-1
$\begingroup$

given two $C^{*}$ algebras $A\subset B $, acting on the same Hilbert space $H$, and $\delta $ is a derivation from $A$ into $B$,(in this case, a derivation is a linear mapping such that $\delta(ab)=a\delta(b)+\delta(a)b,\forall a,b \in A$) and assume it is bounded, then, is there an element $h$ of $A^{-}$(the weak operator closure of $A$), such that $\delta(a)=ha-ah, \forall a\in A$? \

Especially, I want to consider the case when $A,B$ are all commucative $C^{*}$ algebras.In other words, is there no nontrivial bounded derivation?

$\endgroup$
5
  • $\begingroup$ In the case where $A=B$, doesn't this follows from Sakai's theorem that all inner derivations from a von Neumann algebra to itself are inner? $\endgroup$
    – Yemon Choi
    Mar 3, 2011 at 2:58
  • 2
    $\begingroup$ Actually, it seems to me that if the answer to your question were positive, then by taking B=B(H) one would deduce that every derivation from A to B(H) is inner, thus answering Kadison's similarity problem... $\endgroup$
    – Yemon Choi
    Mar 3, 2011 at 3:01
  • 1
    $\begingroup$ ougao, if you have a more specific question you want to ask, you should either ask in the comments to the answer given, or ask a new question. $\endgroup$
    – S. Carnahan
    Mar 3, 2011 at 11:48
  • $\begingroup$ I remarked in the comments to Kate's answer to the old version of this question that the new question is answered by an old paper of Kamowitz (Trans. AMS, 1962): ams.org/mathscinet-getitem?mr=170219 This paper was one of the ingredients that motivated Johnson to define amenability for Banach algebras, and the result I refer to can be generalized to the case of amenable commutative Banach algebras (of which $C(X)$ is a key example). $\endgroup$
    – Yemon Choi
    Mar 3, 2011 at 19:31
  • $\begingroup$ However, Kamowitz' paper may not be the best place to look for a proof of this result. Perhaps I will write up a short argument if I cannot find a self-contained argument; but there is nothing new about it, and it forms part of the basic knowledge needed by people thinking about bounded derivations $\endgroup$
    – Yemon Choi
    Mar 3, 2011 at 19:42

1 Answer 1

6
$\begingroup$

The answer is no. Let $A$ be commutative von Neumann algebra. Let $T\in B(H)$ be such that $Ta-aT\neq 0$ for some $a\in A$. Then $\delta(a)=Ta-aT$ is bounded derivation but there is no element $h\in A^{-}=A$, such that $\delta(a)=ha-ah$ for all $a\in A$.

$\endgroup$
5
  • $\begingroup$ Ah, I missed the fact that the implementing element was meant to lie in the WOT-closure of A. $\endgroup$
    – Yemon Choi
    Mar 3, 2011 at 3:33
  • $\begingroup$ good!but what if $A,B$ are all commutative $C^{*}$ algebras? $\endgroup$
    – Jiang
    Mar 3, 2011 at 4:22
  • $\begingroup$ ougao, every (continuous) derivation from a commutative C*-algebra $A$ into a symmetric $A$-bimodule $X$ is zero; I belive this is due to Guichardet or Kamowitz. Do you know about amenability of Banach algebras? $\endgroup$
    – Yemon Choi
    Mar 3, 2011 at 5:09
  • $\begingroup$ @Kate, sorry for my misfeasance, but I relly want to consider the special case, which I failed to explain clearly in my translation from the original problem faced in differential algebra to this one. $\endgroup$
    – Jiang
    Mar 3, 2011 at 10:00
  • $\begingroup$ @Yemon Choi,thanks for your tip, I think the article written by J.R.Ringrose <Automatic continuity of derivations of operator algebras> is what I want. $\endgroup$
    – Jiang
    Mar 3, 2011 at 10:40

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.