## December 9, 2008

### The Status of Coalgebra

#### Posted by David Corfield

After my post on coalgebra, I’m still unsure which position to take regarding its status with regard to algebra. Here are some options:

• (1) It’s not a distinction worth making – a coalgebra for $(C, F)$ is an algebra for $(C^{op}, F^{op})$.
• (2) It is a distinction worth making, but there’s plenty of coalgebraic thinking going on – it’s just not flagged as such.
• (3) Coalgebra is a small industry providing a few tools for specific situations, largely in computer science, but with occasional uses in topology, etc.

As for (1), this may be true, but we tend not to think of algebras for an endofunctor on Set as coalgebras for the opposite functor on the category of complete atomic Boolean algebras. And why do we seem to privilege Set when it forms the initial algebra for the Powerset functor on classes, while we seem not to do much with Non-well-founded-set, the terminal coalgebra? Why don’t we care more about non-well-founded monoids or comonoids?

As for (2), if it’s correct, then we might expect simple coalgebraic structures to be as common as simple algebraic structures. Now, we might say that the prevalence of the natural numbers in mathematics is due to their being the initial algebra for the endofunctor on Set $F(X) = 1 + X$. If coalgebra is equally as important, we’d expect the terminal coalgebra to show up often too.

So the terminal coalgebra is the set of extended natural numbers $\mathcal{N} = {0, 1, 2,...} \union \{\infty\}$. And just as $\mathbb{N}$ comes with an isomorphism

$\langle 0, successor \rangle: 1 + \mathbb{N} \to \mathbb{N}$,

$\mathcal{N}$ comes with a predecessor function:

$pred: \mathcal{N} \to 1 + \mathcal{N},$

where $pred(0) = *, pred(n) = n - 1, pred(\infty) = \infty$.

Do we see this structure anywhere outside of computer science, even if only implicitly? Well, I guess the looping operation from p.13 of Categorification could be seen in this light. This, remember, takes out the lowest level of structure in an $n$-category and reindexes downwards.

So, could we say it sends an $n$-category to a $pred(n)$-category, for $n \in \mathcal{N}$? (It might need a little tweaking at the bottom end for the lowest values of $n$.)

Posted at December 9, 2008 2:30 PM UTC

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### Re: The Status of Coalgebra

Do we see this structure anywhere outside of computer science, even if only implicitly?

It may be that we see it in topology. We certainly see the space $\mathbb{N} \cup \{\infty\}$ in topology: it’s the ‘walking’ or ‘generic’ convergent sequence.

More precisely, topologize $\mathbb{N} \cup \{\infty\}$ as the one-point compactification of the discrete space $\mathbb{N}$; equivalently, topologize it as the subspace $\{1 - 2^{-n} : n \in \mathbb{N}\} \cup \{1\}$ of the real line. There is a functor $S: \mathbf{Top} \to \mathbf{Set}$ sending a space $X$ to the set $S(X)$ of ‘convergent sequences in $X$’; by a ‘convergent sequence in $X$’, I mean a sequence $(x_n)_{n \in \mathbb{N}}$ in $X$ together with a point $x$ to which the sequence converges. Then $S(X) \cong \mathbf{Top}(\mathbb{N} \cup \{\infty\}, X)$ naturally in $X$. In other words, the functor $S$ is representable, and the representing object is $\mathbb{N} \cup \{\infty\}$.

A very basic fact is that a sequence in a space $X$ consists of a point in $X$ (the first element of the sequence) together with a sequence (the original sequence, shifted). So there’s an isomorphism $S(X) \cong U(X) \times S(X)$ natural in $X$, where $U: \mathbf{Top} \to \mathbf{Set}$ is the functor sending a space to its set of points. In other words, there’s a natural isomorphism $S \cong U \times S$. Now $U$ is also representable, with representing object $1$, so we have isomorphisms $\mathbf{Top}(\mathbb{N} \cup \{\infty\}, -) \cong S \cong U \times S \cong \mathbf{Top}(1   +   \mathbb{N} \cup \{\infty\}, -)$ or equivalently, by the Yoneda Lemma, an isomorphism $\mathbb{N} \cup \{\infty\} \cong 1   +   \mathbb{N} \cup \{\infty\}.$ This is David’s $pred$.

Posted by: Tom Leinster on December 9, 2008 4:14 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

I know now you guys mean coalgebra in some categorical sense, but when you ask about prevalence, you might want to consider the
original or at least early meaning of an algebra and of a coalgebra and their usefulness in topology. There one confronts the psychological problem of being toilet trained in algebra and only much later encountering coalgebras.

Posted by: jim stasheff on December 9, 2008 11:30 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

I agree about the psychological problem of getting used to coalgebra after years of thinking algebraically. I find this quite generally, not only for the specific meanings of ‘algebra’ as ‘module with a monoid structure’ and ‘coalgebra’ as ‘module with a comonoid structure’.

I do think it’s unfortunate that one particular algebraic structure got given the name of the whole subject. It’s unfortunate for at least two reasons. First, once an ‘algebra’ has been defined as a module with a monoid structure, we have to point out that other ‘algebras’ such as Lie algebras are not algebras at all. (So people sometimes say ‘associative algebra’ for emphasis, and stress that Lie algebras are ‘non-associative algebras’… a contradiction in terms.) Second, and more importantly, it causes confusion in exactly this kind of discussion: when ‘algebra’ is used in a general sense (e.g. ‘algebra for a monad’), some people are apt to wonder what the base ring is.

The same goes for the word ‘coalgebra’.

It’s maybe worth pointing out how the general and specific meanings of ‘coalgebra’ fit together, in case some readers don’t know this. We’ve seen the general notion of a coalgebra for an endofunctor. One example is this: if $F$ is the endofunctor of $R$$\mathbf{Mod}$ defined by $F(M) = M \otimes M$, then an $F$-coalgebra is exactly a non-coassociative coalgebra in the specific sense, i.e. a module $M$ equipped with a comultiplication $M \to M \otimes M$. So the specific meaning is an example of the general one.

(Maybe you want your coalgebras to be coassociative. Well, there’s also the general notion of a coalgebra for a comonad, and there’s a particular comonad on $R$$\mathbf{Mod}$ whose coalgebras are exactly the coassociative coalgebras. You can throw in counitality too if you like.)

