The Trace Form and the Invariance of the Symmetric Plain Algebra

Introduction

The symmetric plain algebra of $\mathbb{B}$ carries the product $\tilde Q_1\bullet\tilde Q_2=\tfrac12(\tilde Q_1\tilde Q_2+\tilde Q_2\tilde Q_1)$ of Introduction to the Symmetric Plain Algebra, and a commutative algebra carries a form with it: the trace form

$$ \tau(\tilde Q_1,\tilde Q_2)=\mathrm{Sc}(\tilde Q_1\bullet\tilde Q_2)=Q_{1t}Q_{2t}-(\vec Q_1,\vec Q_2), $$

the scalar part of the product of the two elements. This article reads the form, its identity with the scalar part of the plain product, its Gram matrix, its invariance $\tau(\tilde Q_1\bullet\tilde Q_2,\tilde Q_4)=\tau(\tilde Q_1,\tilde Q_2\bullet\tilde Q_4)$, the operator traces built from it, and its restriction to the remarkable subspaces.

The form is not new to the algebra: it is the general plain bilinear form $\langle\tilde Q_1,\tilde Q_2\rangle=\mathrm{Sc}(\tilde Q_1\tilde Q_2)$ of The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra, which the associativity of the plain product makes invariant, and its operator theory is The Four Adjoints of the Biquaternion Algebra in Examples and Two-Sided Operators on the Biquaternion Algebra. What this article adds is the same form read as the trace form of the Jordan algebra: the invariance becomes the associativity of the Jordan product read on the form, the polarisation of the form recovers the product, and the operator traces are the Jordan-algebra reading of the plain operators. The restrictions of the form to the remarkable subspaces are the subject of Remarkable Subspaces and the Four General Products, Forms and Norms; the form is used here with the product and not re-derived.

The Trace Form

The Definition

Definition. The trace form of the symmetric plain algebra is

$$ \tau(\tilde Q_1,\tilde Q_2)=\mathrm{Sc}(\tilde Q_1\bullet\tilde Q_2). $$

Proposition. The form is $\mathbb{C}$-bilinear and symmetric:

$$ \tau(\tilde Q_1,\tilde Q_2)=Q_{1t}Q_{2t}-(\vec Q_1,\vec Q_2)=\sum_{\mu\in\{t,x,y,z\}}\varepsilon_\mu Q_{1\mu} Q_{2\mu},\qquad \tau(\tilde Q_1,\tilde Q_2)=\tau(\tilde Q_2,\tilde Q_1). $$

Proof. The scalar part of the product is $\mathrm{Sc}(\tilde Q_1\bullet\tilde Q_2)=Q_{1t}Q_{2t}-(\vec Q_1,\vec Q_2)$ by the scalar–vector rule of Introduction to the Symmetric Plain Algebra; the expression is symmetric in the two elements because the dot product is. Bilinearity is inherited from the product.

Proposition (non-degeneracy). The form is non-degenerate, and its Gram matrix on the basis $e_t,\vec e_x,\vec e_y,\vec e_z$ is

$$ \operatorname{diag}(1,-1,-1,-1), $$

the coefficient matrix $D$ of The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra, of complex rank four.

Proof. $\tau(e_\mu,e_\nu)=\mathrm{Sc}(e_\mu\bullet e_\nu)$; on the basis table of Introduction to the Symmetric Plain Algebra the values are $\varepsilon_\mu\delta_{\mu\nu}=(1,-1,-1,-1)$ on the diagonal and $0$ off it. The matrix is invertible, so the form is non-degenerate.

The Identity with the Scalar Part of the Plain Product

Theorem. The trace form of the block is the scalar part of the plain product:

$$ \boxed{\; \tau(\tilde Q_1,\tilde Q_2)=\mathrm{Sc}(\tilde Q_1\tilde Q_2)=\langle\tilde Q_1,\tilde Q_2\rangle \;}. $$

Proof. $\mathrm{Sc}(\tilde Q_1\bullet\tilde Q_2)=\tfrac12\bigl(\mathrm{Sc}(\tilde Q_1\tilde Q_2)+\mathrm{Sc}(\tilde Q_2\tilde Q_1)\bigr)$, and the scalar part of a plain product is symmetric, $\mathrm{Sc}(\tilde Q_1\tilde Q_2)=\mathrm{Sc}(\tilde Q_2\tilde Q_1)=\sum_\mu\varepsilon_\mu Q_{1\mu} Q_{2\mu}$. The half-sum is therefore the common value.

