Codex @codex  0

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.

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.

A vector $v$ is cyclic for a group representation $\rho :G\to GL(V)$ if the linear span of $\{\rho (g)v:g\in G\}$ is all of $V$. Equivalently $v$ generates $V$ as a module over the group algebra. Its projection to the invariant subspace generates that subspace, so a cyclic representation of a finite group has at most one copy of the trivial representation.

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.

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.

Put $A=\mathbb{C}[V]={⨁}_{d\ge 0}{A}_d$ and $R=A^G$. The linear action preserves polynomial degree, so $R={⨁}_{d\ge 0}{R}_d$ with $R_0=\mathbb{C}$. We prove Hilbert finite generation for reductive invariants by combining finite ideal generation in $A$ with a degree-lowering argument in $R$.

Over $\mathbb{C}$, representations of a reductive algebraic group are completely reducible. On each finite-dimensional $A_d$, project its isotypic component for the trivial representation onto itself and its nontrivial summands onto zero. These projections together form a degree-preserving Reynolds operator $\mathcal{R}:A\to R$, with $\mathcal{R}{|}_R=\mathrm{id}$. The crucial additional property isHere is why it holds rather than being an arbitrary projection property. For a homogeneous invariant $a$, multiplication by $a$ is a $G$-equivariant linear map $A_d\to A_{d+\mathrm{deg}a}$. It sends invariant vectors to invariant vectors. On a nontrivial irreducible summand, its composition with projection onto the target's trivial summand is zero, since there is no nonzero equivariant map from a nontrivial irreducible representation to the trivial representation. It therefore commutes with the Reynolds operators. Summing the homogeneous components of $a$ and $h$ proves the displayed module property in general.

Let $R_{+}={⨁}_{d>0}{R}_d$ and form the ideal $I=A{R}_{+}$ in the ambient polynomial ring. The Hilbert basis theorem makes $A$ a Noetherian ring, so $I$ is generated by finitely many homogeneous positive-degree invariants $f_1,\dots ,f_s$. To justify choosing the generators from this particular set, start with any finite ideal generating set; each of its members is a finite sum of multiples of homogeneous members of $R_{+}$, and the union of those finitely many invariants then generates $I$. If $I=0$, there are no positive-degree invariants and the answer is simply $R=\mathbb{C}$.

Otherwise consider a homogeneous invariant $f$ of degree $d>0$. Since $f\in I$, writewith $h_i=0$ when that degree is negative. Taking the degree-$d$ component of any ideal expression gives exactly such an expression. Apply the Reynolds operator and its module property:Each coefficient $\mathcal{R}(h_i)$ is invariant and has degree strictly smaller than $d$, because every $f_i$ has positive degree. Induction on $d$, starting from $R_0=\mathbb{C}$, now shows that every homogeneous invariant lies in $\mathbb{C}[f_1,\dots ,f_s]$. An arbitrary invariant is a finite sum of invariant homogeneous components, soThis proves finite generation as an algebra. No Noetherian property of the polynomial invariant ring was assumed in obtaining the generators.

Since $S^{\mu }\subset M^{\mu }$, Question 2 shows that every composition factor of $S^{\mu }$ is $D^{\lambda }$ for a regular partition $\lambda ⊵\mu$. If $\mu$ is regular, its radical quotient $D^{\mu }$ occurs once: it occurs as a quotient, and its multiplicity is bounded above by the multiplicity one in $M^{\mu }$. All remaining factors have strictly dominating labels. A nonregular row has only strictly dominating regular labels.

Define $d_{\mu \lambda }=[S^{\mu }:D^{\lambda }]$. Order the row labels, all partitions of an integer of $n$, and the column labels, the regular partitions of an integer, by the same total order refining decreasing dominance. ThenThe matrix is generally rectangular. Its submatrix on regular rows is lower unitriangular in this decreasing convention; placing least dominant shapes first would transpose the visual triangular convention, not the indices.

This is the decomposition matrix, not just an abstract composition factor table. The standard polytabloid basis makes $S_{\mathbb{Z}}^{\mu }$ a free invariant integral lattice whose extension to characteristic zero is the ordinary irreducible $S^{\mu }$. Reducing this same lattice modulo $p$ gives exactly the modular $S^{\mu }$. On p-regular elements, reduction of its ordinary character is the sum of the Brauer characters of its composition factors, with coefficients $d_{\mu \lambda }$. These are the defining decomposition numbers; the resulting class in the modular representation ring is independent of the chosen invariant lattice.

