The n-Category Café

Todd,

I have now incorporated your statement that $\Omega^\bullet$ and $\mathrm{Hom}(-,\Omega^\bullet(-))$ form an adjunction.

What you say about how to put this into perspective, $\Omega^\bullet(-)$ being Janusian or ambimorphic etc, I find very intriguing, but haven’t yet incorporated into my notes, since I am not quite sure yet what to make of that. Clearly, I have a lot to learn here.

If and when you might be interested, I’d very much enjoy if you could have a look at the beginning of section 4, where I talk about transgression.

I am pretty sure I am onto something there, but it is also clear that there should be a much better abstract nonsense way to say what I am trying to say.

So here is the point:

usually, transgression is defined as pull-push from right to left through spans of the form $$\array{ &&& hom(par,tar) \otimes par \\ && {}^{p_1}\swarrow && \searrow^{ev} \\ & hom(par,tar) &&&& tar } \,.$$

I am claiming that if what we want to transgress is actually an $n$-functorial thing whose domain is $\mathrm{tar}$, then what we “really want” to regard as its transgression is instead its image under $$\mathrm{hom}(\mathrm{par},-) \,.$$

I know this is what we “really want” by looking at rather large classes of examples.

I might just be content with taking $\mathrm{hom}(\mathrm{par},-)$ by definition to be “my” notion of “good” transgression. But I want to clarify what’s going on, how this relates to the pull-push operation people usually consider.

At the beginning of that section 4 I present a long remark where I give my best attemt at clarifying the situation.

If what I do there is of any value at all, then certainly it is only scratching the surface of something. I have the suspicion you might maybe recognize some abstract nonsense more elegant and more useful which refines my discussion there.

As I said, if and when you are interested, this would be something I’d very much enjoy hearing your comment on.

Todd wrote:

I don’t think this adjunction [between DGCAs and smooth spaces] can be an equivalence,

Yes, I can easily believe that. My vaguely expressed hope that it yields some kind of equivalence was at best too vague.

But let’s see: shouldn’t we be able to obtain some kind of equivalence after passing to cohomologies?

That’s at least what, as far as I understand, the point of Sullivan models is.

For instance theorem 8.1 on page 301 of Sullivan’s old paper states that the cohomology of any DGCA $A$ is isomorphic to the rational polynomial deRham cohomology of the space, which he denotes $\langle A \rangle$, defined by this algebra, $$H^\bullet(A,\mathbb{Q}) \simeq H^\bullet_{pol. deRham}(\langle A \rangle, \mathbb{Q}) \,.$$

And I gather the whole point of the study of Sullivan models is that, in some sense, every space is modeled by a DGCA in rational cohomology in this sense.

I understand that an equivalence in cohomology is much weaker that the full equivalence I seemed to be talking about initially, but I would still be very interested in understanding how this equivalence in cohomology studied in terms of Sullivan models carries over to the context of presheaves over manifolds which we are discussing here.

I wanted to get back to you some time on calculations of internal homs of differential graded cocommutative coalgebras

That would be immensely appreciated!

I haven’t made any progress on this since my attempts on it here and here.

Worse, since I stopped thinking about this, I will have to go back to these discussions to remind myself.

That would be great if we could obtain a good understanding of the internal hom in codifferential coalgebras and/or differential algebras for the relavent cases.

By the way: the fact that, as you told me, codiff. coalgebras are closed, why diff. algebras are not I assume is due to finiteness issues?

Jim Stasheff kindly points me to

M. Mostow, The differential space structrures of Milnor classifying spaces, simplicial complexes, and geometric realizations

which looks like it is very relevant. But I haven’t yet found the time to really study it

For issues of cofree coalgebras, the manual to read might be:
MR2035107 (2005b:16070) Michaelis, Walter Coassociative coalgebras. Handbook of algebra, Vol. 3, 587–788, North-Holland, Amsterdam, 2003. (Reviewer: E. J. Taft) 16W30 (00A20)

I have a pdf copy if anyone is interested.
Both the graded and ungraded cases are considered and quite different or, rather,
if the vector space in degree 0 is not the ground field, watch out.

