The commuting actions of the symmetric group and the general linear group on a tensor power are mutual commutants. The displayed sum ranges over partitions with at most $\dim V$ rows. The Specht modules and Schur modules are the simple factors for the two actions. In particular a primitive Young symmetrizer selects one copy of the matching Schur module.

The commutant of the permutation action on $V^{\otimes n}$ identifies with the symmetric tensors in $\mathrm{End}(V{)}^{\otimes n}$. Polarization spanning of symmetric tensors spans it by $a^{\otimes n}$. Interpolating $(a+tI{)}^{\otimes n}$ at $n+1$ values where $a+tI$ is invertible expresses each such power using invertible ones. Hence the Schur algebra is the linear span of the diagonal general linear action.

A morphism of schemes between integral schemes can have a scheme-theoretic fibre that is neither reduced nor irreducible. For the map of affine lines given by $t=x^2(x-1{)}^2$, the fibre at $t=0$ has ring $k[x]/(x^2(x-1{)}^2)$. The Chinese remainder theorem identifies it with the product of two dual number rings, so it consists of two nonreduced double points. Its nilpotent structure is invisible to the set-theoretic fibre.

The scheme-theoretic fibres over positive real points, negative real points, zero, and nonreal closed points have real coordinate algebras ${\mathbb{R}}^2\times \mathbb{C}$, ${\mathbb{C}}^2$, $\mathbb{R}[x]/(x^4)$, and ${\mathbb{C}}^4$, respectively. Their ordinary irreducible component counts are $3,2,1,4$, so there are four real-scheme isomorphism classes. The last case pulls back an irreducible quadratic to a squarefree degree-eight polynomial, with four conjugate pairs of nonreal roots. The zero fibre retains its nilpotents and is not a reduced point.

Blow up $(0,1,0)$ in ${\mathbb{A}}_{x,y,z}^3$ and compose with $xy:{\mathbb{A}}^3\to {\mathbb{A}}^1$. Nonzero fibres are ${\mathbb{G}}_m\times {\mathbb{A}}^1$. The zero fibre consists of the strict transforms of the two coordinate planes and the exceptional ${\mathbb{P}}^2$, all with multiplicity one. Its plane crossings make it singular; its closed ${\mathbb{P}}^2$ makes it nonaffine. The centre is deliberately on just one component of $xy=0$, keeping the fibre reduced.

An exact $P(k)=A/k^3$ has divergent unsubtracted point variance. Its finite isotropic covariance difference is the displayed logarithm, unchanged by simultaneous dilation of $r,r_0$. Thus the nonzero-mode scaling statement has a natural subtraction interpretation. Exact dilation invariance of a continuous finite-variance isotropic covariance itself would instead force a constant covariance, corresponding only to a zero-mode spectral measure. Cutoffs or a zero-mode subtraction must accompany the nontrivial spectrum.

Give $M\times X$ the diagonal right monoid action. On its equivariant maps into $Y$, define $(ek)(g,x)=e(kg,x)$. The evaluation map of an exponential object is $e(1,x)$ and currying of $h:Z\times X\to Y$ is $\hat{h}(z)(g,x)=h(zg,x)$. Both maps are equivariant, and they are inverse under evaluation. The left multiplication $kg$ in this right action is essential in a noncommutative monoid.

A complex torus with period matrix of a complex torus $\Omega$ admits a Hodge metric precisely when there is a nonsingular integral skew-symmetric matrix $Q$ satisfying the displayed conditions. The entries of $Q$ are the periods of the invariant Kähler form over lattice two-tori. The first condition imposes type $(1,1)$, and the second imposes positivity; the factor $-i$ corresponds to the convention $\omega (v,Jv)>0$.

On $\mathbb{C}/(\mathbb{Z}+\mathbb{Z}i)$, a Hermitian form linear in its first argument has integral imaginary part on the period lattice exactly when its real scalar coefficient is integral. Positivity is equivalent to $t>0$. The associated holomorphic line bundle has degree $t$ because its First Chern class is represented by $(it/2)dz\wedge d\overline{z}$. In particular the oriented Chern pairing is minus the imaginary part under this convention.

