Banach coalgebra

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 $Ban$ of Banach spaces with short linear maps and the projective tensor product. (Recall that a Banach algebra is a monoid object in $Ban$.)

Explicitly, we have:

- a Banach space $A$
- a short linear map, the
**comultiplication**:$\Delta\colon A \to A {\displaystyle\hat{\otimes}_\pi} A$to the projective tensor product;

- a short linear functional, the
**counit**:$\epsilon\colon A \to K ,$where $K$ is the ground field;

- an equation, the
**coassociativity**:$(\Delta {\displaystyle\hat{\otimes}_\pi} \id_A) \Delta x = (\id_A {\displaystyle\hat{\otimes}_\pi} \Delta) \Delta x \in (A {\displaystyle\hat{\otimes}_\pi} A) {\displaystyle\hat{\otimes}_\pi} A \cong A {\displaystyle\hat{\otimes}_\pi} (A {\displaystyle\hat{\otimes}_\pi} A)$for each $x\colon A$;

- an equation, the
**left coidentity**:$(\epsilon {\displaystyle\hat{\otimes}_\pi} \id_A) \Delta x = x \in K {\displaystyle\hat{\otimes}_\pi} A \cong A$for each $x\colon A$;

- and an equation, the
**right coidentity**:$(\id_A {\displaystyle\hat{\otimes}_\pi} \epsilon) \Delta x = x \in A {\displaystyle\hat{\otimes}_\pi} K \cong K$for each $x\colon A$.

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**:

- $\tau \Delta x = \Delta x$ for each $x\colon A$,

where the braiding $\tau\colon A {\displaystyle\hat{\otimes}_\pi} A \to A {\displaystyle\hat{\otimes}_\pi} A$ is generated by $\tau (u \otimes v) = v \otimes u$. Then we have a **cocommutative Banach coalgebra**.

To **freely adjoin a counit** to a non-counital Banach coalgebra $A$, take the Banach space $A \oplus_1 K$ (using the $l^1$-direct sum), let $\Delta_{A \oplus K} (x,c)$ be $(\Delta_A x, 0, 0, c) \in (A {\displaystyle\hat{\otimes}_\pi} A) \oplus_1 A \oplus_1 A \oplus_1 K \cong (A \oplus_1 K) {\displaystyle\hat{\otimes}_\pi} (A \oplus_1 K)$, and let $\epsilon_{A \oplus_1 K} (x,c)$ be $c$. Then $A \oplus_1 K$ is a counital Banach coalgebra. (Freely forcing coassociativity or cocommutativity —or even freely adjoining $\Delta$ in the first place— is harder.)

The category **$Ban Coalg$** of Banach coalgebras has, as objects, Banach coalgebras and, as morphisms, short linear maps $f\colon A \to B$ with equations

$\Delta_B f x = (f {\displaystyle\hat{\otimes}_\pi} f) \Delta_A x$

and (unless we are allowing non-counital coalgebras)

$\epsilon_B f x = \epsilon_A x$

for all $x\colon A$. Warning: the term ‘homomorphism’ is used more generally; see below.

If $A$ and $B$ are Banach coalgebras, then their **projective tensor product** $A {\displaystyle\hat{\otimes}_\pi} B$ is a Banach coalgebra, generated by

$\Delta_{A {\displaystyle\hat{\otimes}_\pi} B} (x \otimes y) = \Delta_A x \otimes \Delta_B y \in (A {\displaystyle\hat{\otimes}_\pi} A) {\displaystyle\hat{\otimes}_\pi} (B {\displaystyle\hat{\otimes}_\pi} B) \cong (A {\displaystyle\hat{\otimes}_\pi} B) {\displaystyle\hat{\otimes}_\pi} (A {\displaystyle\hat{\otimes}_\pi} B)$

and

$\epsilon_{A {\displaystyle\hat{\otimes}_\pi} B} (x \otimes y) = (\epsilon_A x) (\epsilon_B y) \in K .$

Similarly (but more simply), the ground field $K$ is itself a Banach coalgebra, with $\Delta$ and $\epsilon$ both essentially the identity map. In this way, $Ban Coalg$ becomes a symmetric monoidal category.

The full subcategory $Cocomm Ban Coalg$ of cocommutative Banach coalgebras becomes a cartesian monoidal category under the projective tensor product. Actually, $K$ is the terminal object even in $Ban Coalg$ (with the unique coalgebra morphism to $K$ being $\epsilon$ itself), but the pairing

$(f,g)(x) \coloneqq (f {\displaystyle\hat{\otimes}_\pi} g) \Delta x \in A {\displaystyle\hat{\otimes}_\pi} B$

(given $f\colon \Gamma \to A$, $g\colon \Gamma \to B$, and $x\colon \Gamma$) is a morphism of $Ban Coalg$ only when $A$ and $B$ are cocommutative. (I believe that $Ban Coalg$ *does* have a product, but it must be more complicated.)

If $A$ is a Banach coalgebra, then the dual vector space $A^*$ is a Banach algebra. Actually, this is more general than $A^* = [A,K]$; if $B$ is any Banach algebra, then so is $[A,B]$ (the Banach space of bounded linear maps from $A$ to $B$).

This result is nothing special about Banach (co)algebras; it holds in any closed monoidal category. The multiplication operation in $[A,B]$ is given by

$(\lambda \mu) x = m (\lambda {\displaystyle\hat{\otimes}_\pi} \mu) \Delta x ,$

where $m\colon B {\displaystyle\hat{\otimes}_\pi} B \to B$ is (generated by) the multiplication operation on $B$. $[A,B]$ is associative, unital, or commutative if $A$ and $B$ are (with ‘co‑’ in the names of $A$'s properties). In particular, $A^*$ has one of these properties iff $A$ has the corresponding property.

