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This is the supremum of reliable classical bits per use of a memoryless quantum channel when sender and receiver share unlimited prior quantum entanglement. If $|{\psi }_{\rho }{⟩}_{RA}$ purifies the input, let ${\omega }_{RB}=({\mathrm{id}}_R\otimes \mathcal{N})(|{\psi }_{\rho }⟩⟨{\psi }_{\rho }|)$. The single-letter capacity formula is ${\max }_{\rho }[S(\rho )+S(\mathcal{N}(\rho ))-S({\omega }_{RB})]$, the maximum quantum mutual information between reference and output. The preshared resource is independent of the message.

A quantum erasure channel transmits an input state unchanged with probability $1-p$ and replaces it by an orthogonal, recognizable erasure flag with probability $p$. Its output space has one more dimension than the input. This is a quantum channel for $0\le p\le 1$ and preserves all input coherences on the nonerased branch.

In a reflection-symmetric mean-field dynamo model, dipole parity can be represented by an even poloidal-potential amplitude and an odd toroidal-field amplitude; quadrupole parity reverses these assignments. The names describe the reflection classes appropriate to the model, not arbitrary electric multipoles. A complex amplitude $P=A+iB$ allows a phase rotation by $i$ to interchange the two classes when the unperturbed equations have that phase symmetry.

For $k(t)=k_0\sqrt{1+(t/T{)}^2}$, factor resistive damping out of the shearing-mode equations by $R=e^{-D}G$, $S=e^{-D}H$, $D=\eta \int k^{2}dt$. They imply $G^{\prime \prime }={\alpha }^2{k}^{2}G$, so a growing WKB approximation has $G\propto k^{-1/2}{e}^{|\alpha |\int kdt}$ and $H\simeq i\mathrm{sgn}(\alpha )kG$. Its magnetic energy is proportional to $k\exp [2|\alpha |\int kdt-2\eta \int k^{2}dt]$. At long times the positive quadratic exponent is overtaken by negative cubic resistive damping. A well-separated growing interval exists for sufficiently small positive magnetic diffusivity, with $\eta k_0/|\alpha |\ll 1$ at fixed nonzero $|\alpha |k_{0}T$, and the displayed leading peak time. This is transient amplification, not an asymptotically self-sustaining dynamo.

For a nonempty $A\subseteq \{0,1{\}}^n$ of relative size at least $\epsilon \in (0,1]$, its Hamming distance function $F(x)=d(x,A)$ has the bounded differences property with all constants one. If $\mu =\mathbb{E}F$, the concentration bound at $F=0$ gives $\epsilon \le 2e^{-2{\mu }^2/n}$, so $\mu \le t/2$ for $t=\sqrt{2n\log (2/\epsilon )}$. Then $\mathrm{Pr}(F>t)\le 2e^{-2(t-\mu {)}^2/n}\le \epsilon$, proving that the radius-$t$ neighbourhood contains at least $(1-\epsilon )2^n$ cube points.

Vary the antisymmetric auxiliary electromagnetic field tensor and the electromagnetic four-potential independently. The first variation sets $F_{ab}={\nabla }_a{A}_b-{\nabla }_b{A}_a$ algebraically, and variation of the potential gives ${\nabla }_b{F}^{ab}=0$. Substitution of the algebraic equation reduces the first-order action to the usual negative quadratic Maxwell action. Metric variation gives the same stress tensor after eliminating the auxiliary field, because the chain-rule term multiplying its variational equation then vanishes.

For $n$ independent uniform points in a rectangle, condition on one point with normalized coordinates $(x,y)$. It is a maximal external point if all others avoid the upper-right rectangle of area $(1-x)(1-y)$, giving conditional probability $[1-(1-x)(1-y){]}^{n-1}$. Integrating gives $H_n/n$, where $H_n$ is a harmonic number. Linearity of expectation over the $n$ indicator random variables gives $\mathbb{E}{K}_n=H_n$. The same answer holds for independent continuous coordinate distributions, by transforming each coordinate with its distribution function.

The row rank of a matrix is the dimension of the vector space spanned by its rows. Elementary row operations preserve this row space; multiplication on the right by an invertible matrix preserves all linear relations among the rows.

To see why row rank equals column rank, write a reduced row echelon form as $R=EA$, where $E$ is a product of elementary matrices and is an invertible matrix. Reversibility of the row operations gives equal row spaces for $R$ and $A$. Also $Rz=0$ if and only if $Az=0$, so the columns have exactly the same linear relations. If $R$ has $r$ nonzero rows, its pivot columns are the first $r$ standard basis vectors of ${\mathbb{F}}^m$ and span its column space. Its nonzero rows are linearly independent, since their matrix entries in the pivot positions form an identity matrix. Both dimensions therefore equal $r$, and both counts transfer back to $A$.

The column rank of a matrix is the dimension of its image of a linear map, equivalently the dimension of the linear span of its columns. Multiplication on the left by an invertible matrix preserves all column linear relations, and elementary column operations preserve their linear span. In row echelon form the pivot columns form a basis of the column space. Its dimension equals the number of nonzero rows, and hence the row rank.

For composable linear maps, $\mathrm{im}(AB)=A(\mathrm{im}B)$ lies in $\mathrm{im}A$ and has dimension at most that of $\mathrm{im}B$. This proves the bound on the rank of a matrix product. When $A$ is $m\times p$, $B$ is $p\times n$, and $m,n\ge p$, the bound $p$ is attained by $A=\left(\genfrac{}{}{0ex}{}{I_p}{0}\right)$ and $B=(I_{p}0)$.

