For a highest-weight vector $v$ of weight $k$ in an irreducible finite-dimensional sl2 Lie algebra module, the basis $v,Yv,\dots ,Y^{k}v$ diagonalizes $XY$, with eigenvalue $(j+1)(k-j)$ on $Y^{j}v$. Summing those eigenvalues gives $\mathrm{tr}(XY)=k(k+1)(k+2)/6$. The formula $X{Y}^{n}v=n(k-n+1)Y^{n-1}v$ requires $v$ to be a highest-weight vector, not an arbitrary vector of the module.

A differential form of type (p, q) of type $(p,0)$ is holomorphic when its local coefficients in holomorphic coordinates are holomorphic functions, equivalently $\bar{\partial }\alpha =0$. It need not be $d$-closed on every complex manifold. Top-degree holomorphic forms are closed by bidegree. On a compact complex surface, holomorphic forms of all degrees are closed, even without a Kähler metric.

In an adapted smooth unitary frame with input-index connection convention, the Chern connection matrix has blocks $({\theta }_F,B;-{\overline{B}}^t,{\theta }_Q)$, where the second fundamental form of a holomorphic subbundle $B$ has type $(1,0)$. The Cartan curvature equation with input indices gives the displayed formula. If the ambient curvature is zero, ${\Theta }_F(X,\overline{X})=-B(X)B(X{)}^{†}$ is negative semidefinite. Dualizing the determinant line bundle therefore gives semipositive curvature.

The running describes the scale dependence of the scalar spectral index. Differentiating $n_s-1\simeq -6{ϵ}_V+2{\eta }_V$ with the potential slow-roll flow equations gives the displayed second-order slow-roll result. It includes the third potential slow-roll parameter and is generally smaller than the leading spectral tilt.

Write $n=q(p-1)+r$, $0\le r<p-1$. The regular partition labelling the sign representation in characteristic $p>0$ is $\lambda =((q+1{)}^r,q^{p-1-r})$, with zero parts removed. Its conjugate partition is $((p-1{)}^q,r)$, which satisfies the invariant vector criterion for a Specht module. The sign-twisted duality of Specht modules turns this invariant line into a sign quotient of $S^{\lambda }$, necessarily its unique simple head. Conversely, regularity says consecutive column heights differ by less than $p$; combining this with the invariant congruences forces these heights to be $p-1$ except possibly the last. For $p=2$, the label is $(n)$ because sign equals the trivial representation.

For good reduction of an elliptic curve over a complete local field of residue characteristic $p$, smoothness and the Hensel lemma make reduction surjective. Its kernel is a formal group on the maximal ideal, where multiplication by $q^m$, $q\ne p$, is bijective by prime-to-residue-characteristic multiplication on a formal group. Thus there is no nonzero $q$-power torsion in the kernel. Given a lift $P$ of a reduced point killed by $q^m$, choose the unique kernel point $R$ with $q^{m}R=q^{m}P$. Then $P-R$ is a torsion lift. This proves both surjectivity and uniqueness of torsion lifting. For finite residue field these groups are finite.

Let an elliptic curve over $\mathbb{Q}$ have good reduction at distinct primes $p,r$, with point counts $N_p,N_r$. For a torsion point, its $\ell$-primary part injects into every reduction whose residue characteristic differs from $\ell$. Consequently its $\ell$-power order has exponent at most $\min (v_{\ell }(N_p),v_{\ell }(N_r))$ when $\ell \ne p,r$, at most $v_p(N_r)$ when $\ell =p$, and at most $v_r(N_p)$ when $\ell =r$. This precise form avoids assuming injectivity on residue-characteristic torsion. Counts five at two and seven at three force trivial rational torsion.

