The Killing Form of the Antisymmetric Plain Algebra

Introduction

The antisymmetric plain algebra makes the biquaternion space a complex Lie algebra $\mathfrak{g}=(\mathbb{B},\wedge)$ of dimension four, The Lie Algebra of the Antisymmetric Plain Algebra, and every finite-dimensional Lie algebra carries its invariant bilinear form. This article reads the form of the block,

$$ \kappa(\tilde Q_1,\tilde Q_2)=\operatorname{Tr}\bigl(\operatorname{ad}_{\tilde Q_1}\operatorname{ad}_{\tilde Q_2}\bigr), $$

where $\operatorname{ad}_{\tilde Q_1}(\tilde Q_3)=\tilde Q_1\wedge\tilde Q_3$ is the adjoint operator of the bracket. The form is computed on general elements of the algebra and on the units; its invariance and its symmetry are established; its radical, which is the centre $\mathbb{C}_{\mathbb{B}}$, is identified; its relation to the general plain bilinear form on the vector subspace is shown to be a single constant; the Killing data — rank, radical, definite or indefinite character — are read on the basis and on the remarkable subspaces with their isotropic elements; and the Cartan criteria are read on it, together with the Casimir element the form defines.

The bracket and its structure are used from The Lie Algebra of the Antisymmetric Plain Algebra, where the centre, the derived algebra and the identifications $\mathrm{Vect}(\mathbb{B})\cong\mathfrak{sl}(2,\mathbb{C})$, $\mathbb{H}_{\mathbb{B}}\cong\mathbb{R}e_t\oplus\mathfrak{su}(2)$ and $\mathbb{M}_-\cong\mathfrak{u}(2)$ are established. The plain bilinear form $\langle\tilde Q_1,\tilde Q_2\rangle=\operatorname{Sc}(\tilde Q_1\tilde Q_2)$ and its pairings are The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra and The Four Adjoints of the Biquaternion Algebra in Examples; the remarkable subspaces are Introduction to the Remarkable Subspaces; their reading under the general plain form is Remarkable Subspaces and the Four General Products, Forms and Norms; the general theory of the Killing form, of the Cartan criteria and of the Casimir element is The Killing Form Operator, Structure of Lie Algebras and The Casimir Operator; and the operators of the block, which are the adjoints here, are The Adjoint Operators and the Derivations of the Antisymmetric Plain Algebra. The form is a two-variable operation on a module; the squared vector form $\vec Q\cdot\vec Q$ that appears below is a quadratic form on the vector part and not a norm of the algebra.

The Adjoint and the Trace Form

Definition. For $\tilde Q_1\in\mathbb{B}$ the adjoint of $\tilde Q_1$ is the $\mathbb{C}$-linear endomorphism

$$ \operatorname{ad}_{\tilde Q_1}:\mathbb{B}\to\mathbb{B},\qquad \operatorname{ad}_{\tilde Q_1}(\tilde Q_3)=\tilde Q_1\wedge\tilde Q_3, $$

and $\operatorname{ad}:\mathfrak{g}\to\operatorname{End}(\mathbb{B})$, $\tilde Q_1\mapsto\operatorname{ad}_{\tilde Q_1}$, is the adjoint representation of the block.

Proposition (the adjoint is the cross product). For $\tilde Q_1=\sum_\mu Q_{1,\mu} e_\mu$ the adjoint acts by the cross product of the vector part,

$$ \boxed{\; \operatorname{ad}_{\tilde Q_1}(\tilde Q_3)=\vec Q_1\times\vec Q_3 \;}, $$

so its kernel is the complex plane $\mathbb{C}e_t\oplus\mathbb{C}\vec Q_1$ when $\vec Q_1\neq0$ and the whole space when $\vec Q_1=0$, and its rank is $0$ on the centre and $2$ elsewhere. In the basis $(e_t,\vec e_x,\vec e_y,\vec e_z)$ its matrix is block diagonal, with a first column and a first row of zeros and the $3\times3$ matrix of the cross product by $\vec Q_1$ on the vector part.

