Jordan Algebras, Spin Factors, and Alternative Examples

By [F1] and the bilinearity of multiplication in [F2], $\circ$ is bilinear; swapping $a,b$ leaves $ab+ba$ unchanged, so it is commutative. Also $a_{\circ }^2:=a\circ a=aa$ because $2$ is invertible in $k$.

Using associativity from [F2], expand $a\circ (b\circ a_{\circ }^2)=\frac{1}{4}(ab{a}^2+a^{3}b+b{a}^3+a^{2}ba)\mathrm{.}$ Here $a^3=(aa)a=a(aa)$.

In the complex involutive case, $A_{\mathrm{s}\mathrm{a}}$ is closed under addition and real scalar multiplication by conjugate-linearity in [F4], and it contains the unit because $1^{\ast }=1$. For $a,b\in A_{\mathrm{s}\mathrm{a}}$, $(a\circ b{)}^{\ast }=\frac{1}{2}((ab{)}^{\ast }+(ba{)}^{\ast })=\frac{1}{2}(ba+ab)=a\circ b,$ so it is closed under $\circ$.

The other side expands, using [F2], as $(a\circ b)\circ a_{\circ }^2=\frac{1}{4}(ab{a}^2+b{a}^3+a^{3}b+a^{2}ba),$ which contains the same four terms as step 2.1.

Since $A$ is unital, [F1] gives $1\circ a=(1a+a1)\mathrm{/}2=a$ for every $a\in A$. Steps 1.1, 2.1, and 3.1 prove bilinearity, commutativity, and the Jordan identity in [F3]. Thus $A^{+}$ is a unital Jordan algebra.

The real subspace $A_{\mathrm{s}\mathrm{a}}$ inherits commutativity, bilinearity, and the Jordan identity from $A^{+}$, so it is a real Jordan subalgebra.

Formula [F1] is symmetric in $(v,\alpha )$ and $(w,\beta )$ by [F2], and is bilinear. The element $(0,1)$ is a two-sided unit because $(0,1)\circ (v,\alpha )=(v,\alpha )$.

Put $s:=\mathrm{\parallel }v{\mathrm{\parallel }}^2$ and $r:=⟨v,w⟩$. Substituting $x$ twice in [F1] gives $x^2=(2\alpha v,s+{\alpha }^2)$.

Using step 1.2 in [F1] first gives $y\circ x^2=((s+{\alpha }^2)w+2\alpha \beta v,2\alpha r+\beta (s+{\alpha }^2))\mathrm{.}$ A second substitution yields $x\circ (y\circ x^2)=(\alpha (s+{\alpha }^2)w+[2\alpha r+\beta (s+3{\alpha }^2)]v,r(s+3{\alpha }^2)+\beta (3\alpha s+{\alpha }^3))\mathrm{.}$ Each scalar coefficient uses only bilinearity and symmetry in [F2].

First $x\circ y=(\alpha w+\beta v,r+\alpha \beta )$. Applying [F1] to $(x\circ y)\circ x^2$ and using step 1.2 gives $(x\circ y)\circ x^2=(\alpha (s+{\alpha }^2)w+[2\alpha r+\beta (s+3{\alpha }^2)]v,r(s+3{\alpha }^2)+\beta (3\alpha s+{\alpha }^3))\mathrm{.}$ This equals step 2.1 component by component.

Steps 1.1, 2.1, and 3.1 verify the unit, commutativity, bilinearity, and [F3]; hence $S(V)$ is a unital real Jordan algebra.

The real-linear map $\mathrm{\Phi }(v,\alpha )=\bar{v}+\alpha 1$ satisfies $\begin{array}{rl}{\displaystyle \mathrm{\Phi }((v,\alpha )\circ (w,\beta ))}& {\displaystyle =\alpha \bar{w}+\beta \bar{v}+(⟨v,w⟩+\alpha \beta )1,}\\ {\displaystyle \mathrm{\Phi }(v,\alpha )\circ \mathrm{\Phi }(w,\beta )}& {\displaystyle =\frac{1}{2}((\bar{v}+\alpha 1)(\bar{w}+\beta 1)+(\bar{w}+\beta 1)(\bar{v}+\alpha 1))}\\ {\displaystyle }& {\displaystyle =\alpha \bar{w}+\beta \bar{v}+\frac{1}{2}(\bar{v}\bar{w}+\bar{w}\bar{v})+\alpha \beta 1.}\end{array}$ The two expressions agree by [F2]. Thus $\mathrm{\Phi }$ preserves products and sends the unit $(0,1)$ to $1$. Its image is exactly the vector subspace $\mathbb{R}1+\bar{V}$, and product preservation makes that image a Jordan subalgebra.

