The mapping torus of an order-three rotation of a hexagonal flat torus is a closed orientable flat Riemannian manifold with holonomy $C_3$. A screw rotation through $2\pi /3$ with axial displacement $1/3$, together with the planar lattice, describes its deck group. Its first Betti number is one.

The mapping torus of an order-six rotation of a hexagonal flat torus is a closed orientable flat Riemannian manifold with holonomy $C_6$. The screw rotation has angle $\pi /3$ and axial displacement $1/6$. Its first Betti number is one.

The closed orientable flat Riemannian manifold with holonomy $C_2\times C_2$ can be presented by Euclidean screws $\alpha (x,y,z)=(x+1/2,-y,-z)$ and $\beta (x,y,z)=(-x,y+1/2,-z+1/2)$. Their squares and the square of their product generate the unit translation lattice. Each nonidentity holonomy coset has a half-integer displacement along its fixed axis, so acts freely. The holonomy has no common fixed vector, giving first Betti number zero and a finite isometry group for every flat metric.

A section of $\mathcal{O}(1)\oplus \mathcal{O}(1)$ on the complex projective line is a pair of homogeneous linear forms ${\ell }_1=a{Z}_0+b{Z}_1$, ${\ell }_2=c{Z}_0+d{Z}_1$. It vanishes at exactly one projective point precisely when its coefficient matrix has rank one, equivalently $ad-bc=0$ and the matrix is nonzero. Its kernel is then one line in ${\mathbb{C}}^2$. Equivalently, $s=v\ell$ for a constant nonzero vector $v$ and one nonzero linear form $\ell$. Rank two gives no zero, while rank zero gives the identically zero section.

A nowhere-zero smooth section $s$ of a holomorphic line bundle writes its Dolbeault operator as ${\bar{\partial }}_{L}s=a\otimes s$. Its integrability gives $\bar{\partial }a=0$. If $H^{0,1}(X)=0$, then $a=\bar{\partial }f$ globally. The section $e^{-f}s$ is nowhere zero and holomorphic. This argument requires no compactness and explicitly constructs the holomorphic trivialization.

On a compact connected complex manifold, pair nonzero holomorphic sections of a holomorphic line bundle and its holomorphic dual vector bundle. Their product is a global holomorphic function, hence constant by the maximum modulus principle. The identity theorem makes their nonvanishing loci dense, so this constant is nonzero. Both sections are consequently nowhere zero and trivialize their bundles. Conversely, the constant section of a trivial bundle and its dual provide the two sections.

A global holomorphic section of $\mathcal{O}(1)$ on ${\mathbb{CP}}^1$ is a linear homogeneous polynomial in the two projective coordinates. On one patch its coefficient is an entire function $p(\lambda )$; regularity on the other patch says $p(\lambda )/\lambda$ is regular at infinity. Hence $p$ has at most linear growth. Cauchy estimates make its second and higher derivatives zero, so $p=a+b\lambda$. The same argument entrywise applies to matrix-valued sections and makes the logarithmic derivative in the Penrose-Ward correspondence linear in the primed two-component spinor.

Use compactly supported Hamiltonians on open manifolds and zero-mean Hamiltonians on closed manifolds. Because their values straddle zero, $\Vert H_t{\Vert }_{\infty }\le \max H_t-\min H_t\le 2\Vert H_t{\Vert }_{\infty }$. Thus ${\rho }_{\infty }\le d_H\le 2{\rho }_{\infty }$. These are two equivalent, but not identically normalized, conventions for Hofer's metric. A statement about zero distance or zero displacement energy is independent of this choice.

The displacement energy of a subset of a symplectic manifold is the infimum of the energy norms of compactly supported Hamiltonian diffeomorphisms moving it off itself. The infimum is $+\infty$ if no such map exists. One must use the same chosen norm convention in both this definition and any energy lower bound. A compact set can have zero displacement energy even when every nonidentity map has positive norm.

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