For matrices $A\in {\mathbb{F}}^{m\times n}$ and $B\in {\mathbb{F}}^{n\times p}$, their matrix product is the matrix $AB\in {\mathbb{F}}^{m\times p}$ with entries $(AB{)}_{ij}={\sum }_{k=1}^n{A}_{ik}{B}_{kj}$. The operation producing it is matrix multiplication, representing composition of the corresponding linear maps.

One step of symmetric Gaussian elimination removes the pivot vertex and connects its remaining neighbors into a clique. The rule follows from the Schur complement update $a_{ij}\leftarrow a_{ij}-a_{ik}{a}_{kj}/a_{kk}$. Repeating it predicts the generic factor sparsity; exceptional numerical cancellations can remove structurally predicted entries.

For a standard null momentum $(E,0,0,E)$ in signature $(+---)$, define $K_i=M^{0i}$ and $J_i={ϵ}_{ijk}{M}^{jk}/2$. The stabilizing generators are $J_3,N_1,N_2$. The Poincare algebra gives $[J_3,N_1]=i{N}_2$, $[J_3,N_2]=-i{N}_1$ and $[N_1,N_2]=0$, the Lie algebra of planar Euclidean motions. Evaluating the Pauli-Lubanski pseudovector gives $W^2=-E^2(N_1^2+N_2^2)$.

In a finite-dimensional unitary representation of $ISO(2)$, its two commuting Hermitian translation generators have a finite joint spectrum. Conjugating by every planar rotation rotates that spectrum. A finite set invariant under all rotations contains only the origin, so both generators vanish. An irreducible representation is then a one-dimensional rotation character, giving a helicity state. Nontrivial translations instead require the infinite-dimensional continuous-spin representation. The same argument applies to the physical double cover of this little group.

For a future null standard momentum $k$, choose Lorentz transformations $L(p)k=p$ and a unitary irreducible representation $d$ of its little group. The Wigner rotation $R(\Lambda ,p)=L(\Lambda p{)}^{-1}\Lambda L(p)$ acts on the internal state. The resulting induced representation acts on wavefunctions with invariant measure $d^{3}p/(2p^0)$. Trivial little-group translations give fixed helicity; nontrivial translations give a continuous-spin representation with infinitely many polarization states.

For the Z boson vertex $g{\gamma }^{\mu }(c_v-c_a{\gamma }^5)/(2c_W)$, sum over the three initial polarizations and final fermion spins. The symmetric Dirac trace is $4(c_v^2+c_a^2)(k_1^{\mu }{k}_2^{\nu }+k_1^{\nu }{k}_2^{\mu }-g^{\mu \nu }{k}_1\cdot k_2)$. The vector-axial interference is antisymmetric and drops out of the massive vector polarization sum. The longitudinal term drops out by current conservation for massless final particles. Contracting the remaining trace gives the displayed result. The massive-vector spin average divides it by three; two-body Lorentz-invariant phase space then gives $\Gamma =g^2{M}_Z(c_v^2+c_a^2)/(48\pi c_W^2)$ per color. These coefficients are twice those used with a vertex prefactor $g/c_W$.

Each effectively massless active neutrino species contributes the same invisible Z boson partial decay width. Subtracting calculated or measured visible widths from the total width therefore measures the displayed number, assuming no additional invisible channels. The inference concerns particles lighter than $M_Z/2$ with ordinary weak couplings. It does not exclude heavier neutrinos or right-handed neutrinos that are gauge singlets. Early LEP resonance measurements were consistent with three light active species, with statistical uncertainty. See the OPAL report on its 1989 data for the original resonance-width comparison.

For a resting W boson with real spatial polarization vector along $z$, the fermion spin sum gives $\sum |\mathcal{M}{|}^2=g^2{M}_W^2{\sin }^2\theta /2$ for massless charged lepton and antineutrino momenta. The two-body Lorentz-invariant phase space gives the displayed angular decay rate. Since $\int {\sin }^2\theta d\Omega =8\pi /3$, the integrated width is $g^2{M}_W/(48\pi )$, equal to the unpolarized width. The massive-vector spin average changes the angular distribution, not the total width. No initial spin average is taken for a specified pure polarization.

In the leading massless collinear parton approximation, $P^2=0$ and $q^2=-Q^2$ imply $(\xi P+q{)}^2=2\xi P\cdot q-Q^2$. With Bjorken x $x=Q^2/(2P\cdot q)$ and $P\cdot q>0$, the Dirac delta change-of-variable formula yields the displayed identity. The positive-energy branch of the final parton also gives $\delta (E_q+\xi E_P-E_k)/E_k=2\delta ((E_q+\xi E_P{)}^2-E_k^2)$ locally on physical support. Keeping target mass instead introduces ${\xi }^2{M}^2$ and changes the root and Jacobian.

For a surface thermostat, the planes containing $F$ form the projective bundle of $T(SM)/\mathbb{R}F$. The vertical Maslov cycle is the section selecting the image of the vertical vector field of a surface unit tangent bundle. It is a smooth hypersurface. In the local projective chart $\mathbb{R}F+\mathbb{R}(V+rH)$, it is $r=0$; the lifted flow satisfies $\dot{r}=1$ on the cycle, hence crosses it transversely.