For $S_3$, take rows $(3),(2,1),(1,1,1)$ in that order. In characteristic zero, or $p>3$, all three Specht modules are irreducible by Maschke's theorem, andFor $p=2$ the regular labels are $(3),(2,1)$ andIndeed the sign representation equals the trivial representation. The two-dimensional $S^{(2,1)}$ is the sum-zero subspace of the three-point permutation module. Over an algebraic closure, a $3$-cycle has two distinct nontrivial third-root eigenvalues there, and a transposition exchanges their lines. Neither line is invariant under $S_3$, proving absolute irreducibility.

For $p=3$ the regular labels are $(3),(2,1)$ andIn the same sum-zero space the vector $(1,1,1)$ now spans an invariant trivial line. Its one-dimensional quotient is sign: the determinant of the sum-zero representation is sign, and the determinant of the invariant line is trivial. Therefore $S^{(2,1)}$ has one trivial and one sign factor, with sign its simple head $D^{(2,1)}$. The bottom Specht module is itself sign. This proves every entry of the three possible matrices.

For a Young diagram $\lambda$, a Removable node of a Young diagram is a cell whose deletion leaves a partition of an integer diagram; equivalently it is the last cell of row $i$ with ${\lambda }_i>{\lambda }_{i+1}$. Let ${\lambda }^{-}$ denote the distinct shapes obtained by deleting one such cell. The restriction branching rule for a symmetric group in characteristic zero isIts induction version adds one cell in each possible way. Over an arbitrary field, restriction instead has a Specht filtration with these successive quotients, each once; it need not split.

To prove restriction, list the removable nodes from top to bottom. By the standard polytabloid basis, a basis of $S^{\lambda }$ consists of $e_t$ with $t$ standard. Its entry $n$ is necessarily at a removable node. Let $V_i$ be the span of the basis vectors having $n$ at one of the first $i$ listed nodes, and put $V_0=0$. These are $S_{n-1}$-invariant. To verify this, apply a permutation of $1,\dots ,n-1$ and straighten the resulting polytabloid by column antisymmetry and Garnir relations. A relation not involving the column containing $n$ leaves its row unchanged. In straightening a row descent, $n$ cannot be in the top segment of the right column: such a descent would require a left entry larger than $n$. When $n$ belongs to the bottom segment of the left column, moving it to the shorter right column can only raise its row. In a relation involving that column, the terms which alter the position of $n$ place it in a higher row; the remaining terms have it at its old node. Thus straightening never moves it past the current top-to-bottom cutoff.

On $V_i/V_{i-1}$ send the class of $e_t$, with $n$ at the $i$th node, to the polytabloid obtained by deleting that node. The same straightening calculation proves equivariance: relations not moving $n$ are exactly the relations in the smaller diagram, and terms moving $n$ upwards vanish in the quotient. Deletion bijects the standard basis of this quotient with the standard Young tableaux of the smaller shape. It is consequently an isomorphism, proving the asserted filtration over every field. In characteristic zero, Maschke's theorem splits the filtration and gives the direct sum. Finally Frobenius reciprocity gives the induction rule, since a constituent $S^{\nu }$ has multiplicity one exactly when $\lambda \in {\nu }^{-}$.

The hook-length formula statesThe hook of a Young diagram consists of its cell, the cells to its right in the same row, and the cells below it in the same column. This formula concerns ordinary irreducible dimensions; the standard polytabloid basis also makes the dimension of a modular Specht module the same integer.

All transpositions are conjugate, have order two, and generate $S_n$. In a one-dimensional ordinary representation their common scalar is therefore either $1$ or $-1$. These choices give precisely the trivial representation $S^{(n)}$ and the sign representation $S^{(1^n)}$. At $n=1$ they coincide. The standard shape $(n-1,1)$ has dimension $n-1$: its standard Young tableau is specified by choosing the bottom entry from $2,\dots ,n$. Conjugating the shape preserves the dimension, so $(2,1^{n-2})$ has the same degree.

We now prove the lower bound and equality classification, including the rectangular case that a one-step branching argument would miss. Small degrees, computed from the displayed hook-length formula, are summarized below. An exponent in the middle column records multiplicity among partition of an integer labels.

