Using $I_3=\mathrm{diag}(1/2,-1/2,0)$ and flavour $Y=\mathrm{diag}(1/3,1/3,-2/3)$, the three coordinate vectors of the SU(3) group defining group representation have the displayed weights of a representation. The complex-conjugate group representation has their negatives. These coordinates use flavor hypercharge; another Cartan normalization rescales the vertical axis.

For the curve $Y^2=X^3+4p^{2}X$ with $p\equiv 3(\mathrm{mod}4)$, the classes $p$ and $2p$ in two-isogeny descent lead to $N^2=p(U^4+4V^4)$ and $N^2=2p(U^4+V^4)$. A primitive solution forces $p\mid N$ and $p\nmid V$. The resulting congruences make $-1$ a square modulo $p$, a contradiction. Thus these coverings have no rational points.

If $p\equiv 3(\mathrm{mod}8)$, this quartic covering in a two-isogeny descent has no primitive integral solution. Opposite parities of $U,V$ give $N^2\equiv 2(\mathrm{mod}8)$. If both are odd, their fourth powers are one modulo sixteen and $p^2\equiv 9(\mathrm{mod}16)$, giving $N^2\equiv 20(\mathrm{mod}32)$. Neither residue is a square. This excludes the class $2$ on $Y^2=X^3+4p^{2}X$.

The two-isogeny descent quartics associated with square classes $2$ and $-2$ on $Y^2=X^3+24X^2+64X$ have no rational points. A rational solution can be scaled to integral $W,U,V$ with coprime integers $U,V$. If $U$ is odd, reduction modulo sixteen gives $2$ or $10$ for the positive class, and $14$ or $6$ for the negative class; none is a quadratic residue. If $U=2u$ is even, then $V$ is odd and the right side is $32(\pm u^4+3u^2{V}^2\pm V^4)$, with the same sign on the two signed terms. The bracket is odd, so its 2-adic valuation is exactly five, impossible for a nonzero square. These two parity cases exhaust primitive solutions.

At a prime of good reduction of an elliptic curve with residue characteristic not dividing $m$, multiplication by $m$ extends to a finite étale map of the smooth proper elliptic group scheme over the valuation ring. A local rational point extends to a section by properness. Its division fibre is therefore a finite étale torsor, so the Kummer map of an elliptic curve gives an unramified cohomology class. This bounds ramification in the Weak Mordell-Weil theorem.

A direct proof describes the coordinates, not only the cohomology class. Reduce $P$ by properness and choose a point $\overline{Q}$ with $[m]\overline{Q}=\tilde{P}$ over a finite extension of the residue field. Over the corresponding unramified extension of the complete local field, lift $\overline{Q}$ by the Hensel lemma. The error $[m]Q_0-P$ is in the kernel of reduction of an elliptic curve; prime-to-residue-characteristic multiplication on a formal group corrects it uniquely. There is no $m$-torsion in that kernel, even after a ramified field extension, so every division point with that reduction equals this unramified lift. Consequently all division points of an elliptic curve above $P$ have unramified coordinate fields. The torsion case is discussed in J. S. Milne's Elliptic Curves, Chapter V, Proposition 8.1.

Let $E:y^2=(x-\alpha )(x-\beta )(x-\gamma )$ with distinct algebraic integers in a number field $K$. For a rational point $P$, its halving field is unramified outside the finite set $S$ of primes dividing $2(\alpha -\beta )(\alpha -\gamma )(\beta -\gamma )$. At $v\notin S$, the elliptic-curve discriminant is a unit, so there is good reduction. A half of the reduction of $P$ is defined over a finite extension of the residue field. Pass to the corresponding unramified extension of the completion and lift this half using the Hensel lemma. The doubling error lies in the formal kernel of a minimal Weierstrass equation. Since two is a unit, prime-to-residue-characteristic multiplication on a formal group corrects this error uniquely. Every half differs by rational 2-torsion, so all halves are unramified at $v$. The halving cocycle with rational two-torsion has image of order at most four, making the halving field a Galois extension of degree at most four. The bounded-degree extensions with restricted ramification are finite in number; there are only finitely many homomorphisms from their finite Galois groups to $E[2]$. Thus $E(K)/2E(K)$ is finite, without assuming any form of the Mordell-Weil theorem.

