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It seems to be trivial but I am not sure about monotonicity of the norm in the non-commutative case:

Is every C*-algebra a Banach lattice with respect to its natural positive cone?

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If you are asking whether $0\leq a\leq b$ implies $\|a\|\leq \|b\|$, the answer is that that is always true. I could elaborate in an answer, but I'm not sure if that's what you're looking for. Could you please say precisely and explicitly what it is you want to prove about a C*-algebra? –  Jonas Meyer Nov 19 '11 at 17:22

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In order for an ordered vector space to be a Banach lattice, among other things, it is necessary that any two elements of the space have a greatest lower bound and least upper bound in the space. This property usually fails in $C^*$ algebras with their usual $C^*$-algebraic ordering.

Consider for example the $C^*$ algebra $A$ of $2 \times 2$ matrices over the complex numbers (with the operations: matrix addition, matrix multiplication, and the matrix conjugate transpose as the involution). Let $a = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $b = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$. It is easy to see that $a \geq 0$ and $b \geq 0$ in in $A$, and hence that any element $c$ in $A$ satisfying $a \leq c$ or $b \leq c$ must be self-adjoint. Short calculations then show that if $c = \begin{pmatrix} x & y \\ y^* & z \end{pmatrix}$, then $a \leq c$ holds if and only if $x \geq 1$, $z \geq 0$, and $(x - 1) z \geq |y|^2$, and $b \leq c$ holds if and only if $x \geq 0$, $z \geq 1$, and $x(z - 1) \geq |y|^2$.

It follows that the set of upper bounds for $\{a,b\}$ in $A$ is the set $$ U = \left\{\begin{pmatrix} x & y \\ y^* & z \end{pmatrix}: x \geq 1, z \geq 1, xz - \max(x,z) \geq |y|^2\right\}. $$ I claim that $U$ has no least element. Suppose to the contrary that it does; denote it by $\lambda$. There are real numbers $s$ and $t$ and a complex number $u$ satisfying $s \geq 1$, $t \geq 1$, and $st - \max(s,t) \geq |u|^2$ with $\lambda = \begin{pmatrix} s & u \\ u^* & t \end{pmatrix}$. It is clear that the $2 \times 2$ identity matrix $I$ is in $U$, and from $\lambda \leq I$ one easily deduces that $1 - s \geq 0$ and $1 - t \geq 0$. Combining these with the previous constraints on $s$ and $t$ we deduce that $s = t = 1$ and hence $|u|^2 \leq 1 \cdot 1 - \max(1,1) = 0$, so that $u = 0$ and hence $\lambda = I$. But there are elements $d$ of $U$ for which $I \leq d$ does not hold. The matrix $\frac{1}{4} \begin{pmatrix} 5 & 2i \\ -2i & 6 \end{pmatrix}$ is a concrete example but there are many others.

This also shows that $\{-a,-b\}$ has no greatest lower bound in $A$, of course. So there is really no hope of $A$ being a lattice.

(Note that $a$ and $b$ are self-adjoint projections, and if you restrict $\leq$ to the set of self-adjoint projections in $A$, you do get a lattice, which is isomorphic to the lattice of closed subspaces of $\mathbb{C}^2$. The supremum of $a$ and $b$ in this lattice is $I$ and the infimum of $a$ and $b$ is $0$. But the set of projections in $A$ is not a real vector space, let alone a Banach lattice.)

Some general theory that may be of interest:

  • S. Sherman proved (in Order in operator algebras, 1951) that a $C^*$-algebra $A$ is a lattice with respect to its usual ordering if and only if $A$ is commutative.

  • R. V. Kadison proved (in Order properties of bounded self adjoint operators, 1951) that when $H$ is a Hilbert space of dimension $\geq 2$, the $C^*$ algebra $A$ of all bounded operators on $H$ is in some sense "as far from a lattice as you can get" in that if $a$ and $b$ are self-adjoint elements of $A$, then $\{a, b\}$ has a greatest lower bound only if either $a \leq b$ or $b \leq a$. So any pair of non-comparable self-adjoint operators will generate a counterexample like the one I gave above.

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