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For a basepoint-free nef big divisor $D$, this means that the morphism given by its complete system is generically two-to-one onto its image. Equivalently, its general smooth curve is hyperelliptic. A double cover of ${\mathbb{P}}^2$ branched over a smooth sextic, with the pullback of $\mathcal{O}(1)$, gives the degree-two example.

For a basepoint-free nonhyperelliptic nef big divisor $D$, the trigonal condition means that its general smooth curve is a trigonal curve. An elliptic pencil on a K3 surface with $D\cdot E=3$ restricts to its degree-three pencil. On a smooth quartic containing a line $\Gamma$, the hyperplane class $H$ has this property for $E=H-\Gamma$.

For nef big systems on a complex K3 surface, the successive alternatives distinguish a monogonal linear system on a K3 surface from a basepoint-free system; then a degree-two map from a birational map; then a projective ideal generated by quadrics from the trigonal and plane-quintic exceptions. The last alternatives must include the low-genus quartic and quadric-cubic models. The birational morphism can contract exactly the orthogonal $(-2)$-curves; it need not embed the original smooth surface.

An integral Hodge isometry between complex K3 surfaces which carries some Kähler class to a Kähler class is the pullback of a unique isomorphism of the surfaces. A Hodge isometry without that cone condition need not itself be geometric: reflection in an effective $(-2)$-curve reverses its class and cannot be induced by an isomorphism.

Over $\mathbb{C}$, the four partial derivatives show that this projective surface is smooth. Adjunction makes its canonical bundle trivial, and the hypersurface exact sequence gives $H^1({\mathcal{O}}_X)=0$, so it is a K3 surface. Lines given by $x_0=\zeta x_1$, $x_2=\eta x_3$, with ${\zeta }^4={\eta }^4=-1$, lie on it. Choosing both roots differently gives skew lines, one of which is a section of the elliptic pencil defined by planes through the other.

A gas syringe traps a gas in a graduated barrel with a movable plunger, allowing its volume to be varied and read. In an isothermal gas syringe experiment, the relevant total volume also includes any connected tubing and sensor cavity containing the same gas.

For an isothermal collisionless tracer without streaming or an escape cutoff, integrate its three-dimensional Maxwell-Boltzmann velocity distribution over the two transverse velocities. The remaining marginal is the displayed normal distribution. Its width and mean do not depend on position. Integrating along the line of sight therefore only multiplies the marginal by the projected surface density; the normalized observed velocity distribution remains the same Gaussian at every projected radius.

Without an escape-energy cutoff, a galactic distribution function proportional to $\exp [(\Psi -v^2/2)/{\sigma }^2]$ gives $\rho \propto e^{\Psi /{\sigma }^2}$ and a constant one-dimensional velocity dispersion. Since $\Psi ={\Phi }_0-\Phi$, differentiation gives the displayed hydrostatic relation. An ideal gas at constant temperature has pressure $P=\rho k_{B}T/m_0$, so its hydrostatic equilibrium equation is identical at the displayed dispersion-temperature identification. Coupling either relation to the same Poisson equation for Newtonian gravity, with matched central density and boundary data, therefore gives the same density structure. Truncating the stellar velocities at escape changes this equivalence.

The elliptic curves $E_1:y^2=x^3-x$ and $E_2:y^2=x^3-x+1$ over ${\mathbb{F}}_5$ both have eight rational points. The first has four rational 2-torsion points and the second has two, so their point groups are not isomorphic. They are nevertheless degree-two isogenous via $X=x+2/(x-1)$, $Y=y(1-2/(x-1{)}^2)$. The map sends $O$ and $(1,0)$ to $O$; its pole divisor for $X$ is $2O+2(1,0)$, giving degree two. Their j-invariant of an elliptic curve values are respectively three and one, so the curves are not geometrically isomorphic either.

On a level set $f(x,y,z)=c$, implicit differentiation gives the three ratios $-f_y/f_x$, $-f_z/f_y$ and $-f_x/f_z$. Their product is $-1$ wherever all the required coordinate representations are differentiable with nonzero denominators. At coordinate singularities the quotient product need not be defined.

This closed flat Riemannian manifold is the mapping torus of $-I$ on a two-dimensional flat torus. Equivalently, take planar lattice translations and the screw $(v,z)\mapsto (-v,z+1/2)$ in Euclidean three-space. Its square is a unit axial translation, its holonomy is $C_2$, and its first Betti number is one.

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.