Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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

1 Answer 1

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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.