For $x=(v,\alpha )$, [F1] gives $x^2=(2\alpha v,\mathrm{\parallel }v{\mathrm{\parallel }}^2+{\alpha }^2)$. If this square is $(w,t)$, norm homogeneity from [F5] gives $\mathrm{\parallel }w\mathrm{\parallel }=2\mathrm{\mid }\alpha \mathrm{\mid }\mathrm{\parallel }v\mathrm{\parallel }$, while $(\mathrm{\mid }\alpha \mathrm{\mid }-\mathrm{\parallel }v\mathrm{\parallel }{)}^2\ge 0$ gives $2\mathrm{\mid }\alpha \mathrm{\mid }\mathrm{\parallel }v\mathrm{\parallel }\le {\alpha }^2+\mathrm{\parallel }v{\mathrm{\parallel }}^2=t$. Thus every square satisfies $t\ge \mathrm{\parallel }w\mathrm{\parallel }$.

If $\mathrm{\Phi }(v,\alpha )=0$, then $\alpha 1+{\sum }_{i=1}^n{v}_i{\bar{e}}_i=0$ for the coordinates $v={\sum }_i{v}_i{e}_i$. The distinct degree-zero and degree-one PBW basis elements in [F3] are linearly independent, so $\alpha =0$ and every $v_i=0$. Thus $\mathrm{\Phi }$ is injective.

Conversely, suppose $t\ge \mathrm{\parallel }w\mathrm{\parallel }$. If $t=0$, norm separation in [F5] gives $w=0$, so $(w,t)=(0,0{)}^2$. If $t>0$, then $t^2-\mathrm{\parallel }w{\mathrm{\parallel }}^2=(t-\mathrm{\parallel }w\mathrm{\parallel })(t+\mathrm{\parallel }w\mathrm{\parallel })\ge 0$. By [F6], the radicand has a nonnegative square root; set $u=(t+\sqrt{t^2-\mathrm{\parallel }w{\mathrm{\parallel }}^2})\mathrm{/}2>0$. Applying [F6] again gives $\alpha =\sqrt{u}>0$, and set $v=w\mathrm{/}(2\alpha )$. The quadratic equation $u^2-tu+\mathrm{\parallel }w{\mathrm{\parallel }}^2\mathrm{/}4=0$, divided by $u$, gives $u+\mathrm{\parallel }w{\mathrm{\parallel }}^2\mathrm{/}(4u)=t$, while $2\alpha v=w$. Hence [F1] gives $(v,\alpha {)}^2=(w,t)$. Therefore the square cone is exactly $C=\{(w,t):t\ge \mathrm{\parallel }w\mathrm{\parallel }\}$, and its equality locus is $t=\mathrm{\parallel }w\mathrm{\parallel }$, equivalently $t\ge 0$ and $t^2-\mathrm{\parallel }w{\mathrm{\parallel }}^2=0$.

The function $g(w,t)=t-\mathrm{\parallel }w\mathrm{\parallel }$ is continuous in the product topology [F7]: the triangle inequality in [F5] gives $\mathrm{\mid }\mathrm{\parallel }w\mathrm{\parallel }-\mathrm{\parallel }{w}^{\mathrm{\prime }}\mathrm{\parallel }\mathrm{\mid }\le \mathrm{\parallel }w-w^{\mathrm{\prime }}\mathrm{\parallel }$. Continuity makes $\{g>0\}$ and $\{g<0\}$ open; the first lies in $C$ and the second in its complement. At every equality point $(w,t)$ with $g(w,t)=0$, the points $(w,t+\epsilon )$ lie in $C$ and $(w,t-\epsilon )$ lie outside $C$ for every $\epsilon >0$, and both tend to $(w,t)$ as $\epsilon \downarrow 0$. Hence the topological boundary of $C$ is exactly the equality locus identified in step 2.2.

Steps 1.1 and 2.1 establish the injective unital Jordan homomorphism and its image; steps 1.2, 2.2, and 3.1 establish the square cone and its boundary.

Applying the defining conjugation formula twice returns $(a_0,a_1,a_2,a_3)$, so $\overline{\stackrel{ˉ}{a}}=a$. Adding the tuple to its conjugate gives $(2a_0,0,0,0)=\widehat{2a_0}$.

Write $ab=(r_0,r_1,r_2,r_3)$, where $r_0=a_0{b}_0-a_1{b}_1-a_2{b}_2-a_3{b}_3$, $r_1=a_0{b}_1+a_1{b}_0+a_2{b}_3-a_3{b}_2$, $r_2=a_0{b}_2+a_2{b}_0+a_3{b}_1-a_1{b}_3$, and $r_3=a_0{b}_3+a_3{b}_0+a_1{b}_2-a_2{b}_1$. Substituting $(b_0,-b_1,-b_2,-b_3)$ and $(a_0,-a_1,-a_2,-a_3)$ into the same product formula gives $\bar{b}\bar{a}=(r_0,-r_1,-r_2,-r_3)$ coordinate by coordinate. This is $\overline{ab}$ by [F1].

