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.

I am reading Sec33 of Conway's A Course in Operator Theory, according to his definition,

An operator system is a linear manifold $\mathcal{S}$ in a $C^*$-algebra such that $1\in\mathcal{S}$ and $\mathcal{S}=\mathcal{S}^*$.

Then he makes the comment that operator systems have an abundance of positive elements. His argument is that for every hermitian element $a\in\mathcal{S}$, $|a|$ also lies in $\mathcal{S}$ and hence the positive parts and negative parts lie in $\mathcal{S}$ and hence $\mathcal{S}$ is spanned by its positive elements.

However, I do not know why $|a|$ lies in $\mathcal{S}$ since $\mathcal{S}$ is only assumed to be a linear manifold, not an algebra.

Can somebody give a hint? Thanks!

share|improve this question
2  
On page 187, on the same page and very soon after the definition, he shows why $S$ is spanned by its positive elements, with nothing about positive and negative parts. (He uses $\|a\|$ as shorthand for $\|a\|1$, if that is what is confusing you.) –  Jonas Meyer Jul 14 '12 at 21:11
    
@JonasMeyer Yes. I mixed $\|a\|$ with $|a|$. That is what confused me. –  Hui Yu Jul 14 '12 at 21:31
add comment

1 Answer 1

up vote 3 down vote accepted

I assume "linear manifold" means "subspace." I also do not understand the given argument (if $|a|$ denotes the absolute value), but here's another one: the second condition implies that $S$ is spanned by its self-adjoint elements. Let $a$ be such an element. If $\lambda \in \mathbb{R}$ is greater than the spectral radius of $a$ then $a + \lambda$ is positive, so $a = (a + \lambda) - \lambda$ is a difference of positive elements.

share|improve this answer
    
Thanks! Your argument works well. Maybe you already know this but the difference between a linear manifold and a subspace (at least in the literature of operator theory) is that a subspace is closed in $\|\cdot\|$ while there is no such restriction on manifolds. –  Hui Yu Jul 14 '12 at 21:12
2  
@HuiYu: Of course, in other areas of functional analysis, when we say "(linear) subspace" we mean "linear subspace", and when we want to refer to one which is closed, we say "closed (linear) subspace". I think one should be aware that not everyone will follow the terminology of one's particular sample of books –  user16299 Jul 14 '12 at 23:37
    
@Hui: I respectfully disagree. If one wants to talk about a densely defined operator, its domain is certainly not going to be a closed subspace, and there are many situations where one wants to talk about densely defined operators. –  Qiaochu Yuan Jul 15 '12 at 0:50
    
@QiaochuYuan All right. I deleted that comment. I should have said that 'most of the time' we talk about closed subspaces, and we might gain a little efficiency if we could drop the 'closed' before each 'subspace'. –  Hui Yu Jul 15 '12 at 1:18
add comment

Your Answer

 
discard

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.