Jordan Hom-sets under the three unit conventions

Statement

Let $J,K$ be unital Jordan algebras over a fixed field $k$ of characteristic different from $2$ . Under the catalogue conventions for Jord, UJord, and Jord1,

\operatorname{Hom}_{\mathbf{Jord}_{1,k}}(J,K) \subseteq\operatorname{Hom}_{\mathbf{UJord}_k}(J,K) =\operatorname{Hom}_{\mathbf{Jord}_k}(J,K).

The inclusion can be strict. However,

\operatorname{Aut}_{\mathbf{Jord}_{1,k}}(J) =\operatorname{Aut}_{\mathbf{UJord}_k}(J) =\operatorname{Aut}_{\mathbf{Jord}_k}(J).

Proof and strictness

The Hom equality follows because UJord is a full subcategory on unital objects. Jord1 imposes the additional equation $f(1_J)=1_K$ . If $K\ne0$ , the zero homomorphism belongs to the two larger Hom-sets and fails that equation.

If $f:J\to K$ is surjective and preserves the Jordan product, then for every $b=f(a)$ ,

f(1_J)\circ b=f(1_J\circ a)=b.

Thus $f(1_J)$ is the target's unique unit. In particular every invertible Jordan homomorphism is unit-preserving.

A concrete endomorphism comparison

For the real scalar Jordan algebra $\mathbb R$ , every real-linear map is $t\mapsto ct$ . Product preservation says $c^2=c$ , so

\operatorname{End}_{\mathbf{Jord}_{\mathbb R}}(\mathbb R) =\operatorname{End}_{\mathbf{UJord}_{\mathbb R}}(\mathbb R)=\{0,\operatorname{id}\}, \qquad \operatorname{End}_{\mathbf{Jord}_{1,\mathbb R}}(\mathbb R)=\{\operatorname{id}\}.

All three automorphism groups are trivial.