Applying quaternion conjugation twice gives $\overline{\stackrel{‾}{(a,b)}}=(\overline{\stackrel{ˉ}{a}},b)=(a,b)$. Also $(a,b)+\overline{(a,b)}=(a+\bar{a},0)=(\widehat{2a_0},0)$ by [F2].

Using [F1] and reversing quaternion products by [F2], $\overline{(a,b)(c,d)}=(\overline{ac-\stackrel{ˉ}{d}b},-(da+b\bar{c}))=(\bar{c}\bar{a}-\bar{b}d,-da-b\bar{c})\mathrm{.}$ The same pair-product formula gives $\overline{(c,d)}\overline{(a,b)}=(\bar{c},-d)(\bar{a},-b)=(\bar{c}\bar{a}-\bar{b}d,-b\bar{c}-da),$ which is equal because quaternion addition is commutative. Thus $\overline{xy}=\bar{y}\bar{x}$.

Substitution in the pair formula [F1] and the quaternion norm identities [F3] gives $(a,b)(\bar{a},-b)=(a\bar{a}+\bar{b}b,0)=(\widehat{N(a)+N(b)},0),(\bar{a},-b)(a,b)=(\bar{a}a+\bar{b}b,0)=(\widehat{N(a)+N(b)},0)\mathrm{.}$ Hence $x\bar{x}=\bar{x}x=(\widehat{N(a)+N(b)},0)$, completing all claims.

Each quaternion product coordinate is a sum of signed terms $a_p{b}_q$, so the quaternion product is real-bilinear; the coordinate sign-change formula in [F5] makes conjugation real-linear. Therefore every term in the pair formula [F1] is bilinear in $x=(a,b)$ and $y=(c,d)$, and the octonion product is real-bilinear. The quaternion coordinate formula also gives $\widehat{1}a=a\widehat{1}=a$ and $\bar{0}=0$, so [F1] gives $(1,0)(a,b)=(a,b)=(a,b)(1,0)$. Let $t:=2a_0\in \mathbb{R}$. By [F2]–[F3], $a+\bar{a}=\widehat{t}$ is central in $\mathbb{H}$ and $b\bar{b}=\bar{b}b=\widehat{N(b)}$. Squaring $x$ in [F1] gives $x^2=(a^2-\widehat{N(b)},\widehat{t}b)$. Here the second coordinate is $ba+b\bar{a}=b(a+\bar{a})=\widehat{t}b$.

The quaternion table in The quaternions $\mathbb{H}$: real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1,i,j,k$ gives $ij=k$ and $ji=-k$. Formula [F1] gives $(ij)\mathrm{\ell }=(0,k)=k\mathrm{\ell }$ and $i(j\mathrm{\ell })=(0,ji)=(0,-k)=-k\mathrm{\ell }$. The pair $(0,k)$ is nonzero because its quaternion coordinate $k=(0,0,0,1)$ is nonzero. Thus the two bracketings differ, so multiplication is not associative.

Substituting $xy=(ac-\bar{d}b,da+b\bar{c})$ into [F1], expanding with associativity in [F2], and grouping $a+\bar{a}=\widehat{t}$ gives $x(xy)=(a^{2}c-\widehat{N(b)}c-\widehat{t}\bar{d}b,d{a}^2-\widehat{N(b)}d+\widehat{t}b\bar{c})\mathrm{.}$ On the other hand, [F1] and step 1.1 give $x^{2}y=((a^2-\widehat{N(b)})c-\bar{d}(\widehat{t}b),d(a^2-\widehat{N(b)})+(\widehat{t}b)\bar{c}),$ which is the same pair because real scalars are central. Thus $x(xy)=x^{2}y$.

By [F3] and the left alternative law just proved, $\overline{(yx)x}=\bar{x}\overline{yx}=\bar{x}(\bar{x}\bar{y})=({\bar{x}}^2)\bar{y}=\overline{y(x^2)}\mathrm{.}$ The third equality uses the left alternative law on $\bar{x},\bar{y}$; the fourth uses the anti-involution identities in [F3]. Since conjugation is an involution, it is injective, hence $(yx)x=y(x^2)$.

The product is real-bilinear by step 1.1, and steps 2.1 and 3.1 are exactly the two laws in [F4]; step 1.2 proves nonassociativity. Thus this is a unital alternative nonassociative real algebra.