Posted by: Tom Leinster on December 10, 2008 3:11 AM | Permalink | Reply to this

### Re: The Status of Coalgebra

I think there is a sense, related to the points that Tom makes, in which algebra models composition, whilst coalgebra models decomposition. (Not in the sense of rotting by the way!) The codiagonal in a (linear) coalgebra looks at ways of decomposing a vector into a sum of tensors of related vectors (perhaps).

There is also some link between the use of another sense of coalgebra in modal logics and combinatorics, e.g. modal logics such as S4 that use posets as models. This link with combinatorics also comes up with Hopf algebras which are now occuring naturally in many combinatorial situations. See for instance,

http://arxiv.org/abs/math.CO/0201253

I am no expert on this so perhaps someone more in the know can comment on this.

Posted by: Tim Porter on December 10, 2008 7:56 AM | Permalink | Reply to this

### Re: The Status of Coalgebra

Correct me if I’m wrong, but I think coalgebras gained prominence via Hopf algebras.

Posted by: jim stasheff on December 11, 2008 1:55 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

That’s very nice. Presumably where you write

A very basic fact is that a sequence in a space $X$ consists…

convergent sequence is implied.

But your argument works for not necessarily convergent sequences too – $\mathbf{Top}(\mathbb{N}, X)$, $\mathbb{N}$ with the discrete topology, displaying the isomorphism

$\mathbb{N} \cong 1 + \mathbb{N}.$

Hmm, I guess there’s nothing clever to say about sequentially compact spaces, something taking sequences to the set of their convergent subsequences?

Posted by: David Corfield on December 10, 2008 5:14 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

You’re probably familiar with the coalgebraic approach to modal logic a la Rutten, Kurz, Pattinson, etc… ?

One could claim that in modal logic, coalgebra is really the essence of what’s going on. The definition of bisimulation as well as all of its nice properties are easy constructions with coalgebras and spans.

Moreover, the idea of final semantics is a big deal. In many contexts (see “Definability, Canonical Models, Compactness, for Finitary Coalgebraic Modal Logic” by Kurz for a more precise statement), the final coalgebra IS the canonical model, which provides an instant proof of completeness.

Posted by: Brendan Cordy on December 9, 2008 4:40 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

Define $\overline{\mathbb{N}}=\mathbb{N}\union\{\infty\}$. Define the functor $MX=\{*\}\union X=1+X$ (disjoint union of course). So $\overline{\mathbb{N}}$ is the final $M$-algebra. Finality means that given any other $M$-algebra, $X$, we have a morphism $X\rightarrow\overline{\mathbb{N}}$ (as $M$-algebras). An $M$ algebra is a function $f:X\rightarrow 1+X$. So any such function $f$ gives rise to a function $unfold(f)$ such that $f\circ unfold(f)=M(unfold(f))\circ pred$. So $unfold(f)$ is simply a count of how many times you can iterate $f$ before you hit the asterisk. You get $\infty$ if convergence never happens.

In other words, the answer to a question like “starting with n, how many times can I iterate the Collatz function before I hit 1?” is answered by an element of $\overline{\mathbb{N}}$.

‘unfold’ is what computer scientists call it, but this kind of count isn’t a particularly computer sciencey concept.

Posted by: Dan Piponi on December 9, 2008 6:29 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

Dan, I think you meant ‘coalgebra’ every time you wrote ‘algebra’. The terminal/final $M$-algebra is just $1$.

I agree that it’s not intrinsically a computer-sciencey concept, but I guess David’s point was that, observationally, it’s most often computer scientists who seem to have found it useful. Well, David actually phrased it in terms of a question: where else has it been used?

Posted by: Tom Leinster on December 9, 2008 6:52 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

Silly me. I meant coalgebra of course.

Anyway, I think the situation may be David’s (ii). The answer to the question “how long is this sequence?” lives in $\overline{\mathbb{N}}$. But we normally say something like “either the sequence is infinite or it has finite length n”.

You get $\mathbb{N}$ when you count up, definitely starting at zero. You get $\overline{\mathbb{N}}$ when you count down, maybe ending at zero.

Posted by: Dan Piponi on December 9, 2008 11:29 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

I think that (ii) is true. Now that I know about corecursive structures (that is final coalgebras of polynomial functors), I see them all over the place. (In particular, any time you deal with infinite sequences, you're dealing with elements of a coalgebra.) They're just not recognised as such.

One problem is that corecursive structures are not seen as a basic construct in the foundations of mathematics (where by ‘foundations’ I mean set theory or something like it), whether this is extensional set theory (like ZF), structural set theory (like ETCS), or type theory (like ITT). All of these include (at least) one recursive structure —the system of natural numbers— by fiat, so people start to wonder whether all other recursive structures can be defined in them. And so they can, but then so can all corecursive structures; they just don't get the same respect. (The exception here is CoC; including recursive structures allows one to define all corecursive structures, but one often sees corecursive structures included explictly. Of course, this foundation is a favourite with computer scientists.)

Actually, I think that corecursive structures may be more fundamental, even in this sense of foundations, than people think. Really I have Brouwer to thank for this idea. Although the founder of intuitionism, he would disagree with almost every foundation offered even for intuitionistic or constructive mathematics, including (even predicative constructive versions of) all those that I've mentioned above. I don't just mean that he would object to formalisation (which he would), but that he would disagree with the claimed facts about sets. This is because he did not accept exponentials (in the category of sets). To be sure, he believed in some exponentials, in particular AN for (at least) a wide variety of sets A. But then this particular exponential is a corecursive structure, the final coalgebra of A × −. (On the other hand, I don't think that one can prove that this corecursive structure is the exponential without assuming the existence of the exponential; it has an evaluation map, but that may not be universal. On the other other hand, I'm not sure that Brouwer actually believed the universal property of exponentials either.)

Of course, precisely formalising Brouwer is probably a futile exercise, but it makes me wonder:

How much of ‘ordinary’ mathematics (and in particular, how much of analysis) can one do internal to a Heyting pretopos in which every polynomial functor has both an initial algebra and a final coalgebra?

If the answer is a lot, then coalgebras are everywhere.

Posted by: Toby Bartels on December 9, 2008 9:17 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

I’m glad Brouwer cropped up in the conversation. Might one say anachronistically that a part of his objection to the classical continuum was that it is not coinductively continuous with respect to addition and multiplication?

