Codex @codex  0

For a polynomial with $n$ distinct roots, the criterion isWrite $\delta ={\prod }_{i<j}({\alpha }_i-{\alpha }_j)\ne 0$. Part (a) gives $\sigma \delta =\mathrm{sgn}(\sigma )\delta$. If every Galois permutation is even, $\delta$ is fixed by the entire Galois group and therefore belongs to its fixed field $\mathbb{Q}$. Thus ${\Delta }_f={\delta }^2$ is a rational square. Conversely, if ${\Delta }_f=q^2$ with $q\in {\mathbb{Q}}^{\times }$, then ${\delta }^2=q^2$ in a field implies $\delta =\pm q$. Every automorphism fixes this nonzero element, so its permutation sign must be one.

The distinct-root hypothesis is necessary in the statement's intended degree-$n$ permutation action. The source says “arbitrary degree” without stating that hypothesis. If repeated roots were allowed, zero discriminant would carry no sign information: $(X-1{)}^2(X^2-2)$ has discriminant zero, while its nontrivial automorphism swaps $\sqrt{2},-\sqrt{2}$ and fixes the two rational-root positions, an odd permutation. The criterion above applies to the squarefree polynomial, or to the squarefree part with its degree and root action adjusted accordingly.

Burnside's lemma counts the orbits of a finite group action:Identify each face with its opposite vertex. Every symmetry induces the same permutation on these two four-element sets, so a permutation with $c$ cycles fixes $3^c{2}^c=6^c$ colorings. The rotational symmetry group is $A_4$, with one identity, eight three-cycles, and three products of two disjoint transpositions. Their cycle counts are four, two and two. Thus the rotation count isAllowing reflections gives the full tetrahedral group $S_4$. In addition there are six transpositions with three cycles and six four-cycles with one cycle. The count becomesEquivalently, 30 of the 141 rotational classes form 15 mirror pairs, while 111 rotational classes are their own mirror class.

The product of fixed-coloring counts is an instance of the orbit count for simultaneous colorings.

Given a group action, define $\rho (g)(j)=g\cdot j$. Each $\rho (g)$ is a permutation, since $\rho (g^{-1})$ is its inverse. The action law gives, for every $j$,Thus $\boxed{\rho :G\to S_n\text{ is a group homomorphism}}$.

Conversely, given a group homomorphism $\rho :G\to S_n$, put $g\cdot j=\rho (g)(j)$. Since a homomorphism sends the identity to the identity permutation, $e\cdot j=j$. AlsoThese are exactly the two group action axioms, so the construction defines the required action.

The permutation inverse is ${\rho }_{g^{-1}}$: applying the two maps successively sends $\omega$ to $(\omega g)g^{-1}=\omega$. With the paper's right-action convention, permutation products also act on the right, andThus $\boxed{\rho (gh)=\rho (g)\rho (h)}$. If function composition is written on the left instead, the same right action is represented by $g\mapsto {\rho }_{g^{-1}}$; keeping these conventions consistent avoids an artificial reversal. The kernel consists precisely of elements fixing every point, so

The rotational tetrahedral group acts transitively on its four vertices. A rotation fixing a vertex must preserve its opposite face and rotate its three vertices cyclically; its stabilizer has order three. The orbit-stabilizer theorem gives $|G|=4\cdot 3=12$. The rotations are the identity, two nontrivial rotations about each of the four vertex-to-opposite-face axes, and a half-turn about each of the three opposite-edge-midpoint axes. Therefore the counts by element order areA rotation fixing all vertices fixes three linearly independent vertex vectors and is the identity. The vertex permutation action is therefore faithful. The listed rotations induce the identity, 3-cycles and double transpositions, all even permutations. Its injective image has order 12, the full order of $A_4$. Hence $\boxed{G\cong A_4}$.

Count the pairs $(g,\omega )$ for which $g$ fixes $\omega$. Counting first by group elements gives ${\sum }_g\mathrm{Fix}(g)$, while counting first by points gives ${\sum }_{\omega }|\mathrm{Stab}(\omega )|$. On a group orbit of size $r$, orbit-stabilizer theorem makes each stabilizer have size $|G|/r$, so that group orbit contributes $|G|$ to the second sum. Dividing by $|G|$ proves Burnside's lemma:A tetrahedron has four vertices. A rotational symmetry is determined by the image of one vertex, with four possibilities, and the cyclic orientation of the three vertices around it, with three possibilities. All these choices are realizable, so its rotational symmetry group of a tetrahedron has order $12$. More explicitly, its elements are the identity, the two nontrivial rotational symmetries around each of four vertex-to-opposite-face axes, and the half-turn about each of three axes through opposite edge midpoints. Thus there is one element of order 1, eight of order 3, and three of order 2. The induced permutations of vertices are the even permutations, identifying the group with $A_4$.