Proof. The bracket depends only on the vector parts, so $\operatorname{ad}_{\tilde Q_1}(\tilde Q_3)=\vec Q_1\times\vec Q_3$. The kernel of the cross product by $\vec Q_1$ is the line $\mathbb{C}\vec Q_1$ together with the central direction, a complex plane when $\vec Q_1\neq0$; the matrix statement follows because the first coordinate is annihilated. The ranks were verified exactly on general elements and on the basis.

Definition. The trace form of the block is the bilinear form

$$ \kappa(\tilde Q_1,\tilde Q_2)=\operatorname{Tr}\bigl(\operatorname{ad}_{\tilde Q_1}\operatorname{ad}_{\tilde Q_2}\bigr). $$

It is the Killing form of the bracket, computed from the block's own adjoint; this adjoint of the bracket $\wedge$ is written $\operatorname{ad}^{\wedge}_{\tilde Q_1}$ in The Adjoint Operators and the Derivations of the Antisymmetric Plain Algebra, where the commutator $[\tilde Q_1,\tilde Q_2]=\tilde Q_1\tilde Q_2-\tilde Q_2\tilde Q_1=2(\tilde Q_1\wedge\tilde Q_2)$ and its adjoint $[\tilde Q_1,\cdot\,]=2\operatorname{ad}^{\wedge}_{\tilde Q_1}$ are also read. That commutator adjoint has Killing form $4\kappa$ in the complex trace, and the two readings differ by the constant factor of the rescaling of The Lie Algebra of the Antisymmetric Plain Algebra. The form $\kappa$ is the one of the block throughout.

The Form on General Elements

Theorem. For all $\tilde Q_1,\tilde Q_2\in\mathbb{B}$,

$$ \boxed{\; \kappa(\tilde Q_1,\tilde Q_2)=-2\,\vec Q_1\cdot\vec Q_2=-2\,(Q_{1,x}Q_{2,x}+Q_{1,y}Q_{2,y}+Q_{1,z}Q_{2,z}) \;}, $$

and in the basis $(e_t,\vec e_x,\vec e_y,\vec e_z)$

$$ \kappa(e_t,e_t)=0,\qquad \kappa(e_t,\vec e_\nu)=0,\qquad \kappa(\vec e_\mu,\vec e_\nu)=-2\,\delta_{\mu\nu}\quad(\mu,\nu\in\{x,y,z\}), $$

so the Gram matrix of $\kappa$ is $\operatorname{diag}(0,-2,-2,-2)$.

Proof. Let $\vec Q_1$ have coordinates $(Q_{1,x},Q_{1,y},Q_{1,z})$ and let $C_{\vec Q_1}$ be the $3\times3$ matrix of the cross product $\vec X\mapsto\vec Q_1\times\vec X$ on the vector part. The adjoint has block form $\operatorname{diag}(0,C_{\vec Q_1})$ on $\mathbb{C}e_t\oplus\mathrm{Vect}(\mathbb{B})$, so

$$ \operatorname{Tr}\bigl(\operatorname{ad}_{\tilde Q_1}\operatorname{ad}_{\tilde Q_2}\bigr)=\operatorname{Tr}\bigl(C_{\vec Q_1}C_{\vec Q_2}\bigr). $$

For the cross-product matrices one has $C_{\vec Q_1}C_{\vec Q_2}+(\vec Q_1\cdot\vec Q_2)\,\mathrm{I}_3=\vec Q_2\,\vec Q_1^{\mathsf T}$, because $\vec Q_1\times(\vec Q_2\times\vec X)=\vec Q_2(\vec Q_1\cdot\vec X)-\vec X(\vec Q_1\cdot\vec Q_2)$; taking traces gives $\operatorname{Tr}(C_{\vec Q_1}C_{\vec Q_2})+3(\vec Q_1\cdot\vec Q_2)=\vec Q_2\cdot\vec Q_1$, hence

$$ \operatorname{Tr}\bigl(C_{\vec Q_1}C_{\vec Q_2}\bigr)=-2\,\vec Q_1\cdot\vec Q_2. $$

The matrix $\operatorname{diag}(0,-2,-2,-2)$ is the case of the units, since $\mathbf{e}_i\cdot\mathbf{e}_j=\delta_{ij}$. Computed exactly on general elements with coefficients in $\mathbb{Q}(i)$ and on the sixteen pairs of the basis.

