of $M^{(n-1,1)}\cong {\mathbb{C}}^n$ identifies its nontrivial summand as $V^{(n-1,1)}$. Hence $W\cong V^{(n-1,1)}$.

All standard Young tableaux of shape $(n-1,1)$ are $T_j$, $2\le j\le n$, with $j$ below the first cell and the remaining entries increasing along the first row. Their content vectors of standard Young tableaux areAn explicit orthonormal basis realizing these tableau lines isThe sums of their coordinates vanish. Their norms are one, and the inner product of $w_j$ with $w_l$, $j<l$, is zero because the coefficients of $w_j$ sum to zero. Directly summing the action of $(ar)$ gives $X_r{w}_j=c_{T_j}(r)w_j$.

For $s_i=(ii+1)$, $s_1{w}_2=-w_2$. For $2\le i\le n-1$, its only nontrivial two-dimensional block isEvery other $w_j$ is fixed, including all $w_j$ with $j>2$ when $i=1$. These formulas follow by swapping coordinates $i,i+1$ in the displayed vectors, and are the Young orthogonal form with axial distance $i$ for $T_i$.

A hook partition $\lambda =(n-k,1^k)$ has cell contentsIn a standard Young tableau, the unique zero-content cell is $(1,1)$ and contains $1$. The other eigenvalues are precisely the remaining contents, in whatever order the tableau supplies. Their product is independent of that order:The scalar is nonzero over the complex numbers. For the single row or single column, one of the factorials is $0!=1$; the formula still gives $(n-1)!$ or $(-1{)}^{n-1}(n-1)!$, respectively.

The Gelfand–Tsetlin basis spans $V^{\lambda }$, and the scalar computed on its vectors depends only on the shape. Thusfor every $w\in V^{\lambda }$. The product is the sum of the $(n-1)!$ permutations in the conjugacy class of an $n$-cycle. Taking traces therefore gives $(n-1)!{\chi }^{\lambda }((n))$ equal to the displayed scalar times $\dim V^{\lambda }$.

For a hook partition, a standard Young tableau is uniquely determined by the choice of its $k$ entries below the top cell, selected from $\{2,\dots ,n\}$. The column and the remaining row are then forced to increase. Hence $\dim V^{(n-k,1^k)}=\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)$, and cancellation of the factorials yieldsThis uses the central character value of a conjugacy-class sum and tableau counting, without a character rule for removing rim hooks.

The single-column Young diagram has exactly one standard Young tableau, so $V^{(1^n)}$ has dimension one. Consecutive labels are in one column, and the Young seminormal form gives $s_{i}w=-w$ for every adjacent transposition $s_i$. These generators determine the representation, and the sign representation sends every one of them to $-1$. ConsequentlyAt $n=1$ there are no adjacent generators and both modules are the trivial representation.

The conjugate partition is ${\lambda }_j^{\prime }=\#\{i:{\lambda }_i\ge j\}$, obtained by transposing the Young diagram. Transposition sends a standard Young tableau $T$ to a standard tableau $T^{\prime }$ and negates every Content of a Young-diagram cell.

Use the Young orthogonal form. For an admissible pair $R=s_{i}T$ and axial distance $d=c_T(i+1)-c_T(i)$, the action on $(w_T,w_R)$ isIn the transposed pair it is $A_{-d}$, whereas in the tensor product of group representations with the sign representation it is $-A_d$. Their diagonal entries agree, and their off-diagonal entries differ by a sign. Fix a reference tableau and let ${ϵ}_T$ be the sign of the unique label permutation from it to $T$. For every admissible swap ${ϵ}_R=-{ϵ}_T$, sointertwines these two actions. In a nonadmissible pair, transposition exchanges row and column and changes the scalar from $+1$ to $-1$ or conversely, also agreeing with the sign twist. Since adjacent transpositions generate $S_n$, this is an isomorphism:The parity choice is globally well defined because the label permutation is unique; no arbitrary edge-by-edge phase choices are required.

Insert the successive letters of a permutation by row insertion to obtain an insertion tableau $P$, recording the insertion times in the new boxes of $Q$. This is a bijection between permutations and pairs of standard Young tableaux of a common shape. Reverse insertion removes the box marked by the latest recording time. The first row of $P$ has length equal to the longest increasing subsequence.

The polytabloids $e_t$ indexed by standard Young tableaux of shape $\lambda$ form a basis of $S^{\lambda }$ over every field, and a basis of its integral lattice over $\mathbb{Z}$. Column antisymmetry and Garnir relations straighten any polytabloid into this set. Ordering leading tabloids by the rows containing successive entries makes the corresponding coefficient matrix triangular with diagonal entries one, proving independence. In particular the dimension is the standard-tableau count in every characteristic.

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.

Let $T$ be a standard Young tableau, $s_i=(ii+1)$, and $d=c_T(i+1)-c_T(i)$. A suitable Gelfand–Tsetlin basis has, for an admissible pair $R=s_{i}T$ with tableau length increasing,For a nonadmissible swap the scalar is $+1$ in a row and $-1$ in a column. The diagonal coefficient follows from the Young–Jucys–Murphy element relation $s_i{X}_i{s}_i+s_i=X_{i+1}$, and $s_i^2=1$ forces the product of off-diagonal coefficients. One global normalization is $v_T=P_T{\pi }_T{v}_{T_0}$ from the row-reading tableau: a reduced admissible path makes this vector nonzero and gives coefficient one on every length-increasing edge.

The Young lattice is a partially ordered set whose elements are integer partitions, including the empty partition. Each partition of an integer is identified with its diagram, and the order is inclusion of Young diagrams. A cover adds exactly one box. A saturated path from the empty diagram to $\lambda$ is equivalent to a standard Young tableau of shape $\lambda$.

A Young tableau is a filling of a Young diagram with entries, typically positive integers. The shape is the partition of an integer giving its row lengths. We draw Young tableaux in the English convention of Young diagram, with left-aligned rows arranged from top to bottom. A filling need not obey increasing conditions unless these are specified: standard Young tableaux use each of $1,\dots ,n$ exactly once and strictly increase along rows and down columns; Semistandard Young tableaux permit repetitions and require weakly increasing rows and strictly increasing columns.

In the constructions for a symmetric group involving Young symmetrizers, tabloids, and polytabloids, an unqualified numbered Young tableau means a bijective filling by $1,\dots ,n$, with no increasing condition imposed. This narrower convention is compatible with the broader use needed for Semistandard Young tableaux. Historically, Young's work connected these fillings to representation theory of the symmetric group; in particular, the equality between standard Young tableau counts and irreducible representation degrees over the complex numbers predates the modern hook-length formula. See Stanley's Algebraic Combinatorics, historical notes to Chapter 8.