For an isogeny of elliptic curves $\varphi :E\to E^{\prime }$ over a number field $K$, this finite subgroup of $H^1(K,E[\varphi ])$ consists of Galois cohomology classes whose restriction at every place comes from $E^{\prime }(K_v)/\varphi E(K_v)$. It fits intowhere the last group is the kernel of the induced map on Tate–Shafarevich groups. Thus local computations bound the rational-point quotient without necessarily determining it. For multiplication by $n$, this recovers the Selmer group of an elliptic curve.

Four independent complex Dirac solutions in four dimensions span each spinor fibre, because a connection-parallel solution is determined by its value at one point. Their integrability operator must therefore vanish on the whole fibre. Independence of the six Clifford bivectors then gives the displayed constant-curvature tensor and $R_{ab}=-12s{c}^2{g}_{ab}$. When $c>0$, the Ricci tensor is automatically nondegenerate.

Every Lie algebra automorphism $S$ obeys ${\mathrm{ad}}_{SX}=S{\mathrm{ad}}_X{S}^{-1}$. The cyclic property of the trace therefore makes the Killing form invariant under $S$. For a matrix Lie group, $S={\mathrm{Ad}}_g$ is conjugation by $g$, proving the displayed identity even when the group is disconnected. This directly proves gauge invariance of Killing-form contractions of adjoint fields.

For the general linear Lie algebra of all $n\times n$ matrices over a field of characteristic zero, write ${\mathrm{ad}}_X=L_X-R_X$ on the matrix space. The traces of $L_X{L}_Y$ and $R_X{R}_Y$ are $n\mathrm{tr}(XY)$, whereas both mixed traces are $\mathrm{tr}X\mathrm{tr}Y$. Expanding gives the displayed Killing form. Its radical is the scalar matrices because $\kappa (X,Y)=0$ for every $Y$ forces $nX-\mathrm{tr}(X)I=0$. Thus the full general linear algebra is not semisimple, even though its traceless summand is semisimple for $n\ge 2$.

// Target: lie-theory.bigb

The de Rham cohomology class of a Kähler form is a Kähler class. Such classes form the Kähler cone. They have positive integral on every complex curve, and their top self-intersection is positive.

The cup product on the second integral cohomology of a complex K3 surface is an even unimodular lattice of signature $(3,19)$. Here $U$ is the hyperbolic plane and $E_8(-1)$ is the negative-definite form of the E8 lattice. In the M-theory K3 duality, this lattice matches the heterotic electric charge lattice. The self-dual harmonic two-forms of a Ricci flat K3 metric span its positive three-plane.

Locally, a unit-volume Ricci flat metric on a K3 surface determines a positive three-plane in its real second cohomology, with the K3 intersection lattice fixed. The corresponding Grassmannian has dimension $3\cdot 19=57$. Allowing the volume gives one further positive real parameter; large diffeomorphisms identify lattice-equivalent descriptions. This is the same continuous moduli count as the three-torus Narain moduli space plus its coupling.

// Target: supersymmetry.bigb

A primitive effective nef class of square zero on a complex K3 surface defines a basepoint-free pencil with connected genus-one fibres. Its general fibre is a smooth elliptic curve. A section is a smooth rational curve $\Gamma$ with $E\cdot \Gamma =1$ and ${\Gamma }^2=-2$.

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.