An automorphism of a root system $\Phi \subset E$ is an orthogonal transformation of $E$ mapping $\Phi$ onto itself. The Weyl group is a subgroup of this automorphism group; additional automorphisms can act nontrivially on a Dynkin diagram while preserving a root basis.

The $A_2$ root system has equal-length simple roots ${\alpha }_1,{\alpha }_2$ with angle $2\pi /3$, and roots $\pm {\alpha }_1$, $\pm {\alpha }_2$, $\pm ({\alpha }_1+{\alpha }_2)$. Its Cartan matrix is $\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$. The corresponding complex simple Lie algebra is ${\mathfrak{sl}}_3(\mathbb{C})$, of dimension eight.

A reduced root system of rank one is $\Phi =\{\alpha ,-\alpha \}$. Its fundamental weight is $\alpha /2$, and its Weyl group has two elements. Its ordinary Dynkin diagram has one node; adjoining ${\alpha }_0=-\alpha$ gives affine Cartan matrix $\left(\begin{array}{cc}2& -2\\ -2& 2\end{array}\right)$, drawn as a double bond between equal-length roots.

A root is an individual nonzero vector in a root system. For a complex semisimple Lie algebra, it is a nonzero weight of the Adjoint representation restricted to a Cartan subalgebra; its root space consists of vectors $x$ satisfying $[h,x]=\alpha (h)x$ for all $h$ in that subalgebra.

A root vector is a nonzero element of a root space ${\mathfrak{g}}_{\alpha }$ in the root-space decomposition of a semisimple Lie algebra.

The holonomy representation is the natural action of the Riemannian holonomy group on $T_{p}M$. It induces representations on cotangent spaces and tensor powers. Irreducibility means that no nonzero proper real linear subspace of $T_{p}M$ is invariant.

For the retention parameter $p$, the Holevo capacity isEvery output has Von Neumann entropy at least the value on a pure input, $h_2((1+p)/2)$. Two equiprobable orthogonal pure inputs attain this minimum entropy individually and have maximally mixed average output, attaining the bound. Additivity of Holevo capacity makes the unassisted classical capacity of a quantum channel equal to this value. King's depolarizing-channel capacity paper proves additivity with an arbitrary other channel.

A quadratic differential with holomorphic local coefficients. If it is nonzero, its zeros have total order $4g-4$ on a compact genus-$g$ Riemann surface. Away from its zeros, a nonzero differential has flat coordinates whose changes of coordinate have the form $w\mapsto \pm w+c$.

A surface with local Euclidean charts whose changes of coordinate are $w\mapsto \pm w+b$, outside finitely many cone points. A nonzero holomorphic quadratic differential defines such a surface; a zero of order $m$ has cone angle $(m+2)\pi$.

For $q=\varphi (z)d{z}^2$, the length element is $|\varphi (z){|}^{1/2}|dz|$, locally Euclidean in flat coordinates. Zeros give cone points, not punctures. Its infimal length in a homotopy class need not determine the hyperbolic length of that class.

A fundamental wave and its second harmonic satisfy phase matching for a quadratic wave interaction when $\omega (2k)=2\omega (k)$. The square of the fundamental drives the second harmonic, while its product with the conjugate second harmonic drives the fundamental. Projecting these terms onto the resonant modes gives coupled amplitude equations rather than a uniformly valid correction at fixed amplitude.

The three fixed-$a$ estimates below cease to be uniform as the minimum of the phase approaches the endpoint. For the cubic endpoint-to-saddle transition, writeThe Taylor series of the hyperbolic sine gives, for bounded $s$ and fixed $l$,The cubic term controls the tail, so localization of the Laplace integral gives the uniform leading formulaHere $I$ is the cubic Laplace transition integral, equal to $\pi \mathrm{Hi}$ in terms of the Scorer Hi function.

To recover the endpoint regime, let $l\to +\infty$. Scale $s=v/(2^{1/3}l)$ in $I$; the cubic term becomes negligible and $I(-2^{1/3}l)\sim (2^{1/3}l{)}^{-1}$. Thereforewhich agrees with part (i) in the overlap $1\ll l\ll {\nu }^{2/3}$. At $l=0$, the Gamma integral gives $I(0)=3^{-2/3}\Gamma (1/3)$, recovering part (iii).

For $l\to -\infty$, set $M=-2^{1/3}l>0$. The exponent $Ms-s^3/3$ has its maximum at $s=\sqrt{M}$, with second derivative $-2\sqrt{M}$. Thus Laplace's method givesand the transition formula becomesFor $a=1-\eta$ with $\eta \downarrow 0$, part (ii) hasPutting $\eta =(-l){\nu }^{-2/3}$ reproduces both the exponential and its prefactor. For relative agreement of these leading exponentials, one may use the overlap $1\ll -l\ll {\nu }^{4/15}$, which makes $\nu {\eta }^{5/2}\to 0$. Thus the same transition integral connects all three regimes.

Minimize the data-misfit functional $J(x)=\frac{1}{2}\Vert Ax-y{\Vert }^2$. For a perturbation $h$, $DJ(x)[h]=\mathrm{Re}⟨A^{\ast }(Ax-y),h⟩$, so its gradient is $A^{\ast }(Ax-y)$. Gradient descent gives Landweber iteration:On a singular component, the iteration error is multiplied by $1-\tau {\sigma }_j^2$, whose modulus is less than one. Starting from zero preserves orthogonality to $\ker A$ and makes the exact-data iterates converge to $x^{†}$.

For the unit-step formulas and noise bound used below, rescale so that $\tau =1$ and $\Vert A\Vert \le 1$. Finite iteration count bounds amplification of small singular-value components; the stopping count provides the regularization parameter.

Since $(x^2-1)(-{\delta }_0)={\delta }_0$, write $u^{\prime }=-{\delta }_0+w$, where $(x^2-1)w=0$. The polynomial has two simple zeros. The distribution annihilated by a function with simple zeros lemma givesIntegrating with the Heaviside step function yieldsThis includes every solution: its derivative has exactly the required delta terms, and any two primitives differ only by a constant distribution.

The distributional derivative of the Heaviside step function gives $H^{\prime }={\delta }_0$, while $((x-1)H(x-1){)}^{\prime \prime }={\delta }_1$. A particular solution is therefore $H(x)-2(x-1)H(x-1)$.

The difference from any other solution has second derivative zero. Applying a distribution with zero derivative is constant twice shows that difference is affine. ThusThe ramp function is continuous at $x=1$, and its first derivative jumps there, producing the indicated delta source in its second derivative.

The Leibniz rule rewrites the left side as $(xu{)}^{\prime }$. By a distribution with zero derivative is constant,The Dirac delta multiplication identity $x{\delta }_0^{\prime }=-{\delta }_0$ supplies a particular solution, and the preceding division result supplies all solutions of $xu=c$. HenceMultiplying by $x$ and then differentiating verifies the equation. The coordinate-multiplication kernel and the constant-derivative kernel show that no additional solutions are missing.