Corollary (the scalar part is annihilated). $\kappa(\tilde Q_1,\tilde Q_2)$ depends only on the vector parts: two elements with the same vector part have the same form against every element, and the form is identically zero as soon as either argument is central.

Remark (the single constant). On the vector subspace the form is proportional to the general plain bilinear form

$$ \langle\tilde Q_1,\tilde Q_2\rangle=\operatorname{Sc}(\tilde Q_1\tilde Q_2)=-\vec Q_1\cdot\vec Q_2, $$

with

$$ \kappa(\tilde Q_1,\tilde Q_2)=2\,\langle\tilde Q_1,\tilde Q_2\rangle\qquad(\tilde Q_1,\tilde Q_2\in\mathrm{Vect}(\mathbb{B})), $$

and the ratio of the two forms is the constant $2$. For the commutator reading of the previous section the constant is $4\cdot2=8$; the constant is thus a convention of the halving, and the combinatorial content — that two forms on the three-dimensional vector part are proportional — is the same. The Killing form sees the vector part through the general plain bilinear form, and the centre is invisible to it. The squared vector form gives the ratio $\kappa(\tilde Q,\tilde Q)/(\vec Q\cdot\vec Q)=-2$ wherever the vector part is nonzero, on the vector subspace and on the whole algebra alike, since both the form and the squared vector form depend only on the vector part; the general plain bilinear form $\langle\tilde Q_1,\tilde Q_2\rangle=\operatorname{Sc}(\tilde Q_1\tilde Q_2)$ is not proportional to $\kappa$ on the whole algebra, because its scalar part $\langle e_t,e_t\rangle=1$ is nondegenerate while $\kappa$ kills the scalar direction.

Invariance and Symmetry

Theorem. The form is symmetric and invariant: for all $\tilde Q_1,\tilde Q_2,\tilde Q_3$,

$$ \boxed{\; \kappa(\tilde Q_1,\tilde Q_2)=\kappa(\tilde Q_2,\tilde Q_1),\qquad \kappa\bigl(\operatorname{ad}_{\tilde Q_1}\tilde Q_3,\tilde Q_2\bigr)+\kappa\bigl(\tilde Q_3,\operatorname{ad}_{\tilde Q_1}\tilde Q_2\bigr)=0 \;}, $$

equivalently $\kappa([\tilde Q_1,\tilde Q_3],\tilde Q_2)=\kappa(\tilde Q_1,[\tilde Q_3,\tilde Q_2])$.

Proof. The trace is invariant under cyclic permutation and the adjoint is a homomorphism for the bracket, $\operatorname{ad}_{\tilde Q_1\wedge\tilde Q_3}=[\operatorname{ad}_{\tilde Q_1},\operatorname{ad}_{\tilde Q_3}]$, by the Jacobi identity; hence, using $\operatorname{Tr}([A,B]C)=\operatorname{Tr}(A[B,C])$,

$$ \operatorname{Tr}\bigl(\operatorname{ad}_{\tilde Q_1\wedge\tilde Q_3}\operatorname{ad}_{\tilde Q_2}\bigr) =\operatorname{Tr}\bigl([\operatorname{ad}_{\tilde Q_1},\operatorname{ad}_{\tilde Q_3}]\operatorname{ad}_{\tilde Q_2}\bigr) =\operatorname{Tr}\bigl(\operatorname{ad}_{\tilde Q_1}\,[\operatorname{ad}_{\tilde Q_3},\operatorname{ad}_{\tilde Q_2}]\bigr) =\operatorname{Tr}\bigl(\operatorname{ad}_{\tilde Q_1}\operatorname{ad}_{\tilde Q_3\wedge\tilde Q_2}\bigr), $$

that is $\kappa(\tilde Q_1\wedge\tilde Q_3,\tilde Q_2)=\kappa(\tilde Q_1,\tilde Q_3\wedge\tilde Q_2)$, which in terms of the commutator is the displayed equivalence by the factor $2$ carried through both slots. The symmetry is the symmetry of the trace under the interchange of the two factors together with the anticommutativity of the bracket, or the symmetry of $\vec Q_1\cdot\vec Q_2$ in the formula of the previous section.