If the identity is applied with probability $p$ and the three nonidentity Pauli operators each with probability $(1-p)/3$, the Bloch vector retention factor is $(4p-1)/3$. Thus complete depolarization occurs at $p=1/4$, not at zero. The Holevo capacity of a qubit depolarizing channel becomes $1-h_2((2p+1)/3)$ in this identity-probability parametrization.

For nonzero $a,b$ in a number field, the product of their quadratic Hilbert symbols over all places is one. All but finitely many symbols are one. Over $\mathbb{Q}$, this is equivalent to quadratic reciprocity with its supplementary laws. It supplies the compatibility relation between the local invariants of a global quadratic form.

The stocks $S_t=e^{B_t-t/2}$ and $S_t^{\prime }=(1+S_t)/2$ start at one, are true square-integrable zero-rate martingales, and have perfect positive correlation at each positive time. Their bounded multiplicative volatilities are $1$ and $S_t/(1+S_t)$, which differ throughout. The additive intercept is what allows the same correlation but different stochastic-exponential coefficients.

Fix a common Brownian motion $W$ and a finite horizon $T$. Let positive zero-rate stock prices satisfywith predictable processes $\sigma ,{\sigma }^{\prime }$ bounded in absolute value by a deterministic constant. Their stochastic exponentials are martingales with finite second moments.

If $S_T=a{S}_T^{\prime }$ for a deterministic constant $a$, equality of the expected values forces $a=1$. The martingale identity $S_t=\mathbb{E}[S_T\mid {\mathcal{F}}_t]$ then gives $S_t=S_t^{\prime }$ at each $t$. Continuity of the paths makes this equality hold simultaneously for all times outside a single null set.

Subtract the two Itô integrals. The Itô isometry givesSince $S_t>0$, the coefficients are equal almost everywhere for the product measure $dt\otimes \mathbb{P}$. The common driving Brownian motion is part of the statement.

Perfect positive correlation gives an affine terminal relation that may have a nonzero intercept. It does not supply the proportionality used here; the terminal correlation does not identify adapted stock volatility example demonstrates that distinction.

For an implicit Runge-Kutta method, algebraic stability means that $b_i\ge 0$ and the symmetric matrixis positive semidefinite. Here $b_1=b_2=1/2$ and $A+A^T=e{e}^T/2$, soThe method is algebraically stable. This also explains why the property matters for nonlinear problems. If a vector field satisfies $⟨f(v)-f(w),v-w⟩\le 0$, the Runge-Kutta contractivity identity for two solutions, with stage differences $D_i$ and vector-field differences $F_i$, isThe first sum is nonpositive and the second vanishes. Thus, whenever the implicit stage equations have the relevant solutions, their updates are contractive in the Hilbert space norm.

The zero Dirichlet boundary conditions permit an exact finite-interval calculation. Let $T=\mathrm{tridiag}(1,-2,1)$ on the $M$ interior grid points. The Dirichlet discrete Laplacian has an orthogonal eigenvector basisConsequently the update matrix $G=(I-rT{)}^{-1}$ has modal amplification factorsFor every physical $r\ge 0$, all denominators are at least one. Orthogonal diagonalization therefore provesuniformly in $M$, $r$ and $n$. The same bound controls initial perturbations; a forcing increment is propagated by contraction, so successive increments accumulate at most by their sum. The highest-order consistent method is unconditionally stable for all $r\ge 0$. This proof incorporates the boundary conditions, whereas a periodic Fourier mode calculation alone would not do so.

For completeness, if “range” is interpreted algebraically to include negative $r$ on a fixed grid, the exact power-stable range isIndeed, for $r<0$ every denominator is less than one, so $|g_j|\le 1$ requires $1+4r{s}_j\le -1$ for every $j$. The strongest restriction comes from $s_1$. Equality gives a simple amplification factor $-1$ and is allowed because the update is orthogonally diagonalizable. Other negative values either amplify a mode or make the update singular. This extra branch describes backward time stepping on a fixed spatial grid; it is not a positive-time diffusion discretization, and no fixed negative $r$ remains in that branch as $M\to \infty$.