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In the defining representation, the diagonal element of the Cartan subalgebra acts on the four standard basis vectors with eigenvalues $r,s,-r,-s$. ThereforeTheir weight diagram is the four-point cross below.

Let $M_f$ be the matrix of multiplication by $f$ in the basis from Question 1. Define the Coleman norm operator by the norm for a finite free ring extension:The entries lie in $\varphi (R)$, and $\varphi :R\to \varphi (R)$ is an isomorphism, so this determines $Nf$ uniquely. If $f$ is a unit, $M_f$ is invertible and its determinant is a unit. Also $M_{fg}=M_f{M}_g$, giving $N(fg)=NfNg$. Splitting the conjugates, or diagonalizing multiplication after adjoining the roots of unity and inverting $p$, gives the equivalent formulaIn particular, this definition has the required target $R^{\times }$, rather than merely a ring with extra roots of unity in its coefficients.

For the requested ring congruence, observe that the reduction of $\varphi$ is the injective map $\overline{g}(T)\mapsto \overline{g}(T^p)$. If $\varphi (g)\in p^{k}R$, its reduction first gives $g=p{g}_1$. Then $\varphi (g_1)\in p^{k-1}R$; repeat to obtain $g\in p^{k}R$. Applying this to $g=f-1$ provesThis congruence reflection for Frobenius substitution is valid for all $f\in R$, without assuming $f$ is a unit.

A further property needed for the fixed points is Coleman norm contraction. If $g=1+p^{k}h$ with $k\ge 1$, expand the determinant:Here $e_j(M_h)\in \varphi (R)$ is the coefficient of degree $j$ in $\det (I+t{M}_h)$. The trace formula from Question 1 makes the linear term divisible by $p^{k+1}$; every later term is divisible by $p^{2k}$ and $2k\ge k+1$. Reflecting the resulting ring congruence through $\varphi$ givesIf $Nf=f$ and $f\in 1+pR$, iteration gives $f\in 1+p^{k}R$ for every $k$. Since ${\bigcap }_k{p}^{k}R=0$, $f=1$.

To identify all fixed points, we also need $Nf\equiv f(\mathrm{mod}pR)$ for every unit $f$. In characteristic $p$, the extension ${\mathbb{F}}_p[[T]]/{\mathbb{F}}_p[[T^p]]$ has degree $p$ and its norm for a finite free ring extension is $\overline{f}\mapsto {\overline{f}}^p$. Indeed, after passing to fraction fields it is a purely inseparable field extension of degree $p$, so multiplication has only the eigenvalue $\overline{f}$, with multiplicity $p$, after splitting. Its determinant is ${\overline{f}}^p$. Thereforeand congruence reflection for Frobenius substitution gives $Nf\equiv f(\mathrm{mod}pR)$.

Now choose any lift $f_0\in R^{\times }$ of a given $a\in A^{\times }$ and put $f_n=N^n{f}_0$. Multiplicativity givesThus $(f_n)$ is a Cauchy sequence for the finer of the topologies on integral formal power series and converges to a unit $w(a)$ reducing to $a$. The Coleman norm operator is continuous: multiplication matrices and their determinants are continuous, and the finite-free coordinate decomposition and its inverse preserve the $p$-adic filtration defining the finer of the topologies on integral formal power series. Hence $Nw(a)=w(a)$. If two fixed units reduce to $a$, their quotient is fixed and belongs to $1+pR$, so it is $1$ by the preceding argument. This proves uniqueness, independence of the initial lift, and multiplicativity of the Coleman norm-fixed lift. In particular,Every unit $f\in R$ consequently has a unique factorization $f=w(\overline{f})v$ with $v\in 1+pR$, giving the canonical Coleman decomposition of power-series units:The final assertion as printed is false. The original PDF identifies a ring with a product of unit groups. With multiplication, $A$ has a zero and is not a group. With addition, it has exponent $p$, whereas $W\times A^{\times }$ has a nonidentity element of order $2$: $-1\in W$ because $p$ is odd and $N(-1)=(-1{)}^p=-1$. Thus neither interpretation makes the printed assertion true. The two boxed isomorphisms above are the valid natural conclusions; they do not require guessing which symbols the author intended to replace.

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.