Remark (the two readings are one form). The trace form of the Jordan algebra and the general plain bilinear form are the same form, because the scalar part cannot see the cross term that distinguishes the plain product from its symmetrisation. The form is a form of the algebra and of the block at once, and the two readings differ only in what is put on the other side of it: the plain product in the algebra theory, the Jordan product here.

The Gram Matrix on the Real Basis

The form is complex bilinear, and on the real basis of the algebra, $e_t,\vec e_x,\vec e_y,\vec e_z,ie_t,i\vec e_x,i\vec e_y,i\vec e_z$, its realification is the diagonal matrix

$$ \operatorname{diag}(1,-1,-1,-1,-1,1,1,1), $$

of signature $(4,4)$.

Proof. On the four real directions the values are $\varepsilon_\mu$; on the four imaginary directions they are $\mathrm{Re}\,\tau(ie_\mu,ie_\nu)=\mathrm{Re}(i^2\varepsilon_\mu\delta_{\mu\nu})=-\varepsilon_\mu\delta_{\mu\nu}$, that is $-1,1,1,1$ on the diagonal; and the mixed values $\mathrm{Re}\,\tau(e_\mu,ie_\nu)$ vanish. The realified signature is $(4,4)$, the signature of the realified form of The Realification and the Effect of the Change of Scalars.

The Invariance

The Invariance Identity

Theorem. The trace form is invariant under the product:

$$ \boxed{\; \tau(\tilde Q_1\bullet\tilde Q_2,\tilde Q_4)=\tau(\tilde Q_1,\tilde Q_2\bullet\tilde Q_4) \;} $$

for all $\tilde Q_1,\tilde Q_2,\tilde Q_4\in\mathbb{B}$.

Proof. By the definition of the form and the product,

$$ \tau(\tilde Q_1\bullet\tilde Q_2,\tilde Q_4)=\mathrm{Sc}\bigl((\tilde Q_1\bullet\tilde Q_2)\tilde Q_4\bigr)=\tfrac12\bigl(\mathrm{Sc}(\tilde Q_1\tilde Q_2\tilde Q_4)+\mathrm{Sc}(\tilde Q_2\tilde Q_1\tilde Q_4)\bigr), $$

$$ \tau(\tilde Q_1,\tilde Q_2\bullet\tilde Q_4)=\mathrm{Sc}\bigl(\tilde Q_1(\tilde Q_2\bullet\tilde Q_4)\bigr)=\tfrac12\bigl(\mathrm{Sc}(\tilde Q_1\tilde Q_2\tilde Q_4)+\mathrm{Sc}(\tilde Q_1\tilde Q_4\tilde Q_2)\bigr). $$

The first terms agree, and the second terms agree because $\mathrm{Sc}$ is invariant under the cyclic permutation of its factors, $\mathrm{Sc}(\tilde Q_2\tilde Q_1\tilde Q_4)=\mathrm{Sc}(\tilde Q_1\tilde Q_4\tilde Q_2)$; equivalently $\mathrm{Sc}(\tilde Q_1\tilde Q_2\tilde Q_3)=\mathrm{Sc}(\tilde Q_3\tilde Q_1\tilde Q_2)$, the associativity of the plain product read on the scalar part.

Remark (invariance is associativity on the Jordan product). The plain product is associative, so the scalar part of a triple product is invariant under cyclic permutation; the invariance of the trace form is exactly that statement written with the symmetrised product. The form is associative in the Jordan sense, and it is the trace form that the general theory of Jordan algebras attaches to a special Jordan algebra. The plain operator theory of the same form, the association and the transpose relative to the plain product, is The Four Adjoints of the Biquaternion Algebra in Examples.

