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Assume the lower bound $e(L)\ge \gamma (L)/2$ for a closed rational Lagrangian submanifold, with matching energy convention. A nonidentity Hamiltonian map displaces a sufficiently small ball about some moved point. Inside that ball place a Lagrangian torus formed from equal-area circles. Its least positive Liouville period is positive, so the map's norm is at least half this period. This proves positivity of the distance between different maps without appealing to nondegeneracy itself.

The Kähler Laplacian identity makes the space of complex harmonic differential forms of degree $r$ the direct sum of its $(p,q)$ parts with $p+q=r$. Conjugation identifies the $(p,q)$ and $(q,p)$ harmonic spaces. If $r$ is odd these are distinct and pair off, so the Betti number in degree $r$ is even. This is a topological obstruction to existence of a Kähler metric, exhibited by a Hopf surface with first Betti number one.

For a compact Kähler manifold of complex dimension $d$ and $k\le d$, this is the cohomology killed by the indicated power of a Kähler class. The Hodge–Riemann bilinear relations polarize its Hodge structure. On a K3 surface, primitive degree-two cohomology is the orthogonal complement of the chosen positive class.

With $H_1=\mathrm{diag}(1,-1,0)/2$ and $H_2=\mathrm{diag}(1,1,-2)/(2\sqrt{3})$, the simple roots are $(1,0)$ and $(-1/2,\sqrt{3}/2)$. Their coroots give Dynkin labels $a=2p$ and $b=\sqrt{3}q-p$ for Cartesian highest-weight coordinates $(p,q)$. Finite-dimensional irreducible representations require $a,b\in {\mathbb{Z}}_{\ge 0}$. The vertical coordinate is $\sqrt{3}/2$ times flavor hypercharge.

For the one-dimensional Dirichlet second-difference grid, high sine modes have eigenvalues $1-\omega t_k$, with $t_k=1-\cos (k\pi /(n+1))$ and $k\ge \lceil (n+1)/2\rceil$. Let $a$ and $b$ be the smallest and largest of these $t_k$. Convexity gives attenuation $\max (|1-\omega a|,|1-\omega b|)$. Balancing its endpoint values with opposite signs gives the displayed exact minimax choice. For odd $n$, $a=1$; for even $n$, $a=1+\sin (\pi /[2(n+1)])$, and always $b=1+\cos (\pi /(n+1))$. Only in the large-grid or uniform-in-grid limit do these become ${\omega }_{\ast }=2/3$, ${\mu }_{\ast }=1/3$.

For $c\ne 0$ and $d^3\ne 27c$ in characteristic zero, the displayed plane cubic is smooth. Its Hessian determinant is $-6d^2(x^3+y^3+c{z}^3)+(216c-2d^3)xyz$. On the cubic this becomes $8(27c-d^3)xyz$. Hence all nine plane cubic flexes lie on the coordinate axes: $(1:-\zeta :0)$, $(-\sqrt[3]{c}\zeta :0:1)$ and $(0:-\sqrt[3]{c}\zeta :1)$ for ${\zeta }^3=1$.

For a cubic homogeneous polynomial $F$, its Hessian matrix has linear entries, so its determinant is a cubic homogeneous polynomial, possibly zero. On a nonsingular plane cubic in characteristic different from two, its zeros are exactly the flexes, by the Hessian criterion for a flex. If the determinant is nonzero, Bézout's theorem supplies intersections; if it is zero, the local criterion itself supplies flexes.

A holomorphic line bundle is semipositive if it admits a smooth Hermitian metric whose normalized Chern curvature form is semipositive. Tensoring a positive holomorphic line bundle with a nonnegative integer power of a semipositive one remains positive, because their curvature forms add. This smooth-metric property is stronger than simply having a nef numerical class.

Where $\eta =r/v$ decreases outwards, it increases as a surface ray travels inward, so no surface ray turns inside that interval. Penetrating rays may traverse it and turn deeper. The radius is then not a single-valued function of $\eta$, and angular ray data constrain combined branch contributions rather than the single function inverted by Herglotz–Wiechert inversion. Additional structural or observational information is needed to locate the hidden profile.

For a fixed rational point $R$ on an elliptic curve over a number field, the addition formula gives $h_x(P-R)\le 2h_x(P)+O_R(1)$. The duplication map has degree four, so $h_x(2Q)=4h_x(Q)+O(1)$. Combining them gives the displayed contraction. A finite set of representatives modulo twice the point group makes the constants uniform. Points above a fixed height threshold descend by a definite amount; Northcott theorem makes the terminal bounded-height set finite. This converts the Weak Mordell-Weil theorem into the full finite-generation theorem.

For $\bar{\nu }=Ar$ with $A>0$, use $x=2\sqrt{r/(3A)}$ and $g=A{r}^{3/2}\Sigma$. The Keplerian viscous diffusion equation becomes $g_t=g_{xx}$. The zero-torque inner boundary condition is $g(0,t)=0$, and the positive solution $g=Cx{t}^{-3/2}{e}^{-x^2/(4t)}$ gives $\Sigma \propto r^{-1}{t}^{-3/2}{e}^{-r/(3At)}$. Its mass is proportional to $t^{-1/2}$, angular momentum is constant and inward mass flow at the origin is proportional to $t^{-3/2}$. Its radial velocity is $r/t-3A/2$, exhibiting inner accretion and outer spreading.

// Target: astrophysics.bigb

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.