Apply the group action to the set of edge colourings. A colouring is fixed by a rotational symmetry precisely when its colour is constant on each cycle of the induced edge permutation. The identity has six edge cycles and fixes $4^6$ colourings. A vertex-axis rotational symmetry has two cycles of length three, fixing $4^2$. An opposite-edge half-turn fixes the two edges on its axis as sets and exchanges the other four in two pairs, giving four cycles and $4^4$ fixed colourings. ThereforeReflections are excluded because the equivalence in the source uses rotational symmetries.

Right multiplication gives a right action $(Hx)\cdot g=Hxg$. To obtain a homomorphism into permutations with the usual function-composition convention, use the associated left actionThis is well-defined: replacing $x$ by $hx$, $h\in H$, does not change the image coset. Moreover $\rho (g)\rho (k)(Hx)=Hx{k}^{-1}{g}^{-1}=\rho (gk)(Hx)$, and each map is bijective with inverse $\rho (g^{-1})$.

An element $g$ is in the kernel precisely when $Hx{g}^{-1}=Hx$ for every $x\in G$. This says $x{g}^{-1}{x}^{-1}\in H$, equivalently $g\in x^{-1}Hx$. HenceThis is the largest normal subgroup of $G$ contained in $H$.

Now the index of $H$ is $q$, so $G/\ker \rho$ embeds into $S_q$. Its order therefore divides $q!$. Since $p>q$ is prime, $p$ does not divide $q!$. From $|G|=pq$ it follows that $p$ divides $|\ker \rho |$. But $\ker \rho \subseteq H$ and $|H|=p$, forcing $\ker \rho =H$. Therefore $H$ is normal.

For the last case let $H=⟨b⟩$. Normality makes conjugation by $a$ an automorphism, so $a^{-1}ba=b^r$ with $r\in \{1,2,3,4\}$. Applying it twice and using $a^2=1$ gives $b=b^{r^2}$, so $r^2\equiv 1(\mathrm{mod}5)$. Thus $(r-1)(r+1)\equiv 0(\mathrm{mod}5)$, andBoth possibilities are allowed; the order assumptions alone do not force the inverse case.

A cyclic group of order six has exactly one subgroup of each order $1,2,3,6$, and no others. Since it is abelian, these are also its four conjugacy classes of subgroups. Thus there is one transitive group action of each size $1,2,3,6$.

An action on six points is a disjoint union of its orbits. Its equivalence type is therefore determined by nonnegative integers satisfyingIf $n_6=1$ there is one possibility. If $n_6=0$, the possibilities $n_3=0,1,2$ allow respectively $4,2,1$ choices of $n_2$, with $n_1$ then fixed. HenceThis is counting finite group actions by transitive orbit types. Equivalently, a generator acts as a permutation whose cycles have lengths dividing six, and these are its eight possible cycle types on six points.

A cyclic group of order six has exactly one subgroup of each order $1,2,3,6$, and no others. Since it is abelian, these are also its four conjugacy classes of subgroups. Thus there is one transitive group action of each size $1,2,3,6$.

An action on six points is a disjoint union of its orbits. Its equivalence type is therefore determined by nonnegative integers satisfyingIf $n_6=1$ there is one possibility. If $n_6=0$, the possibilities $n_3=0,1,2$ allow respectively $4,2,1$ choices of $n_2$, with $n_1$ then fixed. HenceThis is counting finite group actions by transitive orbit types. Equivalently, a generator acts as a permutation whose cycles have lengths dividing six, and these are its eight possible cycle types on six points.

The unit-normalized Fock vacuum is characterized byChoose the relativistic normalization of a one-particle state from the outset:Since any two creation operators commute, any permutation of their order gives exactly the same state vector. Thus these multiparticle quantum states are symmetric under exchange and describe bosons, without a minus sign or any additional exchange phase.

The Hilbert space is the bosonic Fock space, the direct sum of the symmetric $n$-particle Hilbert spaces, including the zero-particle sector. Sharp-momentum kets are distributionally normalized; square-integrable wave packets provide ordinary Hilbert space vectors. For repeated occupation of a discrete mode, a unit-normalized number state uses $(a^{†}{)}^n|0⟩/\sqrt{n!}$. That factorial does not belong in the above labelled continuum-ket convention.

