The arm consists of the cells strictly to the right of $(i,j)$ in its row of the Young diagram; its length is $a_{\lambda }(i,j)={\lambda }_i-j$. It excludes the anchor cell. At $(1,2)$ in $(3,2)$ the arm contains only $(1,3)$, so its length is one. Together with the leg of a Young-diagram cell and the anchor, it forms the hook of a Young diagram.

Take independent ensembles for two quantum channels and their Cartesian product ensemble. The Von Neumann entropy of a tensor product is the sum of the two entropies, both for its average output and for each labelled output. Thus its Holevo quantity is the sum of the two individual quantities. Taking ensembles arbitrarily close to each supremum proves the displayed inequality without requiring maximizing ensembles to exist. It does not assert additivity for arbitrary channels.

For $dX=dB+a{X}^{-1}dt$, put $\gamma =1-2a\in (0,1)$ and $v_R(x)=(R^{2-\gamma }{x}^{\gamma }-x^2)/(1+2a)$. It is nonnegative on $[0,R]$, vanishes at both endpoints and solves $\frac{1}{2}{v}_R^{\prime \prime }+(a/x)v_R^{\prime }=-1$ on $(0,R)$. Optional stopping at $T_{\epsilon }\wedge T_R$ gives $\mathbb{E}(T_{\epsilon }\wedge T_R)\le v_R(x)$ uniformly as $\epsilon \downarrow 0$. Thus the limiting exit to zero or $R$ occurs in finite time, not merely as an asymptotic approach to zero. Scale probabilities give ${\mathbb{P}}_x(T_0<T_R)=1-(x/R{)}^{\gamma }$; letting $R\to \infty$ gives almost sure finite hitting of zero.

Choose $\chi (z)=\sqrt{z^2-a^2}\sim z$ at infinity, cut on the finite crack. Its upper crack value is $i\sqrt{a^2-t^2}$, so $H=\chi G$ has jump $-2i\sqrt{a^2-t^2}p(t)$. The Sokhotski–Plemelj theorem gives the displayed Cauchy integral solution. It decays as $z^{-2}$ for integrable weighted traction. A possible homogeneous term $C/\chi$ would give a logarithmic displacement period; it is removed for a dislocation-free finite crack with no extra remote loading.

For two semi-infinite cracks separated by a ligament, choose $\chi (z)=\sqrt{a^2-z^2}>0$ on the ligament. Its upper crack values are $-i\sqrt{t^2-a^2}$ on the right and $+i\sqrt{t^2-a^2}$ on the left. Multiplication by $\chi$ converts the Hilbert problem for an antiplane crack to an additive jump; the Sokhotski–Plemelj theorem gives the displayed solution, assuming convergence. A homogeneous term $C/\chi$ transfers an extra resultant $\pi C$ through the ligament and has logarithmic displacement at infinity. The usual correction with no additional remote resultant excludes it. The far-field normalization is essential to uniqueness.

For the moving semi-infinite crack, take the branch cut of $\sqrt{z}$ on the negative real axis and $\sqrt{x}>0$ ahead of the tip. With steadily moving antiplane shear, face tractions ${\sigma }_{23}^{\pm }=-f(x)$ imply $\mathrm{Im}{F}^{\prime \pm }=f/\mu$. The antisymmetric displacement response gives $F^{\prime -}=-\overline{F^{\prime +}}$. Hence $G=\sqrt{z}{F}^{\prime }$ has jump $G^{+}-G^{-}=-2\sqrt{-x}f(x)/\mu$. The Sokhotski–Plemelj theorem gives the displayed Cauchy integral, assuming convergence and no extra remote homogeneous loading. For $x>0$ it is purely imaginary, so ${\sigma }_{23}={\int }_{-\infty }^0\sqrt{-s}f(s)/(x-s)ds/(\pi \sqrt{x})$. At fixed co-moving loading this ahead-of-tip stress is independent of speed; the displacement still has the factor $1/\beta$.