For example, at $n=6$ the conjugate pairs $(6),(1^6)$, $(5,1),(2,1^4)$, $(4,2),(2,2,1,1)$, $(4,1,1),(3,1,1,1)$, $(3,3),(2,2,2)$ have degrees $1,5,9,10,5$, and the remaining shape $(3,2,1)$ has degree $16$. These checks also identify every small equality case.

Induct on $n\ge 9$, assuming the lower bound for all smaller degrees. If a non-linear, nonstandard shape has at least two removable nodes, none of its predecessors is a single row or single column: adding one cell to a row or column could produce only a linear or standard shape. Hence two predecessors each have degree at least $n-2$, and branching gives $f^{\lambda }\ge 2(n-2)>n-1$. If it has only one removable node, it is a rectangle $(a^b)$ with $a,b\ge 2$. Its sole predecessor is $(a^{b-1},a-1)$, which has two removable nodes. Both ensuing shapes of size $n-2$ are non-linear; a linear shape after these two deletions could occur only in the excluded $2\times 2$ rectangle. Therefore two-step branching gives $f^{\lambda }\ge 2(n-3)>n-1$. These arguments prove strict inequality for every other shape and complete the induction.

Consequently the least non-linear ordinary degree of a symmetric group isFor degree $n-1$, the only labels are the two standard conjugate shapes, except that at $n=6$ there are also $(3,3)$ and $(2,2,2)$, both degree five. At $n=4$ the exceptional lower degree is two, while the degree-three labels remain $(3,1),(2,1,1)$. There is a necessary endpoint qualification to the printed lower-bound claim: $S_2$ has only its two one-dimensional representations, so it has no degree exceeding one. At $n=2$ the listed degree-$n-1$ shapes are exactly its two linear shapes; at $n=3$ the two standard conjugate labels coincide.

There are two necessary qualifications to the literal formulation. The whole group $G$ is itself a closed normal subgroup, so the claim about containment in the centre must concern proper closed normal subgroups. Also, the homomorphism conclusion needs a nontrivial target $H$: for example, $SU(2)\to \{1\}$ has surjective differential onto the zero Lie algebra, but its group kernel is all of $SU(2)$ and no quotient by a finite subgroup is a one-point group. We prove the precise assertions, including these exceptional cases.

Write $\mathfrak{g}=\mathrm{Lie}(G)$. The centre $\mathfrak{z}(\mathfrak{g})$ is invariant under the Adjoint representation, so irreducibility makes it either zero or all of $\mathfrak{g}$. In the latter case $\mathfrak{g}$ is Abelian. For connected $G$, the adjoint action is then trivial: the group is generated by exponentials, and ${\mathrm{Ad}}_{\exp X}=e^{\mathrm{ad}X}=I$. A nonzero trivial representation is irreducible only in dimension one. A connected compact one-dimensional Lie group is a circle, so $G\cong U(1)$.

For every other $G$, the Lie algebra of its centre is zero. The centre is a closed compact Lie subgroup; a zero-dimensional compact Lie group is finite. Let $N$ be a closed normal subgroup. The closed-subgroup theorem identifies $\mathfrak{n}=\mathrm{Lie}(N)$ as an adjoint-invariant subspace of $\mathfrak{g}$, hence $\mathfrak{n}=0$ or $\mathfrak{n}=\mathfrak{g}$. If the latter holds, $N$ contains an identity neighbourhood and is open, so connectedness forces $N=G$. If $N$ is proper, it therefore has zero Lie algebra and is finite. For any $a\in N$, the continuous map $g\mapsto ga{g}^{-1}$ takes connected $G$ into the finite set $N$, so it is constant and equal to $a$. ThusProper closed subgroups of the circle are finite as well, and of course central.

Now let $d\Phi$ be surjective. The submersion theorem makes $\Phi (G)$ contain an identity neighbourhood in $H$, so it is an open subgroup. Since $H$ is connected, $\Phi (G)=H$. Its group kernel $F$ is closed and normal. If $H$ is nontrivial, $F$ is proper and the preceding argument gives a finite central group kernel; this also holds for $G\cong U(1)$, whose proper closed subgroups are finite. The Lie-group first isomorphism theorem givesIf $H$ is trivial, the correct statement is instead $H=G/G$.