Use base-two logarithms for Von Neumann entropy, so the units are bits, and adopt $0{\log }_{2}0=0$. All systems here may be taken finite-dimensional; the entropy arguments also apply when the displayed Von Neumann entropies are finite. The density operator of the joint pure state is ${\rho }_{AB}=|{\Psi }_{AB}⟩⟨{\Psi }_{AB}|$. It is a rank-one orthogonal projector, whose only nonzero eigenvalue is one. ConsequentlyThe reduced density matrices are ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$ and ${\rho }_B={\mathrm{Tr}}_A{\rho }_{AB}$. The quantum conditional entropy is defined by the entropy differenceFor the pure state under consideration this becomes $S(B|A)=-S({\rho }_A)$. To determine when it is negative, write the Schmidt decompositionThe eigenvalues of ${\rho }_A$ are the ${\lambda }_j$, together with possible zeros. Each contribution $-{\lambda }_j{\log }_2{\lambda }_j$ is nonnegative, and their sum vanishes exactly when one eigenvalue equals one. Equivalently the Schmidt rank is one and $|{\Psi }_{AB}⟩$ is a product state. If the pure state is entangled, then $r\ge 2$, every nonzero eigenvalue is strictly below one, and the Von Neumann entropy is strictly positive. Thus the pure-state negative conditional entropy criterion givesThe purity assumption matters: for mixed density operators, negative quantum conditional entropy is not necessary for entanglement.

Let a density operator on a Hilbert space ${\mathcal{H}}_S$ have the spectral decompositionIntroduce an auxiliary Hilbert space ${\mathcal{H}}_R$ with an orthonormal basis containing $r$ vectors $|j{⟩}_R$, and setThis vector has norm one. Its partial trace isThis explicitly constructs a purification of a density operator. The same construction works for an arbitrary trace-class density operator: its nonzero eigenvalues form a finite or countable summable family, and the displayed vector converges in the Hilbert space norm. The entropy inequality below is understood with finite Von Neumann entropies, avoiding undefined differences of infinities.

Apply this construction with $S=AB$ to obtain a joint pure state on $ABC$. The Schmidt decomposition across $AB:C$, $A:BC$ and $B:AC$ respectively givesNow apply Subadditivity of Von Neumann entropy to the reduced density matrices on $BC$ and $AC$:These inequalities say $S(A)-S(B)\le S(C)$ and $S(B)-S(A)\le S(C)$. Replacing $S(C)$ by $S(AB)$ proves the Araki–Lieb inequality:The subadditivity used here concerns the actual reduced density operators, rather than statistical independence. In particular it follows from nonnegativity of quantum relative entropy, sinceand similarly for $AC$. Thus no independence assumption on the purification of a density operator is needed.

Diagonalize the reduced density matrix on Alice's Hilbert space:For every positive eigenvalue, define a vector on Bob's Hilbert space byThe definition of the partial trace givesThus the $|b_j⟩$ are orthonormal. Extend the $|a_j⟩$ to an orthonormal basis of Alice's Hilbert space. If $|a⟩$ lies in the zero-eigenvalue subspace of ${\rho }_A$, thenThere is therefore no component of $|{\Psi }_{AB}⟩$ in those additional basis directions. Expansion in Alice's orthonormal basis yields the Schmidt decompositionIts positive Schmidt coefficients are $s_j=\sqrt{{\lambda }_j}$ and its Schmidt rank is $r$. Taking the two partial traces and using the orthonormality of both families givesHence the two reduced density matrices have exactly the same nonzero eigenvalues, including multiplicities. Their numbers of zero eigenvalues may differ when the local Hilbert spaces have different dimensions. In particular their Von Neumann entropies agree.

Factor the coefficients into a tensor product:This is already a Schmidt decomposition with one term. ConsequentlyEach reduced density matrix is the rank-one orthogonal projector onto $(|0⟩-|1⟩)/\sqrt{2}$, so its other eigenvalue is zero. The joint pure state is a product state, with no entanglement.

The coefficient matrix in the computational orthonormal bases, with Alice's index labelling rows, isSince the entries are real, the reduced density matrices areBoth have trace one and determinant $1/9$. Their characteristic polynomial is ${\lambda }^2-\lambda +1/9$, so the nonzero eigenvalues areBoth are positive. Thus the Schmidt rank, or the Schmidt number of this pure state, isIf “Schmidt numbers” denotes the individual positive Schmidt coefficients instead, those areTheir squares are ${\lambda }_{\pm }$; giving both the Schmidt rank and the Schmidt coefficients removes this terminology ambiguity. The pure state is entangled because its Schmidt rank exceeds one.

Choose $m\ge \ell (\lambda )$, pad the partition of an integer with zeros, and define its beta numbers byThey form a strictly decreasing set of nonnegative integers. Conversely, from its decreasing enumeration recover ${\lambda }_i={\beta }_i-m+i$. With $m=\ell (\lambda )$ these are exactly the first-column hook lengths, since $h_{i1}={\lambda }_i+{\lambda }_1^{\prime }-i={\lambda }_i+m-i$. Padding changes the numbers: increasing $m$ by one replaces $B$ by $\{0\}\cup (B+1)$, and leaves the represented partition of an integer unchanged.