Let $H$ be a Cartan subalgebra of a finite-dimensional complex Lie algebra $L$. In the adjoint generalized-weight decomposition for a nilpotent Lie algebra, its zero summand satisfies $L^0=H$. Nilpotence gives $H\subseteq L^0$. If $L^0/H$ were nonzero, all adjoint operators of $H$ on this quotient would be nilpotent, so the Engel theorem would supply a nonzero coset $x+H$ with $[H,x]\subseteq H$. That contradicts self-normalization. The derivation rule ensures that $L^0$ is a subalgebra, so the quotient action is well-defined.

Composition of endomorphisms and wedge product of differential forms give the class $[{\Theta }^k]\in H^k(M,{\Omega }_M^k\otimes \mathrm{End}E)$. Even total degree of vector-bundle curvature and the noncommutative telescoping identity give ${\Theta }_1^k-{\Theta }_0^k=\bar{\partial }{\sum }_{j=0}^{k-1}{\Theta }_1^{j}a{\Theta }_0^{k-1-j}$ whenever ${\Theta }_1-{\Theta }_0=\bar{\partial }a$. Thus all powers are metric independent; no commutativity of endomorphisms is assumed.

The flat polar metric $d{r}^2+r^{2}d{z}^2$ has a circle isometry away from $r=0$. Its local Buscher rules give the displayed dual metric and dilaton. The dual Ricci tensor has components $R_{rr}=-2/r^2$ and $R_{\tilde{z}\tilde{z}}=-2/r^4$, while the dilaton Hessian has components $1/r^2$ and $1/r^4$. Thus the leading metric-dilaton Weyl condition $R_{ab}+2{\nabla }_a{\nabla }_b\Phi =0$ holds despite nonzero curvature. Flat spectators complete a critical target; the scalar anomaly condition also holds because $|\nabla \Phi {|}^2=1/r^2$, $\square \Phi =2/r^2$ and $R=-4/r^2$. The construction is local on $r>0$, excluding the degenerating circle.

For a Hermitian holomorphic bundle over a Kähler manifold, let $\Theta$ act by exterior multiplication. The bundle-valued Kähler identities give $[\Lambda ,{\Delta }_{\bar{\partial }}]=-i(\Theta \wedge -{)}^{\ast }$, and taking adjoints yields $[{\Delta }_{\bar{\partial }},L]=i\Theta \wedge -$. This measures the failure of the bundle-valued Dolbeault Laplacian to commute with the Lefschetz operator.

This elliptic curve has torsion subgroup $\{O,(0,0),(2,0),(10,0)\}\cong (\mathbb{Z}/2\mathbb{Z}{)}^2$ and the infinite-order point $(1,3)$. Good-reduction point counts at three and eleven are four and sixteen, which bound the torsion order by four using reduction of torsion points on an elliptic curve. Doubling $(1,3)$ gives $(361/36,-323/216)$; after the shift $X=x-4$ to $y^2=X^3-28X-48$, its nonintegral coordinate proves nontorsion by the Nagell–Lutz theorem. The two-isogeny formula gives the partner $Y^2=X^3+24X^2+64X$. The first two-torsion square-class homomorphism has image $\{1,2,5,10\}$, while the second has image $\{1,-1\}$ by the signed divisor-two obstruction for an isogeny covering. The two-isogeny rank formula therefore gives $2^r=4\cdot 2/4=2$, so the rank is one. The argument does not determine the index of the subgroup generated by $(1,3)$.

The vector-length multiset of a rank-two Euclidean lattice determines it up to an orthogonal map. Let $a$ be the shortest nonzero length. Removing two copies of each positive integer multiple of $a$ removes one primitive lattice line; the shortest remaining length $b$ is the shortest independent-vector length. Those two vectors form a lattice basis: otherwise a nonzero point in the centred basis parallelogram would be an independent lattice vector of length less than $b$. Lattice-point density determines the covolume $A$. The basis Gram matrix then has diagonal entries $a^2,b^2$ and absolute off-diagonal entry $\sqrt{a^2{b}^2-A^2}$. This proves two-dimensional spectral rigidity for a flat torus through its dual lattice.