Under $\mathrm{diag}(t,t^{-1})$, acting by inverse substitution, $X^{d-j}{Y}^j$ has weight $2j-d$. If $r$ is the smallest active exponent of $Y$, the multiplicity at $[1:0]$ is $r$ and the Hilbert-Mumford value is $d-2r$. Conjugating replaces this point by any projective point. Thus strict positivity is exactly the displayed condition; replacing the strict inequality by $m_p\le d/2$ gives semistability. A binary quartic is stable precisely when it has no repeated projective root.

The two three-generator components on $\{a,b,c\}$ and $\{c,d,a\}$ are iterated HNN extensions of infinite cyclic groups. The subgroup on $\{a,c\}$ is free of rank two in each component by Britton's lemma. Amalgamating over these free subgroups gives the displayed infinite finitely presented group. Its cyclic squaring relations forbid every nontrivial finite quotient by the least-prime multiplicative-order argument.

Every finite-dimensional representation of a complex semisimple Lie algebra is a direct sum of irreducible representations.

For a complex semisimple Lie algebra and a choice of positive roots, every finite-dimensional Irreducible Lie algebra representation has a unique highest weight, which is a dominant integral weight. Conversely each dominant integral weight labels exactly one such irreducible module up to isomorphism. Its highest-weight vector spans the highest-weight space and generates the module. This classification and Weyl complete reducibility theorem reduce decomposing any finite-dimensional module to its irreducible highest-weight summands.

The pseudocomplement of $a$ is the greatest element disjoint from $a$. It need not satisfy $a\vee \neg a=1$. In a frame, double pseudocomplementation defines the Booleanization of a locale; its fixed elements are regular open elements.

The Hermitian part of a square matrix is half its sum with its conjugate transpose. It is a Hermitian matrix and satisfies $\mathrm{Re}(x^{\ast }Ax)=x^{\ast }Hx$. Its largest eigenvalue is the numerical abscissa, controlling instantaneous growth of the Euclidean norm under $x^{\prime }=Ax$.

A bounded Hermitian operator determines a Hermitian form $b_T(x,y)=⟨Tx,y⟩$ and its real-valued diagonal expression $q_T(x)=b_T(x,x)$. On a complex Hilbert space,Thus it is a quadratic form on the underlying real vector space; it is generally not a quadratic form over the complex field, whose scaling rule would involve ${\alpha }^2$. This distinction explains why replacing $x^{\ast }$ by $x^{\mathrm{T}}$ loses the geometry of a complex inner product.

For a finite-dimensional Hermitian operator with an orthonormal basis of eigenvectors $e_j$ and corresponding eigenvalues ${\lambda }_j$, the spectral theorem givesFor a unit vector, this is an average of the eigenvalues with nonnegative weights summing to one. It is also the Rayleigh quotient, so its minimum and maximum are the smallest and largest eigenvalues.

In the Hermitian operator diagram, take $x(\theta )=(\cos \theta ,\sin \theta {)}^{\mathrm{T}}$ for $D,P$, and $x_H(\theta )=\cos \theta e_1+\sin \theta e_3$ for $H$. The computed curves areThese paths illustrate the extremal values; the spectral theorem supplies the bounds for every complex unit vector, including those outside the plotted paths.

The real matrix space defined by $C+C^{\ast }=\mathrm{tr}(C)I$ consists exactly of $C=aI+i\mathbf{v}\cdot \mathbf{\sigma }$, with $a$ and the three components of $\mathbf{v}$ real. The Pauli matrices have $\mathrm{tr}({\sigma }_i{\sigma }_j)=2{\delta }_{ij}$, giving the displayed real symmetric bilinear form of signature $(3,1)$. The real isomorphism $C\mapsto aI-\mathbf{v}\cdot \mathbf{\sigma }$ to Hermitian matrices preserves this form when the latter have the polarized negative-determinant form $[\mathrm{tr}(AB)-\mathrm{tr}A\mathrm{tr}B]/2$.

For the $n=1$, $m=1$ Hermite vertical mode of an isothermal disk, the local dispersion relation is $[(\omega -{\Omega }_K{)}^2-{\Omega }_K^2{]}^2=k^2{c}_s^2(\omega -{\Omega }_K{)}^2$. Its low inertial-frame frequency branches satisfy $\omega \simeq \pm k{c}_s/2$. These waves transport a changing tilt of successive annuli, or a warp, with group velocity $\pm c_s/2$. The vertical velocity is approximately independent of height. A rigid tilt is the zero-wavenumber, zero-pattern-frequency limit; the gas still executes vertical orbital motion at frequency ${\Omega }_K$ in its rotating frame.