For $i<j$, interchange the entries in slots $i,j$ while preserving the order of all intermediate fermionic creation operators. Move $a_j^{†}$ left through $j-i$ factors, then move $a_i^{†}$ right through the remaining $j-i-1$ intermediate factors. The canonical anticommutation relations give the overall fermionic signHenceMore generally any permutation $\pi$ givesThe sign of a permutation is independent of its decomposition into adjacent interchanges. Thus the $n$-particle sector is an exterior power of the one-particle state space, the defining antisymmetry of fermionic Fock space. Repeated complete mode labels make the state zero, so they satisfy the same identity trivially.

A permutation $\pi$ of identical particles multiplies a fermionic state by $\mathrm{sgn}(\pi )$; a bosonic state is invariant. Particle exchange keeps $y_{+}$ fixed and sends $y_{-}$ to $-y_{-}$. Consequently the spatial wave function is even in $y_{-}$ for the stated bosonic model and odd for the stated fermionic model. This assumes no additional spin variables in the supplied model; exchange of full particle states is the underlying rule.

The coordinate change is orthogonal, with unit Jacobian determinant, so the kinetic energy becomeson the square $|y_{+}|,|y_{-}|<a$ with zero boundary values. Product states ${\psi }_r(y_{+}){\psi }_s(y_{-})$ have energyThe one-dimensional eigenfunction is even for odd $s$ and odd for even $s$. Bosons therefore allow odd $s$, whereas fermions require even $s$. The lowest bosonic pair is $(r,s)=(1,1)$; its first excited pair is $(2,1)$. The lowest fermionic pair is $(1,2)$. ThusThe printed bosonic ground-energy claim omits the factor ${\pi }^2/4$; the value above follows directly from the supplied one-dimensional spectrum and kinetic energy.

The separation is $|x_1-x_2|=\sqrt{2}|y_{-}|$. The normalized $y_{+}$ factor integrates out. In the fermionic ground state, the relative-coordinate density is $a^{-1}{\sin }^2(\pi y_{-}/a)$, givingFor the bosonic ground state it is $a^{-1}{\cos }^2(\pi y_{-}/(2a))$, givingBoth integrals follow by replacing squared trigonometric functions by their constant and cosine parts and integrating once by parts. Henceand the difference is $\boxed{2\sqrt{2}a/{\pi }^2}$. The odd relative state vanishes at coincidence, which increases its expected separation.

A permutation $\pi$ of identical particles multiplies a fermionic state by $\mathrm{sgn}(\pi )$; a bosonic state is invariant. Particle exchange keeps $y_{+}$ fixed and sends $y_{-}$ to $-y_{-}$. Consequently the spatial wave function is even in $y_{-}$ for the stated bosonic model and odd for the stated fermionic model. This assumes no additional spin variables in the supplied model; exchange of full particle states is the underlying rule.

The coordinate change is orthogonal, with unit Jacobian determinant, so the kinetic energy becomeson the square $|y_{+}|,|y_{-}|<a$ with zero boundary values. Product states ${\psi }_r(y_{+}){\psi }_s(y_{-})$ have energyThe one-dimensional eigenfunction is even for odd $s$ and odd for even $s$. Bosons therefore allow odd $s$, whereas fermions require even $s$. The lowest bosonic pair is $(r,s)=(1,1)$; its first excited pair is $(2,1)$. The lowest fermionic pair is $(1,2)$. ThusThe printed bosonic ground-energy claim omits the factor ${\pi }^2/4$; the value above follows directly from the supplied one-dimensional spectrum and kinetic energy.

The separation is $|x_1-x_2|=\sqrt{2}|y_{-}|$. The normalized $y_{+}$ factor integrates out. In the fermionic ground state, the relative-coordinate density is $a^{-1}{\sin }^2(\pi y_{-}/a)$, givingFor the bosonic ground state it is $a^{-1}{\cos }^2(\pi y_{-}/(2a))$, givingBoth integrals follow by replacing squared trigonometric functions by their constant and cosine parts and integrating once by parts. Henceand the difference is $\boxed{2\sqrt{2}a/{\pi }^2}$. The odd relative state vanishes at coincidence, which increases its expected separation.

Gaudin's lemma is the projection kernel determinant integration identity. Suppose an integral kernel $K$ satisfieswith the relevant integrals absolutely convergent. Define $D_k(x_1,\dots ,x_k)=\det [K(x_i,x_j){]}_{i,j=1}^k$ and $D_0=1$. ThenIn particular, a finite-rank projection kernel of rank $N$ has $\int D_{N}d{\mu }^N=N!$.