For an elliptic curve over $\mathbb{Q}$, reduction at a good prime $p$ is injective on the $\ell$-primary torsion subgroup when $\ell \ne p$: its kernel is a formal group, where multiplication by $\ell$ is bijective. Hence a rational point of order $\ell$ forces $\ell \mid \#\tilde{E}({\mathbb{F}}_p)$. If for every possible prime $\ell$ some good $p\ne \ell$ has a point count not divisible by $\ell$, rational torsion is trivial. In particular two distinct good primes $p,r$ with coprime counts, $p\nmid \#\tilde{E}({\mathbb{F}}_r)$ and $r\nmid \#\tilde{E}({\mathbb{F}}_p)$ suffice: the coprime counts exclude other primes, and the two cross conditions exclude $p,r$. This argument does not incorrectly apply prime-to-$p$ injectivity to $p$-power torsion.

Choose a graph vertex $x=(n,x_2,\dots ,x_d)$ on the boundary of a cubic lattice box. Translation and reflection invariance identify the probabilities of $0\leftrightarrow x$ and $x\leftrightarrow 2n{e}_1$. Their intersection implies $0\leftrightarrow 2n{e}_1$, so positive association of random variables gives $C_{2n}\ge \varphi (0\leftrightarrow x{)}^2$. A union bound and cubic lattice symmetries choose such an $x$ with probability at least the box-boundary connection probability divided by the number of boundary graph vertices. This polynomial factor disappears in the exponential rate, identifying boundary and axial inverse correlation lengths.

This quartic has no primitive integral solution. If $5\nmid U$, reduction modulo five makes $W^2\equiv 2$, impossible. If $5\mid U$, coprimality gives $5\nmid V$ and $5\mid W$; dividing by twenty-five then gives $(W/5{)}^2\equiv 2V^4\equiv 2(\mathrm{mod}5)$, again impossible. It excludes the square class two in two-isogeny descent on $Y^2=X^3+100X$. The point $(5,25)$ supplies class five, so the subgroup property excludes class ten as well.

For $E_5^{\prime }:y^2=x^3+100x$, the two-isogeny descent classes $2$ and $10$ have no rational covering points. Their primitive integral quartics are respectively $N^2=2U^4+50V^4$ and $N^2=10(U^4+V^4)$, with $\gcd (U,V)=1$.

For the first equation, reduction modulo $5$ forces $5\mid U$ and $5\mid N$, since $2$ is not a square modulo $5$ and a nonzero fourth power is $1$. Put $U=5U_1,N=5N_1$. Dividing by $25$ gives $N_1^2=50U_1^4+2V^4$, again a nonsquare modulo $5$ because $5\nmid V$. For the second equation, a primitive pair has $U^4+V^4$ nonzero modulo $5$: it is $1$ or $2$. The right side has five-adic valuation exactly $1$, impossible for a square. Since $(5,25)$ lies on $E_5^{\prime }$, its square-class image is exactly $\{1,5\}$. Combined with the rank bound for a prime congruent-number elliptic curve computation, the two-isogeny rank formula yields rank exactly $1$ for $E_5:y^2=x^3-25x$.

For $y^2=x^3+a{x}^2+bx$ with $a\equiv 8$ and $b\equiv 9(\mathrm{mod}16)$, the square class $-1$ in a two-isogeny descent would require coprime integers $u,v$ and an integer $N$ with $N^2=-u^4+a{u}^2{v}^2-b{v}^4$. If $u$ is odd and $v$ even, the right side is $15$ modulo $16$; if $u$ is even and $v$ odd, it is $7$; if both are odd, it is $14$. None is a square modulo $16$. Therefore $-1$ does not occur in the two-torsion square-class homomorphism image.

This quartic covering in a two-isogeny descent has no solution with coprime integers $U,V$ not both zero. If precisely one is odd, the right side is $3$ modulo four. If both are odd, their fourth powers are one modulo sixteen and $14U^2{V}^2=14$ modulo sixteen, so the right side is twelve modulo sixteen. Neither residue is a square. Coprimality excludes the remaining parity case. Consequently the negative unit square class is absent from the two-torsion square-class homomorphism for $Y^2=X^3+14X^2+X$.

