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The time-ordered exponential solves $i\hslash {\partial }_{t}U=H(t)U$ with initial condition $U(t_0,t_0)=I$. Its expansion is the Dyson series. For bounded continuous Hamiltonian operators the series converges, as in differentiation of the Dyson time-ordered exponential. If the generators commute at all times, time ordering is unnecessary and an ordinary operator exponential suffices.

A majority Boolean function outputs one when more than half its input bits are one. An odd number of inputs avoids a tie convention. For three bits its unique multilinear polynomial is $x_1{x}_2+x_1{x}_3+x_2{x}_3-2x_1{x}_2{x}_3$, so the polynomial method for quantum query lower bounds requires at least two exact queries.

For covers branched over three real points with a real base point, choose the two generating loops so complex conjugation reverses each. A sheet permutation satisfying the displayed identities lifts that conjugation to an anticonformal equivalence of the covers, including the filled branch points. For closed hyperbolic Riemann surfaces this is a Riemannian isometry. Such an isometry need not come from conjugating the subgroups inside the original deck group, so a Gassmann equivalent pair can be isometric despite nonconjugacy there.

For a liquid film heated below and cooled above, the quasistatic temperature of a conductively cooled film makes a thicker region colder at the surface. If surface tension decreases with temperature, Marangoni stress draws fluid towards thicker regions, competing with smoothing by capillary pressure. Writing ${\gamma }_s={\gamma }_0+bh$ gives $a_c={\gamma }_0/(3\mu )$, $a_m=b/(2\mu )$. A balanced long-wave scaling gives $H_{\tau }+(H^2{H}_X{)}_X+(H^3{H}_{XXX}{)}_X=0$. The dispersion relation $s=k^2-k^4$ has unstable band $0<|k|<1$ and fastest normal mode at $|k|=1/\sqrt{2}$. This thermal mechanism does not require surfactant.

If a charged-particle rate coefficient has $\lambda \propto {\eta }^2{e}^{-\eta }$ with $\eta \propto T^{-1/3}$, its logarithmic temperature derivative is $(\eta -2)/3$. This is a local slope evaluated at fixed composition. It increases as temperature falls, so extrapolating a single power law across a large temperature range can be inaccurate.

For two equal-frequency quantum harmonic oscillators, the ordinary thermal trace is $r/(1-r{)}^2$ and the trace with the parity operator is $r/(1+r{)}^2$, where $r=e^{-\beta \omega }$. Total excitation $N$ has degeneracy $N+1$ and parity $(-1{)}^N$. Therefore the traces expand as ${\sum }_{n\ge 1}(\pm 1{)}^{n-1}n{r}^n$. The alternating weights come from parity insertion in a bosonic state space, not fermionic statistics.

A zero-energy mode lies in the kernel of the quadratic elastic filament operator. For pure bending, zero curvature makes the displacement affine. Free-free endpoints therefore leave translation and tilt modes, while slope-constrained, force-free endpoints leave only translation. Both clamped-clamped and hinged-hinged endpoints remove these modes. An unrestricted modal amplitude has constant Boltzmann distribution weight, so no normalizable canonical ensemble or finite displacement variance exists. Fixing the rigid degrees of freedom allows the equipartition theorem on the positive complement.

For a positively oriented basis $\mathbf{a},\mathbf{b},\mathbf{c}$, put $\tau =\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})>0$. The center of an interior sphere at distance $r$ from each of the three tetrahedron face planes through the origin has the displayed expression. The three inward unit vectors are the normalized cross products $\mathbf{b}\times \mathbf{c}$, $\mathbf{c}\times \mathbf{a}$, $\mathbf{a}\times \mathbf{b}$. Taking inner products with these vectors solves three independent signed-distance equations. Reversing some signs instead produces centers on exterior sides. Existence of a sphere contained in the full tetrahedron also requires clearance from the fourth face.

The tensor-product weight diagram has six distinct nonzero differences of the three defining weights and a zero weight of multiplicity three. Identify the tensor product with endomorphisms of ${\mathbb{C}}^3$ under conjugation. Its scalar line is a singlet, while the traceless matrices give the irreducible adjoint octet, with two zero weights. The root commutators connect all of the traceless matrix units, proving irreducibility rather than inferring it solely from a drawing.

