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For the sum-zero standard representation of the symmetric group, the vectors $v_j=e_1+\cdots +e_j-j{e}_{j+1}$ form a Gelfand–Tsetlin basis. The Young–Jucys–Murphy element $X_i$ has eigenvalue $i-1$ for $i\le j$, $-1$ for $i=j+1$, and $i-2$ for $i>j+1$. Normalizing by $\sqrt{j(j+1)}$ gives an orthonormal basis corresponding to tableaux of shape $(n-1,1)$ whose second-row entry is $j+1$.

For the standard representation of the symmetric group $S_3$, the unramified local factor of an Artin L-function, with $t=p^{-s}$, is respectively the displayed factor for identity, transposition, and three-cycle Frobenius cycle types. These follow by removing the trivial summand from the three-point permutation representation. If the inertia group contains a three-cycle, its invariant subspace in the standard representation is zero, so the ramified local factor is one.

Order the conjugacy classes of the symmetric group $S_4$ by representatives $1,(12),(12)(34),(123),(1234)$; their sizes are $1,6,3,8,6$. For a three-dimensional rotation through an angle $\theta$, the eigenvalues are $1,e^{i\theta },e^{-i\theta }$, so the trace of a three-dimensional rotation is $1+2\cos \theta$.

Under the identification with the rotational symmetry group of a cube, elements of order two give half-turns, elements of order three give turns through $\pm 2\pi /3$, and elements of order four give turns through $\pm \pi /2$. A half-turn can have an axis through opposite edges or opposite faces, but both kinds have trace $-1$. ThusIts character inner product with itself isBy character orthogonality, a nonzero complex representation has character inner product one with itself exactly when it is an irreducible representation. Therefore $C$ is an irreducible representation of dimension three.

For comparison useful in part (e), the standard representation of the symmetric group has character equal to the number of fixed letters minus one, namely $(3,1,-1,0,-1)$ in this order. Multiplication by the sign representation changes this character to $(3,-1,-1,0,1)$. Thus $C$ is the tensor product of the standard representation of the symmetric group with the sign representation.

Realize the two-dimensional irreducible representation $\rho$ as the standard representation of the symmetric group onwith $S_3$ permuting coordinates. It is the complement of the trivial summand in the three-point permutation representation. For each prime the local factor of an Artin L-function iswhere $I_p$ is the inertia group and Frobenius means any lift of the residue-field Frobenius automorphism. On $V^{I_p}$ the lift is independent of that choice. At an unramified prime, put $t=p^{-s}$. The Local Artin factors of the standard representation of S3 areThe polynomial discriminant of $X^3-3$ is $-243=-3^5$, so only $3$ can ramify in its splitting field. For $p\ne 3$, the squarefree factorization of $X^3-3$ modulo $p$ determines the Frobenius cycle type on its roots.

At $p=2$,The quadratic is irreducible, so Frobenius is a transposition andThus $a_2=0$, $a_4=1$ and $a_8=0$. At $p=5$,The quadratic discriminant is $3(\mathrm{mod}5)$, a nonsquare, so this is again a transposition. Consequently $L_5(\rho ,s)=(1-5^{-2s}{)}^{-1}$ and $a_5=0$.

At $p=7$, the possible cubes are $0,1,-1$, so $X^3-3$ has no root modulo $7$. Being cubic, it is irreducible; Frobenius is a three-cycle. Thereforeand $a_7=-1$.

The ramified factor at $3$ must be computed on inertia group invariants. Write $\alpha =3^{1/3}$ and set $\pi =({\zeta }_3-1)/\alpha$. Using $({\zeta }_3-1{)}^2=-3{\zeta }_3$ givesThe Eisenstein polynomial $X^6+3$ over ${\mathbb{Q}}_3$ has degree six, so ${\mathbb{Q}}_3(\pi )$ is a totally ramified degree-six subfield of ${\mathbb{Q}}_3({\zeta }_3,\alpha )$. The latter has degree at most $[F:\mathbb{Q}]=6$, so the fields agree. This Eisenstein sextic presentation of the splitting field of X3 minus 3 over Q3 shows $D_3=I_3=S_3$. Since the only permutation-invariant vectors in ${\mathbb{C}}^3$ have all coordinates equal, their intersection with $V$ is zero. ThusIn particular $a_3=a_9=0$; an unramified degree-two factor at $3$ would give the wrong coefficients.

