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Embed SU(2) into SU(3) by $U\mapsto \mathrm{diag}(U,1)$. Its defining representation decomposes as an irreducible doublet and a trivial representation. The Schur lemma forces every commuting matrix to be $\mathrm{diag}(z{I}_2,w)$. Unitarity and unit determinant imply $|z|=1$ and $w=z^{-2}$, proving the formula. With Hermitian generator $X=\mathrm{diag}(1,1,-2)$, the defining representation restricts as $2_1\oplus 1_{-2}$ and the Adjoint representation as $3_0\oplus 2_3\oplus 2_{-3}\oplus 1_0$. The U(1) charges in the Adjoint representation are differences of defining charges. The generated connected subgroup is $(SU(2)\times U(1))/{\mathbb{Z}}_2$, since the two factors meet in $\mathrm{diag}(-I_2,1)$.

For a finite group, the regular representation contains $d_{\rho }$ copies of each irreducible representation $\rho$. Its character of a representation is $|G|$ at the identity and zero elsewhere. Thus ${\sum }_{\rho }{d}_{\rho }{\chi }_{\rho }(x)=|G|1_{\{x=e\}}$, including the trivial representation. This formula converts independent uniform expectations into constrained configuration counts.

Use the standard polytabloids for tableaux with rows $12\mid 34$ and $13\mid 24$, as an ordered basis $e_1,e_2$. Adjacent transpositions act byIn characteristic three their common fixed line is $F(e_1+e_2)$. On its quotient every transposition acts by minus one, giving the sign representation. Thus the displayed sequence has the trivial representation and sign as its composition factors, each once. It is nonsplit: the minus-one eigenspace of $\rho (12)$ is $F{e}_2$, which is not preserved by $\rho (23)$.

The natural tabloid form restricts to Gram matrix $\left(\begin{array}{cc}4& 2\\ 2& 4\end{array}\right)$. Its radical in characteristic three is exactly that fixed line. Therefore $D^{(2,2)}$ is the sign quotient, while the trivial factor is $D^{(4)}$. This gives an explicit small failure of semisimplicity in modular symmetric-group representation theory.

A colour singlet transforms in the trivial representation of the SU(3) colour group. Global colour rotations leave it invariant, and its colour generators annihilate it. A three-quark colour singlet uses the completely antisymmetric colour tensor; a quark–antiquark singlet contracts their colour indices.

Four quark flavours in the defining representation of flavour SU(4) and their antiquarks give sixteen meson flavour states. The trace of $M^i{}_j=q^i{\bar{q}}_j$ is the trivial representation; its traceless part is the Adjoint representation. Under light-flavour SU(3), the full content is $8\oplus 3\oplus \overline{3}\oplus 21$. Relative to the light-flavour nonet, the new states are $u\bar{c},d\bar{c},s\bar{c},c\bar{u},c\bar{d},c\bar{s},c\bar{c}$. Inside the traceless fifteen-dimensional multiplet the additional singlet is instead $(u\bar{u}+d\bar{d}+s\bar{s}-3c\bar{c})/\sqrt{12}$. The pure $c\bar{c}$ state is a linear combination of this state and the full SU(4) singlet. The classification concerns flavour; it does not imply degenerate physical masses or fix spin.

The trivial representation of a molecular point group: every symmetry operation is represented by $1$. Labels such as $A_1$ or $A_{1g}$ depend on the group and its labelling convention.

A field is a gauge singlet if it transforms in the trivial representation of the entire gauge group: every gauge transformation leaves it unchanged. A singlet of only one factor need not be a gauge singlet of the product group. For example a right-handed neutrino with representation $(1,1{)}_0$ is a Standard Model gauge singlet, whereas a color-singlet charged lepton is still acted on by the electroweak group. A singlet can have ordinary derivatives in its kinetic term and gauge-invariant masses, subject to Lorentz invariance and any additional global symmetries.

A group representation is a group homomorphism from a group to the general linear group of a vector space. It realizes abstract group elements as invertible linear transformations.

The defining identities are $\rho (e)=I$ and $\rho (gh)=\rho (g)\rho (h)$, so an inverse element is represented by a matrix inverse. A choice of basis gives a matrix representation, but the underlying action does not depend on that choice. The action need not distinguish every group element: the trivial representation sends all of them to $I$.

