The spin factor and the plus-quadratic Clifford convention

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.