Zero-product Jordan algebra R^1

The zero-product Jordan algebra on $\mathbb R^{1}$ is the vector space $\mathbb R^{1}$ , with its usual addition and scalar multiplication, and with

x\circ y=0\qquad\text{for all }x,y\in \mathbb R^{1}.

This bilinear commutative product satisfies the Jordan identity because both sides are zero. Its dimension over $\mathbb R$ is $1$ .

No unit

The vector space has a nonzero vector $x$ , whereas $e\circ x=0$ for every $e$ . Thus no element acts as a multiplicative unit. This is an object of $\mathbf{Jord}$ over the stated field, but is not an object of either $\mathbf{UJord}$ or $\mathbf{Jord}_1$ . Positivity of the dimension matters: the zero-dimensional algebra is outside this family.

Complete endomorphism and automorphism descriptions

Every $\mathbb R$ -linear map $T$ preserves the product, since

T(x\circ y)=T(0)=0=T(x)\circ T(y).

Conversely, a Jordan homomorphism over $\mathbb R$ is required to be $\mathbb R$ -linear. Therefore, in $\mathbf{Jord}_{R}$ ,

\operatorname{End}(\mathbb R^{1},0)=M_{1}(\mathbb R), \qquad \operatorname{Aut}(\mathbb R^{1},0)=\operatorname{GL}_{1}(\mathbb R).

These identifications use the standard coordinate basis. On endomorphisms, composition becomes ordinary matrix multiplication and pointwise addition gives the matrix algebra. On automorphisms, composition gives the general linear group $\operatorname{GL}_{1}(\mathbb R)$ . Invertibility is sufficient because the inverse of an invertible linear map also preserves the zero product.

The resulting matrix algebra has its ordinary associative product, and is a different object from this zero-product Jordan algebra.