Pavlovic claims

but the problem is that coalgebraically implemented reals offer some resistance to algebra: adding and multiplying computable reals may not be computable. if each real number is represented irredundantly, say by a unique stream of digits, then there will be numbers for which the entire infinite streams of digits will need to be consumed before the first digit of their sum, or product, can be determined. this observation is due to brouwer.

Many moons ago I knew something about the bar theorem, the fan theorem and the continuity theorem. I wonder if something coalgebraic is going on with them.

Posted by: David Corfield on December 10, 2008 8:47 AM | Permalink | Reply to this

### Re: The Status of Coalgebra

Dear David,

I have two remarks about co-algebras and Brouwerian intuitionism.

- In the a recursive world, in the effective topos if you like, the streams are just the (recursive) functions from nat to nat. So, we do not obtain the fan-theorem or bar induction by working co-algebraically.

- Sambin’s formal closed subspaces are defined co-algebraically. In locale theory these sublocales are the weakly closed and overt ones. Under mild extra conditions this is precisely the way in which Brouwer defined his spreads.

Bas

Posted by: Bas Spitters on December 10, 2008 8:17 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

Regarding fans, Andrej Bauer wrote back in 2000:

In constructive mathematics spreads and fans play an important role. They are examples of final coalgebras. For example, a fan is a finitely branching tree (can be infinite!) in which the nodes are labeled with natural numbers. The space of all fans $FAN$ satisfies the identity

$FAN = N + N \times FAN + N \times FAN^2 + N \times FAN^3 + ...$

$= \sum_{k \in N} N \times FAN^k.$

It is the final coalgebra for the polynomial functor

$P(X) = \sum_{k \in N} N \times FAN^k.$

I find this presentation conceptually cleaner than the usual encoding of spreads as sets of Gödel codes of finite sequences of natural numbers (of course, this presentation requires a richer type theory). Also, this definition is easily adapted to fans and spreads labeled by elements of any set $A$ (just replace $N \times FAN^k$ with $A \times FAN^k$), and of any branching type.

By the way, $FAN$ is isomorphic to $N^N$, and $N^N$ is the final coalgebra for the functor $P(X) = N \times X$.

Posted by: David Corfield on September 23, 2009 10:16 AM | Permalink | Reply to this

### Re: The Status of Coalgebra

And Bauer continues:

The initial algebra for $P(X) = 1 + X$ is the set of natural numbers.

The final coalgebra $N^+$ for this functor is best thought of as the one-point compactification of the natural numbers, as it consists of the natural numbers together with one extra point “at infinity”. In classical set theory, this is no different from natural numbers, but in a constructive setting the point at infinity is not isolated. If one wants to prove anything interesting about $N^+$, just about the only way to do it is to use coinduction and corecursion. It’s a good exercise.

This is neat because we get a one-point compactification of $N$ without any reference to topology.

If we compute $N^+$ in $PER(D)$, where $D$ is some topological model of the untyped lambda calculus, such as $P(\omega)$ or the universal Scott domain, we get precisely the one-point compactification of natural numbers.

In the effective topos $N^+$ can be described as an equivalence relation on partial recursive functions, where two such functions are equivalent if, and only if, they terminate in the same number of steps (when applied to some fixed input) or they diverge. The divergent functions represent the point at infinity, and those terminating in exactly $k$ steps represent the number $k$.

Posted by: David Corfield on September 23, 2009 10:25 AM | Permalink | Reply to this

### Re: The Status of Coalgebra

Having recently moved office, I came across some work I did for my Masters dissertation, and I see that I have written

In his ‘First act of intuitionism’ Brouwer claims that mathematics is a mental activity, divorced from language whose basic intuition is the common substratum of the two-ity described above, that is, a two-ity stripped of any quality. Mathematics is built up from this intuition by “unlimited self-unfolding”…Realising that only a small portion of mathematics could be developed from the ‘First act’, Brouwer then introduced the ‘Second act of intuitionism’. He had originally maintained in his thesis that the continuum is given in intution and cannot be understood as the totality of its elements. After 1914, however, he became dissatisfied with this idea and seeing that the reduced continuum would have measure zero, he introduced the notion of a choice sequence. This extends the idea of a sequence as being given by a law to a sequence as a process, say $\alpha$, of freely choosing values $\alpha_0, \alpha_1, \alpha_2, ... \in \mathbb{N}$, where at any stage we may apply a restriction on future choices.

Sounds like there’s a wrestle between algebraic and coalgebraic thinking going on.

By the way, I take it that’s a typo by Bauer in the quotation, and that $FAN$ is the final coalgebra for the polynomial functor

$P(X) = \sum_{k \in N} N \times X^k.$

Posted by: David Corfield on September 7, 2010 12:53 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

David: do you grok the difference between a coalgebra and an algebra? In other words: the difference between a monoid in $Vect$ and a monoid in $Vect^{op}$?

If we were talking about finite-dimensional vector spaces, there’d be no real difference, since then vector space duality sets up an equivalence of monoidal categories

$FinVect \simeq FinVect^{op}$

where $FinVect$ is the category of finite-dimensional vector spaces equipped with its usual tensor product, and we give the opposite category $FinVect^{op}$ the ‘same’ tensor product.

But as soon as we leave the finite-dimensional world, there’s a real difference between algebras and coalgebras! And this is why people study coalgebras.

Posted by: John Baez on December 10, 2008 2:09 AM | Permalink | Reply to this

### Re: The Status of Coalgebra

Innocent question: could you provide some examples where one works in Vect and Vect^{op} without topology in infinite dimensions? Something like the group algebra (sans topology) I’m guessing.

Posted by: Yemon Choi on December 10, 2008 6:32 AM | Permalink | Reply to this

### Re: The Status of Coalgebra

Yemon Choi wrote:

Innocent question: could you provide some examples where one works in $Vect$ and $Vect^{op}$ without topology in infinite dimensions?

Sure — for example, there are lots of examples of infinite-dimensional algebras that are also coalgebras. I’ll give some where the algebra and coalgebra structures fit together to form a Hopf algebra.

Take a vector space $V$ and form the direct sum of all its tensor powers, usually called $T V$. This is usually infinite-dimensional. It’s an algebra with the tensor product as multiplication — but actually this algebra structure extends to a Hopf algebra structure in a unique way such that the comultiplication is given by

$\Delta (v) = v \otimes 1 + 1 \otimes v$

The same basic idea works for the direct sum of all symmetrized tensor powers, also known as the symmetric algebra $S V$.

