Positive element of a C-star algebra

An element $a$ of a C-star algebra is positive when $a=b^{\ast }b$ for some $b$, equivalently when $a=a^{\ast }$ and $\sigma (a)\subseteq [0,\infty )$. Continuous functional calculus gives it a unique positive square root $a^{1/2}$.

The positive elements of a C-star algebra form a norm-closed proper convex cone. For $p,q\ge 0$ and $s=\Vert p\Vert +\Vert q\Vert$, continuous functional calculus gives $\Vert s1-p-q\Vert \le s$, forcing the real spectrum of $p+q$ to be nonnegative. The same bound with one fixed upper norm bound proves norm closedness under limits; scaling by nonnegative real scalars preserves positivity. If both $h$ and $-h$ are positive, their spectra lie in $\{0\}$, and spectral radius norm equality for normal elements forces $h=0$. This cone defines the order $a\le b$ precisely when $b-a\in A_{+}$.

In $M_2(\mathbb{C})$ take $a=\mathrm{diag}(1,2)$ and $b=\left(\begin{array}{cc}2& 1\\ 1& 3\end{array}\right)$. Then $a-1\ge 0$ and $b-a=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\ge 0$, but $b^2-a^2=\left(\begin{array}{cc}4& 5\\ 5& 6\end{array}\right)$ has determinant $-1$ and a negative eigenvalue. In contrast, for commuting positive elements $b^2-a^2=(b-a)(b+a)$ is positive in the commutative function algebra supplied by continuous functional calculus. The obstruction is noncommutativity, not a failure of the scalar inequality.

Here strictly positive means positive and invertible. Set $k=a^{-1/2}b{a}^{-1/2}\ge 1$ using congruence preserves positivity in a C-star algebra. The continuous functional calculus gives $k^{-1}\le 1$, since $1-t^{-1}\ge 0$ for $t\ge 1$. Then $a^{-1}-b^{-1}=a^{-1/2}(1-k^{-1})a^{-1/2}\ge 0$. The proof works without commutativity. If $1\le a$, functional calculus also gives $a^{-1}\le 1$.

For a positive element $a$, its positive square root in a C-star algebra gives $c^{\ast }ac=(a^{1/2}c{)}^{\ast }(a^{1/2}c)$, which is positive. Applying this to $b-a$ shows that $a\le b$ implies $c^{\ast }ac\le c^{\ast }bc$. The map need not preserve multiplication, but it preserves order; if $c$ is invertible, the inverse congruence also preserves order.

Define positivity spectrally: a self-adjoint element has nonnegative spectrum. First, addition preserves it. For $p,q\ge 0$ and $s=\Vert p\Vert +\Vert q\Vert$, functional calculus gives $\Vert s1-p-q\Vert \le s$, forcing every real spectral value of $p+q$ to be nonnegative. The cone is proper because a self-adjoint element with spectrum contained in $\{0\}$ has norm zero.

For $h=a^{\ast }a$, let $h_{-}=\max (-h,0)$ by the continuous functional calculus, and set $c=a{h}_{-}^{1/2}$. Then $c^{\ast }c=-h_{-}^2$ is nonpositive. The nonzero spectra of products in opposite orders imply that $c{c}^{\ast }$ is nonpositive too. Write $c=u+iv$ with $u,v$ self-adjoint. Then $c^{\ast }c+c{c}^{\ast }=2(u^2+v^2)$ is positive, since self-adjoint squares have nonnegative spectrum. It is also nonpositive, hence zero. It follows that $c^{\ast }c=-c{c}^{\ast }$ is both positive and nonpositive, hence zero. The C-star identity gives $c=0$, so $h_{-}^2=0$ and $h_{-}=0$. Thus $h\ge 0$, without assuming positivity of adjoint products in an earlier step.

Every Positive element of a C-star algebra $a$ has a unique positive element $a^{1/2}$ satisfying $(a^{1/2}{)}^2=a$. It is obtained by applying the continuous functional calculus to $t\mapsto \sqrt{t}$ on $\sigma (a)\subseteq [0,\infty )$.

An element $a$ of a C-star algebra is positive if and only if it is Hermitian and $\sigma (a)\subseteq [0,\infty )$. This characterization transfers the pointwise order on continuous functions to an abstract C-star algebra through the continuous functional calculus.