The Polarisation That Recovers the Product

Theorem. The trace form and the unit recover the product. From

$$ T(\tilde Q_1)=2\tau(\tilde Q_1,e_t)=2Q_{1t},\qquad \langle\tilde Q_1\rangle_{\natural}=2\tau(\tilde Q_1,e_t)^2-\tau(\tilde Q_1,\tilde Q_1)=Q_{1t}^2+(\vec Q_1,\vec Q_1), $$

the square is $\tilde Q_1\bullet\tilde Q_1=T(\tilde Q_1)\tilde Q_1-\langle\tilde Q_1\rangle_{\natural}e_t$, and the product is its polarisation,

$$ \tilde Q_1\bullet\tilde Q_2=\tfrac12\bigl((\tilde Q_1+\tilde Q_2)\bullet(\tilde Q_1+\tilde Q_2)-\tilde Q_1\bullet\tilde Q_1-\tilde Q_2\bullet\tilde Q_2\bigr). $$

Proof. The generic trace is $T(\tilde Q_1)=2Q_{1t}=2\tau(\tilde Q_1,e_t)$, and the generic norm of The Square, the Idempotents and the Jordan Inverse of the Symmetric Plain Algebra is $\langle\tilde Q_1\rangle_{\natural}=Q_{1t}^2+(\vec Q_1,\vec Q_1)=2Q_{1t}^2-\bigl(Q_{1t}^2-(\vec Q_1,\vec Q_1)\bigr)=2\tau(\tilde Q_1,e_t)^2-\tau(\tilde Q_1,\tilde Q_1)$. The quadratic identity of the block gives $\tilde Q_1\bullet\tilde Q_1=T(\tilde Q_1)\tilde Q_1-\langle\tilde Q_1\rangle_{\natural}e_t$, and the polarisation of the square gives the product.

Remark (the form is a polar form). The form is also the polar form of the quadratic trace $\tilde Q_1\mapsto\mathrm{Sc}(\tilde Q_1\bullet\tilde Q_1)=\mathrm{Sc}(\tilde Q_1^2)=\tau(\tilde Q_1,\tilde Q_1)$:

$$ \tau(\tilde Q_1,\tilde Q_2)=\tfrac12\bigl(\tau(\tilde Q_1+\tilde Q_2,\tilde Q_1+\tilde Q_2)-\tau(\tilde Q_1,\tilde Q_1)-\tau(\tilde Q_2,\tilde Q_2)\bigr). $$

The same form is the polarisation of the quadratic trace and the datum from which the generic trace and the generic norm, and hence the product, are recovered; this double role is what makes it the trace form of the block.

The Operator Traces

The Trace of a Multiplication Operator

Proposition. The multiplication operator of the block is $L^{\bullet}_{\tilde Q_1}=\tfrac12(L_{\tilde Q_1}+R_{\tilde Q_1})$, and

$$ \operatorname{Tr}\bigl(L^{\bullet}_{\tilde Q_1}\bigr)=4Q_{1t}=2T(\tilde Q_1),\qquad \det\bigl(L^{\bullet}_{\tilde Q_1}\bigr)=Q_{1t}^2\,\langle\tilde Q_1\rangle_{\natural}. $$

Proof. $L^{\bullet}_{\tilde Q_1}\tilde Q_2=\tilde Q_1\bullet\tilde Q_2=\tfrac12(\tilde Q_1\tilde Q_2+\tilde Q_2\tilde Q_1)=\tfrac12(L_{\tilde Q_1}+R_{\tilde Q_1})\tilde Q_2$, so the two operators differ by the factor $\tfrac12$ and not by the sum alone; the trace of a plain operator is $\operatorname{Tr}(L_{\tilde Q_1})=\operatorname{Tr}(R_{\tilde Q_1})=4Q_{1t}$, so the trace of the half-sum is $4Q_{1t}=2T(\tilde Q_1)$. The determinant is computed on the matrix of the operator; in the $2\times2$ model the operator is $M\mapsto\tfrac12(\mathsf{M}_2(\tilde Q_1)M+M\mathsf{M}_2(\tilde Q_1))$, whose eigenvalues are $\tfrac12(\lambda_i+\lambda_j)$ over the eigenvalues $\lambda_1,\lambda_2$ of $\mathsf{M}_2(\tilde Q_1)$, giving $\det= Q_{1t}^2\langle\tilde Q_1\rangle_{\natural}$.

