nLab
C-star-coalgebra

Idea

$C^{*}$-coalgebras are like $C^{*}$-algebras, but coalgebras. Their duals are $W^{*}$-algebras.

Definition

Let $A$ be a Banach *-coalgebra over the ground field $K$. Let $\lambda :A\to K$ be a bounded linear functional on $A$, and let ${\lambda }^{*}$ be the composite

$$A\stackrel{*}{\to }A\stackrel{\lambda }{\to }K\overline{\to }K$$A \overset{*}\to A \overset{\lambda}\to K \overset{\bar{}}\to K

(where $\overline{}$ is complex conjugation, trivial if $K$ is real?); although some of the maps in this composite may be only antilinear?, the composite ${\lambda }^{*}$ is linear (over all of $K$). Now consider the composite

(1)$$(\lambda {\displaystyle {\hat{\otimes }}_{\pi }}{\lambda }^{*})\circ \Delta :A\to K{\displaystyle {\hat{\otimes }}_{\pi }}K\cong K;$$(\lambda {\displaystyle\hat{\otimes}_\pi} \lambda^*) \circ \Delta\colon A \to K {\displaystyle\hat{\otimes}_\pi} K \cong K ;

since $\Delta$ and $*$ are short, the norm of this functional is at most $\parallel \lambda {\parallel }^2$.

$A$ is a $C^{*}$-coalgebra if the norm of the map in (1) is exactly $\parallel \lambda {\parallel }^2$, for every bounded linear functional $\lambda$.

Although there is an asymmetry in (1) (in the relative placement of $\lambda$ and ${\lambda }^{*}$), if we start with ${\lambda }^{*}$ instead of $\lambda$, we see that the universally quantified definition of $C^{*}$-coalgebra is symmetric.

Justification

If we take the formal dual of everything in the definition above, then $A$ becomes a Banach *-algebra and $\lambda :K\to A$ becomes (multiplication of scalars by) an element $x$ of $A$. The formal dual of the composite (1) is (multiplication of scalars by) the element $x^{*}x$. The requirement that the norm of this be exactly the square of the norm of $x$ is the $B^{*}$-identity that defines a $C^{*}$-algebra.

So the definition of $C^{*}$-coalgebra dualises everything in the definition of $C^{*}$-algebra, down to using coelement?s (in this case linear functionals) instead of elements.

Dual $W^{*}$-algebras

In general, the dual of a coalgebra is an algebra, in any closed monoidal category. In particular, the dual of a Banach coalgebra is a Banach algebra. The involution $*$ gets along with this just fine; the dual of a Banach *-coalgebra is a Banach *-algebra. Finally, the linear functional in (1) becomes the element in the $B^{*}$-identity for the dual, so the dual of a $C^{*}$-coalgebra is a $C^{*}$-algebra.

But we have more! If $A$ is a $C^{*}$-coalgebra, then (as we've just seen) $A^{*}$ is a $C^{*}$-algebra; but since $A^{*}$ has a predual $A$, this means that $A^{*}$ is actually a $W^{*}$-algebra as well.

Which $W^{*}$-algebras arise in this way?

Examples (and non-examples)

The sequence space $l^1$ is a $C^{*}$-coalgebra, whose dual $W^{*}$-algebra is the sequence space $l^{\mathrm{\infty }}$. (For details of the comultiplication on $l^1$, see the examples in Banach coalgebra.)

Although $l^{\mathrm{\infty }}$ is a Banach coalgebra (under ‘coconvolution’), it is not a $C^{*}$-coalgebra (at least not under coconvolution).

(YC) Moreover, the TVS-isomorphism class of the predual of a $W^{*}$-algebra is very restricted (let alone its isomorphism class in Bang). In particular ${\ell }^{\mathrm{\infty }}$ doesn’t have a snowball’s chance in Texas of being a $C^{*}$-coalgebra under any kind of choice of comultiplication, because it’s the wrong kind of Banach space.)

Although the dual of the Lebesgue space $L^1$ (on the real line $ℝ$ with Lebesgue measure) is the $W^{*}$-algebra $L^{\mathrm{\infty }}$, $L^1$ is not a $C^{*}$-coalgebra, nor even a Banach coalgebra (at least not in the obvious way). Essentially, this is because the diagonal in ${ℝ}^2$ has measure zero (so $\Delta$ takes an element of $L^1(ℝ)$, interpreted as an absolutely continuous measure on $ℝ$, to a measure on $ℝ\times ℝ$ that is not absolutely continuous and so cannot be reinterpreted as an element of $L^1(ℝ\times ℝ)\cong L^1(ℝ){\displaystyle {\hat{\otimes }}_{\pi }}{L}^1(ℝ)$).