An invariant tensor is fixed by every element of the represented group, so it spans or belongs to a trivial group representation. The special unitary group preserves its identity tensor and the upper and lower volume tensors. Such tensors make tensor contraction equivariant and permit invariant inner products and real structures.

A $(j,k)$ tensor has one fundamental representation factor for every upper index and a dual factor for every lower index. Under $A\in SU(n)$, upper factors transform by $A$ and lower factors by $A^{-1}$; unitarity identifies the latter with conjugate fundamental matrices. Thus complex conjugation exchanges $(j,k)$ and $(k,j)$. Index permutations and contraction with an invariant tensor commute with the action, producing invariant subspaces and decomposition maps.

When the two indicated groups are finite, their size ratio is multiplicative on short exact sequences. It is one for finite modules and therefore unchanged by finite-index modifications. A trivial integral module $\mathbb{Z}$ has quotient $|G|$, while an induced regular lattice ${\mathbb{Z}}_p[G]$ has both Tate groups zero and quotient one. These facts permit norm-index calculations without explicitly finding every norm.

A tangent vector is a linear map on germs of smooth functions satisfying $D(ab)=a(x)D(b)+b(x)D(a)$. In a manifold chart, first-order expansion shows $Dh={\sum }_{i}D(u^i){\partial }_{i}h(x)$. Thus the coordinate derivations form a basis, and the dimension equals the manifold dimension. Composition of germs with a smooth map defines its differential of a smooth map without choosing charts.

A conservation law flux $F(U)$ transports the conserved density $U$ across a surface. In one dimension $U_t+{\partial }_{x}F(U)=0$; a moving discontinuity satisfies $s[U]=[F]$. For a stationary shock wave, each conserved flux has the same value on both sides.

A total is a set of five edge-disjoint synthemes covering all fifteen duads of a six-point set. There are six totals; each syntheme belongs to two and every two totals share exactly one syntheme. Each total contains exactly one syntheme containing any prescribed duad. Totals are one-factorizations of the complete graph.

For constant $P=w\rho$ and velocity potential $v_i=i{k}_i\theta$, stress-energy conservation gives the displayed density equation in synchronous gauge in cosmology. Here $h$ is the trace of the spatial metric perturbation, and primes denote conformal time. In particular cold baryons obey ${\delta }_b^{\prime }=k^2{\theta }_b-h^{\prime }/2$. A positive sign in front of $k^2{\theta }_b$ on the left contradicts this potential convention.

If each daughter subsequently conserves its own spin while contracting at fixed outer separation, its next critical radius is $a/8$. A local repetition of the synchronous fission model for binary formation gives the displayed inner-to-outer separation ratio. Continuous tidal locking throughout the contraction would instead keep the spin frequency fixed and decrease its centrifugal-to-gravity ratio as the cube of its radius; the repeated fission would not follow. Spin-orbit torque assumptions must therefore be explicit.

Counting successive symplectic pairs givesFor each nonzero first vector there are $q^{2m-1}$ partners with pairing one, after which its nondegenerate plane splits off. The exact defining-characteristic part is $q^{m^2}$, realized by an upper symplectic unitriangular subgroup.

The stabilizer of a totally isotropic $k$-space in a finite symplectic space also stabilizes its orthogonal complement. Its block decomposition has a unipotent normal subgroup $Q$ and a block-diagonal complement $L\cong {\mathrm{GL}}_k(q)\times {\mathrm{Sp}}_{2m-2k}(q)$. The paired $k$-space carries the contragredient action. A row-vector convention gives upper-triangular blocks; a column-vector convention transposes the arrangement. The unipotent radical count for a symplectic parabolic subgroup yields its order.

An invertible real matrix preserving the standard symplectic form. With $J=\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right)$, it satisfies $A^{T}JA=J$, equivalently $AJ{A}^T=J$. The first identity describes preservation of the form; the second describes preservation of coordinate Poisson brackets.