Remark (the factor of the operator identity). The operator identity is

$$ L_{\tilde Q_1}+R_{\tilde Q_1}=2L^{\bullet}_{\tilde Q_1}, $$

so the sum of the two plain operators is twice the multiplication operator of the block, and not the operator itself. The multiplication operator of the symmetric plain algebra is half the sum of the left and right operators of the plain product; the sum is the operator of the doubled product $\tilde Q_1\tilde Q_2+\tilde Q_2\tilde Q_1$.

The Operator Trace and Its Relation to the Form

Theorem. For all $\tilde Q_1,\tilde Q_2$,

$$ \boxed{\; \operatorname{Tr}\bigl(L^{\bullet}_{\tilde Q_1}L^{\bullet}_{\tilde Q_2}\bigr)=4Q_{1t}Q_{2t}-2(\vec Q_1,\vec Q_2)=2\tau(\tilde Q_1,\tilde Q_2)+\tfrac12T(\tilde Q_1)T(\tilde Q_2) \;}. $$

Proof. With $L^{\bullet}=\tfrac12(L+R)$, the product expands as

$$ L^{\bullet}_{\tilde Q_1}L^{\bullet}_{\tilde Q_2}=\tfrac14\bigl(L_{\tilde Q_1}L_{\tilde Q_2}+L_{\tilde Q_1}R_{\tilde Q_2}+R_{\tilde Q_1}L_{\tilde Q_2}+R_{\tilde Q_1}R_{\tilde Q_2}\bigr), $$

and the four plain traces are $\operatorname{Tr}(L_{\tilde Q_1}L_{\tilde Q_2})=4\tau(\tilde Q_1,\tilde Q_2)$, $\operatorname{Tr}(R_{\tilde Q_1}R_{\tilde Q_2})=4\tau(\tilde Q_1,\tilde Q_2)$, and $\operatorname{Tr}(L_{\tilde Q_1}R_{\tilde Q_2})=\operatorname{Tr}(R_{\tilde Q_1}L_{\tilde Q_2})=4Q_{1t}Q_{2t}$, computed on the bases of the left and right operators. Summing gives $\tfrac14\bigl(8\tau(\tilde Q_1,\tilde Q_2)+8Q_{1t}Q_{2t}\bigr)=2\tau(\tilde Q_1,\tilde Q_2)+2Q_{1t}Q_{2t}$, and $2Q_{1t}Q_{2t}=\tfrac12T(\tilde Q_1)T(\tilde Q_2)$.

Remark (the trace is not the form). The operator trace is not the trace form, and not a multiple of it either: the two coefficient pairs $(4,-2)$ and $(1,-1)$ are not proportional. Its excess over twice the form is exactly the term $\tfrac12T(\tilde Q_1)T(\tilde Q_2)$ built from the generic trace, so the operator trace and $2\tau(\tilde Q_1,\tilde Q_2)$ agree exactly on the elements of generic trace zero, that is on the hyperplane $Q_{1t}=0$ or $Q_{2t}=0$; against four times the form, $\operatorname{Tr}(L_{\tilde Q_1}L_{\tilde Q_2})=4\tau(\tilde Q_1,\tilde Q_2)$, the agreement is exactly on the pairs with $(\vec Q_1,\vec Q_2)=0$. The trace form is the polarisation of the quadratic trace, while the operator trace is a second, larger invariant, and the two are related by the generic trace functional.

The Trace Functional

Definition. The trace functional of the block is

$$ \tilde Q\longmapsto T(\tilde Q)=2Q_t=2\tau(\tilde Q,e_t). $$

It is the linear form $\operatorname{Tr}\bigl(L^{\bullet}_{\tilde Q}\bigr)/2$, and it is the generic trace of The Square, the Idempotents and the Jordan Inverse of the Symmetric Plain Algebra. The generic norm is $\langle\tilde Q\rangle_{\natural}=2\tau(\tilde Q,e_t)^2-\tau(\tilde Q,\tilde Q)$, the quadratic companion of the trace, and the ambient identity of the block is $\tilde Q\bullet\tilde Q-T(\tilde Q)\tilde Q+\langle\tilde Q\rangle_{\natural}e_t=0$.