Set $\epsilon =i{\tau }_2$, whose Pauli matrix identity is $\epsilon {\tau }_i^{\ast }=-{\tau }_i\epsilon$. Exponentiation gives $\epsilon U^{\ast }=U\epsilon$ for $U\in SU(2)$. Hence complex conjugating a Higgs doublet and multiplying by $\epsilon$ restores the fundamental representation of $SU(2)$, while reversing hypercharge. A Higgs field of hypercharge $1/2$ yields a conjugate doublet of hypercharge $-1/2$. Its product ${\tilde{\varphi }}^{†}L$ with a lepton doublet $L$ of hypercharge $-1/2$ transforms in the trivial representation of the whole electroweak gauge group. At vacuum $(0,v{)}^T/\sqrt{2}$, this composite is $v{\nu }_L/\sqrt{2}$.

For any partition of an integer $\lambda ⊢n$, the number $f^{\lambda }$ of standard Young tableaux of shape $\lambda$ isThe numerator is a factorial, and the denominator is the hook product of a partition, using every cell's hook length once. The empty shape has one empty standard Young tableau, and the formula still holds with $0!=1$ and an empty product equal to one. By the standard polytabloid basis, $f^{\lambda }$ is also the dimension of the Specht module $S^{\lambda }$ over the complex numbers, or equivalently the degree ${\chi }^{\lambda }(1)$ of its irreducible character. For a teaching derivation of this dimension formula, see Etingof et al., Introduction to Representation Theory, §5.17, Theorem 5.17.1.

In the Young diagram of $(3,2)$, the hook length array $[[4,3,1],[2,1]]$ givesThis agrees with the five explicit fillings in standard Young tableau. Its conjugate partition $(2,2,1)$ has the same hook lengths after transposition, hence also gives dimension five. At the extremes, for $n\ge 1$, a single row $(n)$ and a single column $(1^n)$ each have hook product $n!$ and exactly one standard Young tableau; their Specht modules over the complex numbers are the trivial representation and sign representation, respectively.

Why the formula is beautiful: global compatibility of all the row and column inequalities reduces to a product of local hook lengths. It is useful when enumeration is already expensive: computing one hook length per cell replaces testing up to $n!$ numbered Young tableaux. For $(3,2)$, the proportion of the $120$ bijective fillings that are standard Young tableaux is $5/120=1/24$, the reciprocal of the hook product. This product does not justify treating overlapping hook conditions as independent; Grinberg explains that tempting but invalid argument in §1.3 of his teaching slides.

Historically, the modern hook-product statement was published by J. S. Frame, G. de B. Robinson and R. M. Thrall in 1954: ^[ref]. Their manuscript was received on 1 September 1953, and their proof reorganizes an earlier irreducible character degree formula into the product of the hook lengths. The theorem unifies the counting and representation viewpoints developed earlier by Young. For proof routes within this wiki, see trace computation of Specht module dimension, determinant formula for Specht module dimension, and Greene–Nijenhuis–Wilf hook walk.

A horizontal strip is a skew Young diagram containing at most one cell in each column. Equivalently, $\lambda /\mu$ is a horizontal strip when ${\lambda }_i\ge {\mu }_i\ge {\lambda }_{i+1}$ for all $i$. The multiplicity of the trivial representation in the associated skew representation of a symmetric group is one for horizontal strips and zero otherwise. A column with two cells supplies a cyclic tableau vector on which an adjacent transposition acts by $-1$, excluding invariants.

With $s=(12)$, the idempotents $e_{\pm }=(1\pm s)/2$ split the right regular module into two three-dimensional projective modules. Their tops are respectively the trivial representation and sign representation. The nilpotent ideal $J(k{S}_3)$ ensures that any nonzero direct summand has nonzero top, so the one-dimensional tops make these projective modules indecomposable. Their successive radical series of a module factors are trivial, sign, trivial and sign, trivial, sign. This decomposition concerns right modules; the block of S3 in characteristic three remains a single algebra.

For a subgroup $H\le G$ and a left group algebra $k[H]$-module $V$, its induced representation is ${\mathrm{Ind}}_H^{G}V=k[G]{\otimes }_{k[H]}V$, with $G$ acting by left multiplication on the first tensor factor. If $H$ has finite index of a subgroup and $V$ is a finite-dimensional vector space, the dimension of the induced representation is $[G:H]\dim V$.

Inducing the trivial representation gives the permutation representation on cosets $G/H$. Frobenius reciprocity characterizes the construction: intertwining operators from the induced representation to $W$ correspond to $H$-intertwining operators from $V$ to the restriction of a representation $W$. Thus induction of representations extends a smaller group action while imposing only the identifications forced by that subgroup.