Conjugation of the $(i,j)$ entry of $XY$ gives $\overline{(XY{)}_{ji}}={\sum }_r\overline{X_{jr}{Y}_{ri}}={\sum }_r\overline{Y_{ri}}\overline{X_{jr}}={\sum }_r{Y}_{ir}{X}_{rj}=(YX{)}_{ij}\mathrm{.}$ Thus $XY+YX$ is Hermitian off the diagonal. Its $i$th diagonal entry is ${\sum }_r(X_{ir}\overline{Y_{ir}}+Y_{ir}\overline{X_{ir}})\mathrm{.}$ Each summand is fixed by conjugation, since its conjugate is the same sum in reverse order; by [F2] the fixed elements are real. Therefore $m(X,Y)$ is in $H_3(\mathbb{O})$. For bilinearity, [F5] shows that quaternion multiplication is real-bilinear and quaternion conjugation is real-linear. The pair formula in Real octonions by Cayley–Dickson doubling therefore makes octonion multiplication real-bilinear. Each entry of the matrix product in [F1] is a sum of three octonion products, so matrix multiplication is real-bilinear; consequently $D(X,Y)=XY+YX$ and $m(X,Y)=D(X,Y)\mathrm{/}2$ are bilinear. The formula for $m$ makes it commutative. The quaternion coordinate formula gives $\widehat{1}a=a\widehat{1}=a$ and $\bar{0}=0$, so the pair formula gives $(1,0)(a,b)=(\widehat{1}a-\bar{b}0,b\widehat{1}+0\bar{a})=(a,b)$ and $(a,b)(1,0)=(a\widehat{1}-\bar{0}b,0a+b\bar{\widehat{1}})=(a,b)$; hence $1_{\mathbb{O}}=(1,0)$ is a two-sided unit. The identity matrix is in $H_3(\mathbb{O})$ and is a two-sided unit for the matrix product, hence $m(I,X)=X$.

Let $1,i,j,k,\mathrm{\ell },i\mathrm{\ell },j\mathrm{\ell },k\mathrm{\ell }$ be the ordered real basis of $\mathbb{O}=\mathbb{H}\times \mathbb{H}$ specified by the pair formula. The three diagonal matrix units, together with the eight Hermitian matrices having one of these octonions in the $(1,2)$, $(2,3)$, or $(3,1)$ slot and its conjugate in the transposed slot, form a real basis of $H_3(\mathbb{O})$: the three real diagonal coordinates and the three independent octonion entries are recovered uniquely from any Hermitian matrix. There are $3+3\cdot 8=27$ basis elements.

Set $D(X,Y):=XY+YX$. With $m=D\mathrm{/}2$, bilinearity gives $8(m(X,m(Y,m(X,X)))-m(m(X,Y),m(X,X)))=D(X,D(Y,D(X,X)))-D(D(X,Y),D(X,X))\mathrm{.}$ The right side is a polynomial homogeneous of degree three in $X$ and degree one in $Y$, with coefficients in the real basis of $H_3(\mathbb{O})$ fixed here.

Write $X={\sum }_i{x}_i{E}_i$ and $Y={\sum }_{\mathrm{\ell }}{y}_{\mathrm{\ell }}{E}_{\mathrm{\ell }}$ in this basis. The coefficient of $x_i{x}_j{x}_k{y}_{\mathrm{\ell }}$ in the degree-$(3,1)$ expression in step 2.1 is the sum over the distinct permutations $(p,q,r)$ of $(i,j,k)$ of $D(E_r,D(E_{\mathrm{\ell }},D(E_p,E_q)))-D(D(E_r,E_{\mathrm{\ell }}),D(E_p,E_q))\mathrm{.}$ This follows by distributing each bilinear $D$; when indices repeat, distinct permutations count the distinct placements in the cubic monomial exactly once. The script in [F4] computes all ${27}^2=729$ ordered basis-pair products to encode $D$, rejecting any non-Hermitian result, then checks these coefficients for every sorted triple $0\le i\le j\le k<27$ and every $0\le \mathrm{\ell }<27$. It checks $\left(\genfrac{}{}{0px}{}{29}{3}\right)\cdot 27=\mathrm{98,658}$ monomial coefficients, or $\mathrm{98,658}\cdot 27=\mathrm{2,663,766}$ output coordinates. Its signed integer arithmetic returns zero in every coordinate, with the convention witnesses $ij=k$, $(ij)\mathrm{\ell }=k\mathrm{\ell }$, and $i(j\mathrm{\ell })=-k\mathrm{\ell }$ all true. Since the basis coordinates of the polynomial are exactly the coefficients enumerated, the polynomial vanishes for every $X,Y$.

step 1.1 and step 3.1 establish closure, commutativity, bilinearity, a unit, and the Jordan identity in [F3]. Hence $H_3(\mathbb{O})$ is a unital real Jordan algebra.