11
$\begingroup$

Can anyone give me a relatively simple proof or Some reference for the following fact.(I know that there is a proof of this theorem in Gerard J. Murphy'book: "$C^*$-Algebras and Operator Theory", but I'm sure that there should be a simple proof of this.

Every hereditary C*-subalgebra of a simple $C^*$-algebra is also simple!

Maybe this is easy for someone, but it makes me confused for a long time. I am a novice!

$\endgroup$
3
  • 4
    $\begingroup$ You should unaccept the irrelevant answer and accept the relevant answer. $\endgroup$
    – Yemon Choi
    May 1, 2012 at 9:46
  • $\begingroup$ I agree with Captain Oates. $\endgroup$ Jan 21, 2014 at 4:44
  • $\begingroup$ This is Theorem 3.2.8 in Murphy's book $\endgroup$
    – Muddana
    Mar 13 at 11:50

2 Answers 2

0
$\begingroup$

I don't know much about C* algebras, but I would guess that there shouldn't be a simple proof of the theorem you state. Here's an algebraic example which seems to be a counterexample to the theorem once the modifier C* is removed from the statement. Let A be the Weyl algebra, i.e. the algebra generated by x,y with the relation xy-yx=1. It isn't hard to show this algebra is simple and that it has global dimension 1. The subalgebra generated by x also has global dimension 1, but of course it is far from simple.

$\endgroup$
3
  • 6
    $\begingroup$ Peter, I don't think the hereditary that you use in algebra is the same as the hereditary that people use for $C^\ast$-algebras. The latter has nothing to do with homological dimension in the Tor-Ext sense, or probably in any other sense come to think of it. $\endgroup$
    – Yemon Choi
    Feb 20, 2012 at 1:31
  • $\begingroup$ Thank you for your interest in my question! Yes, the hereditary that people define in $C^*$-algebra is not the same as the hereditary that use for general algebra. $\endgroup$
    – Aviv
    Feb 20, 2012 at 2:25
  • 2
    $\begingroup$ Ah, that's unfortunate terminology, then. I guess this is evidence that the first phrase in my first sentence is correct :-) $\endgroup$ Feb 20, 2012 at 3:32
15
$\begingroup$

Here is a direct argument which may not differ much in its essence from the one in the book that you mention:

Say $A$ is simple and $B$ is a closed hereditary subalgebra of $A$. This means that if $a\in A$ and $b_1,b_2\in B$ then $b_1ab_2\in B$. Let $x\in B$ be non-zero and let us show that it generates $B$ as a closed two sided ideal. Since $x$ generates $A$ as a closed two-sided ideal, the finite sums of elements of the form $axa'$, with $a,a'\in A$, form a dense subset in $A$. In particular, if $y\in B$ and $\epsilon>0$ then there exists an element of the form $$ \sum_{i=1}^{n} a_ixa'_i $$ within a distance $\epsilon$ of $y$. The problem with this is that the $a_{i}$s and $a'_i$s are not in $B$. They are only in $A$. This is fixed using that in a C*-algebra one always has that $|c^*|^{1/n}c|c|^{1/n}\to c$ for any $c$ (alternatively, you can use an approximate unit for $B$). Then for $k$ large enough the element $$ \sum_{i=1}^{n} (|y^*|^{1/k}a_i|x^*|^{1/k}) x (|x|^{1/k}a'_i |y|^{1/k} ) $$ is also within a distance of $\epsilon$ of $y$.

$\endgroup$
3
  • $\begingroup$ Thank you very much Leonel! I have to say that this proof may be what I expect! I think your method is similar to the proof of the following fact: Let $A$ be a simple $C^*$-algebra and $q \in A$ a projection. Then $qAq$ is a simple $C^*$-algebra. I once tried to imitate the proof of the above easy result, But I didn't get it. Thank you again! $\endgroup$
    – Aviv
    Feb 21, 2012 at 1:25
  • 2
    $\begingroup$ You're welcome. The same argument with very little modification shows also that every closed two-sided ideal of the hereditary subalgebra is the intersection of an ideal of the algebra with the hereditary subalgebra. $\endgroup$ Feb 22, 2012 at 0:32
  • $\begingroup$ Yes! Let $B$ be a hereditary subalgebra of a simple $C^*$-algebra $A$, essentially, as in Gerard J. Murphy'book, and also your argument, let $I$ be any closed ideal of $B$ and $J$ be the closed ideal of $A$ generated by $I$, i.e., $J = \overline{(AIA)}$. Then, [J \cap B = \overline{(BJB)}=I.] then my question follows.(one direction is trivial.) $\endgroup$
    – Aviv
    Feb 22, 2012 at 1:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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