Is the cone of squares in a Jordan algebra a cone?

Question

Let $A$ be a finite-dimensional Jordan algebra over $\mathbb{R}$, i.e. a finite-dimensional real vector space with a commutative bilinear product $\circ: A \times A \rightarrow A$ satisfying $(a^2 \circ b) \circ a = a^2 \circ (b \circ a)$ for all $a, b \in A$. Let $Q = \{ a^2 \mid a \in A\}$ be the cone of squares.

I want to show: $Q$ is cone (i.e. closed under addition and multiplication with non-negative reals and $Q \cap -Q = \{0\}$).

The only thing I don't know how to show: If $a, b \in Q$, then $a + b = c^2$ for some $c \in Q$. How can I show this?

Answer 1

It is likely this is false for a general Jordan algebra, even finite dimensional over $\mathbb R.$ I downloaded the review of Faraut and Kuranyi by Kenneth Gross, and he emphasizes in a footnote he says:

What is now called a symmetric cone in older terminology was referred to as a domain of positivity, and what is now known as a Euclidean Jordan algebra was called a formally real Jordan algebra.

Now, from page 3 of McCrimmon's book, we find

A Jordan algebra is formally real if

$$ x_1^2 + \cdots + x_n^2 = 0 \; \Longrightarrow \; x_1 = \cdots = x_n = 0. $$

This is A Taste of Jordan Algebras by Kevin McCrimmon, about 2003.

So, my advice is to see if you can prove your cone property with formal reality as an axiom. As a matter of taste in terminology, I don't see how you can call something a domain of positivity if the sum of strictly positive elements can be zero. If it does not work out, you have several people to ask for a counterexample, that is showing what goes wrong when not formally real.

P S I think you will like the description of Euclidean Jordan algebras in Newton's Algorithm in Euclidean Jordan Algebras, with Applications to Robotics by Uwe Helmke, Sandra Ricardo, Shintaro Yoshizawa.