Mackey theory analyzes induced representations by combining restriction of a representation, conjugate subgroup representations, and double cosets. For subgroups $H,K\le G$ of a finite group, the restriction of a representation induced from $H$ to $K$ decomposes into terms indexed by the double cosets $K\backslash G/H$.

A double coset representative $g$ contributes an induced representation from the subgroup $K\cap gH{g}^{-1}$, using the appropriate conjugate subgroup representation and its restriction of a representation. For the trivial representation of $H$, this is the decomposition of the coset permutation representation into its $K$-orbits. The intersection subgroups are stabilizers of those orbits, explaining geometrically why double cosets occur. This framework computes spaces of intertwining operators and tests whether induced representations are irreducible representations.

A mean-zero function has expectation zero with respect to the specified probability measure. On a finite set with uniform measure this means ${\sum }_{x}f(x)=0$. A balanced subset indicator is an example. In Fourier analysis on a finite group, this is equivalent to vanishing of the component at the trivial representation.

A one-dimensional complex group representation acts on a one-dimensional complex vector space, and is equivalently a group homomorphism into the multiplicative group ${\mathbb{C}}^{\times }$. It is automatically an irreducible representation. Its character of a representation is its scalar value, a one-dimensional character. A change of basis cannot change those scalar values, so two such group representations are equivalent exactly when their homomorphisms agree. For a symmetric group $S_n$ with $n\ge 2$, all transpositions are conjugate, square to the identity and generate the group. Their common scalar image is therefore $1$ or $-1$, giving exactly the trivial representation and the sign representation.

The additive cyclic group ${\mathbb{Z}}_n$ is generated by $1$, with $n\cdot 1=0$. Every complex irreducible representation is one-dimensional by a. If $R(1)=z$, the defining relation requires $z^n=1$. Conversely each such root of unity defines a group homomorphism. Hence the complete list isChanging the representative of $g$ changes the exponent by an integer multiple of $2\pi i$, so the formula is well-defined. These representations are pairwise inequivalent because their values on the generator differ. The case $k=0$ is the trivial representation.

The matrix $r$ has distinct eigenvalues $\alpha ,{\alpha }^2$. Any one-dimensional invariant subspace for the defining representation would therefore have to be one of its two coordinate eigenspaces. But $s$ exchanges those lines, so neither is invariant under the whole group. Hence $R_2$ is an irreducible representation over the complex numbers.

There are three complex irreducible representations up to equivalence of group representations, because there are three conjugacy classes. The trivial representation and the sign representation already give two inequivalent one-dimensional representations. Together with $R_2$, their dimensions satisfy the sum of squares of irreducible representation dimensions:Thus there are no further irreducible representations. Alternatively the character of $R_2$ is $(2,-1,0)$, and its character inner product with itself is $(4+2)/6=1$, independently confirming irreducibility.

The irreducible representations can all be specified on the generators:The first is the trivial representation and the second is the sign representation. To see directly that these exhaust the one-dimensional representations, their scalar images commute, so the relation $srs=r^{-1}$ forces $r=r^{-1}$. Combining $r^2=1$ with $r^3=1$ gives $r=1$, and then $s$ may be either sign.

Taking traces gives the character table, in the conjugacy class order of solution:Here $\alpha +{\alpha }^2=-1$ gives the middle entry in the last row. The weighted character inner products of distinct rows vanish, and the norm of each row is one. Together with solution, this proves completeness of the table.

The first coordinate is fixed by the embedded SU(5), while the last five transform in its defining representation. Therefore $6{\downarrow }_{SU(5)}=1\oplus 5$, and taking the dual representation gives $\overline{6}{\downarrow }_{SU(5)}=1\oplus \overline{5}$.

For the symmetric square, the fixed-fixed component is one trivial representation, the fixed-five mixed components give $5$, and the symmetric five-five block is $15$. For the exterior square, the fixed-fixed component vanishes, the mixed components give $5$, and the antisymmetric five-five block is $10$. Finally restrict $35=6\otimes \overline{6}-1$, using $5\otimes \overline{5}=24\oplus 1$.

The five SU6 tensor branchings to SU5 are thusThe symbols $15$ on the left and right refer to different groups: the left-hand one is the SU(6) exterior square, whereas the right-hand one is the SU(5) symmetric square. The dimension checks are $6=5+1$, $21=15+5+1$, $15=10+5$ and $35=24+5+5+1$.

Explicitly, in the Adjoint representation the lower-right traceless five-by-five block gives $24$; the lower-left column and upper-right row give $5$ and $\overline{5}$; and the diagonal direction $\mathrm{diag}(-5,1,1,1,1,1)$ gives the remaining trivial representation.