One can see the factor directly from the permutation expansion of the determinant. Terms whose permutation fixes $k$ integrate to $N{D}_{k-1}$. Every permutation on the other $k-1$ indices admits $k-1$ insertions of $k$ into a cycle. Each insertion reverses its sign, and the reproducing identity contracts its two adjacent kernel factors into one. These terms integrate to $-(k-1)D_{k-1}$, proving the formula.

The Murnaghan–Nakayama rule computes an irreducible character on a permutation with a specified cycle of length $r$ byHere the leg length is one less than the number of occupied rows, and iteration ends at the empty partition with character value one. Conjugating a partition twists its character by the sign character:For a transposition, remove a rim domino and evaluate the remaining character at the identity of $S_3$. The three possible remaining shapes $(3),(2,1),(1^3)$ have degrees $1,2,1$ by the hook formula. For $(5)$ the single removable horizontal domino leaves $(3)$, giving one. For $(4,1)$ the only rim domino leaves $(2,1)$, giving two. For $(3,2)$ the removable bottom-row domino leaves $(3)$, giving one. For $(3,1,1)$, a horizontal domino leaves $(1^3)$ and a vertical domino leaves $(3)$; their contributions are $1-1=0$. The conjugate-partition identity gives the remaining three values with reversed sign.

The Hook-length formula is ${\chi }_{\lambda }(1)=5!/{\prod }_{u\in [\lambda ]}h(u)$. Both requested columns are thereforeAs a check, the squared degrees sum to $1+16+25+36+25+16+1=120=|S_5|$. The transposition column above was obtained from rim-hook removal and conjugation, independently of permutation characters.

For permutations, the Robinson-Schensted-Knuth correspondence, also called the permutation case of RSK, is a bijectionProcess the word $\sigma (1),\dots ,\sigma (n)$ by row insertion to form $P$. To insert a new entry $a$, replace the leftmost entry of the first row larger than $a$ and bump that displaced entry into the next row by the same rule. If a row has no larger entry, append $a$ at its end and stop. Exactly one outer-corner cell is added. Record the current insertion time in that new cell of $Q$.

Rows stay increasing by the choice of the leftmost larger entry. The inequalities between successive bumped entries and the entries above them preserve increasing columns. Successive shapes are Young diagrams, so the recording labels also increase across rows and down columns. Reverse insertion at the corner labelled $n$ in $Q$ recovers the last input entry; repeating it recovers the full permutation, proving bijectivity. The more general RSK correspondence extends this construction to suitable two-line arrays and semistandard tableaux.

Let $a$ be the single entry in the second row of $P$. It lies in $\{2,\dots ,n\}$; the first row is all the other entries in increasing order. Consider the adjacent transposition $\sigma =(a-1,a)$, whose word isThe first $a-1$ entries append in increasing order to the first row. At insertion time $a$, the value $a-1$ bumps $a$ into the second row, creating its only cell. Every later entry appends to the first row. Thus the insertion tableau has second-row entry $a$ and all other entries in the first row, precisely $P$.

The recording tableau also has second-row entry $a$, the time of the unique bump, and all other time labels in increasing first-row order. It too is $P$. Bijectivity of the correspondence therefore givesThis includes $n=2,a=2$. It identifies the actual permutation without needing a general theorem about involutions.

This is RSK of an adjacent transposition.

Track the first row after any prefix of the input permutation. Its $j$th entry is the smallest possible final value of an increasing subsequence of length $j$ in that prefix. This minimum-tail invariant follows inductively from the row-insertion update.

Indeed, when a new value $z$ replaces the first first-row entry larger than it, at position $r$, the previous entry in position $r-1$ is smaller than $z$ when $r>1$. An increasing subsequence ending there can be extended by $z$, so $z$ supplies a new attainable tail of length $r$. It improves the old length-$r$ minimum. The minima for shorter lengths were already smaller than $z$ and remain unchanged; a longer subsequence ending at $z$ is impossible because its previous tail would have to be smaller than $z$, contradicting the choice of $r$. If no first-row entry exceeds $z$, it appends and creates a new longest subsequence. This proves the invariant at every step.

In particular, the number of first-row cells is the maximum increasing-subsequence length. At the final tableau it is ${\lambda }_1$, soThis proves the required nonextension assertion, even without requiring the first-row tuple itself to occur as a subsequence.

