Operator algebra

algebraic quantum field theory (perturbative, on curved spacetimes, homotopical)


field theory: classical, pre-quantum, quantum, perturbative quantum

Lagrangian field theory


quantum mechanical system

free field quantization

gauge theories

interacting field quantization


States and observables

Operator algebra

Local QFT

Perturbative QFT

C *C^*-coalgebras


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


Let AA be a Banach *-coalgebra over the ground field KK. Let λ:AK\lambda\colon A \to K be a bounded linear functional on AA, and let λ *\lambda^* be the composite

A*AλK¯K A \overset{*}\to A \overset{\lambda}\to K \overset{\bar{}}\to K

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

(1)(λ^ πλ *)Δ:AK^ πKK; (\lambda {\displaystyle\hat{\otimes}_\pi} \lambda^*) \circ \Delta\colon A \to K {\displaystyle\hat{\otimes}_\pi} K \cong K ;

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

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

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


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

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

Dual W *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 *B^*-identity for the dual, so the dual of a C *C^*-coalgebra is a C *C^*-algebra.

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

Which W *W^*-algebras arise in this way?

Examples (and non-examples)

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

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

(YC) Moreover, the TVS-isomorphism class of the predual of a W *W^*-algebra is very restricted (let alone its isomorphism class in Bang). In particular \ell^\infty doesn’t have a snowball’s chance in Texas of being a C *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 1L^1 (on the real line \mathbb{R} with Lebesgue measure) is the W *W^*-algebra L L^\infty, L 1L^1 is not a C *C^*-coalgebra, nor even a Banach coalgebra (at least not in the obvious way). Essentially, this is because the diagonal in 2\mathbb{R}^2 has measure zero (so Δ\Delta takes an element of L 1()L^1(\mathbb{R}), interpreted as an absolutely continuous measure on \mathbb{R}, to a measure on ×\mathbb{R} \times \mathbb{R} that is not absolutely continuous and so cannot be reinterpreted as an element of L 1(×)L 1()^ πL 1()L^1(\mathbb{R} \times \mathbb{R}) \cong L^1(\mathbb{R}) {\displaystyle\hat{\otimes}_\pi} L^1(\mathbb{R})).

Revised on August 28, 2012 05:16:53 by Yemon Choi (