Finding a $C^*$-subalgebra of $B(H)$ for $c_0$?

Question

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.

Answer 1

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.

Answer 2

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.