Note that $[B,A]$ (or even $B^*$) is *not*, in general, a Banach coalgebra. (That's because $Ban$ is closed, not coclosed?.)

Let $A$ and $B$ be Banach coalgebras.

Of course, $A$ and $B$ are Banach spaces, so we may consider the whole panoply of linear operators from $A$ to $B$. In general, a linear operator is only a partial function, defined on a linear subspace of $A$ (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 $A$), the linear mappings (each defined on all of $A$), the bounded operators (each defined on all of $A$ and bounded or equivalently continuous), and the short operators (each bounded with a norm at most $1$).

A **comultiplicative linear operator** from $A$ to $B$ is a linear operator $T\colon A \to B$ such that the following hold for all $x \in \dom T$:

- $\Delta x \in (\dom T) {\displaystyle\hat{\otimes}_\pi} (\dom T)$,
- $\Delta_B T x = (T {\displaystyle\hat{\otimes}_\pi} T) \Delta_A x$ (which exists by the previous line), and
- $\epsilon_B T x = \epsilon_A x$ (which always exists).

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 $Ban Coalg$; 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 $Ban$: a Banach space $A$ equipped with the structures of both a Banach algebra and a Banach coalgebra, such that $\Delta$ and $\epsilon$ 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:

- $\Delta (x y) = (\Delta x) (\Delta y)$ (with the induced multiplication on $A {\displaystyle\hat{\otimes}_\pi} A$),
- $\Delta 1 = 1 \otimes 1$ (which is the identity in $A {\displaystyle\hat{\otimes}_\pi} A$),
- $\epsilon (x y) = (\epsilon x) (\epsilon y)$ (with the multiplication on the right in $K$), and
- $\epsilon 1 = 1$ (with the $1$ on the right in $K$).

The category **$Ban Bialg$** of Banach bialgebras has, as objects, Banach bialgebras and, as morphisms, short linear maps that are morphisms of both Banach algebras and Banach coalgebras.

A **Banach Hopf algebra** is a Hopf object? in $Ban$: a Banach bialgebra $A$ with a (necessarily unique) short linear map (the **antipode**) $S\colon A \to A$ such that

$m (S {\displaystyle\hat{\otimes}_\pi} \id) \Delta x, m (\id {\displaystyle\hat{\otimes}_\pi} S) \Delta x = 1 \epsilon x$

(where $m$ is the multiplication with identity $1$) for all $x\colon A$.

The category **$Ban Hopf Alg$** of Banach Hopf algebras has, as objects, Banach Hopf algebras and, as morphisms, short linear maps $f\colon A \to B$ that are morphisms of Banach bialgebras and preserve antipodes:

$f (S_A x) = S_B (f x)$

for all $x\colon A$.

A **Banach $*$-coalgebra** is a $*$-monoid object? in $Ban$: a Banach coalgebra $A$ equipped with an antilinear map (the **adjoint**) $x \mapsto x^*\colon A \to A$ such that

$\Delta x^* = (\tau \Delta x)^*$

(where $\tau$ again is the braiding and $*$ on $A {\displaystyle\hat{\otimes}_\pi} A$ is generated by $(x \otimes y)^* = x^* \otimes y^*$) for all $x\colon A$ and

$\epsilon x^* = \overline {\epsilon x}$

(where a bar indicates complex conjugation) for all $x\colon A$.

The category **$Ban {*} Coalg$** of Banach $*$-coalgebras has, as objects, Banach $*$-coalgebras and, as morphisms, short linear maps $f\colon A \to B$ that are morphisms of Banach bialgebras and preserve adjoints:

$f(x^*) = f(x)^*$

for all $x\colon A$.

There are also $C^*$-coalgebras, which have their own page.

It’s well known that the sequence space $l^1$ of absolutely summable infinite sequences, thought of as $l^1(\mathbb{Z})$ (where $\mathbb{Z}$ 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 $l^1$ is a Banach coalgebra, its dual space $l^\infty$ (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 $l^\infty$ is also a Banach coalgebra, and indeed a Banach Hopf $*$-algebra too.

Explicitly: The projective tensor square $l^1 {\displaystyle\hat{\otimes}_\pi} l^1$ is the space of absolutely summable infinite matrices; convolution takes the matrix $(a_{i,j})_{i,j}$ to the sequence

$(\sum_{i + j = k} a_{i,j})_k$

(summing along antidiagonals); comultiplication takes $(a_k)_k$ to the diagonal matrix?

$\Delta a = (\sum_{i = j = k} a_k)_{i,j}$

(which is not quite the origin of the symbol ‘$\Delta$’ but might as well be). The tensor square $l^\infty {\displaystyle\hat{\otimes}_\pi} l^\infty$ 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 $l^\infty$ takes the matrix $(a_{i,j})_{i,j}$ to the sequence

$(\sum_{i = j = k} a_{i,j})_k = (a_{k,k})_k$

of its diagonal entries; the dual comultiplication (which part of me wants to call ‘nvolution’, but let's say coconvolution instead) takes $(a_k)_k$ to

$(\sum_{i + j = k} a_k)_{i,j} = (a_{i + j})_{i,j}$

(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 $l^\infty$, but (at least in classical mathematics) $l^1$ is only a subspace of that. So from this perspective, what’s lucky is that $l^1$ is closed under convolution. (In dream mathematics, $l^1$ is the entire dual of $l^\infty$, so no luck is required.) Of course, $l^1$ *is* the dual of $c_0$ (the space of sequences with limit $0$), but $c_0$ is *not* closed (coclosed?) under coconvolution (try any non-zero example), so we are still lucky.

Revised on August 27, 2012 20:24:06
by Toby Bartels
(98.19.40.130)