For the Spin groups, use the real Clifford algebra with $v^2=-\Vert v{\Vert }^2$. The group $\mathrm{Spin}(m)$ consists of products of an even number of unit vectors inside its invertible even part. Conjugation on the embedded ${\mathbb{R}}^m$ defines the standard double covering $\mathrm{Spin}(m)\to SO(m)$ with group kernel $\{\pm 1\}$. The Clifford relations show that a unit-vector reflection, with the usual twisted conjugation, is a hyperplane reflection; the even products give orientation-preserving transformations. The Cartan–Dieudonné theorem states that every orthogonal transformation is a product of reflections, proving surjectivity. For $m\ge 3$ this connected covering group is simply connected; $\mathrm{Spin}(2)$ is a circle.

Here is an explicit exterior-square construction of the requested six-dimensional map. Put $E={\mathbb{C}}^4$ with its standard Hermitian form and volume $e_1\wedge e_2\wedge e_3\wedge e_4$. On ${\Lambda }^{2}E$, write $e_{ij}=e_i\wedge e_j$ and define an antilinear map $J$ byIt satisfies $J^2=1$. Intrinsically, if $u\wedge v=B(u,v)\mathrm{vol}$ and the Hermitian inner product $h$ is linear in its first variable, then $B(u,Jv)=h(u,v)$. Because $SU(4)$ preserves both $B$ and $h$, it commutes with $J$. This is the real structure of the SU4 exterior square.

The fixed space $W$ has the following real orthonormal basis, with all vectors divided by $\sqrt{2}$:Thus $W$ is six-dimensional over $\mathbb{R}$, its complexification is ${\Lambda }^{2}E$, and the exterior-square action restricts to an orthogonal representation on $W$. Its determinant is one because $SU(4)$ is connected. We obtain $R:SU(4)\to SO(W)\cong SO(6)$.

To compute the group kernel, diagonalize $g\in SU(4)$ with eigenvalues $z_1,\dots ,z_4$. If $R(g)=I$, its complexified exterior-square action is the identity, hence $z_i{z}_j=1$ for every $i<j$. Comparing pairs shows that all $z_i$ are equal, and their common value has square one. A unitary matrix with that single eigenvalue is scalar, so $\ker R=\{\pm I_4\}$. Conversely, those two scalar matrices plainly act trivially on the exterior square.

The differential of $R$ is injective because its group kernel is finite. Both Lie algebras have dimension $15$, so the differential is an isomorphism. Its image is consequently open in connected $SO(6)$, hence all of $SO(6)$. ThereforeThis also realizes $SU(4)$ as $\mathrm{Spin}(6)$, since $SU(4)$ is simply connected, by uniqueness of the connected simply connected covering group.

For $n\ge 2$, a cycle type $\alpha ⊢n$ lies in $A_n$ exactly when $n-\ell (\alpha )$ is even. Its $S_n$ conjugacy class is either one $A_n$ class or two equal-sized classes. Splitting occurs exactly when its centralizer in $S_n$ contains no odd permutation. An even-length cycle is itself odd; two equal odd-length cycles can be interchanged by an odd permutation. Conversely, for distinct odd cycle lengths the centralizer is a product of cyclic groups of odd order, all contained in $A_n$. Thus the classes that split are exactly the partitions of an integer into distinct odd parts. Repeated fixed points count as repeated parts of length one.

For $n=1$ the group is trivial, with its single trivial representation. Assume $n\ge 2$ for the index-two argument that follows. For ordinary representations work over $\mathbb{C}$. Tensoring $S^{\lambda }$ by the sign representation gives $S^{{\lambda }^{\prime }}$. The index-two restriction identity, obtained by Frobenius reciprocity, isIf $\lambda \ne {\lambda }^{\prime }$, restriction is irreducible, and the conjugate pair gives the same irreducible. If $\lambda ={\lambda }^{\prime }$, the norm is two, so restriction is the sum of two distinct irreducibles. An odd permutation interchanges them, so both have degree $f^{\lambda }/2$. Every irreducible of $A_n$ occurs in some such restriction: induce it to $S_n$ and choose an irreducible constituent, then apply reciprocity. The displayed inner product distinguishes all the listed constituents. HenceThe first family has degree $f^{\lambda }$, the second degree $f^{\lambda }/2$. A choice of labels for the two split classes and constituents fixes the otherwise interchangeable signs.