In coordinates $h(x,y,z)=\left(\begin{array}{ccc}1& x& y\\ 0& 1& z\\ 0& 0& 1\end{array}\right)$, the Heisenberg group law is $h(x,y,z)h(a,b,c)=h(x+a,y+b+xc,z+c)$. Commutation for every $(a,b,c)$ forces $xc=az$, hence $x=z=0$. The center of a group is therefore the $y$-axis. The surjective group homomorphism $h(x,y,z)\mapsto (x,z)$ has exactly this kernel of a group homomorphism, giving the stated quotient group by the first isomorphism theorem.

For an admissible pointed Heegaard diagram, $C{F}^{-}$ is a free $\mathbb{Z}[U]$ chain complex with one generator for each point of ${\mathbb{T}}_{\alpha }\cap {\mathbb{T}}_{\beta }$. Each generator is an unordered selection containing exactly one intersection on each alpha circle and exactly one on each beta circle. After choosing coherent orientations,Here $\hat{\mathcal{M}}$ is the moduli space modulo strip translation, $\mu$ is the Maslov index, and $n_z$ is multiplicity at the basepoint. Holomorphic representatives have nonnegative domain multiplicities. The differential lowers grading by one and $U$ lowers it by two. Boundary strata of one-dimensional moduli spaces give the cancellation proving ${\partial }^2=0$. Its homology is $H{F}^{-}$.

Use the convention that $C{F}^{-}$ is generated by $[\mathbf{x},i]$ with $i<0$, and $U[\mathbf{x},i]=[\mathbf{x},i-1]$. The chain complexes areThe last complex is generated by $[\mathbf{x},0]$. Equivalently it is $C{F}^{-}/UC{F}^{-}$ with its absolute grading shifted up by two. The short exact sequence $0\to C{F}^{-}\to C{F}^{\infty }\to C{F}^{+}\to 0$ gives the fundamental long exact sequence. For a rational homology sphere in each Spin-c structure, $H{F}^{\infty }\cong \mathbb{Z}[U,U^{-1}]$. The plus tower is ${\mathcal{T}}_d^{+}=\mathbb{Z}[U,U^{-1}]/U\mathbb{Z}[U]$, with bottom degree $d$; its corresponding minus tower is a free $\mathbb{Z}[U]$ module with top degree $d-2$.

A smooth oriented cobordism $W:Y_1\to Y_2$ with a Spin-c structure induces $U$-equivariant maps $F_{W,\mathfrak{s}}^{\circ }$ on the corresponding Heegaard Floer homology groups. A handle decomposition defines the maps using generator selection for one- and three-handles and holomorphic triangle counts for two-handles. Composition sums over extensions of the prescribed Spin-c structures. When the boundary determinant classes are torsion, the absolute grading shift isThe square is defined using the rational relative lift. If $b_2^{+}(W)>0$, the infinity map vanishes. Between rational homology spheres, when $b_1(W)=b_2^{+}(W)=0$, the infinity map is an isomorphism of Laurent towers.

For a knot $K$ in an integral homology three-sphere and $n>0$, fix a Spin-c structure ${\mathfrak{t}}_k$ on the $n$-surgery. Write ${\mathfrak{s}}_i$ for structures on zero-surgery with $⟨c_1({\mathfrak{s}}_i),[\hat{F}]⟩=2i$, using a capped Seifert surface. In compatible labels the completed-minus triangle isOne can equally use the plus theory without completing. The first map is the congruence-restricted zero-surgery trace map; the last is the sum of reverse, orientation-reversed $n$-surgery trace maps. The middle map uses a Heegaard triple with an auxiliary lens-space generator. For $n=1$ that lens space is $S^3$, giving the usual three two-ended cobordism maps. For $n>1$ it is important to retain the auxiliary boundary and its fixed Spin-c structure. The U-adic completion of Heegaard Floer homology makes the sums of maps well-defined.