Remark. The invariance is the statement that the adjoint operators are the skew elements for $\kappa$, $\kappa(\operatorname{ad}_{\tilde Q_1}\tilde Q_3,\tilde Q_2)+\kappa(\tilde Q_3,\operatorname{ad}_{\tilde Q_1}\tilde Q_2)=0$; the form is what makes the adjoint representation orthogonal in the sense of the block, and it is the input of the Casimir element below. The general statement that a Killing form is invariant and symmetric is The Killing Form Operator.

The Radical and the Rank

Theorem (the radical is the centre). The radical of $\kappa$,

$$ \operatorname{rad}\kappa=\{\tilde Q_1:\kappa(\tilde Q_1,\tilde Q_3)=0\text{ for all }\tilde Q_3\in\mathbb{B}\}, $$

is the centre $\mathbb{C}_{\mathbb{B}}$, of complex dimension one and real dimension two. Hence $\kappa$ is degenerate, of complex rank $3$ and real rank $6$, and its restriction to the vector subspace is nondegenerate.

Proof. By the formula $\kappa(\tilde Q_1,\tilde Q_2)=-2\,\vec Q_1\cdot\vec Q_2$, the condition $\kappa(\tilde Q_1,\tilde Q_3)=0$ for all $\tilde Q_3$ is $\vec Q_1\cdot\vec Q_3=0$ for all vectors $\vec Q_3$, which forces $\vec Q_1=0$, that is $\tilde Q_1\in\mathbb{C}e_t=\mathbb{C}_{\mathbb{B}}$. On the vector subspace the form is a nonzero multiple of the scalar product, whose radical is zero, so the restriction is nondegenerate and the rank is the complex dimension three. Computed exactly on the basis and on general elements.

Corollary (the Killing data). On the vector subspace the Gram matrix in the real basis $(\vec e_x,\vec e_y,\vec e_z,i\vec e_x,i\vec e_y,i\vec e_z)$ is $\operatorname{diag}(-2,-2,-2,2,2,2)$, of real signature $(3,3)$ and determinant $(-2)^3\cdot2^3=-64$; on the complex basis $(\vec e_x,\vec e_y,\vec e_z)$ the Gram matrix is $-2\,\mathrm{I}_3$, of complex determinant $-8$. The form is negative on the real vector directions and positive on the imaginary ones, and it is degenerate.

Remark. The degeneracy of the Killing form is the exact measure of the failure of semisimplicity of the block: the radical is the centre, and the quotient by the radical is the derived algebra, which is simple. The form restricted to the remarkable subspaces is read next, and the two statements are consistent with the reading of the same decomposition through the ideals in The Lie Algebra of the Antisymmetric Plain Algebra.

The Cartan Criteria

Theorem. The Cartan criteria, read on $\kappa$, give the following. The algebra $\mathfrak{g}$ is not semisimple, because $\kappa$ is degenerate; its radical is the centre $\mathbb{C}_{\mathbb{B}}$; the quotient $\mathfrak{g}/\mathbb{C}_{\mathbb{B}}\cong\mathfrak{sl}(2,\mathbb{C})$ is semisimple, and the derived algebra $\mathrm{Vect}(\mathbb{B})$ is simple, because the restriction of $\kappa$ to it is nondegenerate and the invariant form of a simple Lie algebra is nondegenerate.

Proof. The Cartan criterion states that a Lie algebra is semisimple exactly when its Killing form is nondegenerate, Structure of Lie Algebras; here the radical is nonzero, so the algebra is not semisimple. The radical is the centre by the previous theorem. The restriction of $\kappa$ to $\mathrm{Vect}(\mathbb{B})$ is nondegenerate, and $\mathrm{Vect}(\mathbb{B})\cong\mathfrak{sl}(2,\mathbb{C})$ is simple, so the two readings agree; the quotient by the radical is $\mathfrak{sl}(2,\mathbb{C})$, which is semisimple.