A continuum field level line is specified by the harmonic boundary heights it creates and the conditional zero-boundary field remaining on each side. Since a Continuum Gaussian free field is a random distribution, this is not the pointwise set of its zeros. For the height jump appropriate to the Green-function normalization, the zero-height line between opposite Dirichlet heights is chordal ${\mathrm{SLE}}_4$.

Let $M$ be a continuous local martingale on $[0,T]$, with deterministic $M_0=m$, and let $V$ be a continuous nonnegative adapted finite-variation process with deterministic $V_0=v<\infty$. Suppose $d[M{]}_t=-d{V}_t$ and $V_T=0$ almost surely. For each real $u$, the Itô formula showsis a complex local martingale: its two finite-variation terms cancel. Its absolute value is at most one, so it is a true bounded martingale. Consequentlywhich identifies the entire normal distribution $N(m,v)$ by its characteristic function.

For example, $M_t=B_t$ and $V_t=T-t$ give $B_T\sim N(0,T)$. The residual-variance condition $V_T=0$ is essential for this conclusion. With $M_t=0$ and $V_t=1$ throughout, the displayed exponential is still a bounded martingale, but its terminal mean variable is the constant zero, not $N(0,1)$. In a field-revelation construction, $M_t$ represents a conditional mean and $V_t$ the remaining conditional variance; only after the latter has vanished does the martingale expectation identify the terminal sample law.

For a line bundle $M$ on $C\times T$, with $C$ a smooth projective curve, this function is locally constant. Twist both $M$ and $M^{-1}$ by a large power of an ample line bundle on $C$. Near a chosen fiber their first sheaf cohomology groups vanish. The Riemann-Roch theorem and upper semicontinuity of their zeroth sheaf cohomology groups then bound the degree from above and below by the degree on that fiber. Connectedness of $T$ is necessary for a single constant across all fibers.

The standard-tableau right-ideal decomposition of $\mathbb{C}{S}_n$, tensored over that algebra with $V^{\otimes n}$, gives this direct sum of general linear group modules. With $e_t=h_t/H_{\lambda }$, the map $e_t\mathbb{C}{S}_n\otimes T\to e_{t}T$ is an isomorphism. Individual summands generally need not be symmetric group submodules; instead each is a Schur module for the commuting action.

If $\sigma$ has ${\alpha }_r$ cycles of length $r$, contraction of matrix entries around its cycles gives $\mathrm{tr}(\sigma {\xi }^{\otimes n})={\prod }_r\mathrm{tr}({\xi }^r{)}^{{\alpha }_r}$. The formula holds for every endomorphism, not only diagonalizable ones. The Schur–Weyl duality decomposition equates it with a sum of products of Specht and Schur characters.

A column of length greater than $m$ antisymmetrizes more than $m$ vectors, giving zero. When there are at most $m$ rows, place the $i$th basis vector in every tensor position of row $i$. Row symmetrization multiplies by a nonzero factorial product, and column antisymmetrization is nonzero because the vectors in every column are distinct. This proves the precise nonvanishing criterion for a Schur module.

For a bounded operator on a real Hilbert space, $⟨Lv,v⟩=⟨Sv,v⟩$ with $S=(L+L^{\ast })/2$. Consequently the Euler-Lagrange equation of $⟨Lv,v⟩-2⟨f,v⟩$ is $Sv=f$. It is $Lv=f$ only when $L$ is self-adjoint or when the relevant solution also annihilates the skew part. For example $L=\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)$ has positive quadratic form $\Vert v{\Vert }^2$, but that form contains no information about its skew part.

At the critical field,The quasiparticle gap closes and the low-energy branch is linear, unlike a massive excitation. This is the Ising-chain quantum critical point of the Transverse-field Ising model; the cusp in the nonnegative dispersion represents the two opposite propagation directions.