Finally the Euler product makes its Dirichlet series coefficients multiplicative for coprime indices. Its constant coefficient is $a_1=1$, and $a_6=a_2{a}_3=0$, $a_{10}=a_2{a}_5=0$. Every integer up to ten is now covered, giving

The permutation representation on ${\mathbb{C}}^X$ sends the basis vector $e_x$ to $e_{gx}$. Its character is ${\pi }_X(g)=|X^g|$, because precisely the fixed basis vectors contribute to the trace. Burnside lemma says that the number of orbits is $|G{|}^{-1}{\sum }_g|X^g|$. To prove it, double-count $(g,x)$ with $gx=x$: the count is ${\sum }_g|X^g|={\sum }_x|G_x|$. Each orbit contributes $|Gx||G_x|=|G|$ by the orbit-stabilizer theorem.

A pair is fixed under the diagonal action exactly when both entries are fixed. Since permutation characters are real,by Burnside lemma on the product.

The general linear group ${\mathrm{GL}}_2(q)$ acts on the $q+1$ lines of ${\mathbb{F}}_q^2$. It is doubly transitive: choosing nonzero representatives of any two distinct lines gives a basis, and a linear map can send that basis to representatives of any other ordered pair. Thus there are two orbits on ordered pairs, the diagonal and its complement. The character inner product of this permutation character with itself is two, while the multiplicity of the trivial character is one by transitivity. The sum of squares of irreducible multiplicities is therefore two; the remaining summand is one irreducible representation, of dimension $(q+1)-1$. Its dimension is $\boxed{q}$.

For $S_n$ acting on two-element subsets, an ordered pair of subsets is classified by intersection size $0,1,2$. All three occur for $n\ge 4$, and a permutation carries any pair of a given type to any other. Hence $⟨{\pi }_Z,{\pi }_Z⟩=3$. The natural point permutation representation is the trivial representation plus the standard representation of the symmetric group, of dimension $n-1$: its irreducibility also follows from double transitivity and inner product two.

Embed this point representation into ${\mathbb{C}}^Z$ by $e_i\mapsto {\sum }_{j\ne i}{e}_{\{i,j\}}$. The map is equivariant, and its kernel consists of vectors with $c_i+c_j=0$ for every distinct $i,j$. Three distinct indices force every $c_i=0$, so it is injective for $n\ge 3$. Thus the trivial and standard representations each occur in ${\mathbb{C}}^Z$. The norm three leaves exactly one further irreducible, with multiplicity one. It is the orthogonal complement of that embedded subspace, equivalently the vectors $a_{ij}=a_{ji}$ satisfying ${\sum }_{j\ne i}{a}_{ij}=0$ for every $i$. Its dimension isThe constituent dimensions are $\boxed{1,n-1,n(n-3)/2}$, each once. For $n=3$, complementation identifies two-element subsets with points, so there are only the trivial and 2-dimensional standard constituents.

Take $m=2$ and $\lambda =(2,1)$, so $\lambda +\delta =(3,1)$. A transposition in $S_3$ has one singleton cycle and one two-cycle. The Frobenius alternant character formula therefore givesIn the coordinate permutation representation on ${\mathbb{C}}^3$, a transposition fixes one basis vector, so its character is $1$. This representation is the direct sum of the invariant line of constant vectors and the standard representation of the symmetric group, whose vectors have coordinate sum zero. Subtracting the trivial character gives ${\chi }_V((12))=1-1=0$, as required.

Let $e_1,\dots ,e_n$ be the coordinate vectors and chooseThese vectors lie in the sum-zero standard representation of the symmetric group. They are mutually orthogonal for the usual Hermitian inner product, with $\Vert v_j{\Vert }^2=j(j+1)$, so they form a basis of its $(n-1)$-dimensional space.

If $i\le j$, every transposition in $X_i$ exchanges two coordinates equal to $1$, so each fixes $v_j$. If $i=j+1$, summing the $j$ transpositions gives coordinate $-1$ in each of positions $1,\dots ,j$ and coordinate $j$ in position $j+1$, hence $X_{j+1}{v}_j=-v_j$. Finally, if $i>j+1$, the $i$th coordinate of $v_j$ is zero. In the sum of the $i-1$ transpositions, any coordinate in its support retains its original value in $i-2$ terms and becomes zero in the remaining term; the $i$th output coordinate is the sum of the original first $i-1$ coordinates, which is zero.

Thus all actions are diagonal in this basis:This supplies an explicit Gelfand–Tsetlin basis of the standard representation.

Assume $n\ge 2$. The standard Young tableaux of shape $(n-1,1)$ are $T_k$, for $2\le k\le n$: the sole second-row entry is $k$, and the first row isThis lists all of them, since $1$ must occupy the first box and the rest of the first row is forced to increase. Their contents areWrite $w_k=w_{T_k}$. The only nontrivial adjacent swaps exchange the lower entry with $k-1$ or $k+1$. Substituting the content differences into Young orthogonal form yields all generator actions:In the first line at $k=2$, the square-root coefficient is zero and the action is $-w_2$, so no vector $w_1$ is needed. The second line is used only when $k<n$, since the generator $s_n$ does not exist.

