I have a $C^\*$-algebra $\mathcal{A}$, and would like to make use of the spectral order $\preceq$ coming from (the self-adjoint part of) its enveloping von Neumann algebra $\mathcal{A}^{\*\*}$.


I am most interested in checking that the spectral join/meet of finite subsets of $\mathcal{A}^\text{sa}$ are contained in $\mathcal{A}^\text{sa}$. It suffices to check for pairs only.


Given self-adjoint $a, b \in \mathcal{A}^{\*\*}$, the spectral join $a \vee b$ turns out to be the self-adjoint element of $\mathcal{A}^{\*\*}$ whose spectral projections satisfy
$$\chi\_{(\lambda, \infty)}(a \vee b) = \chi\_{(\lambda, \infty)}(a) \vee \chi\_{(\lambda, \infty)}(b),$$
where $\chi\_S$ denotes the indicator function of a subset $S \subset \mathbb{R}$, and the $\vee$ on the right is the standard join for projections of a von Neumann algebra.


If we restrict $a$ and $b$ to be elements of $\mathcal{A}$, then the two projections on the right above are open (in the sense described by Akemann in *The general Stone-Weierstrass problem*, 1969), and hence so is the one on the left (which is necessary for inclusion of $a \vee b$ in $\mathcal{A}$). However, I am not sure how to guarantee that $a \vee b \in \mathcal{A}$.


Would anyone have a reference or proof relevant to my objective? If it happens that $a \vee b$ is not an element of $\mathcal{A}$, might $a$ and $b$ still have a larger join in the smaller poset $(\mathcal{A}, \preceq|\_{\mathcal{A}})$? I am fairly new to the operator algebra world, so I apologise if this is a standard result that I just haven't been able to track down in the literature.



Spectral join in a $C^*$-algebra relative to its enveloping von Neumann algebra
Sorry, the answer is no. In general the spectral join of two positive operators $a$ and $b$ is the strong operator limit of $(a^k + b^k)^{1/k}$ as $k \to \infty$ (Corollary 10 of [this](https://www.ams.org/journals/proc/1996-124-11/S0002-9939-96-03474-0/S0002-9939-96-03474-0.pdf) old old paper of mine). But this need not belong to the C${}^\*$-algebra generated by $a$ and $b$. For a counterexample, work in $l^\infty(\mathbb{N}, M\_2)$ acting on the $l^2$ direct sum $\mathbb{C}^2 \oplus \mathbb{C}^2 \oplus \cdots$ and let $p$ and $q$ be the projections with $p(n) = \left[\begin{matrix}1&0\cr 0&0\end{matrix}\right]$ and $q(n) = \left[\begin{matrix}\cos^2(1/n)&\cos(1/n)\sin(1/n)\cr \cos(1/n)\sin(1/n)&\sin^2(1/n)\end{matrix}\right]$ for all $n$. The spectral join of two projections is just their join as projections, i.e., the projection onto the join of their ranges, and the join of the ranges of $p$ and $q$ contains each $\mathbb{C}^2$ summand, so it is the whole space and thus $p \vee q$ is the identity. But every element $r$ of the C\*-algebra generated by $p$ and $q$ has $(2,2)$ entries going to zero as $n \to \infty$ (since this is true of any monomial in $p$ and $q$), so this C${}^\*$-algebra doesn't contain the identity.


Any self-adjoint operator that is spectrally larger than the identity has null spectral projection for the interval $(-\infty, 1/2]$ and is therefore invertible, so it can't belong to the C${}^\*$ algebra generated by $p$ and $q$. So that falsifies your final question too.


