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I believe that the answer is affirmative and I would be grateful to any comments on my attempt (see below) of proving this.

Let $A$ be a C*-algebra and denote by $Z(A)$ the centre of $A$.

  1. First of all, it is clear that $Z(A)$ is a subalgebra of $A$.
  2. Furthermore, it is easy to see that $Z(A)$ is closed under the $*$-operation.
  3. All elements of $Z(A)$ satisfy the so called C*-identity, since $Z(A)$ is a subset of $A$.
  4. $Z(A)$ is norm-closed.

The only step that I don't feel comfortable with is step 4. Here is how I would attempt to prove it:

Take a sequence $\{c_n\}_{n\in \mathbb{N}}$ of elements from $Z(A)$ converging to $c\in A$. We want to show that $c\in Z(A)$.

Let $a\in A$ be arbitrary. Clearly $c_n a - a c_n = 0$ holds for all $n\in \mathbb{N}$. By assumption $\lim_{n\to \infty} ||c-c_n|| = 0$.

Now,

$|| ac-ca || = || ac-ca + c_n a - a c_n || = || a(c-c_n) + (c_n-c)a) || \leq 2 ||a|| \cdot ||c-c_n|| $

Therefore $||ac-ca||=0$ and hence $ac=ca$. Right?

(Somehow, the last two lines give me a bad feeling in my stomach that I can't explain.)

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1 Answer 1

up vote 11 down vote accepted

Edit: I forgot to answer the question in the title: yes the center of a $C^{\ast}$-algebra is indeed a $C^{\ast}$-algebra.


As with the other steps, also your argument for step 4. is fine. If $c_n \in Z(A)$ and $c_n \to c \in A$ then $c \in Z(A)$ because for every $a \in A$ we have $\|ac -ca\| = \|a(c-c_n) - (c-c_n)a\|$ which is equal to the left hand side below and can be estimated as $$\|a(c-c_n) + (c_n-c)a\| \leq \|a\|\,\|c-c_n\|+ \|c-c_n\|\|a\| = 2\|a\|\,\|c-c_n\| \;\xrightarrow{n\to\infty}\; 0,$$ hence $\|ac-ca\| = 0$ and thus $c$ commutes with every $a \in A$.

Here's how I would argue (only writing your argument slightly differently and avoiding explicit use of sequences): For a fixed $a \in A$ consider the map $x \mapsto [a,x] = ax - xa$. This is clearly continuous and linear.

Now $\ker{[a,\cdot}]$ is a closed subspace of $A$ and it consists of the elements $x$ commuting with $a$. The center of $A$ is thus $Z(A) = \bigcap\limits_{a \in A} \ker{[a,\cdot]}$, hence it is closed as an intersection of closed sets.

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