The corresponding Specht module is the sum-zero standard representation of the symmetric group:A concrete orthonormal realization isPermuting coordinates gives exactly the actions above, linking this orthogonal tableau basis with the Gelfand–Tsetlin basis found in Question 1. Here the printed $s_j$ are the adjacent Coxeter generators, rather than products of all generators.

The implication is false. ChooseThe inequalities $0\le 4<9$ and the relation $13=1\cdot 9+4$ hold. At cell $(1,4)$, the hook length is $12-4+1=9=a{p}^k$.

For the partition of an integer $(12,1)$, the Hook-length formula giveswhich is divisible by $3$. This also agrees with the degree of the standard representation of the symmetric group $S_{13}$.

The obstruction is visible in the core of a partition: removing the horizontal rim hook of length nine leaves $\gamma =(3,1)$, which has no hook length nine and hence is the $9$-core. Its degree is ${\chi }^{(3,1)}(1)=3$. The large hook ensures the required weight of a partition here, but the core degree is not coprime to $p$.

The normis a positive integer, because it is the sum of the squares of the multiplicities of the irreducible constituents. The character restriction norm bound and $[G:K]=3$ make it at most three. It is therefore $1$, $2$, or $3$.

All three values occur:

Permuting the gaps, or equivalently relabelling the particles, leaves both quadratic forms invariant. The three-dimensional permutation representation splits as the rigid-rotation line spanned by $(1,1,1)$ and the two-dimensional sum-zero standard representation of the symmetric group.

The two displayed vibrational vectors merely choose a basis of that sum-zero plane. A permutation generally mixes them, but their common frequency leaves the whole plane invariant. This is precisely the degenerate normal modes from permutation symmetry required by the physical symmetry.

Definitions and labels. A complex group representation on a vector space $V$ is a group homomorphism $\rho :G\to \mathrm{GL}(V)$, so $\rho (gh)=\rho (g)\rho (h)$ and $\rho (e)=I$. It is an irreducible representation if $V\ne 0$ and its only invariant subspaces are $0$ and $V$.

Read permutation matrices by their action on column basis vectors. The labels in the original PDF areIn particular, the original PDF gives the last row of $C$ as $(1,0,0)$; the provided TeX transcription's $(0,0,1)$ is incorrect and would not give an invertible permutation matrix.

Multiplication and conjugacy. Matrix products act on columns with the right factor first. For example $A^2=B$, $AD=C$, and $DA=E$. The complete multiplication table, with row factor multiplied by column factor, isConjugating a permutation relabels its cycles. Thus the conjugacy classes areof sizes one, two and three. These are respectively the identity, the three-cycles and the transpositions in the symmetric group on three letters.

Dimensions. The sum of squares of irreducible representation dimensions equals the group order, which here is six. A three-dimensional irreducible representation would already contribute nine, so the displayed representation is reducible. There are three inequivalent complex irreducible representations, one for each conjugacy class, including the trivial representation of dimension one. If their dimensions are $1,d_2,d_3$, then $d_2^2+d_3^2=5$, whose positive integer solution is $\{d_2,d_3\}=\{1,2\}$. HenceOne can also see reducibility directly: every permutation matrix fixes $(1,1,1{)}^T$.

Characters. The character of a representation is the trace of its representing matrix, constant on each conjugacy class. The given permutation representation has character equal to the number of fixed basis vectors:in class order $(I;A,B;C,D,E)$. A trivial representation has character $(1,1,1)$. For the second one-dimensional character, a transposition has value $\pm 1$ because its square is the identity. The two three-cycles are conjugate and inverse, so their common value $a$ obeys both $a^2=1$ and $a^3=1$, forcing $a=1$. Character orthogonality with the trivial character then requires $1+2+3b=0$, giving $b=-1$. This determines the sign representation without constructing its matrices.

Write the two-dimensional character as $(2,x,y)$. Character orthogonality with the first two rows givesso $x=-1$, $y=0$. Its squared norm is $(4+2)/6=1$, as required. The complete character table is thereforeThe last row belongs to the standard representation of the symmetric group.

To decompose the given representation, use the character inner productIts multiplicities areThusConcretely the invariant line is spanned by $(1,1,1{)}^T$ and its invariant complement is $\{(x,y,z):x+y+z=0\}$. Their characters are exactly the first and third rows, agreeing with the decomposition obtained entirely from character orthogonality.