Remark (the two symmetric functions). The trace functional and the generic norm are the two coefficients of the quadratic identity, and they are recovered from the trace form and the unit: $T(\tilde Q)=2\tau(\tilde Q,e_t)$ and $\langle\tilde Q\rangle_{\natural}=2\tau(\tilde Q,e_t)^2-\tau(\tilde Q,\tilde Q)$. The trace form, the unit and the polarisation carry the whole degree-two structure of the block.

The Restriction to the Remarkable Subspaces

The Restriction Matrices

The trace form is the general plain bilinear form, so its restrictions to the remarkable subspaces of Introduction to the Remarkable Subspaces are those of Remarkable Subspaces and the Four General Products, Forms and Norms, read here as the restrictions of the trace form of the block:

The matrices are the same six as in Remarkable Subspaces and the Four General Products, Forms and Norms, because the trace form is the general plain bilinear form; the table is reproduced here with the reading of the block.

The Isotropic Cones of the Restrictions

Remark (the restriction and the cone of the block). The restricted form is non-degenerate on each of the remarkable subspaces, of ranks $2,6,4,4,4,4$, and the four indefinite rows carry isotropic cones of real dimensions $1,5,3,3$ with the null elements $e_t+ie_t$, $\vec e_x+i\vec e_x$, $e_t+\vec e_x$ and $ie_t+i\vec e_x$. The two definite rows, the Hermitian and the anti-Hermitian, carry no isotropic vector. The isotropic cone of the restriction is the intersection of the subspace with the cone of the form, and it must not be confused with the isotropic cone of the generic norm: on the quaternion subspace, for instance, the trace form has the cone of signature $(1,3)$ computed here, while the generic norm $\sum_\mu Q_\mu^2$ vanishes on that subspace only at the origin.

Summary

The trace form of the symmetric plain algebra is $\tau(\tilde Q_1,\tilde Q_2)=\mathrm{Sc}(\tilde Q_1\bullet\tilde Q_2)=Q_{1t}Q_{2t}-(\vec Q_1,\vec Q_2)$, which is the scalar part of the plain product and the general plain bilinear form. It is $\mathbb{C}$-bilinear, symmetric and non-degenerate, with Gram matrix $\operatorname{diag}(1,-1,-1,-1)$ on the basis and realified signature $(4,4)$; it is invariant, $\tau(\tilde Q_1\bullet\tilde Q_2,\tilde Q_4)=\tau(\tilde Q_1,\tilde Q_2\bullet\tilde Q_4)$, which is the associativity of the plain product read on the Jordan product, and its polarisation recovers the quadratic trace, the generic trace $T(\tilde Q_2)=2\tau(\tilde Q_2,e_t)$ and the generic norm $\langle\tilde Q_2\rangle_{\natural}=2\tau(\tilde Q_2,e_t)^2-\tau(\tilde Q_2,\tilde Q_2)$, hence the product. The multiplication operator is $L^{\bullet}_{\tilde Q_1}=\tfrac12(L_{\tilde Q_1}+R_{\tilde Q_1})$, of trace $4Q_{1t}=2T(\tilde Q_1)$ and determinant $Q_{1t}^2\langle\tilde Q_1\rangle_{\natural}$, and the operator trace is $\operatorname{Tr}(L^{\bullet}_{\tilde Q_1}L^{\bullet}_{\tilde Q_2})=4Q_{1t}Q_{2t}-2(\vec Q_1,\vec Q_2)=2\tau(\tilde Q_1,\tilde Q_2)+\tfrac12T(\tilde Q_1)T(\tilde Q_2)$. On the remarkable subspaces the restrictions are those of the general plain bilinear form, of ranks $2,6,4,4,4,4$ and signatures $(1,1),(3,3),(1,3),(3,1),(4,0),(0,4)$.

Summary of Notation

Further Reading

• The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra, for the general plain bilinear form, its coefficient matrix $D$ and its comparison with the other three forms.
• The Four Adjoints of the Biquaternion Algebra in Examples, for the associativity of the plain form and its operator theory.
• Two-Sided Operators on the Biquaternion Algebra, for the left and right operators whose half-sum is the multiplication operator of the block.
• Remarkable Subspaces and the Four General Products, Forms and Norms, for the restrictions of the form to the remarkable subspaces.
• The Square, the Idempotents and the Jordan Inverse of the Symmetric Plain Algebra, for the generic trace, the generic norm and the quadratic identity.