Caveat: The author unfortunately DEFINES tensor coagebra to be the cofree thing, which in the general case is NOT the tensor space with the deconcantenation diagonal.

Urs,

A couple of weeks ago (eons in Café-blogging time!), I promised you I’d say a little more about internal homs of cocommutative coalgebras. Now that you’re returning your attention to some related issues (e.g., possible internal homs in $DGCA^{op}$, as for example here), I thought I’d make good on that promise.

To review: back here, I gave a construction for internal homs of (dg) cocommutative coalgebras (the category of which I had, regrettably, denoted $CoDGA$ – I’ll use $Coc$ instead). This depended crucially on cofree cocommutative coalgebras, which are a little trickier to construct than one might think at first. I began to say something about that here, but then there were various interruptions. So now let me pick up where I left off.

The upshot of my previous comment was that general cocommutative coalgebras form a category dual not to commutative algebras, but rather to the category of commutative algebras of pro-finite-type (to adopt the terminology suggested by David Roberts). This rests on two observations:

• The category of finite-dimensional cocommutative coalgebras is dual to the category of finite-dimensional commutative algebras (by taking linear duals, which is a contravariant equivalence).
• A general cococommutative coalgebra is a canonical filtered colimit (viz., the union) of its finite-dimensional subcoalgebras.

The linear dual is a contravariant functor

$$(-)^\prime: Coc \to DGCA$$

which takes colimits in $Coc$ to limits in $DGCA$. Hence, by the two observations above, each cocommutative coalgebra is taken to a commutative algebra which is the cofiltered limit of its finite-dimensional quotients, that is to say, a pro-finite-type commutative algebra.

Pro-finite-type algebras may be regarded as having TVS (topological vector space) structures, as inverse limits of finite-dimensional spaces with their usual topologies, and the linear dual or transpose of a coalgebra map is a continuous algebra map. It is asserted by Getzler and Goerss that the linear dual

$$(-)^\prime: Coc \to ProfAlg_{cont}$$

is a contravariant equivalence; a quasi-inverse map going the other way is gotten by taking the continuous dual.

(Urs also surmised, and I was inclined to agree, that the algebras $\Omega^\bullet(X)$ were not themselves of profinite type. I seem to remember that the usual TVS structure on $\Omega^\bullet(X)$ is a structure of Montel space.)

According to Getzler and Goerss, the cofree cocommutative coalgebra generated by a finite-dimensional chain complex $V$ may be constructed as the unique coalgebra whose linear dual is the profinite symmetric algebra generated by the the linear dual $V^\prime$ (i.e., take the algebra of polynomial functions on $V$, and pass to the inverse limit of its finite-dimensional quotient algebras to get the profinite symmetric algebra). The cofree cocommutative coalgebra is therefore the continuous dual of the profinite completion of the symmetric algebra.

This description looks somewhat abstract and forbidding; it would be nice to have a more user-friendly description. A natural first instinct is that the cofree cocommutative coalgebra is just the product of $S_n$-invariants

$$C(V) = \prod_n (V^{\otimes n})^{S_n},$$

but recall that’s not right because this product isn’t a coalgebra in any obvious way (it would be correct if tensor products respected countable products, but they don’t). But, the cofree coalgebra is embedded inside this product, and can be considered the best coalgebra approximation to this guess: it’s the union of all finite-dimensional coalgebras “naturally embedded” inside $C(V)$.