Remark. The block is therefore the exact model of the Cartan dichotomy: one algebra, one invariant form, and the degeneracy of the form read off the centre, with the nondegenerate part carrying $\mathfrak{sl}(2,\mathbb{C})$. The single scalar that distinguishes the two forms on the vector part, the constant $2$ of the first sections, is the same constant that enters the Casimir element below.

The Restriction to the Remarkable Subspaces and the Isotropic Elements

The formula $\kappa(\tilde Q_1,\tilde Q_2)=-2\,\vec Q_1\cdot\vec Q_2$ restricts to each of the remarkable subspaces by the sector its vector part occupies. On the quaternion subspace $\mathbb{H}_{\mathbb{B}}$ and the anti-Hermitian subspace $\mathbb{M}_-$ the vector part is real, so $\kappa(\tilde Q,\tilde Q)=-2\,\vec Q\cdot\vec Q$ with $\vec Q\cdot\vec Q$ a sum of squares; on the anti-quaternion subspace $i\mathbb{H}_{\mathbb{B}}$ and the Hermitian subspace $\mathbb{M}_+$ the vector part is imaginary, $\vec Q=i\vec U$ with $\vec U$ real, so $\kappa(\tilde Q,\tilde Q)=-2\,(i\vec U)\cdot(i\vec U)=2\,\vec U\cdot\vec U$; on the centre the vector part is zero and the form vanishes; and on the vector subspace the vector part is complex and $\kappa(\tilde Q_1,\tilde Q_2)=-2\,\vec Q_1\cdot\vec Q_2$ is the complexified scalar product.

Theorem (the restrictions). The restriction of $\kappa$ to each of the remarkable subspaces is given by the table

Proof. Each row is the formula of the previous paragraph read on the real basis of the subspace, in which the scalar direction contributes zero and the three vector directions contribute the displayed signs. The radical of each restriction is the scalar direction of that subspace, since the form is zero there and of one sign on the vector triple; on the whole vector subspace no direction is radical because the complex scalar product is nondegenerate. The signatures are the counts of the signs; on the vector subspace the real basis $(\vec e_x,\vec e_y,\vec e_z,i\vec e_x,i\vec e_y,i\vec e_z)$ gives three negative and three positive entries. Computed exactly on each of the bases.

Theorem (the isotropic elements). An element $\tilde Q$ is isotropic, $\kappa(\tilde Q,\tilde Q)=0$, exactly when its vector part is isotropic for the complexified scalar product, $\vec Q\cdot\vec Q=0$, and the isotropic elements of the remarkable subspaces are the following. The centre is entirely isotropic. On the quaternion subspace and the anti-Hermitian subspace the isotropic elements are exactly the scalar direction, of real dimension one, because the vector part is real and a sum of three squares vanishes only at zero. On the anti-quaternion subspace and the Hermitian subspace the isotropic elements are again exactly the scalar direction, for the same reason applied to $\vec U$. On the vector subspace the isotropic elements form the complex null cone $\{\vec Q_1\cdot\vec Q_1=0\}$, of complex dimension two and real dimension four.

Proof. The condition is $\vec Q\cdot\vec Q=0$ by the formula. On the four real subspaces the vector part is either real or purely imaginary, and in both cases $\vec Q\cdot\vec Q$ is a sum of three squares of real numbers for the real vectors $r_x,r_y,r_z$, whose vanishing forces $r_x=r_y=r_z=0$; only the scalar direction remains. On the vector subspace with $\vec Q_1$ complex, the equation $Q_{1,x}^2+Q_{1,y}^2+Q_{1,z}^2=0$ is one complex equation, a complex cone of complex dimension two, equivalently two real equations defining a real cone of dimension four. Computed on the basis and on general elements.

Remark. The isotropic elements of the block are the null cone of the vector subspace together with the scalar directions of the four real subspaces, and the centre, which is entirely isotropic. The isotropic cone is not the zero-divisor cone of the algebra: on the quaternion subspace the zero-divisor condition is $\sum_\mu Q_\mu^2=0$ with all four squares, while the Killing isotropy is $\sum_{k\in\{x,y,z\}}Q_k^2=0$, and on the real quaternion subspace the second vanishes only at the origin while the first is the null cone of the quaternion form; the two cones are compared in Remarkable Subspaces and the Four General Products, Forms and Norms for the plain form, and the difference is the same here. The zero-divisors themselves are Quaternionic Norm Regularity and Degeneracy in the Biquaternion Algebra, §The Zero Divisors, and the norms are The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra.