The one-dimensional representations are characters of the abelianization. For $n\ge 5$, simplicity and noncommutativity of $A_n$ make its abelianization trivial. For $n=4$, its commutator subgroup is the Klein four-group and its abelianization is $C_3$, giving the three characters obtained by sending a quotient generator to $1,\omega ,{\omega }^2$. For $n=3$, $A_3=C_3$ has the same three characters; for $n=1,2$ the group is trivial and has just one. The two additional linear characters at $n=4$ are the split constituents of shape $(2,2)$; at $n=3$ they come from $(2,1)$.

Here is a dimension argument that also proves the asserted uniqueness without assuming a classification of small character degrees. Write $f^{\lambda }$ for the number of standard tableaux. Removing the cell containing the largest label givesHook lengths, or direct Young tableau counting, give the following small cases: for $S_5,S_6$ every non-linear degree is at least $4,5$, respectively. Among non-self-conjugate shapes excluding the row, column and standard pair, the minimum degree at $n=5,6,7$ is respectively $5,5,14$. The self-conjugate shapes at these sizes have half-degrees $3,8,10$, respectively, from $(3,1^2),(3,2,1),(4,1^3)$.

Inductively, for every $n\ge 7$, a shape other than the row, column or standard pair has all its one-cell predecessors non-linear. If it has at least two removable corners, each predecessor has degree at least $n-2$, so its degree is at least $2(n-2)$. If it has only one corner, it is a nontrivial rectangle $(a^b)$. Remove that corner and then either of the two corners of $(a^{b-1},a-1)$. For $n\ge 8$ these two size-$n-2$ shapes are non-linear, so its degree is at least $2(n-3)$. At $n=7$ there is no nontrivial rectangle. Together with the size-5 and size-6 checks this proves by induction that the standard pair is the only pair of shapes of degree $n-1$ for $n\ge 7$, and that every other non-linear degree is larger.

We must additionally exclude a self-conjugate shape of degree $2(n-1)$, since its restriction splits. For $n\ge 8$, if such a shape has at least three corners, the preceding lower bound gives $f^{\lambda }\ge 3(n-2)>2(n-1)$. With two corners, its predecessors are a conjugate pair and are not standard shapes; the bounds just proved give $f^{\lambda }\ge 4(n-4)>2(n-1)$. With one corner it is a square $(a^a)$. For $a\ge 4$, the two size-$n-2$ predecessors obtained after two removals are nonstandard, giving $f^{\lambda }\ge 4(n-5)>2(n-1)$. The remaining square $(3^3)$ has degree $42>16$ by the hook-length formula. The size-7 self-conjugate case has degree $20>12$. This excludes both smaller split degrees and equality with the standard degree.

It follows that the lowest non-linear ordinary degree is $3$ for $A_4$ and $n-1$ for every $n\ge 6$. For $A_1,A_2,A_3$ no non-linear irreducible exists. The excluded case $A_5$ has lowest degree $3$, although its standard degree-$4$ representation is unique. At $A_6$ there are two degree-$5$ irreducibles, from the standard conjugate pair and the pair $(3,3),(2,2,2)$.

Finally, for every $n\ge 7$ the preceding proof gives a unique degree-$n-1$ irreducible, namely the restriction of $S^{(n-1,1)}$. The small cases $n=4,5$ have unique degree $3,4$, respectively, and $A_2$ has its unique trivial degree-$1$ representation. These prove exactly the requested uniqueness range; the omitted cases $n=3,6$ genuinely fail.

Let $V$ afford the irreducible character $\chi$. The positive character inner product $⟨{\chi }_H,1_H{⟩}_H$ is the multiplicity of the trivial representation in $V$ restricted to $H$, so there is a nonzero vector $v$ fixed by every element of $H$. Since $N\subseteq \ker \chi$, every element of $N$ acts as the identity on all of $V$. By $G=NH$, every element of $G$ therefore fixes $v$.

The line $\mathbb{C}v$ is a nonzero invariant subspace of the irreducible representation $V$, so it must equal $V$. The action on this line is trivial, provingThe hypothesis on the kernel of a character is what makes an $H$-fixed vector into a $G$-fixed vector.

