The hook length of cell $(i,j)$ in the Young diagram of a partition of an integer $\lambda$ is the number of cells in its Young-diagram hook:Here ${\lambda }^{\prime }$ is the conjugate partition. There is one hook length for every cell, so the multiset of hook lengths has cardinality $|\lambda |$.

For $\lambda =(3,2)$ and ${\lambda }^{\prime }=(2,2,1)$, compute $h_{1,1}=(3-1)+(2-1)+1=4$ and $h_{1,2}=(3-2)+(2-1)+1=3$. Continuing cell by cell gives the hook graph of a partitionor the row array $[[4,3,1],[2,1]]$. Its hook product is $4\cdot 3\cdot 1\cdot 2\cdot 1=24$. A cell has hook length one exactly when it is a Removable node of a Young diagram: it has no cell to its right or below. Here these are $(1,3)$ and $(2,2)$. Transposing the Young diagram gives the array $[[4,2],[3,1],[1]]$ with the same hook product; this explains geometrically why conjugate shapes have equal counts in the hook-length formula.

For a code with product state inputs to $n$ uses of a memoryless quantum channel, Subadditivity of Von Neumann entropy bounds its output Holevo quantity by $n{\chi }^{\ast }(\Lambda )$. The Holevo bound and Fano's inequality then give, for asymptotic rate $R$, average and hence maximum error at least $1-{\chi }^{\ast }(\Lambda )/R$ in the limit inferior. In particular, no code above that rate can have vanishing maximum error, even with a collective decoder measurement.

A channel's Holevo capacity is additive with another channel when their joint capacity equals the sum of the individual capacities. If this holds with every other channel, the channel's tensor powers have capacity $n{\chi }^{\ast }(\Lambda )$ and regularization does not increase its unassisted classical capacity of a quantum channel. Additivity is a special property, not a general rule for quantum channels.

A finite-dimensional irreducible representation of a complex semisimple Lie algebra is minuscule if its weights form one Weyl group orbit. Every weight multiplicity is then one, because multiplicities are invariant under the Weyl group and the highest weight has multiplicity one.

If a dominant integral weight $\lambda$ lies in the root lattice, its finite-dimensional irreducible highest-weight representation contains zero as a weight. First $\lambda$ is a nonnegative integral combination of simple roots: a negative part would have negative inner product with $\lambda$, contradicting dominance. For any nonzero weight $\mu =\sum m_i{\alpha }_i$ with $m_i\ge 0$, the identity $(\mu ,\mu )=\sum m_i(\mu ,{\alpha }_i)>0$ supplies an index with $m_i>0$ and $⟨\mu ,{\alpha }_i^{\vee }⟩>0$. By Injectivity of sl2 lowering above weight zero, applying the corresponding lowering operator produces a nonzero vector of weight $\mu -{\alpha }_i$. Repetition reaches zero.

For $m_H>2m_Z$, the Higgs boson coupling to Z bosons gives $\mathcal{M}=(2m_Z^2/v){ϵ}_1^{\ast }\cdot {ϵ}_2^{\ast }$. The massive vector polarization sum gives $\sum |\mathcal{M}{|}^2=(m_H^4/v^2)(1-4x+12x^2)$, with $x=m_Z^2/m_H^2$. The two-body decay phase space and identical final-state symmetry factor $1/2$ giveThe contraction uses $k_1\cdot k_2=(m_H^2-2m_Z^2)/2$. The on-shell rate vanishes at threshold; below threshold an off-shell calculation is required.

For a finite-index subgroup $L\le {\mathbb{Z}}^2$, its projection to the first coordinate is $a\mathbb{Z}$ and its intersection with the second axis is $\{0\}\times d\mathbb{Z}$, with $a,d>0$. Choose $(a,b)\in L$, reducing $b$ modulo $d$. Together with $(0,d)$ it is a basis of this row lattice: subtracting a multiple of $(a,b)$ from any vector leaves a vector on the second axis. These parameters are unique, and reduction of the two coordinates shows $[{\mathbb{Z}}^2:L]=ad$. Consequently an integral matrix of positive determinant $n$ has a unique representative of this form under left multiplication by $S{L}_2(\mathbb{Z})$, with $ad=n$. This proves the lattice facts underlying determinant-n matrix representatives for Hecke operators.

The row lattice of an integer matrix with $m$ rows is the set of their integer linear combinations. It is an Euclidean lattice in its real span, since it is contained in the discrete set ${\mathbb{Z}}^n$ and spans that real vector space. For a nonsingular two-by-two integer matrix, the row Hermite normal form in rank two gives a canonical basis and proves that its index in ${\mathbb{Z}}^2$ is the absolute value of its determinant.

The upper numbering is obtained by applying the inverse Herbrand function to the index in the lower ramification numbering. It is compatible with quotient Galois groups, whereas lower numbering is compatible with subgroups.

For multiplication $(x,y,z)(a,b,c)=(x+a,y+b,z+c+xb)$, the left-invariant frame isIts Lie brackets of vector fields satisfy $[E_1,E_2]=E_3$ and the other basis brackets vanish.