Arrange positions $0,1,2,\dots$ on $p$ runners, putting position $r+pk$ on runner $r$ at height $k$. A bead occupies each position in the beta set. This is the abacus of a partition. The hook criterion in a beta set says that a bead $b$ and a gap $a<b$ correspond to a hook of length $b-a$; replacing $b$ by $a$ removes the corresponding rim hook. In particular removing a rim $p$-hook moves a bead from $b$ to the vacant position $b-p$ immediately above it on the same runner. Its leg length counts the beads in the intervening positions of the full linear numbering.

For an explicit example, take $\lambda =(4,2,1)$, $p=3$, and $m=3$. Its beta set is $\{6,3,1\}$. Runner zero has beads at $3,6$, runner one has a bead at $1$, and runner two is empty. Packing upwards gives $\{3,1,0\}$, which represents $(1)$; thus the three-core is $(1)$ and the weight is two.

Every bead slide reduces the sum of occupied positions, so repeated slides terminate. The number $n_r$ of beads on each runner is unchanged. At termination they must occupy exactlyon runner $r$, since any gap above a bead would allow a further slide. Hence the final beta set is uniquely determined by the initial runner counts, independently of the order of hook removals. Increasing the number of beads by $p$ replaces $B$ by $\{0,\dots ,p-1\}\cup (B+p)$, adds one bead to each runner, and gives exactly the same represented core after packing. General changes in padding follow by applying $B\mapsto \{0\}\cup (B+1)$; packing commutes with this operation up to the cyclic relabelling of runners. Therefore the result is independent of both removal order and padding:The weight of a partition is $w_p(\lambda )=(n-|C_p(\lambda )|)/p$.

A block of a group algebra is an indecomposable two-sided summand $F{S}_{n}e$, where $e$ is a primitive central idempotent. The identity decomposes into pairwise orthogonal such idempotents; a module belongs to a block when its idempotent acts as identity and the other block idempotents act as zero. For ordinary characters, use a splitting p-modular system $(K,\mathcal{O},F)$, or extend the residue field first: the block idempotents lift to $\mathcal{O}{S}_n$, and determine which ordinary irreducible $K{S}_n$-modules lie in each $p$-block. The symmetric group is split over its prime field, so extending $F$ introduces no change to the resulting partition of an integer labels.

The Nakayama block theorem, traditionally called the Nakayama Conjecture for symmetric groups, isThe simple modules in that block are exactly the $D^{\nu }$ with regular labels having that core. At fixed $n$, the common core also fixes the weight. Here is an essay proof route through central characters, specifying the main algebraic and combinatorial steps and why each is needed.

First introduce the commuting Young–Jucys–Murphy elementsTheir commutators vanish because contributions supported on each triple of letters cancel. Symmetric polynomials in the $J_i$ are central: for adjacent transpositions $s_i$, the relations $s_i{J}_i=J_{i+1}{s}_i-1$, $s_i{J}_{i+1}=J_i{s}_i+1$ show that $s_i$ commutes with symmetric expressions in $J_i,J_{i+1}$, and it commutes with the other $J_j$. Since these transpositions generate $S_n$, the symmetric expressions commute with the whole group algebra.

The substantial generation ingredient is the integral Jucys–Murphy description of the center of a symmetric-group algebra: every conjugacy-class sum is an integral symmetric polynomial in the $J_i$. Its proof expands symmetric monomials in these sums of transpositions and organizes the resulting class sums by the number of moved letters and cycle filtration; triangular elimination supplies integral spanning, rather than merely rational spanning. A useful first identity in this construction isIndeed, when adjoining the letter $n$, either choose $t$ and leave it as a singleton cycle, or choose a transposition $(jn)$ and insert it into the cycle containing $j$; this gives every permutation uniquely. The full generation step distinguishes the individual class sums which this first identity groups together. Its integrality is essential: a description only over $\mathbb{Q}$ could use denominators divisible by $p$ and would not prove generation of the modular center. Since class sums form the integral basis of the center, this integral theorem reduces to the same generation statement over $F$.

Next determine their scalar action. In the characteristic-zero tableau basis of $S^{\lambda }$, $J_i$ is diagonal with entry the Content of a Young-diagram cell occupied by $i$, namely column minus row. This follows recursively along the multiplicity-free branching chain: the commuting operators preserve each path line, and $J_{i+1}=s_i{J}_i{s}_i+s_i$ determines the successive eigenvalues. Consequently a symmetric polynomial $f(J_1,\dots ,J_n)$ acts on $S^{\lambda }$ byThis is independent of the ordering of the cells. The scalar identity holds on the integral Specht lattice, so after modular reduction it depends only on the multiset of contents modulo $p$. Every composition factor has this same central character of a block.