The continuous functional calculus of a bounded normal operator $T$ is a representation of $C(K)$, with $K=\sigma (T)$. The Riesz-Markov-Kakutani representation theorem represents $f\mapsto ⟨f(T)v,v⟩$ by a positive regular measure ${\mu }_v$. The assignment $f\mapsto f(T)v$ is isometric for the $L^2({\mu }_v)$ norm, and extends to a unitary map onto the closed cyclic subspace it generates. Multiplication by $z$ and $\overline{z}$ represent $T$ and $T^{\ast }$. Orthogonal decomposition into cyclic reducing subspaces, allowing an uncountable family, constructs the Borel functional calculus for a normal operator without requiring separability.

With $m$ beta numbers, the staircase has set $\{0,2,\dots ,2m-2\}$. Lowering a bead by a positive even number either hits an occupied bead or a negative position, so no even-size rim hook can be removed. The adjoint of multiplication by $p_r$ under the Hall inner product of symmetric functions is $r\partial /\partial p_r$. The power-sum border-strip multiplication formula therefore gives zero derivative with respect to every even power sum. Since the rational symmetric-function algebra is freely generated by the power sums, the stated conclusion follows.

Read a word of type $b$ from the left. Every $1$ is good; an $i+1$ is good when the number of previous good $i$'s exceeds the number of previous good $(i+1)$'s. The set $s(a,b)$ contains words with at least $a_i$ good $i$'s for each $i$. Their matched chains yield independent marked-column polytabloids; column antisymmetrization and the add-or-raise recursion give $\dim S^{a,b}=|s(a,b)|$ over every field.

If a projective complete intersection $Z\subseteq {\mathbf{P}}^n$ is cut out by a homogeneous regular sequence of degrees $d_1,\dots ,d_c$, then $\mathcal{I}/{\mathcal{I}}^2\cong {⨁}_j{\mathcal{O}}_Z(-d_j)$ and ${\mathcal{N}}_{Z/{\mathbf{P}}^n}\cong {⨁}_j{\mathcal{O}}_Z(d_j)$. The classes of the defining equations form the conormal basis: the first syzygies of a regular sequence are the Koszul complex relations, whose coefficients vanish modulo the defining ideal. Thus the generator classes have no residual linear relations. Dualizing gives the stated normal sheaf.

In the induction ring of ordinary symmetric-group characters, $[\lambda ][r]$ is the sum, with multiplicity one, of $[\nu ]$ for which $\nu /\lambda$ is a horizontal strip of $r$ cells. Here $[r]$ is the trivial representation of $S_r$. In the Littlewood–Richardson rule, content $(r)$ means every skew entry is $1$; strict column increase is possible exactly when there is at most one new cell in every column. This proves the rule.

The nonnegative integer $c_{\alpha \beta }^{\nu }$ is the multiplicity of $S^{\nu }$ in ${\mathrm{Ind}}_{S_{|\alpha |}\times S_{|\beta |}}^{S_{|\nu |}}(S^{\alpha }⊠S^{\beta })$ in characteristic zero. The Littlewood–Richardson rule counts it by Semistandard Young tableaux of skew shape $\nu /\alpha$, content $\beta$, and lattice word in right-to-left, top-to-bottom reading order. For example $c_{(3,2,1),(3,2)}^{(5,3,2,1)}=3$: the two new top cells must be ones, and the remaining reading letters can be $122,212,221$. This coefficient also multiplies Schur functions in the induction ring of symmetric-group characters.

The Kostant partition function counts nonnegative integer combinations of the positive roots of a root system, without ordering the summands. Multiply the formal geometric series ${\sum }_{k\ge 0}{e}^{-k\alpha }$ for every positive root. The coefficient of $e^{-\nu }$ counts precisely these combinations. Every coefficient is finite because a linear functional strictly positive on the positive roots bounds every participating integer coefficient. The sign of each exponent in the cancelling product must agree with that of this series.