Let $J$ be an unital Jordan algebra (over $\mathbb{R}$) - recall that this means that $J$ is an unital $\mathbb{R}$-algebra (whose product we denote by $\bullet$) satisfying $x\bullet y=y\bullet x$ and $(x^{\bullet 2}\bullet y)\bullet x=x^{\bullet 2}\bullet(y\bullet x)$ (where $x^{\bullet 2}=x\bullet x$) for all $x,y\in J$. We assume that:

  1. $J$ is special, i.e. $J\subset A$ is a $\mathbb{R}$-vector subspace of some associative unital $\mathbb{R}$-algebra $A$ such that $x\bullet y=\frac{1}{2}(xy+yx)$ for all $x,y\in J$, where $xy$ denotes the associative product of $x$ and $y$ in $A$, and the units of $A$ and $J$ coincide.

  2. $A$ as above may be chosen to be a $*$-algebra - that is, a $\mathbb{C}$-algebra endowed with an involution $*:A\mapsto A^*$ (an anti-linear map satisfying $(xy)^*=y^*x^*$ and $(x^*)^*=x$ for all $x,y\in A$) - such that $x^*=x$ for all $x\in J$. In other words, $J$ is a Jordan subalgebra of the Jordan algebra of the Hermitian (i.e. self-adjoint) elements of $A$.

  3. $J$ is formally real, i.e. if $x_1,\ldots,x_n\in J$ satisfy $x_1^{\bullet 2}+\cdots+x_n^{\bullet 2}=0$, then $x_1=\cdots=x_n=0$. If $A$ as in (1.-2.) above is finite-dimensional (and therefore so is $J$), then 3. follows immediately from 2. due to the spectral theorem.

As a consequence of the formal reality of $J$, we see that $-1$ cannot be a sum of squares of elements of $J$, that is, $-1\neq x_1^{\bullet 2}+\cdots+x_n^{\bullet 2}$ for all $x_1,\ldots,x_n\in J$, $n\in\mathbb{N}$ - otherwise, we would have $x_1^{\bullet 2}+\cdots+x_n^{\bullet 2}+1^{\bullet 2}=0$ for some choice of $x_1,\ldots,x_n\in J$.

We say that a unital *-algebra $A$ containing $J$ as in (1.-2.) above is an associative $*$-envelope of $J$ if, in addition, $A$ coincides with the $*$-subalgebra of $A$ generated by $J$. Even if $J$ is formally real, it is still possible for such an $A$ to have nilpotent elements of order two, therefore $A$ need not be formally real. On the other hand,

Question: If an unital *-algebra $A$ is an associative $*$-envelope of an unital, formally real Jordan algebra $J$, is $-1$ a sum of Hermitian squares in $A$? In other words, are there $x_1,\ldots,x_n\in A$ such that $x_1^*x_1+\cdots+x_n^*x_n=-1$?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.