With $x=2\sqrt{r/(3A)}$ and $g=A{r}^{3/2}\Sigma$, the half-line heat equation solution $g=C{t}^{-1/2}{e}^{-x^2/(4t)}$ has $g_x(0)=0$ but $g(0)>0$. It gives $\Sigma \propto r^{-3/2}{t}^{-1/2}{e}^{-r/(3At)}$, constant disk mass and angular momentum proportional to $t^{1/2}$. Its radial velocity is $r/t$, and a central torque proportional to $t^{-1/2}$ supplies the increasing angular momentum. It is an outward-spreading, torque-driven disk, not a zero-torque accreting solution.

// Target: astrophysical-fluid-dynamics.bigb

An irreducible complex $G^F$-module is cuspidal if its Harish-Chandra restriction is zero for every proper rational parabolic subgroup. By adjunction, it occurs in no Harish-Chandra induction from a proper rational Levi subgroup. This is a finite-group notion, distinct from a cuspidal automorphic representation.

For the sphere bundle of an oriented rank-$r$ real vector bundle, the integration map in the Gysin sequence lowers degree by $r-1$. Its compatibility with the cup product is the displayed projection formula. The construction is the connecting map of the disk/sphere pair followed by the inverse Thom isomorphism theorem; the connecting map has its graded module sign, and the degree sign in the conventional fibre-integration map gives the displayed left-module formula.

Let $E$ be an oriented real vector bundle of positive even rank $r$, with zero Euler class, and suppose the integral cohomology of $B$ is torsion-free. The Gysin sequence gives $0\to H^j(B)\to H^j(S(E))\to H^{j-r+1}(B)\to 0$. Choose an odd-degree class $a\in H^{r-1}(S(E))$ with ${\pi }_{!}a=1$. The projection formula for sphere bundle integration identifies the cohomology additively with ${\pi }^{\ast }{H}^{\ast }(B)\oplus {\pi }^{\ast }{H}^{\ast }(B)a$. It is torsion-free, so graded commutativity of the cup product gives $a^2=0$ from $2a^2=0$. These facts establish the displayed ring isomorphism, not just a decomposition of groups.

Choose a temporal polarization vector, two transverse polarization vectors and a spatial longitudinal polarization vector. The Gupta-Bleuler quantization condition is $(a^0-a^3)|\chi ⟩=0$. Applied to ${\sum }_{\lambda }{c}_{\lambda }{a}^{\lambda †}|0⟩$, it gives $c_3=-c_0$. The remaining temporal-longitudinal excitation $a^{0†}-a^{3†}$ has zero norm and is orthogonal to every constrained state. Its field wavefunction is proportional to the null four-momentum, hence is pure gauge. Quotienting this direction leaves the two positive transverse photon states. Use wave packets or a box normalization to avoid treating continuum momentum eigenstates as normalizable vectors.

For an open clamped spline knot sequence and degree $d\ge 1$, the Greville abscissae ${\xi }_i=d^{-1}{\sum }_{r=1}^d{t}_{i+r}$ reproduce the parameter. The B-spline differentiation formula gives derivative coefficients $d({\xi }_i-{\xi }_{i-1})/(t_{i+d}-t_i)=1$. Their lower-degree partition of unity makes the derivative of the weighted sum equal to one. The left endpoint value is the left parameter endpoint, fixing the integration constant to zero. This proves linear precision even with unequal knot intervals.

Over an algebraically closed field, every regular function on a nonempty irreducible projective variety is constant. Regard the function as a morphism to ${\mathbb{P}}^1$ avoiding infinity. Its graph is closed; the closedness of projection from projective space makes its image closed in ${\mathbb{P}}^1$. A proper closed subset there is finite, and an irreducible finite image is a singleton. Irreducibility is the usual variety convention; on a disconnected projective algebraic set different components may have different constants.

Over an algebraically closed field, choose a point $P$ on the smooth projective curve. The Riemann-Roch theorem gives $\dim L(P)=2$: the complementary canonical divisor term has negative degree. A nonconstant function in $L(P)$ has exactly one simple pole, so it defines a degree-one morphism of algebraic curves to the projective line. A degree-one map between smooth projective curves is an isomorphism. The existence of a rational point is essential over a field which is not algebraically closed.

For $F=X+\lambda V$, the canonical fields $X,V$ preserve Liouville volume of a surface geodesic flow, so the divergence is $V\lambda$. For a Gaussian thermostat, $\lambda =⟨E,iv⟩$ and $V(iv)=-v$, giving the stated sign. If $E=\nabla f$, then the density $e^{f\circ \pi }$ makes this divergence zero.

Normalize the target in the function field of the generically finite Gauss map of a theta divisor; its finite map has an intrinsic branch locus. For a nonhyperelliptic smooth projective curve, this locus is the projective dual variety of the canonical curve. For a hyperelliptic curve of genus at least three, it is the union of the dual hypersurface of its canonical rational normal curve and the $2g+2$ hyperplanes corresponding to branch points. Locally, ramification exchanges two roots of a hyperplane section, or exchanges the two lifts of a root passing through a branch point. Biduality then recovers the curve or its double-cover data.