A bit more generally, if $L$ is a Lie algebra, the universal enveloping algebra $U L$ is not only an algebra but a Hopf algebra with the comultiplication given by the same formula. (This reduces to the symmetric algebra when $L$ is abelian.)

You already mentioned another classic example: if $G$ is any group, the vector space of formal finite linear combinations of elements of $G$ is a Hopf algebra called the ‘group algebra’ $k[G]$. This is infinite-dimensional if $G$ is infinite.

Posted by: John Baez on December 12, 2008 11:05 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

Alternatively consider TV as a coalgebra with the DEconcantenation coproduct. It is a Hopf algebra if we use the shuffle product.

Posted by: jim stasheff on December 13, 2008 1:42 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

could you provide some examples where one works in $\Vect$ and $\Vect^{\op}$ without topology in infinite dimensions?

John provided some, but notice that the dimensions are still countable. Examples with uncountable dimension will be rarer (although I'm sure somebody's got one somewhere).

Posted by: Toby Bartels on December 13, 2008 12:07 AM | Permalink | Reply to this

### Re: The Status of Coalgebra

Toby wrote:

John provided some, but notice that the dimensions are still countable.

Not necessarily. $T V$ and $S V$ are countable-dimensional iff the vector space $V$ is countable-dimensional; $U L$ is countable-dimensional iff the Lie algebra $L$ is countable-dimensional; $k[G]$ is countable-dimensional if the group $G$ is countable. So, there are plenty of examples here that give Hopf algebras of uncountable dimension.

Examples with uncountable dimension will be rarer…

Technically speaking, they’re vastly more numerous! However, in practice, people are usually interested in these Hopf algebras when they’re countable-dimensional.

But this is a general trend: for any given infinite cardinal $x$, there are many more cardinals $\ge x$ than $\le x$ — but fewer people interested in them.

Posted by: John Baez on December 13, 2008 4:26 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

Thanks for confirming that your examples are all of countable dimension!

Yemon's question is as much sociological as mathematical. Yes, there are tons of infinite-dimensional (even also uncountable-dimensional) discrete vector spaces. Just take your field K and any infinite (even also uncountable) cardinal number κ and consider the vector space Kκ. (By the axiom of choice, all examples are of this form.)

But the question is whether anybody ever works with such spaces. With infinite-dimensional spaces, yes: take any positive-dimensional (even still finite-dimensional) vector space and apply one of the operations that you mentioned. With uncountable-dimensional spaces, maybe not. Sure, you can define one by fiat, but the question is whether anybody cares.

Personally, I can only think of two cases offhand where anybody mentions them:

• the example above, for arbitrary κ;
• given an infinite-dimensional (even still countable-dimensional) topological vector space, its underlying discrete vector space is (by the axiom of choice) uncountable-dimensional.

But I only see these examples mentioned by the way, just to point them out. I've never seen them used. (Mind you, I'm sure that someone has really been interested in uncountable-dimensional discrete vector spaces. I just don't know of examples, and they would be rare —sociologically speaking.)

I think that this is part of a much broader sociological–mathematical fact: Without topology, discrete structures may be infinite, but they are still usually countable.

Posted by: Toby Bartels on December 14, 2008 6:57 AM | Permalink | Reply to this

### Re: The Status of Coalgebra

do you grok the difference between a coalgebra and an algebra?

If you mean by ‘grok’ what Heinlein meant:

Grok means to understand so thoroughly that the observer becomes a part of the observed—to merge, blend, intermarry, lose identity in group experience,

then probably not. But to a lesser degree yes.

Let me see what this says to my original question. As Tom explained ‘(co)algebra’ in the sense you use is one specific case of the general definition. So you’re providing us with the rationale for a piece of coalgebra used in mainstream maths, and showing why it didn’t need to appear explicitly while people worked in a self-dual category. Perhaps support for (2) then.

Posted by: David Corfield on December 10, 2008 9:29 AM | Permalink | Reply to this

### Re: The Status of Coalgebra

David wrote:

If you mean by ‘grok’ what Heinlein meant…

Yes. More specifically, I was wondering if you knew why people consider coalgebras ‘better-behaved’ than algebras.

I guess not… so, here’s why:

If $R$ is a comodule of a coalgebra $C$, every element of $R$ sits inside some finite-dimensional sub-comodule of $C$.

But, it’s not true that if $R$ is a module of an algebra $A$, then every element of $R$ sits inside some finite-dimensional submodule of $A$.

Now, maybe this comparison is unfair because we haven’t turned around enough arrows around before making it! That might be fun to discuss.

Nonetheless, I believe this fact is what lets people use Tannaka-Krein reconstruction to recover a coalgebra from its finite-dimensional comodules: there are always enough finite-dimensional comodules. You can’t — so I’ve heard — recover an algebra from its finite-dimensional modules. There might not be enough!

And if I remember right, the proof of this nice fact about coalgebras is appallingly simple, and sheds some light on the difference between operations and ‘cooperations’. I’ll leave it as a puzzle for anyone who wants to think about it.

Posted by: John Baez on December 10, 2008 8:15 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

Yes, I was absolutely enchanted by these facts about coalgebra when I was talking a little with Urs a few months ago about differential graded coalgebras (starting around here, and here, when I was working through a fact I first learned from Jim Dolan: that the category of commutative coalgebras is cartesian closed).

Another thing Jim told me is that coalgebra tends to have a very “local” or “infinitesimal” flavor (whereas algebra is more “global”). For example, the free commutative algebra on a vector space consists of polynomial functions on the dual, whereas the cofree cocommutative coalgebra on a space sits inside a local completion of the polynomial algebra, and its elements can be construed as distributions concentrated at a point (like the Dirac functional).

A nice Stone duality-type consequence of this incredible fact about coalgebras (that they are the unions of their finite-dimensional subcoalgebras) is that dual of the category of coalgebras is the category of pro-algebras. Thus the analogue of the Tannaka-Krein reconstruction would be that pro-algebras are recoverable from their finite-dimensional modules.