A group representation on a vector space $W$ is a group homomorphism $\rho :G\to GL(W)$: $\rho (gh)=\rho (g)\rho (h)$ and $\rho (e)=I$. An invariant subspace $U$ obeys $\rho (g)U\subseteq U$ for every $g$. A nonzero representation is irreducible if its only invariant subspaces are $0$ and $W$. It is a faithful representation if it is injective, equivalently if its group kernel is $\{e\}$. Its character is ${\chi }_{\rho }(g)=\mathrm{tr}\rho (g)$. For inequivalent complex irreducible representations of a finite group, character orthogonality is

Use right-to-left composition of permutations and assign the matrix labels consistently byThe resulting Cayley table, with row multiplying column, isConjugation relabels the entries of a cycle, preserving its length. Thus the conjugacy classes are

Define the permutation representation by $P_{\sigma }{e}_j=e_{\sigma (j)}$, or $(P_{\sigma }{)}_{ij}={\delta }_{i,\sigma (j)}$. Applying it to a basis vector proves $P_{\sigma }{P}_{\tau }=P_{\sigma \tau }$. The six assignments above produce exactly the listed matrices. They are distinct, and $P_{\sigma }=I$ forces $\sigma (j)=j$ for every $j$, so the representation is faithful.

Every permutation fixes $(1,1,1)$, hence fixes every vector $(c,c,c)$. Therefore $V_0$ is a one-dimensional invariant subspace carrying the trivial representation. Its orthogonal complement issince the inner product with $(c,c,c)$ is $c(v_1+v_2+v_3)$. Permuting coordinates preserves their sum, so this plane is invariant too. Every real vector has the unique decompositionThus the permutation representation is the direct sum of group representations on the line and the plane.

To find the eigenvectors of $A$, note that $A(v_1,v_2,v_3)=(v_3,v_1,v_2)$. With $\zeta =e^{2\pi i/3}$, its three eigenvalues and eigenspaces areThe only real eigenvalue is one, whose eigenline is $V_0$, so $A$ has no nonzero real eigenvector in $V_0^{\perp }$. A proper nonzero real invariant subspace of that two-dimensional plane would be a line, giving such an eigenvector. Hence the plane representation is irreducible over $\mathbb{R}$.

For the later complex dimension-count theorem one should also establish irreducibility over $\mathbb{C}$; real irreducibility alone would not imply it in general. In the complexified plane, the only candidate $A$-invariant lines are its $\zeta$ and ${\zeta }^2$ eigenlines. But $C$ exchanges them:Neither is invariant under the whole group, proving complex irreducibility as well. The required decomposition is thereforewhere $W$ is the standard representation of the symmetric group for three letters. Its character is obtained by subtracting one from the number of fixed coordinates: on the three classes it is $(2,-1,0)$.

Finally, the sign representation gives a second one-dimensional irreducible, with value $+1$ on $I,A,B$ and $-1$ on the transpositions. These are exactly the two one-dimensional possibilities: if $r=A,s=C$, a one-dimensional representation sends them to commuting nonzero scalars $v,u$. The relations $r^3=s^2=e$ and $srs=r^{-1}$ imply $v^3=u^2=1$ and $v=v^{-1}$. Hence $v=1$ and $u=\pm 1$, the trivial and sign choices. Together with the two-dimensional $W$, these giveThe sum of squares of irreducible representation dimensions leaves room for no other irreducible. Thus there are exactly two one-dimensional and one two-dimensional complex irreducible representations.

Realize the standard representation of the symmetric group on $W=\{x\in {\mathbb{C}}^4:{\sum }_i{x}_i=0\}$. The symmetric power sums $p_k={\sum }_i{x}_i^k$, for $k=2,3,4$, freely generate its polynomial invariant ring. To justify the description, average any polynomial lift from $W$ to ${\mathbb{C}}^4$ over the symmetric group; the resulting symmetric lift is a polynomial in the elementary symmetric polynomials $e_1,e_2,e_3,e_4$, and restriction sets $e_1=0$. The remaining three generators are algebraically independent. The Newton identities express them equivalently through $p_2,p_3,p_4$ over the complex numbers. In coordinates $x=(a,b,c,-a-b-c)$, one has $p_2=a^2+b^2+c^2+(a+b+c{)}^2$ and $p_3=-3(a+b)(b+c)(c+a)$.

The standard representation of the symmetric group $S_4$ restricts irreducibly to $A_4$. Its character has value $3$ at the identity and $-1$ on each nonidentity element of the normal Klein four-group. Restricted to that subgroup, it is the sum of the three nontrivial linear characters.