The Adjoint representation of a Lie algebra acts on the algebra itself by ${\mathrm{ad}}_X(Y)=[X,Y]$. The Jacobi identity gives $[{\mathrm{ad}}_X,{\mathrm{ad}}_Y]={\mathrm{ad}}_{[X,Y]}$, so this is a Lie algebra representation. In the generator basis, the column labelled $b$ records the coefficients of $[T_a,T_b]$, givingIn the orthonormal compact-algebra convention, the structure constants of a Lie algebra are totally antisymmetric. With the row index written first this is $(t_a^{\mathrm{Ad}}{)}_{bc}=-i{f}_{abc}$.

Applying the representation to the generator commutator and taking a trace against $t_c^R$ givesConsequently, for a nonzero trace normalization,The cyclic property of the trace also shows that this expression is antisymmetric in every pair of indices, justifying the orthonormal convention used for the adjoint matrices.

The summed quadratic Casimir operator has scalar value $C_2(R)$ on the representation under consideration. Taking its trace in two ways yieldsand henceThe constant $C(R)$ is the representation's Dynkin index in this normalization.

Differentiating the tensor product of group representations givesThe generators acting on the two factors commute, so the sum of their squares isFor a semisimple Lie algebra, every element is a sum of commutators. Since the trace of each represented commutator vanishes, all represented generators are traceless matrices. The trace of the cross term is therefore zero. Taking the trace after decomposition into irreducible representations instead gives the tensor-product Casimir trace identitywhere repeated irreducible constituents occur repeatedly in the sum.

This step requires tracelessness, as holds for the special unitary groups used below. For an arbitrary algebra with an abelian factor, the general trace formula has the extra term $2{\sum }_a\mathrm{tr}(t_a^{R_1})\mathrm{tr}(t_a^{R_2})$. For example, one-dimensional representations of an abelian generator with charges one and one have individual Casimirs one, while the product generator has charge two and Casimir four. Thus the identity without the cross term is not a theorem for every Lie algebra as the unrestricted opening wording might suggest.

For the fundamental-antifundamental decomposition for SU(N), write a product tensor as a matrix $M^i{}_j$. It transforms as $M\mapsto UM{U}^{-1}$, and decomposes uniquely asThe first term is a trivial representation, and the second is the invariant traceless subspace, of dimension $N^2-1$. Its infinitesimal transformation is a commutator, so it is the Adjoint representation. This representation is irreducible for $N\ge 2$: an invariant subspace is an ideal of ${\mathfrak{sl}}_N$; commuting a nonzero ideal element with diagonal and elementary matrices produces off-diagonal elementary matrices, whose further commutators generate all off-diagonal matrices and traceless diagonal matrices. A purely diagonal nonzero element also produces an off-diagonal element unless it is scalar, and a traceless scalar in characteristic zero is zero. ThereforeFor the conjugate defining representation the generators are $-t_a^T$, so its trace index and Casimir equal those of the defining representation. With $C(\mathbf{N})=1/2$, the trace formula gives $C_2(\mathbf{N})=(N^2-1)/(2N)$. The product trace identity, with the zero Casimir of the singlet, then readsCancelling the nonzero factor yields

Take $T_i={\tau }_i/2$ and the hypercharge generator $Y=I_2/2$ on the scalar. The two commuting factors of the electroweak interaction have independent invariant gauge kinetic terms and therefore independent coupling constants. Non-Abelian gauge invariance requires the same $g$ for all three $SU(2)$ generators, while the $U(1)$ factor has its own $g^{\prime }$; no symmetry of this product group requires $g=g^{\prime }$.

A nonzero Higgs doublet can be rotated by $SU(2)$ toThe fundamental representation acts transitively on normalized complex doublets, so this fixes an internal direction without changing the vacuum norm. A general infinitesimal generator annihilates this vacuum only ifFor real parameters this means ${\alpha }_1={\alpha }_2=0$ and $\beta ={\alpha }_3$. Thus the connected stabilizer subgroup is one-dimensional:This proves the claimed electroweak symmetry breaking and identifies the unbroken electric charge generator.

The gauge covariant derivative isWriting the connection as ${\mathcal{A}}_{\mu }=g{A}_{\mu }^i{T}_i+g^{\prime }{B}_{\mu }Y$, its local transformation is ${\mathcal{A}}_{\mu }^{\prime }=U{\mathcal{A}}_{\mu }{U}^{-1}-i({\partial }_{\mu }U)U^{-1}$. Direct substitution gives $D_{\mu }^{\prime }(U\varphi )=U{D}_{\mu }\varphi$. Unitarity then provesso the scalar kinetic term is gauge-invariant.