In a gas syringe, the graduated barrel reading $V_{\mathrm{read}}$ can omit a fixed connecting or sensor volume $V_d$ occupied by the same sealed gas. For a common settled temperature, Boyle's law then requiresThis second relation makes a graph of $1/p$ against $V_{\mathrm{read}}$ a straight line; its intercept divided by its slope estimates $V_d$ under these assumptions. Unequal temperatures between the barrel and the connecting space require a different correction, analogous to the gas thermometer dead-space correction.

For an original hypothetical example, let $V_d=1\mathrm{c}{\mathrm{m}}^3$, $V_{\mathrm{read},1}=20\mathrm{c}{\mathrm{m}}^3$, $p_1=100\mathrm{kPa}$, and $V_{\mathrm{read},2}=10\mathrm{c}{\mathrm{m}}^3$. The total volume falls from $21$ to $11\mathrm{c}{\mathrm{m}}^3$, so $p_2=100(21/11)\mathrm{kPa}\approx 190.91\mathrm{kPa}$, rather than $200\mathrm{kPa}$. The omitted volume makes the total compression ratio smaller than the barrel-reading ratio. These stipulated values are independent of the manufacturer's apparatus correction and are not measurements.

The tensor product of Lie algebra representations $3\otimes 3$ splits into the symmetric $6$ and antisymmetric $\overline{3}$. Then $6\otimes 3=10\oplus 8$ and $\overline{3}\otimes 3=8\oplus 1$, as highest-weight character subtraction verifies. The decuplet is fully symmetric under quark permutations, the singlet fully antisymmetric, and the two octets constitute the two-dimensional mixed permutation multiplicity space. Applying the Pauli constraint on three-quark flavour multiplets gives one ground-state spin-one-half octet and one spin-three-halves decuplet.

// Target: physics.bigb

Split the two dual factors into antisymmetric and symmetric tensors. The invariant Levi-Civita symbol identifies the antisymmetric pair with $3$, giving $6\oplus \overline{3}$ after multiplying by the remaining triplet. On the symmetric part, contraction $T^a{}_{ba}$ maps onto $\overline{3}$; the equivariant splitting is $T^a{}_{bc}=({\delta }_b^a{v}_c+{\delta }_c^a{v}_b)/4$. Its trace-free kernel has dimension $18-3=15$ and highest weight with Dynkin labels $(1,2)$. The Weyl dimension formula gives dimension $15$, so this is exactly the irreducible kernel. It is conjugate to the other fifteen-dimensional module with labels $(2,1)$; dimension alone does not distinguish the two.

In an adapted real orthonormal coframe, the fundamental form of a Hermitian manifold is $\omega ={\sum }_j{e}^j\wedge f^j$ and the Riemannian volume form is ${\omega }^n/n!$. The complex-linear Hodge star sends each summand to the product of all the other oriented two-dimensional factors. Summation gives the displayed formula, valid for every Hermitian manifold, without a Kähler assumption.

For a directed set of Scott continuous maps $f_i:D\to E$, its least upper bound is $f(x)={\sup }_i{f}_i(x)$. The two directed suprema over $i$ and over inputs commute, proving that $f$ remains Scott continuous. Thus the function space of complete partial orders is a complete partial order.

In the unique decomposition $f={\sum }_{i=0}^{p-1}(1+T{)}^i\phi (f_i)$, define $\psi (f)=f_0$. Averaging translates over the p-th roots of unity annihilates every nonzero basis index and retains the zeroth, proving the displayed identity and its integrality. Thus $\psi \phi =1$ and $\psi (\phi (g)f)=g\psi (f)$. Under the Amice transform, $\psi =0$ characterizes measures supported on ${\mathbb{Z}}_p^{\times }$.

For $R={\mathbb{Z}}_p[[T]]$ and $\varphi (f)=f((1+T{)}^p-1)$, reduction modulo $p$ is the injective substitution $\overline{f}(T)\mapsto \overline{f}(T^p)$. Thus $\varphi (f)\in pR$ implies $f\in pR$. Dividing by $p$ and repeating proves the displayed equality for every $k\ge 1$. In particular, a ring congruence of Frobenius substitutions on cyclotomic power series reflects back to the original formal power series.

For an elliptic curve over ${\mathbb{F}}_q$,

If an isogeny of elliptic curves $\pi :E_1\to E_2$ satisfies $\pi {\varphi }_1=-{\varphi }_2\pi$ for their Frobenius isogenies, then $\pi (1-{\varphi }_1)=(1+{\varphi }_2)\pi$. Multiplicativity of the degree of an isogeny gives $\mathrm{deg}(1-{\varphi }_1)=\mathrm{deg}(1+{\varphi }_2)$. The degree parallelogram law then gives the displayed point-count identity, since $\mathrm{deg}(1-{\varphi }_i)=\#E_i({\mathbb{F}}_q)$ and $\mathrm{deg}{\varphi }_i=q$.

For an integral one-dimensional formal group law over a finite extension of the p-adic field, the inverse of ${\partial }_{Y}F(T,0)$ has integral coefficients because its constant term is one. Integrating this invariant differential gives ${\log }_F(T)=T+{\sum }_{n\ge 2}{c}_n{T}^n$ with $n{c}_n$ integral. For $v_K(\pi )=1$ and $e=v_K(p)$ the displayed bound follows. The estimate $v_p(n)\le (n-1)/(p-1)$ makes the nonlinear part strictly contracting on ${\pi }^s{\mathcal{O}}_K$ for $s>e/(p-1)$, yielding the deep logarithm subgroup of a formal group.