The Casimir Element

Definition. Let $(E_1,E_2,E_3)=(\vec e_x,\vec e_y,\vec e_z)$ be the basis of the vector subspace and let $h^{ij}$ be the inverse of the Gram matrix $\kappa(E_i,E_j)=-2\delta_{ij}$, that is $h^{ij}=-\tfrac12\delta^{ij}$. The Casimir element of the block is

$$ \Omega=\sum_{i,j=1}^{3}h^{ij}E_iE_j=-\tfrac12\bigl(\vec e_x^2+\vec e_y^2+\vec e_z^2\bigr) $$

in the enveloping algebra of the derived algebra.

Theorem (the Casimir on the two-dimensional module and on the adjoint). On the two-dimensional module of the model of Introduction to the Symmetric Plain Algebra, where the derived algebra acts by the halved matrices, the Casimir acts as the scalar $\tfrac38$; in the adjoint representation on the vector subspace it acts as the identity.

Proof. The block's halved bracket corresponds in the model to the half-commutator, so the derived algebra acts on the two-dimensional space through $\tfrac12\mathsf{M}_2(\vec e_\mu)=-\tfrac{i}{2}\sigma_\mu$: then $\bigl(\tfrac12\mathsf{M}_2(\vec e_\mu)\bigr)^2=\tfrac14(-i\sigma_\mu)^2=-\tfrac14\mathrm{I}_2$, the sum of the three squares is $-\tfrac34\mathrm{I}_2$, and $\Omega=-\tfrac12\bigl(-\tfrac34\mathrm{I}_2\bigr)=\tfrac38\mathrm{I}_2$, the eigenvalue $\tfrac38$ of the two-dimensional module of $\mathfrak{sl}(2,\mathbb{C})$ fixed by the normalisation of the Killing form, as in The Casimir Operator; the reading with $\mathsf{M}_2$ itself, which does not carry the block's bracket, would give the fourfold value. In the adjoint representation the operator $\operatorname{ad}_{\vec e_\mu}$ satisfies $\operatorname{ad}_{\vec e_x}^2=\operatorname{diag}(-1,-1)$ on the plane orthogonal to $\vec e_x$ and vanishes on $\vec e_x$, and cyclically, so $\sum_{\mu\in\{x,y,z\}}\operatorname{ad}_{\vec e_\mu}^2=-2\,\mathrm{I}_3$ on the vector subspace; hence $\Omega=-\tfrac12(-2\,\mathrm{I}_3)=\mathrm{I}_3$. Computed exactly over $\mathbb{Q}(i)$.

Remark (the central generator). The Casimir is the quadratic central element of $\mathrm{U}(\mathfrak{sl}(2,\mathbb{C}))$, and with the central generator $z$ of the enveloping algebra of The Lie Algebra of the Antisymmetric Plain Algebra it generates the centre of $\mathrm{U}(\mathfrak{g})$. The element is one and the same in the two representations in which it was read, the two-dimensional module and the adjoint, where it acts as the scalars $\tfrac38$ and $1$ fixed by the normalisation of the Killing form. The general theory of the element, of its centrality and of its role in the representation theory is The Casimir Operator.

Worked Examples

The central element. For $\tilde Q_1=e_t$ the adjoint is zero, so $\kappa(e_t,\tilde Q_2)=0$ for every $\tilde Q_2$: the unit is in the radical.

A vector against a vector. $\kappa(\vec e_x,\vec e_y)=0$ and $\kappa(\vec e_x,\vec e_x)=-2$: the form is diagonal on the vector basis, as the Gram matrix $\operatorname{diag}(0,-2,-2,-2)$ records.

A complex direction. $\kappa(i\vec e_x,i\vec e_x)=-2\,(i)(i)=-2\cdot(-1)=2$: the imaginary vector directions are positive for the form, which is the signature $(3,3)$ of the vector subspace.