A Heegaard diagram marks the boundaries of compressing disks from the two sides of a Heegaard surface. Attaching two-handles along the two collections, and capping resulting spherical boundary components, reconstructs the three-manifold.

In a constant-volume gravity-current box model without settling, $\varphi ={\varphi }_0$ and $g^{\prime }=g{\varphi }_0(\gamma -\beta t)$. A constant Froude number then gives finite runout length of a gravity current at the neutral-buoyancy time $\gamma /\beta$. Further motion requires a different physical model once the stable density stratification is lost.

For a time-independent strictly positive Dirichlet realization of an elliptic operator and $\psi \in L^2(U)$, the eigenfunction expansionsolves $u_t+Lu=0$ with homogeneous Dirichlet boundary conditions. For $t\ge \epsilon >0$, every power ${\lambda }_m^N$ times the exponential is bounded, so the series and all its time derivatives converge in all the Sobolev domains of powers of an elliptic Dirichlet operator. Elliptic regularity and the Sobolev embedding theorem give smoothness up to the spatial boundary for positive time. Parseval identity and the dominated convergence theorem give $u(t)\to \psi$ in $L^2$ as $t\downarrow 0$.

Under a product measure on a finite Boolean lattice, an increasing event $E$ and a decreasing event $G$ obey the displayed inequality. The complement $G^c$ is an increasing event, so Harris' inequality gives $\mathbb{P}(E\cap G^c)\ge \mathbb{P}(E)\mathbb{P}(G^c)$. Subtract from $\mathbb{P}(E)$ to obtain the result. This implication includes degenerate Bernoulli distribution parameters $0$ and $1$.

For $\varphi =A\cos (q_{0}z)$, the gradient-quartic Brazovskii model has volume-averaged free-energy densityThe averages are $⟨{\cos }^2⟩=⟨{\sin }^2⟩=1/2$ and $⟨{\cos }^2{\sin }^2⟩=1/8$. For $B>0$, the stable amplitude is zero for $a\ge a_c$ and satisfies $A^2=(a_c-a)/(B{q}_0^2)$ below it. The mean-field approximation therefore predicts a continuous phase transition.

For $H=\int [a{\varphi }^2/2+\kappa |\nabla \varphi {|}^2/2+\gamma ({\nabla }^2\varphi {)}^2/2+B{\varphi }^2|\nabla \varphi {|}^2]$ and an even positive trial kernel $J(q)$, define $S_0={\sum }_q^{+}1/J(q)$ and $S_2={\sum }_q^{+}{q}^2/J(q)$ using a positive-wavevector sum for a real field. Isserlis theorem gives $⟨H_4{⟩}_0=4B{S}_0{S}_2/V$. The Feynman-Bogoliubov inequality therefore has stationary kernelsThus both the mass and gradient coefficient shift. In the thermodynamic limit, their self-consistency equations areA coarse-graining ultraviolet cutoff $\Lambda$ is necessary for the first integral in three dimensions: its large-$k$ radial integrand tends to a constant. Physical solutions require $J(q)>0$.

If $\alpha =a+bi$ is a Gaussian integer and $P$ is a nonzero prime ideal of $R=\mathbb{Z}[\alpha ]$, then $R/P$ is a finite field. Complex conjugation preserves $R$, so a nonzero $\beta \in P$ gives the positive integer $\beta \bar{\beta }\in P$. Primality supplies a rational prime $p\in P$. Since $\alpha$ satisfies the monic polynomial $t^2-2at+a^2+b^2$, the ring $R/P$ is a quotient of ${\mathbb{F}}_p[t]/(\bar{F})$ and has at most $p^2$ elements. It is a finite integral domain, hence a field. The proof does not require $R$ to be a principal ideal domain.

For a nearly circular stellar orbit of radius $a_b$ in gas moving at $(1-\eta )$ times the local circular speed, the averaged tangential gas drag on the centre of mass is $-\eta {\kappa }_b{v}_K$. Orbit-averaged drag work givesIf $m_2/m_1=ϵ\ll 1$ and aerodynamic stopping time scales as $m^k$, $0<k<1$, then $\tau /{\tau }_b\simeq \eta {ϵ}^k\ll 1$. Internal contraction can outpace stellar migration. Tightly gas-coupled drift requires solving radial and azimuthal motion together instead of this weak-drag approximation.

For a weakly perturbed circular binary planetesimal, the constant differential headwind does zero work averaged over the unperturbed orbit. Orbit-averaged drag work givesWith constant aerodynamic stopping times, $a(t)=a_0{e}^{-t/\tau }$. This averaging assumes drag is slow compared with the orbital period and does not appreciably distort the circular orbit within one period.

On a bounded domain of positive measure, the constant functions $u_k\equiv k$ satisfy $\mathrm{TV}(u_k)=0$ while $\Vert u_k{\Vert }_{BV}=k|\Omega |\to \infty$. Thus the total variation seminorm on a domain is neither coercive in the $BV$ norm nor strictly convex. Additional control of the mean can remove this obstruction; the Poincaré inequality for total variation supplies the needed bound on suitable connected domains.