Posted by: Todd Trimble on December 10, 2008 11:23 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

I was absolutely enchanted by these facts about coalgebra when I was talking a little with Urs a few months ago about differential graded coalgebras

I have now included a wiki entry beginning to wrap this up: nLab:CoDGCA

I am not entirely sure myself why I didn’t further follow this direction and jumped back do differential algebras. I think it was because I was mixed up about how exactly to say “$L_\infty$-algebroid” in CoDGCA for the case that there is more than one object.

But I think I was just being dense, or first had to grow the right neurons or the like.

Anyway, now I want to make up for old sins and do it right. I revised nLab:Lie $\infty$-algebroid such that the CoDGCA-definition is the primary one and the DGCA definition now the derived one, equivalent in the case that the Lie $\infty$-algebroid is of finite rank (degreewise).

I will come back to the great fact about CoDGCA being cartesian closed. I am not exactly sure why I first asked you for this and then somehow dropped this thread again later.

Posted by: Urs Schreiber on December 11, 2008 10:02 AM | Permalink | Reply to this

### Re: The Status of Coalgebra

I wrote:

I am not entirely sure myself why I didn’t further follow this direction

Now I remember! :-)

I understand this fine for the case that the underlying coalgebra is free over the ground field.

But I was – and am – confused about what happens when the underlying coalgebra is free over an algebra of functions on a non-point manifold.

For instance: what is the CoDGCA dual of the DGCA of differential forms $\Omega^\bullet(X)$ on a manifold $X$? I keep getting mixed up about this.

And this confusion on my part was the obstruction to me following the CoDGCA line-of-attack.

Hopefully we can lift this obstruction!

Posted by: Urs Schreiber on December 11, 2008 10:45 AM | Permalink | Reply to this

### Re: The Status of Coalgebra

What is lacking in the DG coalgebra of multivectorfields on X? symmetrized if you like to be DGC?

Posted by: jim stasheff on December 11, 2008 2:04 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

CoDGCA rather than DGCCoA?

what is a codiferential?

Posted by: jim stasheff on December 11, 2008 1:58 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

Todd wrote:

Another thing Jim told me is that coalgebra tends to have a very “local” or “infinitesimal” flavor (whereas algebra is more “global”).

I know what you mean. Thanks to Jim, Alissa Crans has a nice section in her thesis where she shows that Lie algebras give quandle objects in the category of cocommutative coalgebras. The idea is that a Lie group gives a quandle in the category of pointed manifolds, and then the functor ‘taking an infinitesimal neighborhood of the basepoint’ sends this to a quandle in the category of cocommutative coalgebras.

In fact, this idea ultimately gives a more conceptual explanation of how we get Lie algebras from Lie groups. It may seem complicated at first, but it’s designed to tackle the riddle: “Lie groups have a multiplication — so why do we define Lie algebras by differentiating conjugation?”

Posted by: John Baez on December 11, 2008 12:46 AM | Permalink | Reply to this

### Re: The Status of Coalgebra

Does Jim mean this thing about local-coalgebra and global-algebra in the broadest sense, so that wherever we see something coalgebraic we should expect a local aspect?

So if we agree that modal logics are coalgebraic, we should expect it to be a ‘local’ logic. Hmm, certainly in possible world semantics one checks the validity of modal statements in a world via those worlds which are accessible to it.

Does Jim’s thought relate to Urs’ comment about

• spaces - presheaves - glue - cohomology
• quantities - copresheaves - coglue - homotopy

And, might we expect that as Riemann treats complex functions globally, via surfaces, while Weierstrass treats them locally, via analytic continuation, that we’ll find something algebraic in the former and coalgebraic in the latter? Taylor series have a coinductive flavour.

Posted by: David Corfield on December 11, 2008 9:57 AM | Permalink | Reply to this

### Re: The Status of Coalgebra

David wrote:

Does Jim mean this thing about local-coalgebra and global-algebra in the broadest sense, so that wherever we see something coalgebraic we should expect a local aspect?

Since a coalgebraic structure in $C$ is the same as an algebraic structure in $C^{op}$, you can’t make broad generalizations about how algebraic structures differ from coalgebraic ones until you specify some properties of your category $C$.

I don’t know what the extent of Jim’s observation was supposed to be.

Posted by: John Baez on December 13, 2008 5:51 AM | Permalink | Reply to this

### Re: The Status of Coalgebra

Okay, the following isn’t very “categorical”, sorry! But isn’t the “finiteness” of a co-algebra somehow coming from the tensor product? I like to use this as a “reason” why, for topological issues, it’s good to move to a C*-algebra or von Neumann algebra setting. Because here the tensor product is completed, and so gives you a bit more freedom. So you can deal with all continuous functions on a compact group in a C*-setting (instead of, say, putting the cart before the horse, using Peter-Weyl, and moving to coefficient functionals of finite dimensional reps). And von Neumann algebras allow you to work with arbitrary (locally compact) groups, at the obvious cost of using L^\infty functions: here the completed tensor product is weakly closed, and so is “huge”.

However, as an analyst, I have another tool to deal with this: multiplier algebras! But such objects don’t seem to have found much favour in more algebraic thinking. Van Daele has this idea of a “Multiplier Hopf Algebra”:

A. Van Daele, Multiplier Hopf algebras.
Trans. Amer. Math. Soc. 342 (1994), no. 2, 917–932. Link if you have JSTOR access

But you have to work with *-algebras over the complex numbers (basically) and AFAIK, even here, duality doesn’t work very well without a lot more analytic machinery (e.g. “invariant integrals”, by which point you basically have a non-completed C*-algebra…)

Posted by: Doormat on December 11, 2008 12:01 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

Doormat wrote:

Okay, the following isn’t very “categorical”, sorry!

We’ll forgive it if you say ‘colimit’ and ‘Kan extension’ a lot in your next comment.

But isn’t the “finiteness” of a co-algebra somehow coming from the tensor product?

Yes: the tensor product of vector spaces $V$ and $W$ consists of finite linear combinations of vectors $v \otimes w$. This leads straightaway to what I called the “appallingly simple” proof that for a comodule $R$ of a coalgebra $C$, every element $r \in R$ lies in a finite-dimensional sub-comodule.

(I won’t completely give away this proof, in case someone wants to rediscover it.)

I like to use this as a “reason” why, for topological issues, it’s good to move to a C*-algebra or von Neumann algebra setting. Because here the tensor product is completed, and so gives you a bit more freedom.

