On ${\bigwedge }^2{\mathbb{C}}^4$, the invariant Hermitian form and volume tensor define the displayed antilinear map with square one, commuting with SU(4). Its fixed space satisfies $T^{12}=\overline{T^{34}}$, $T^{13}=-\overline{T^{24}}$, and $T^{14}=\overline{T^{23}}$. Three complex choices give six real coordinates, and this fixed space has the original exterior square as its complexification. It is therefore a real representation of dimension six, rather than merely the twelve-dimensional realification of an arbitrary six-dimensional complex representation.

For distinct integral roots $\alpha ,\beta ,\gamma$ with $m=\max (|\alpha |,|\beta |,|\gamma |)$, the finite set in unramified halving fields for a split cubic contains only primes at most $2m$. A quadratic Galois character unramified away from $s$ primes is represented by a signed square-free integer supported on those primes, giving at most $2^{s+1}$ characters. The halving cocycle with rational two-torsion has two independent coordinates in $\mathbb{Z}/2\mathbb{Z}$, so the quotient has at most $2^{2s+2}$ elements. Since $m\ge 1$ and there are at most $m$ primes at most $2m$, this gives $2^{2m+2}$. The bound is deliberately crude but depends only on the root size.

The Lefschetz operator commutes with the bundle-valued Dolbeault Laplacian on the entire smooth form space exactly when the Chern connection is flat. The Lefschetz-Dolbeault curvature commutator gives one direction. Conversely test its curvature-wedge operator on degree-zero sections with arbitrary prescribed values at any point to force every vector-bundle curvature endomorphism to vanish.

At a prime of good reduction of an elliptic curve, the reduction map is a group homomorphism on all local points. If $N$ annihilates the reduced finite group, every $NP$ reduces to the identity at infinity. A finite point with integral affine coordinates reduces to an affine point, so a nonidentity $NP$ cannot have integral affine coordinates. The identity has no affine coordinates and must be treated separately.

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.

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 ${\mathrm{SLE}}_4$.

If a continuous mean martingale $M$ and a nonnegative finite-variation process $V$ satisfy $d⟨M⟩=-dV$, then the Itô formula shows $\exp (iM-V/2)$ is a bounded complex martingale. Its terminal expectation establishes an entire Gaussian distribution with the initial mean and variance. This is stronger than matching only the first two moments in a random-domain field construction.

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.