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Reducing a minimal integral Weierstrass equation modulo the maximal ideal gives a projective cubic over the residue field. Reducing primitive projective coordinates defines the reduction map on rational points.

Over the complex numbers, $A_i=(J_i+i{K}_i)/2$ and $B_i=(J_i-i{K}_i)/2$ generate commuting copies of ${\mathfrak{sl}}_2(\mathbb{C})$, givingFinite-dimensional irreducible representations are labelled $(j_L,j_R)$.

Here descent is an arithmetic calculation with square classes and an isogeny of elliptic curves, rather than the gluing of objects along a covering.

For $E:y^2=x^3+a{x}^2+bx$ and its two-isogenous curve $E^{\prime }:y^2=x^3-2a{x}^2+(a^2-4b)x$, square-class maps on $E(\mathbb{Q})$ and $E^{\prime }(\mathbb{Q})$ determine the Mordell-Weil rank.

Let $n$ be prime to $\mathrm{char}K$ and suppose ${\mu }_n\subset K$. For a finite Galois extension $L/K$, the multiplicative Kummer pairing has domainand sends $(\sigma ,[a])$ to $\sigma (\alpha )/\alpha$, where $\alpha \in L$ and ${\alpha }^n=a$. The restriction to $V_L$ ensures that the chosen root actually lies in the domain of $\sigma$. Changing that root or the representative of $[a]$ leaves the value unchanged; multiplication in either argument multiplies the values. If $L$ is generated by these roots, this is the perfect duality of finite Kummer theory.^[ref]

For an elliptic curve, choose $Q$ with $nQ=P$. The analogous expression is $\sigma Q-Q\in E[n]$. In general it is a Galois cohomology cocycle, satisfying $c(\sigma \tau )=c(\sigma )+\sigma c(\tau )$, rather than a group homomorphism. If all $n$-torsion is rational, the action on $E[n]$ is trivial and it becomes a homomorphism. This is the connecting construction in the Kummer map of an elliptic curve.

On the complex special linear Lie algebra ${\mathfrak{sl}}_n(\mathbb{C})$, the Killing form isThus its intrinsic definition through the Adjoint representation becomes an ordinary matrix trace calculation.

To derive the coefficient, act first on the full space of $n\times n$ matrices. Write $L_X(A)=XA$ and $R_X(A)=AX$. The Adjoint representation acts by ${\mathrm{ad}}_X=L_X-R_X$. In the basis of matrix units, direct calculation givesHere $\mathrm{Tr}$ is the trace on the $n^2$-dimensional space of matrices. Expanding the product yieldsFor traceless $X,Y$, the last term vanishes. The full matrix space splits as ${\mathfrak{sl}}_n(\mathbb{C})\oplus \mathbb{C}I$, and every commutator acts trivially on the scalar summand. Removing that summand therefore leaves the trace unchanged, giving the Killing form on ${\mathfrak{sl}}_n(\mathbb{C})$.

Every invariant bilinear form on a Lie algebra $B$ on a finite-dimensional complex simple Lie algebra $\mathfrak{g}$ is a scalar multiple of its Killing form $\kappa$. Since $\kappa$ is nondegenerate, there is a unique endomorphism $T$ with $B(x,y)=\kappa (Tx,y)$, even when $B$ itself is degenerate. Invariance of both forms gives $T[z,x]=[z,Tx]$. Thus $T$ intertwines the Adjoint representation of a Lie algebra, which is an irreducible representation because its invariant subspaces are ideals of a Lie algebra. The Schur lemma gives $T=cI$, hence $B=c\kappa$. In particular, every nonzero such form is automatically nondegenerate and symmetric.

For a finite-dimensional Lie algebra over a field of characteristic zero, the radical ${\mathfrak{g}}^{\perp }=\{x:\kappa (x,\mathfrak{g})=0\}$ of the Killing form is a solvable ideal. Invariance makes it an ideal, and the Cartan solvability criterion applied to its adjoint image proves solvability.

The Dolbeault Laplacian is ${\Delta }_{\bar{\partial }}=\bar{\partial }{\bar{\partial }}^{\ast }+{\bar{\partial }}^{\ast }\bar{\partial }$. On a Kähler manifold, the Kähler identities imply

The Witt group of isometric structures identifies two isometric structures when their orthogonal difference is metabolic.