Two simple modules are in the same block exactly when their central characters agree. The center of a block is a commutative local finite-dimensional algebra; its semisimple quotient gives one scalar character, while the orthogonal block idempotents separate different blocks. By the generation theorem, equality on all symmetric polynomials in the $J_i$ is enough. Conversely, different residue multisets are detected by elementary symmetric polynomials: the monic polynomial ${\prod }_{(i,j)\in \lambda }(z-(j-i))$ over $F$ recovers the multiplicity of each residue by unique factorization, including multiplicities larger than $p$. Thus block equivalence is exactly equality of the integersThis gives the residue-content criterion for symmetric-group blocks without assuming that the sum of contents alone distinguishes blocks.

Finally identify residue content with the core. A removable rim $p$-hook has consecutive contents modulo $p$, hence removes one cell of each residue. Therefore shapes with the same core and the same size have the same residue counts. For the converse choose $m$ divisible by $p$ and at least $n$, and write $q_r=n_r-m/p$ for the runner charges. Starting from the empty partition of an integer gives $q_r=0$. Adding a cell of residue $r$ moves a beta-number bead from residue $r-1$ to residue $r$, increasing $q_r$ and decreasing $q_{r-1}$. The same changes occur in $c_s-c_{s+1}$ when $c_r$ increases by one. Induction on the cells therefore givesEqual residue counts give equal charges and hence equal packed runner configurations, so they give the same core. This completes both directions of the block criterion. The argument exposes the three essential bridges: integral center generation, its scalar content evaluation, and the abacus identification of residue data with a unique core.

Write $T=\left(\begin{array}{cc}P& Q\\ R& S\end{array}\right)$ and $J=\left(\begin{array}{cc}0& I_l\\ -I_l& 0\end{array}\right)$. Block multiplication givesIts vanishing is equivalent toThis identifies $L$ with the symplectic Lie algebra ${\mathfrak{sp}}_{2l}(\mathbb{C})$.

We now supply the requested root-space decomposition, including explicit vectors. Let $E_{ab}$ be the matrix units, put $h_i=E_{ii}-E_{l+i,l+i}$, and define the linear functionals ${\epsilon }_i$ on the diagonal Cartan subalgebra by ${\epsilon }_i(h)={\lambda }_i$. Then $H={⨁}_i\mathbb{C}{h}_i$. The Cn root system and a corresponding matrix root basis of the symplectic Lie algebra areEach listed vector satisfies the block conditions just proved. For a diagonal matrix $h$, the matrix unit calculation $[h,E_{ab}]=(h_{aa}-h_{bb})E_{ab}$ gives all the requested eigenvalues explicitly:The diagonal vectors and difference-root vectors span every possible $P$ block. The sum-root and doubled-root vectors span all symmetric $Q$ and $R$ blocks. They are linearly independent, as is evident from their disjoint independent block entries. Equivalently there are $l+2l^2=l(2l+1)$ of them, exactly the dimension $l^2+2l(l+1)/2$ of $L$. The zero weight space is precisely $H$: a zero weight forces $P$ diagonal and every entry of $Q,R$ to vanish. Hence every nonzero root space is one-dimensional and

Put $D_1=X_1-Y_1$, $D_2=X_2-Y_2$ andThe difference vector is $D=BZ$. Its mean is zero because the two sons have the same mean vector. Its covariance matrix isFor example, $\mathrm{Cov}(D_1,D_2)=b-c-c+b=2(b-c)$. ThusThe positive definiteness assumption supplies $a-b>0$ and $a+b-2c>0$: these are eigenvalues of the original matrix on within-son contrast and between-son difference directions. In particular the resulting difference covariance is positive definite.

Extend the selected ${\ell }_1$ to an orthonormal basis of eigenvectors, writing it as $v_1$. For a unit vector orthogonal to it, $a_1=0$, and thereforeA unit eigenvector $v_2$ attains equality. Thus ${\ell }_2$ is a unit eigenvector for the largest eigenvalue on ${\ell }_1^{\perp }$, and the maximum is ${\lambda }_2$. If the largest eigenvalue is repeated, a second orthogonal vector in that eigenspace is allowed and ${\lambda }_2={\lambda }_1$; there is no uniquely determined first pair of directions.

Let ${\lambda }_1\ge \cdots \ge {\lambda }_p\ge 0$ be the eigenvalues of the symmetric covariance matrix $V$, and choose an orthonormal basis of eigenvectors $v_1,\dots ,v_p$. If $\ell ={\sum }_j{a}_j{v}_j$ has unit length, then ${\sum }_j{a}_j^2=1$ andEquality is attained by $v_1$, and precisely by unit vectors in the top eigenspace. Hence choose ${\ell }_1$ to be any unit eigenvector for the largest eigenvalue; the maximum variance is ${\lambda }_1$. This proves the maximizing property of the Rayleigh quotient, rather than merely specifying a stationary point.