Jordan maps from the scalar algebra are idempotents

Statement

Let $k$ be a field of characteristic different from $2$ , viewed as a Jordan algebra with ordinary multiplication, and let $J$ be a Jordan $k$ -algebra. Evaluation at $1$ is a bijection

\operatorname{Hom}_{\mathbf{Jord}_k}(k,J) \longleftrightarrow \{e\in J:e\circ e=e\}, \qquad f\longmapsto f(1).

The inverse sends an idempotent $e$ to $t\mapsto te$ . If $J$ is unital, the same description holds in UJord, whereas Jord1 contains exactly the map $t\mapsto t1_J$ .

Proof

Linearity forces $f(t)=tf(1)$ . Preservation of $1\circ1=1$ requires $e\circ e=e$ . Conversely, if $e$ is idempotent, bilinearity gives $(se)\circ(te)=st e$ , so the map preserves products. It preserves units precisely when $e=1_J$ .

The complex qubit algebra

For $J=\operatorname{Herm}_2(\mathbb C)$ , as a real Jordan algebra, idempotents are Hermitian projection matrices. They are $0$ , $I_2$ , and rank-one projections

p=uu^*,\qquad u\in\mathbb C^2,\qquad u^*u=1.

Vectors differing by a unit complex scalar give the same projection. Thus the real Jordan Hom-set is parametrized by two isolated points together with $\mathbb{CP}^1$ ; only $I_2$ gives a unit-preserving map. This is a complete classification of these maps, not of all endomorphisms of the qubit algebra.