Banach coalgebras (or cogebras) are like Banach algebras, but coalgebras. The dual of a Banach coalgebra is a Banach algebra (but not conversely). We can also consider Banach bialgebras (or bigebras).
A Banach coalgebra, or Banach cogebra, is a comonoid object in the monoidal category of Banach spaces with short linear maps and the projective tensor product. (Recall that a Banach algebra is a monoid object in .)
Explicitly, we have:
to the projective tensor product;
where is the ground field;
for each ;
for each ;
for each .
Technically, we've defined a counital coassociative Banach coalgebra. We can leave out (3,5,6) to get a non-counital Banach coalgebra, and (also) leave out (4) to get a non-coassociative Banach coalgebra. Warning: these terms are examples of the red herring principle. Note that (3) is a property-like structure (and 4–6 are obviously just properties).
On the other hand, we can add the property of cocommutativity:
where the braiding is generated by . Then we have a cocommutative Banach coalgebra.
To freely adjoin a counit to a non-counital Banach coalgebra , take the Banach space (using the -direct sum), let be , and let be . Then is a counital Banach coalgebra. (Freely forcing coassociativity or cocommutativity —or even freely adjoining in the first place— is harder.)
and (unless we are allowing non-counital coalgebras)
for all . Warning: the term ‘homomorphism’ is used more generally; see below.
If and are Banach coalgebras, then their projective tensor product is a Banach coalgebra, generated by
The full subcategory of cocommutative Banach coalgebras becomes a cartesian monoidal category under the projective tensor product. Actually, is the terminal object even in (with the unique coalgebra morphism to being itself), but the pairing
(given , , and ) is a morphism of only when and are cocommutative. (I believe that does have a product, but it must be more complicated.)
If is a Banach coalgebra, then the dual vector space is a Banach algebra. Actually, this is more general than ; if is any Banach algebra, then so is (the Banach space of bounded linear maps from to ).
This result is nothing special about Banach (co)algebras; it holds in any closed monoidal category. The multiplication operation in is given by
where is (generated by) the multiplication operation on . is associative, unital, or commutative if and are (with ‘co‑’ in the names of 's properties). In particular, has one of these properties iff has the corresponding property.
Note that (or even ) is not, in general, a Banach coalgebra. (That's because is closed, not coclosed?.)
Let and be Banach coalgebras.
Of course, and are Banach spaces, so we may consider the whole panoply of linear operators from to . In general, a linear operator is only a partial function, defined on a linear subspace of (and otherwise only required to be a linear map); but in particular we consider the densely-defined operator?s (each defined on a dense subspace of ), the linear mappings (each defined on all of ), the bounded operators (each defined on all of and bounded or equivalently continuous), and the short operators (each bounded with a norm at most ).
A comultiplicative linear operator from to is a linear operator such that the following hold for all :
We can also consider densely-defined comultplicative linear operators. A coalgebra homomorphism, or cohomomorphism, is a comultiplicative linear mapping; we can also consider bounded homomorphisms and short homomorphisms. The last of these are, as above, the morphisms in ; of course, any of these classes of operators (except the densely-defined ones, which are not closed under composition) could be taken to be morphisms of a different category with the same objects, but then we would have isomorphisms that are not isometries. (See also isomorphism of Banach spaces?.)
A Banach bialgebra, or Banach bigebra, is a bimonoid in : a Banach space equipped with the structures of both a Banach algebra and a Banach coalgebra, such that and are both morphisms of Banach algebras, or equivalently such that the multiplication and unit of the Banach algebra are both morphisms of Banach coalgebras. Explicitly, this requirement is:
The category of Banach bialgebras has, as objects, Banach bialgebras and, as morphisms, short linear maps that are morphisms of both Banach algebras and Banach coalgebras.
(where is the multiplication with identity ) for all .
The category of Banach Hopf algebras has, as objects, Banach Hopf algebras and, as morphisms, short linear maps that are morphisms of Banach bialgebras and preserve antipodes:
for all .
(where again is the braiding and on is generated by ) for all and
(where a bar indicates complex conjugation) for all .
The category of Banach -coalgebras has, as objects, Banach -coalgebras and, as morphisms, short linear maps that are morphisms of Banach bialgebras and preserve adjoints:
for all .
There are also -coalgebras, which have their own page.
It’s well known that the sequence space of absolutely summable infinite sequences, thought of as (where is the abelian group of integers under addition), is a Banach algebra under convolution; however, it is also a Banach coalgebra, and these structures together make it a Banach bialgebra, in fact a Banach Hopf -algebra. Since is a Banach coalgebra, its dual space (the sequence space of absolutely bounded sequences) is a Banach algebra (which is also well known); and although there is no guarantee that it should work, in this case is also a Banach coalgebra, and indeed a Banach Hopf -algebra too.
Explicitly: The projective tensor square is the space of absolutely summable infinite matrices; convolution takes the matrix to the sequence
(summing along antidiagonals); comultiplication takes to the diagonal matrix?
(which is not quite the origin of the symbol ‘’ but might as well be). The tensor square is the space of infinite matrices with absolutely bounded entries;
(comment added 26-08-2012 by YC: I am not convinced; Grothendieck’s inequality, anyone?)
the dual multiplication on takes the matrix to the sequence
of its diagonal entries; the dual comultiplication (which part of me wants to call ‘nvolution’, but let's say coconvolution instead) takes to
(so each antidiagonal is constant).
We are lucky that coconvolution exists, since the dual of a Banach algebra need not be a Banach coalgebra; but arguably coconvolution is easier to describe than convolution, so let us shift perspective and take coconvolution as basic. Then convolution necessarily exists on the dual of , but (at least in classical mathematics) is only a subspace of that. So from this perspective, what’s lucky is that is closed under convolution. (In dream mathematics, is the entire dual of , so no luck is required.) Of course, is the dual of (the space of sequences with limit ), but is not closed (coclosed?) under coconvolution (try any non-zero example), so we are still lucky.