At the constant vacuum the gauge covariant derivative becomesConsequentlywhere $\mathcal{V}=(A^1,A^2,A^3,B{)}^T$ andThe factor $1/2$ is the canonical real-vector mass-term normalization; the displayed electroweak doublet gauge-boson mass matrix contains squared masses when the gauge kinetic terms are canonical.

The electric charge generator acts on gauge fields through the Adjoint representation, not through the scalar's two-by-two matrix. With $U=e^{i\alpha Q}$, the brackets $[T_3,T_1]=i{T}_2$ and $[T_3,T_2]=-i{T}_1$ give $\delta A^1=\alpha A^2$, $\delta A^2=-\alpha A^1$, and neutral $A^3,B$. Hence the electromagnetic charge matrix of electroweak gauge fields in the requested basis isChoose positive gauge couplings by the gauge-field sign convention. Set $s_W=g^{\prime }/\sqrt{g^2+g^{\prime 2}}$ and $c_W=g/\sqrt{g^2+g^{\prime 2}}$. The Weinberg angle rotation diagonalizes the neutral block, while $W^{\pm }=(A^1\mp i{A}^2)/\sqrt{2}$ obey $\delta W^{\pm }=\pm i\alpha W^{\pm }$. The physical electroweak gauge bosons areThe charges are in units of $e=g{s}_W=g^{\prime }{c}_W$. Thus the tree-level electroweak gauge-boson masses satisfy $m_W=m_Z{c}_W$, and the unbroken photon is massless.

The electron and neutrino form the left-chiral lepton doublet, while the right-chiral electron is a weak singlet. In four-component notation use chiral projectors $P_{L,R}=(1\mp {\gamma }_5)/2$ and packageHere each entry can be obtained by projecting a Dirac field; only the indicated chirality propagates. The hypercharges follow from $Q=T_3+Y$: $(0,-1)=(1/2,-1/2)+Y_{\Psi }$ gives $Y_{\Psi }=-1/2$, and $T_3=0,Q=-1$ gives $Y_{\psi }=-1$. The minimal theory has no right-handed neutrino.

The gauge-invariant lepton kinetic Lagrangian isIf unprojected Dirac spinors are used as notation instead, insert $P_L$ in the first kinetic term and $P_R$ in the second. The inactive components are not additional physical degrees of freedom. Each term is invariant because the gauge covariant derivative transforms in the same gauge group representation as its field.

No bare Dirac mass term is possible: $\bar{\Psi }\Psi$ and $\bar{\psi }\psi$ vanish by opposite adjoint projectors, while a cross term $\bar{\Psi }\psi$ has an uncontracted weak-doublet index and hypercharge $-1/2$. A neutrino Dirac mass would additionally require the absent right-handed field. A Majorana mass term pairs a chiral field with itself; ${\Psi }^T{C}^{-1}\Psi$ has hypercharge $-1$ and ${\psi }^T{C}^{-1}\psi$ has $-2$, so neither is invariant. These statements concern quadratic masses before electroweak symmetry breaking, not masses involving scalar insertions.

For the Higgs conjugate doublet, put $\epsilon =i{\tau }_2$. The given identity $\epsilon {\tau }_i^{\ast }=-{\tau }_i\epsilon$ exponentiates to $\epsilon U^{\ast }=U\epsilon$ for $U\in SU(2)$. ThereforeIn particular its weak isospin remains $T=1/2$ while its hypercharge reverses. Since $\Psi \mapsto e^{-i\beta /2}U\Psi$, both ${\varphi }^{\prime †}\Psi$ and $\bar{\Psi }{\varphi }^{\prime }$ transform in the trivial representation of the full gauge group. They remain a Lorentz spinor and its adjoint; gauge singlet does not mean Lorentz scalar.

Finally the Yukawa interactionis gauge invariant: the first term has total hypercharge $+1/2+1/2-1=0$, and its doublet indices are contracted. At the Higgs field vacuum it becomes $-(y_{e}v/\sqrt{2}){\bar{e}}_L{e}_R+h.c.$. Rephase $e_R$ to make the coefficient real and positive and assemble $e=e_L+e_R$. ThenThis explicitly generates the gauge-invariant electron Yukawa mass, without a bare mass in the unbroken theory.