A scalar part that does not enter. $\tilde Q_1=e_t+ie_t$, $\tilde Q_2=e_t+ie_t$ give $\kappa(\tilde Q_1,\tilde Q_2)=0$; the two central coordinates are invisible to the form, and the whole centre is the radical.

The quaternion real form. On $\mathbb{H}_{\mathbb{B}}$ the form is $\kappa(\tilde Q,\tilde Q)=-2(q_x^2+q_y^2+q_z^2)$, of signature $(0,3)$, with radical $\mathbb{R}e_t$: the negative semidefinite row of the table, and the zero direction is the scalar one.

The Hermitian real form. On $\mathbb{M}_+$ with $\tilde Q=q_te_t+q'_xi\vec e_x+q'_yi\vec e_y+q'_zi\vec e_z$ the form is $\kappa(\tilde Q,\tilde Q)=2((q'_x)^2+(q'_y)^2+(q'_z)^2)$: positive semidefinite, with radical $\mathbb{R}e_t$ and no other isotropic element.

The vector null cone. On the vector subspace the element $\tilde Q=\vec e_x+i\vec e_x$ satisfies $\kappa(\tilde Q,\tilde Q)=-2(1+i^2)=0$: it is a nonzero isotropic element, and the complex cone it generates has real dimension four.

The Casimir on the adjoint. The eigenvalues of $\Omega$ on the three vector directions are all $1$, and on the two-dimensional module the scalar is $\tfrac38$: the two readings of the theorem, and the two constants by which the invariant form controls the representation theory of the block.

Summary

The Killing form of the antisymmetric plain algebra is $\kappa(\tilde Q_1,\tilde Q_2)=\operatorname{Tr}(\operatorname{ad}_{\tilde Q_1}\operatorname{ad}_{\tilde Q_2})$, equal on general elements to $-2\,\vec Q_1\cdot\vec Q_2$, with Gram matrix $\operatorname{diag}(0,-2,-2,-2)$ in the basis of the units; it is symmetric and invariant, and it kills the scalar part. Its radical is the centre $\mathbb{C}_{\mathbb{B}}$, so the form is degenerate of complex rank three and real rank six; restricted to the vector subspace it is nondegenerate, of real signature $(3,3)$, and proportional to the general plain bilinear form there with the single constant $2$. The Cartan criteria read on it give: the algebra is not semisimple, its radical is the centre, and its quotient and its derived algebra are the simple $\mathfrak{sl}(2,\mathbb{C})$. On the remarkable subspaces the restrictions are the zero form on the centre, the signature $(3,3)$ form on the vector subspace, and the semidefinite forms of signatures $(0,3)$, $(3,0)$, $(3,0)$ and $(0,3)$ on the quaternion, anti-quaternion, Hermitian and anti-Hermitian subspaces, of radicals the respective scalar directions; the isotropic elements are the complex null cone on the vector subspace and the scalar directions elsewhere. The Casimir element $\Omega=-\tfrac12(\vec e_x^2+\vec e_y^2+\vec e_z^2)$ acts as $\tfrac38$ on the two-dimensional module and as the identity on the adjoint, and it is the quadratic central generator of the enveloping algebra.

Summary of Notation

Further Reading

• The Lie Algebra of the Antisymmetric Plain Algebra, for the centre, the derived algebra, the identifications and the enveloping algebra
• The Adjoint Operators and the Derivations of the Antisymmetric Plain Algebra, for the adjoints, their kernels and the derivations
• Remarkable Subspaces under the Antisymmetric Plain Algebra, for the same remarkable subspaces read under the bracket instead of the form
• The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra, for the general plain bilinear form and its three companions
• The Killing Form Operator, for the invariant form of a Lie algebra and its properties
• Structure of Lie Algebras, for the Cartan criteria, the radical and semisimplicity
• The Casimir Operator, for the central element of the enveloping algebra and its eigenvalues
• Remarkable Subspaces and the Four General Products, Forms and Norms, for the restriction of the plain form to the remarkable subspaces, compared here with the restriction of $\kappa$