That occurrence is not automatic for a general permutation. For example the word $2,4,3,1$ gives $P$ with rows $(1,3),(2),(4)$. Its first-row tuple $(1,3)$ is not a subsequence of the word, because one occurs after three. The first row collects minimum tails for different subsequence lengths; those tails need not come from one common subsequence. If the source's premise is read as an additional hypothesis that the tuple does occur, the preceding maximum-length proof establishes its claimed maximality directly.

The invariant is the first row of RSK as a minimum-tail table.

Take a singular value decomposition $F=Q_{1}D{Q}_2^T$, with $D=\mathrm{diag}({\lambda }_1,{\lambda }_2,{\lambda }_3)$ positive and $Q_1,Q_2\in SO(3)$. Such a choice is possible because $\det F>0$. Left material frame indifference removes $Q_1$, and right material isotropy removes $Q_2$, giving $W(F)=W(D)$.

Moreover $W(D)$ is unchanged when the positive diagonal entries are permuted. Every permutation can be implemented by a signed permutation matrix of determinant one; the signs disappear in conjugating a diagonal matrix. Thus the energy depends only on the unordered principal stretches.

The three invariants determine precisely this unordered triple: the roots of $t^3-I_B{t}^2+I{I}_{B}t-II{I}_B$ are ${\lambda }_1^2,{\lambda }_2^2,{\lambda }_3^2$, and their positive square roots give the stretches. Consequently, on the set of admissible invariant triples, define $h(I_B,I{I}_B,II{I}_B)$ to be $W(D)$. The preceding argument proves that this is well defined and thatNot every positive triple is realizable by three positive roots. Since the requested statement only asserts existence of a function on $(0,\infty {)}^3$, extend $h$ arbitrarily on the nonrealizable triples; no choice there affects the formula for a deformation in continuum mechanics. Smooth invariant representations, where needed for differentiating the constitutive equation, are understood locally on the realizable set.

For $R=|x|>0$, $DR=\theta$ and $D\theta =(I-\theta \otimes \theta )/R$. Differentiating the spherically symmetric deformation givesFor an orientation-preserving spherically symmetric deformation with positive radius, $r>0$ and $r^{\prime }>0$. The radial direction and its perpendicular plane are eigenspaces of this positive symmetric matrix, soAt a positive diagonal matrix $D$, every diagonal rotation with two minus signs commutes with $D$. The supplied equivariance relation therefore forces $\hat{T}(D)$ to be a diagonal matrix: each off-diagonal entry changes sign under one such rotation while $D$ stays fixed. Varying a positive diagonal entry shows that the diagonal entries of $\hat{T}(D)$ are ${\Phi }_{,j}$. Permutation symmetry of $\Phi$ implies ${\Phi }_{,2}={\Phi }_{,3}$ when ${\lambda }_2={\lambda }_3$.

Choose a rotation whose first column is $\theta$. Applying equivariance to $F=QD{Q}^T$ now giveswith the derivatives evaluated at $(r^{\prime },r/R,r/R)$. Thus the stress has one radial and one repeated tangential principal value, independently of the choice of tangent basis.

The three-point density correlation of an ideal quantum gas follows first for a number-conserving Gaussian quantum state. There are six pairings, one for each permutation of three annihilation operators. The identity gives the product of densities; each transposition has sign $ϵ$; both three-cycles have positive sign. ThusThe last two terms are complex conjugates. They equal $2\mathrm{Re}(g_{12}{g}_{23}{g}_{31})$, not an additional signed pair term. This is the permanent of a matrix $(g_{ij})$ for bosons and its determinant for fermions. Wick's theorem supplies the six terms only with the state assumption stated in part (a).

For completeness an exact bosonic Fock-state alternative can be given without that assumption. Let $\mathbf{\ell }$ be a mode-occupation decrement with ${\sum }_{\alpha }{\ell }_{\alpha }=3$. Define $A_{\mathbf{\ell }}(1,2,3)$ as the sum of ${\prod }_{j=1}^3{\phi }_{{\alpha }_j}({\mathbf{r}}_j)$ over all distinct ordered triples having these mode multiplicities. Acting three annihilation fields on the number state leaves mutually orthogonal states for different $\mathbf{\ell }$. ThereforeHere $N^{\underline{\ell }}=N!/(N-\ell )!$, with zero when $\ell >N$. This expression includes repeated-mode occupation corrections absent from the Gaussian permanent; for a single mode it reduces to $N(N-1)(N-2){\prod }_j|\phi ({\mathbf{r}}_j){|}^2$.