Question about the determinant of the Jordan algebra

Question

I'm recently reading the book Analysis on Symmetric Cones and meet with a question. In Chapter 2, the author asserts that for given $x\in V$, where $V$ is a Jordan algebra, the function $q=\det (e,x,x^2 ,x^3 ,\dotsb ,x^{r-1} ,e_{r+1} ,\dotsb ,e_n )$, where $n$ is the dimension of $V$ and $r$ is the rank of $V$, is a polynomial. I guess at first that it is a homogeneous function and so it is called a "polynomial", but when I read further, I find that it must be actually a polynomial. I just don't know why it could be a polynomial since it is a function from $V$ to the scalar field $\mathbb{R} $.

Answer 1

A polynomial on a finite-dimensional vector space $V$ is defined to be a polynomial function of any choice of linear coordinates $x_1, \dots, x_n$ on that vector space.

In more detail:

Given any basis $e_1, \dots, e_n$ of a finite-dimensional vector space $V$, we can write any vector $x \in V$ as

$$ x = \sum_{i=1}^n x_i e_i $$

for some unique choice of numbers $x_i \in \mathbb{R}$. These numbers $x_i$ are thus functions of $x$, called (linear) coordinate functions.

Any function

$$ f \colon V \to \mathbb{R} $$

that is a polynomial in the linear coordinate functions is called a polynomial on $V$.

In particular, the determinant function on a Jordan algebra is a polynomial on that Jordan algebra, in this sense.

The usual determinant of an $n \times n$ matrix is likewise a polynomial function on the space of $n \times n$ matrices.