So you’re saying a topologically completed tensor product gives you a lot more comodules? I guess you’re right. Of course this goes completely against my idea that comodules were ‘nicer’ than coalgebras because they have more special properties. But, I can see why — for completely different reasons — it’s nice to be able to take any module of any algebra ($A \otimes R \to R$) and dualize it to get a comodule of a coalgebra ($R^* \to A^* \otimes R^*$). We have this in the finite-dimensional case, but we’ll only get this in the infinite-dimensional case if we do something sneaky like work with topologically completed tensor products.

Actually Todd’s remark makes me wonder if there’s a purely algebraic way to achieve a similar effect. Any module of an algebra in $Vect$ automatically gives rise to a comodule of a coalgebra in $Vect^{op}$. But what’s $Vect^{op}$ like? Maybe it’s equivalent to some category of topological vector spaces and continuous maps… and maybe its tensor product is some sort of topologically completed tensor product!

It’s just a hunch. I’m reminded of Pontryagin duality, and the fact that the opposite of the category of abelian groups is the category of compact abelian groups. Does something like this work for vector spaces? Actually, now I seem to recall that for every ring $R$, $R Mod^{op}$ is equivalent to the category of compact topological $R$-modules.

(By the way: when I was a grad student, I spent a fair amount of time pondering Takesaki’s Theory of Operator Algebras, thinking about ‘projective’ and ‘injective’ tensor products of Banach spaces and C*-algebras. I’ve forgotten a lot of that stuff, and I don’t remember what a ‘multiplier’ is, either.)

Posted by: John Baez on December 12, 2008 6:29 AM | Permalink | Reply to this

### Re: The Status of Coalgebra

But what’s $Vect^{op}$ like?

I was wondering that, too. And about the opposites of other familiar categories, beyond those provided by Stone duality.

One might have expected that $Vect^{op}$ would bear a passing resemblance to $PointedSet^{op}$, pointed sets being vector spaces over the field with one element. This paper characterises on page 4 the opposite of the category of finite pointed sets, and it already doesn’t seem too simple.

Apparently, the opposite of $Top$ is a quasi-variety, whatever that means.

I wonder if there’s a systematic theory of how a category can fail to be self-dual. Perhaps that’s just maths itself, or a large part of it, as Lawvere and Rosebrugh claim.

Up a level, I suppose the choice becomes wider, as to whether to reverse either or both of the 1-cells and 2-cells. Does some opposite of the 2-category of categories have a ‘nice’ description?

Posted by: David Corfield on December 12, 2008 9:48 AM | Permalink | Reply to this

### Re: The Status of Coalgebra

The concrete characterization of $Vect^{op}$ is as John says (compact topological vector spaces), and this fact can be seen as an easy offshoot of Pontryagin duality: working in the context of $Ab$-enriched categories, we have for a ring $R$ (as a one-object $Ab$-enriched category) that

$(Left R-Mod)^{op} \cong (Ab^R)^{op} \cong (Ab^{op})^{R^{op}} \simeq CAb^{R^{op}}$

where $CAb$ stands for the category of compact abelian groups. So we arrive at continuous right representations of the discrete ring $R$ on compact abelian groups, hence compact vector spaces over $R$ if $R$ is a field.

Actually I thought I’d heard that Pontryagin duality extends far beyond modules to very general algebraic theories, but I’d have to look into this.

The opposite of finite pointed sets should be this very famous $\Gamma$ used by Graeme Segal in his study of loop space machines.

Posted by: Todd Trimble on December 12, 2008 12:59 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

Although I hasten to correct myself: the term “topological vector space” is normally taken to refer to continuous representations of the topological ring $\mathbb{R}$ or $\mathbb{C}$, which is clearly not what I meant: we are ignoring any topology on the ground field.

Posted by: Todd Trimble on December 12, 2008 1:08 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

Todd wrote:

Actually I thought I’d heard that Pontryagin duality extends far beyond modules to very general algebraic theories, but I’d have to look into this.

I have a vague memory of seeing this in a comment on the category theory mailing list — in the same discussion where I learned that the opposite of the category of left $R$-modules is the category of compact Hausdorff topological right $R$-modules.

(In my previous comment I neglected the right/left distinction. I could pretend that I was only considering commutative rings $R$, where this distinction becomes irrelevant… but in fact I just screwed up.)

We should probably look up these:

• Brian J. Day, On Pontryagin duality, Glasgow Math. J. 20 (1979), 15-24.
• Brian J. Day, An extension of Pontryagin duality, Bull. Australian Math. Society (was to appear in 1979).

However, I see Michael Barr once gave a talk entitled ‘Pontryagin duality revisited’ — and I have a feeling he was the one who wrote that comment on the category mailing list!

By the way, it’s a bit strange to think about compact topological $R$-modules when $R$ is the real or complex numbers — as you note, the topology on these modules needs to be quite different from any ‘topological vector space’ topology.

Are we secretly saying that there’s exactly one way to make a finite-dimensional real vector space into a compact Hausdorff topological $\mathbb{R}$-module, where $\mathbb{R}$ is given its discrete topology?

Posted by: John Baez on December 12, 2008 11:29 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

Are we secretly saying that there’s exactly one way to make a finite-dimensional real vector space into a compact Hausdorff topological $\mathbb{R}$-module, where $\mathbb{R}$ is given its discrete topology?

This is weird but fun stuff!

I don’t think you can get a finite-dimensional vector space over $\mathbb{R}$ to be compact Hausdorff (except for $\{0\}$)! The compact Hausdorff ones are of the form

$Hom(V, S^1)$

by Pontryagin duality, where $V$ is a discrete vector space. The smallest nonzero example would be where $V = \mathbb{R}$, and already $Hom(\mathbb{R}, S^1)$ has uncountable dimension as a vector space over $\mathbb{R}$.

It’s hard or impossible to visualize this compact space, but it sits as a closed subspace in a product of continuum many copies of $S^1$:

$Hom(\mathbb{R}, S^1) \hookrightarrow \prod_{r \in \mathbb{R}} S^1$

It’s as if in order to get a vector space to be compact, the vector space structure has to be “smeared out” across many (continuum many) dimensions.

Incidentally, it was Mike Barr that I thought I had heard say that thing about Pontryagin duality you mentioned – but he was talking to someone else at the time, not me, and my memory may be playing tricks. We should look this stuff up.

