At equally spaced distinct times, a mean profile is affine exactly when its consecutive second differences vanish. These form $p-2$ independent contrasts. At unequal times, compare adjacent slopes instead: $({\mu }_{i+1}-{\mu }_i)/(t_{i+1}-t_i)-({\mu }_i-{\mu }_{i-1})/(t_i-t_{i-1})$. A Hotelling test of linear hypotheses tests these restrictions jointly with unknown covariance.

In a local trivialization, a vector-bundle connection with matrix $A$ defines the displayed complement of the vertical tangent space. The frame-change law makes these complements independent of the trivialization. Their dependence on the fiber coordinate is linear; an arbitrary horizontal complement need not satisfy this condition and need not define a connection on a vector bundle. Horizontal lifts satisfy the ordinary differential equation defining parallel transport.

The holonomy of a connection along a closed curve is the group element relating the endpoints of its horizontal lift. A flat connection can have nontrivial holonomy around a noncontractible loop.

For $M$ binary units and desired patterns ${\xi }^{\mu }$, use these symmetric weights and set $w_{ii}=0$. For one pattern, the field at that pattern is $(M-1){\xi }_i/M$, so it is stable. Several patterns add cross-talk terms; storage by this rule alone does not guarantee stability of every arbitrary collection of patterns.

Writing $m_{\mu }=M^{-1}{\sum }_j{\xi }_j^{\mu }{x}_j$ givesfor $P$ stored patterns. The subtraction removes the forbidden self-couplings. It makes explicit how the desired association competes with all other stored patterns. An unintended stable state can therefore arise even though every update obeys asynchronous Hopfield energy descent.

For symmetric zero-diagonal weights and zero thresholds, update one binary unit to the sign of its current field, retaining it at a zero field. The displayed energy change is negative on every actual flip. There are finitely many states, so a fair asynchronous update schedule eventually reaches a fixed point that is a minimum against single-unit flips. Synchronous updates can instead cycle, and local minima need not be global minima.

To obtain the displayed identity, subtract the two energies. Terms not involving $x_k$ cancel, and symmetry combines the two copies of each pair. With thresholds included the result is $\Delta E=-(x_k^{\prime }-x_k)h_k$. The zero-threshold expression above is its specialization. At $h_k=0$, keeping the old state rules out motion along equal-energy states. Finiteness then excludes infinitely many strict decreases, while fairness ensures that an unstable unit cannot be ignored forever. A local minimum here means no single flip lowers energy; equality directions can remain.

The asynchronous condition matters even for two units: with $w_{12}=w_{21}=1$ and zero thresholds, simultaneous updates alternate $(1,-1)$ and $(-1,1)$ forever. Either single-unit update instead reaches an aligned stable state.

The leg consists of the cells strictly below $(i,j)$ in its column of the Young diagram; its length is $l_{\lambda }(i,j)={\lambda }_j^{\prime }-i$, where ${\lambda }^{\prime }$ is the conjugate partition. It excludes the anchor cell. At $(1,2)$ in $(3,2)$ the leg contains only $(2,2)$, so its length is one. Transposing the Young diagram exchanges the leg and the arm of a Young-diagram cell.

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 $[\mathrm{tr}(AB)-\mathrm{tr}A\mathrm{tr}B]/2$.