5
$\begingroup$

Let $c_0$ denote the set of all complex sequences that converge to zero.

We can show that $c_0$ is a $C^*$-algebra with the $*$-involution defined as complex conjugate and norm $$\|x\| = \max_j |x_j|$$ for every $x \in c_0$.

I know that every $C^*$-algebra is $*$-isomorphic to a $C^*$-subalgebra of $B(H)$ for some Hilbert space $H$.

How do I go about finding this $C^*$-subalgebra of $B(H)$ for $c_0$?

Any help pointing me in the right direction would be much appreciated.

$\endgroup$
5
$\begingroup$

The other answer is obviously the canonical one. I just wanted to mention that it can be obtained via the GNS construction, if one uses its generalized form for weights (and not states).

Indeed, we can define a faithful weight $\varphi:c_0\to\mathbb C$ by $$ \varphi(x)=\sum_jx_j $$ (recall that a weight is a priori only defined on positive elements, and it can be infinite). If we do GNS for this weight, no quotient is necessary (because $\varphi$ is faithful) and the inner product is given by $$ \langle x,y\rangle=\varphi(y^*x)=\sum_j x_j\overline{y_j}. $$ This inner product is defined on a dense subspace of $c_0$ (the sequences with finitely many nonzero elements) and its completion is $\ell^2(\mathbb N)$, with $c_0$ acting as multiplication operators.

$\endgroup$
  • $\begingroup$ If one does not care that much about $\ell^2(\mathbb{N})$ then the state $\varphi(x) = \sum_{j} \frac{x_j}{2^{j}}$ would also do. $\endgroup$ – Mateusz Wasilewski Oct 12 '17 at 20:01
  • $\begingroup$ Of course. But I don't see that one could get an easy description of what the completion of $c_0$ is with the norm induced by $\varphi$. $\endgroup$ – Martin Argerami Oct 12 '17 at 20:22
  • $\begingroup$ Well, it would be just weighted $\ell^2$. I agree, however, that it is more natural to work with the usual $\ell^2$. $\endgroup$ – Mateusz Wasilewski Oct 13 '17 at 7:13
4
$\begingroup$

Let's begin with $\ell_\infty$: every bounded sequence induces an operator on the sequence space $\ell_2$ by multiplication. That is, $x\in \ell_\infty$ corresponds to the multiplication operator $M_x(y) = (x_ky_k)_{k=1}^\infty$ on $\ell_2$. These are all diagonal operators with respect to the standard basis of $\ell_2$.

Restricting to $c_0$, we get only operators with diagonal entries that tend to $0$. These are precisely compact diagonal operators. Indeed, being compact is equivalent to being the limit of finite-rank operators; on the other hand, being in $c_0$ is equivalent to being the limit of sequences that have finitely many nonzero terms.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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