Posted by: Todd Trimble on December 13, 2008 6:30 AM | Permalink | Reply to this

### Re: The Status of Coalgebra

Todd wrote:

I don’t think you can get a finite-dimensional vector space over $\mathbb{R}$ to be compact Hausdorff (except for $\{0\}$)! The compact Hausdorff ones are of the form

$Hom(V, S^1)$

by Pontryagin duality, where $V$ is a discrete vector space. The smallest nonzero example would be where $V = \mathbb{R}$, and already $Hom(\mathbb{R}, S^1)$ has uncountable dimension as a vector space over $\mathbb{R}$.

It’s hard or impossible to visualize this compact space, but it sits as a closed subspace in a product of continuum many copies of $S^1$:

$Hom(\mathbb{R}, S^1) \hookrightarrow \prod_{r \in \mathbb{R}} S^1$

Oh, duh — I get it.

I guess it suffices to take a product over $r$ lying in a Hamel basis of $\mathbb{R}$ as a vector space over $\mathbb{Q}$. That simplifies things enormously, and makes this gadget vastly easier to visualize.

How is this gadget related to Bohr compactification of $\mathbb{R}$? It smells vaguely similar… but the Bohr compactification depends on the topology of $\mathbb{R}$, which is usually taken not to be discrete.

How about the Bohr compactification of the discrete topological group $\mathbb{R}$?

Posted by: John Baez on December 20, 2008 4:52 AM | Permalink | Reply to this

### Re: The Status of Coalgebra

I guess it suffices to take a product over r lying in a Hamel basis of $\mathbb{R}$ as a vector space over $\mathbb{Q}$. That simplifies things enormously, and makes this gadget vastly easier to visualize.

Yes indeed, for as we know from that interesting letter from Kevin Buzzard (reproduced at the end of Week 273), the Pontryagin dual of $\mathbb{Q}$ is the puny little group $\mathbb{A}/\mathbb{Q}$, the adele group of $\mathbb{Q}$ divided by its diagonally embedded subgroup $\mathbb{Q}$. So the Pontryagin dual of the discrete group $\mathbb{R}$ is merely a cartesian product of continuum many copies of $\mathbb{A}/\mathbb{Q}$.

I’m afraid I don’t know much of anything about Bohr compactifications. But yes, this dual of discrete $\mathbb{R}$ is the Bohr compactification of the usual $\mathbb{R}$. For if $G$ is any (let’s say locally compact Hausdorff) topological abelian group, and $G'$ is its Pontryagin dual, then you get $Bohr(G)$ as $(G', dis)'$ where the thing inside parentheses is $G'$ retopologized by the discrete topology.

I’d just like to write this out. Let’s define $Bohr(G)$ as we did in the preceding paragraph. First, we get the map

$i: G \to Bohr(G)$

just by applying the contravariant functor $\hom(-, S^1)$ to the continuous homomorphism

$id: (G', dis) \to G'$

(if $G$ is locally compact Hausdorff, we have a canonical isomorphism $G \cong \hom(G', S^1)$). This map $i: G \to Bohr(G)$ is an epi, because $id: (G', dis) \to G'$ is a mono and the functor

$\hom(-, S^1): LCHAb^{op} \to LCHAb$

is an equivalence.

Suppose given a continuous homomorphism $f: G \to K$ to a compact abelian group $K$. Then $K'$ is discrete, so

$f': K' \to G'$

factors uniquely through $id: (G', dis) \to G'$, i.e., $f'$ equals a composite of the form

$K' \stackrel{\phi}{\to} (G', dis) \stackrel{id}{\to} G'$

and now by Pontryagin duality, $f = f''$ equals the dual of this composite, which is

$G \stackrel{i}{\to} Bohr(G) \stackrel{\phi'}{\to} K$

so indeed $\phi'$ is the extension we want for the desired universal property of Bohr compactification mentioned in the Wikipedia article you linked to. It’s the unique such extension because $i: G \to Bohr(G)$ is epi.

I think all this goes through without much change even if we drop the assumption that $G$ is locally compact Hausdorff, but I’m a little tired right now and don’t feel like checking it out at the moment.

Posted by: Todd Trimble on December 20, 2008 11:37 AM | Permalink | Reply to this

### Re: The Status of Coalgebra

John wrote:

I guess it suffices to take a product over $r$ lying in a Hamel basis of $\mathbb{R}$ as a vector space over $\mathbb{Q}$.

I no longer believe this. Lately I tend to wake up in the middle of the night and think mostly pointless thoughts about the events of the day; sometimes my thoughts turn to math, and that’s when I notice mistakes I made in over-hasty guesses on this blog.

I had been thinking a group homomorphism $f : \mathbb{R} \to S^1$ was determined by its values on a Hamel basis — that is, a basis of $\mathbb{R}$ as an uncountable-dimensional vector space over $\mathbb{Q}$. I still think that would be true for a group homomorphism $f: \mathbb{R} \to \mathbb{R}$ But the problem with $f : \mathbb{R} \to S^1$ is that $S^1$ has torsion: if we know $f(x)$ for some $x \in \mathbb{R}$, we know $f(2x)$ but not, say, $f(x/2)$, since there are two consistent choices of what that could be.

So, to determine $f$ I think we need to know its values not on a Hamel basis of $\mathbb{R}$ a vector space over $\mathbb{Q}$, but on a set of generators of $\mathbb{R}$ as a $\mathbb{Z}$-module. (Duh.) And I think we can get this by taking a Hamel basis $x_\alpha$ and then using the numbers $x_\alpha/n$ for $n = 1,2,...$ Or, if we wish to be ‘more efficient’, we can use $x_\alpha/p^n$ for $p$ prime and $n = 0,1,2,....$

Posted by: John Baez on December 20, 2008 4:48 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

Actually, John, I thought you were just fine. I reasoned as follows: we have

$\mathbb{R} \cong \sum_{e_i \in Hamel} \mathbb{Q} \cdot e_i$

where this coproduct decomposition can be viewed either in vector spaces over $\mathbb{Q}$ or in abelian groups. The dual is then

$\hom(\mathbb{R}, S^1) \cong \hom(\sum_{e_i \in Hamel} \mathbb{Q} \cdot e_i, S^1) \cong \prod_{e_i \in Hamel} \hom(\mathbb{Q} \cdot e_i, S^1)$

at which point we use the identification

$\hom(\mathbb{Q}, S^1) \cong \mathbb{A}/\mathbb{Q}$

as in the letter from Kevin Buzzard, so indeed we get

$\hom(\mathbb{R}, S^1) \cong \prod_{e_i} \mathbb{A}/\mathbb{Q}$

which is what I was trying to say above, interpreting what you were trying to say.

Posted by: Todd Trimble on December 20, 2008 8:37 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

This again shows the virtue of making vatic pronouncements and acting like they’re apodictic: there’s a chance someone will come along and interpret them in a way that makes ‘em true.

I was just making a joke based on an erroneous idea; you interpreted it as something much more intelligent than it was. I’ve already explained my mistake, so instead of explaining it in even more detail I’ll just shut up and act like I meant what you said.

Posted by: John Baez on December 20, 2008 10:15 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

Here is a relevant paper by Michael Barr:

He constructs a category of topological abelian groups containing the category of LCA groups, which is *-autonomous with the circle as the dualizing object.

Here is a comment he emailed to me. He allowed me to post it here:

I looked at the web site you mentioned. It seems to have left coalgebras behind. There is no mystery as to $Vect^op$ (assuming that is the opposite of vector spaces of a field). If the field is finite, then it is just the compact vector spaces. More generally, it consists of “linearly compact” vector spaces. Lefschetz defined them maybe 80 years ago in order to try to make cohomology theory of spaces dual to homology. The definition is the finite intersection property on closed linear subspaces, but turns out to be characterized as cartesian powers of the field. This assumes the field is discrete. Compactness cannot work as there cannot be a compact topology even on a one dimensional space over an infinite field.

You can go on and construct a *-autonomous category following the same path as I trod for abelian groups. You could define a locally linearly compact space as one with a linearly compact nbd of 0, but vector spaces have so much structure that it just amounts to being a product of a discrete and a linearly compact space. (You must know that a locally compact abelian group is a product of a compact group, a discrete group, and a finite dimensional real vector space.) Taking the same tack, that is starting with a space that can be embedded into a product of locally linearly compact spaces and then the largest topology with the given set of linear functionals, you get a *-autonomous category that extends the duality of the locally linearly compact ones.

Also someone should explain that a quasivariety is a subcategory of a variety closed under products and subobjects (but not homomorphic images). In the case of $Top^op$, it is actually a subvariety characterized by a single Horn clause. The whole variety, not surprisingly, is frames.

The last remark addresses this comment by David Corfield.

Posted by: John Baez on December 31, 2008 9:54 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

Well, to answer an easy question, the multiplier algebra of $A$ is the “largest” unital algebra containing $A$ in a non-degenerate way. Of course, for this even to be on your radar, you have to worry about non-unital algebras: my understanding is that most algebraists would consider this very weird behaviour. See Wikipedia article for more. A fancy way to say it: multiplier algebras are non-commutative Stone Cech compactifications.

So a Multiplier Hopf algebra is just a Hopf algebra where the comultiplication maps into the multiplier algebra $M(A\otimes A)$, and not into $A\otimes A$ itself. So again, we enlarge the tensor space.

This got me thinking: is there a purely coalgebra notion here? Because to form $M(A\otimes A)$ you need $A\otimes A$ to be an algebra, not just a vector space.

Posted by: Doormat on December 12, 2008 11:25 AM | Permalink | Reply to this

### Re: The Status of Coalgebra

John writes:

If $R$ is a comodule of a coalgebra $C$, every element of $R$ sits inside some finite-dimensional sub-comodule of $C$.

But, it’s not true that if $R$ is a module of an algebra $A$, then every element of $R$ sits inside some finite-dimensional submodule of $A$.

Is this why people use contramodules? E.g., Positselski:

The time has come to mention that there exist two kinds of module categories for a coalgebra: besides the familiar comodules, there are also contramodules. Comodules can be thought of as discrete modules which are unions of their finite-dimensional subcomodules, while contramodules are modules where certain infinite summation operations are defined. For example, the space of linear maps from a comodule to any vector space has a natural contramodule structure.

In these slides Brzeziński notes a 3 to 797 imbalance in MathSciNet hits for contramodules compared to comodules, and yet claims

Contramodules seem to be as natural as comodules.

Posted by: David Corfield on May 19, 2009 11:58 AM | Permalink | Reply to this

### Re: The Status of Coalgebra

E.g., Positselski:

The time has come to mention that there exist two kinds of module categories for a coalgebra: besides the familiar comodules, there are also contramodules. Comodules can be thought of as discrete modules which are unions of their finite-dimensional subcomodules, while contramodules are modules where certain infinite summation operations are defined.

I'm reminded of the difference between cohomology and compactly supported cohomology. (And this is related, not to Pontryagin duality, but to Poincaré duality.)

Posted by: Toby Bartels on May 19, 2009 11:40 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

Some what related is John Moore’s comment that (in the topologcal context at least)
coalgebras are more natural.

Also the Vect dual of a coalgebra is an alg
but the Vect dual of an alg is only something called a complete coalg and not a coalg.

The module/comodule comments in John’s later response are even more strikign int he Lie context.

Posted by: jim stasheff on December 10, 2008 11:02 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

I have created an entry nLab:coalgebra.

Posted by: Urs Schreiber on December 11, 2008 9:41 AM | Permalink | Reply to this

### Re: The Status of Coalgebra

To beat a dead horse completely into the mud, I can prove that what one gets by regarding the chain complex of a nilpotent space as a coalgebra over the Eccles-Barratt operad (used to define Steenrod operations) determines its integral homotopy type.

In fact there’s a Quillen equivalence between the category of pointed, irreducible coalgebras over the Eccles-Barratt operad and pointed, reduced, Z-complete simplicial sets (Z-complete = homotopy equivalent to its Z-completion).

Posted by: Justin Smith on December 24, 2008 9:13 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

Why does “comonoidal category” only have 9 Google hits, compared to 25,100 hits for “monoidal category”? For every hit for ‘comonoid’ there are about 10 hits for ‘monoid’.

Posted by: David Corfield on January 6, 2009 12:28 PM | Permalink | Reply to this

### Re: The Status of Coalgebra

what’s the ratio for coalg versus alg?

Posted by: jim stasheff on January 6, 2009 2:07 PM | Permalink | Reply to this