By “naturally embedded”, I mean this. The symmetric algebra $S(V^\prime)$ is dense in its profinite-type completion $S_{prof}(V^\prime)$, so a continuous algebra map $S_{prof}(V^\prime) \to \mathbb{R}$ is uniquely determined by its restriction to $S(V^\prime)$. Now the space of linear functionals $S(V^\prime) \to \mathbb{R}$ is $C(V)$ if $V$ is finite-dimensional. So the cofree coalgebra must be embedded in the ordinary linear dual $S(V^\prime)^\prime = C(V)$. By a “natural embedding” of a finite-dimensional coalgebra $i: C \hookrightarrow C(V)$, I just mean the linear transpose of an algebra quotient $q: S(V^\prime) \to C^\prime$.

This still doesn’t seem particularly useful, but perhaps we can develop some intuition. We are taking the continuous linear dual of an algebra of polynomial functions (with a TVS structure specified by a uniformity defined by ideals of finite codimension). It is natural to think of the elements of the dual as measures of some sort, and (therefore) the cofree coalgebra as a coalgebra of measures. For example, if $V$ is 1-dimensional (and here I’m ignoring dg structure), we can think of $C(V) = \prod_n \mathbb{R}$ as the space of formal power series $\mathbb{R}[[x]]$, or (better) as a space of formal linear combinations

$$\sum_n \frac{a_n \delta_{0}^{(n)}}{n!}$$

where $\delta_0$ is the Dirac functional supported at $0$ and $\delta_{0}^{(n)}$ is its $n^{th}$ distributional derivative. We are trying to cut back on this space of measures supported at $0$ to get the cofree coalgebra, where the comultiplication $\Delta$ on the “good” measures $m$ should satisfy

$$\langle \Delta(m), f \otimes g \rangle = \langle m, f g \rangle.$$

(Urs, I’m going to stop here for the time being; this is to be continued, but I want/need to do some calculations first, and also think about what Jim Stasheff said here, which may be very useful. Also, Jim has mentioned the name Walter Michaelis a few times, who has apparently written a handbook on coalgebras with some information on cofree stuff – with some luck, I can get access to that.)

PS: I haven’t forgotten that you are contemplating internal homs for DCGA’s, based on passing back and forth between this category and the topos of presheaves on smooth manifolds. Hopefully I can think about what you are saying here.

Wait, wait! Did you quasi-internal hom use the algebra structure of the source? how?

For such interest as it may have, I’ll continue with the theme on cofree cocommutative coalgebras, as discussed last time here. This time I’ll calculate a basic example: working within the category of vector spaces over a field (say $\mathbb{C}$), I’ll compute the cofree cocommutative coalgebra $C$ generated by a 1-dimensional space. It’s simpler than I thought it would be!

(And though this may seem very much to be a ‘toy’ example, I believe it helps me see my way clear to the more general case, and also to the dg case; more on this later.)

As explained last time, the cofree coalgebra $C$ on one cogenerator $x$ is the best coalgebra approximation to the dual of the free algebra,

$$C \hookrightarrow \mathbb{C}[x]^\prime \cong \mathbb{C}[[x]].$$

More precisely, $C$ is formed as the union or filtered colimit (in $Vect$) of subspace inclusions

$$i = q^\prime: A^\prime \hookrightarrow \mathbb{C}[x]^\prime$$

that are linear transposes of projections to finite-dimensional quotient algebras $A$:

$$q: \mathbb{C}[x] \to \mathbb{C}[x]/I = A.$$

The $A^\prime$ here carry coalgebra structures dual to the algebra structures $A$, and since the underlying functor

$$Coc \to Vect$$

is easily seen to create colimits (even if we don’t yet assume the theorem that it is comonadic), we get an induced coalgebra structure $C$ on the filtered colimit as calculated in $Vect$. This $C$ is the cofree coalgebra.

By a kind of “calculus of symbols”, I will show that

• $C$ is naturally isomorphic to the localization of $\mathbb{C}[x]$ with respect to the ideal $(x)$

where the localization $\mathbb{C}[x]_{(x)}$ is naturally embedded in the local completion $\mathbb{C}[[x]]$. I will also compute the coalgebra structure on $\mathbb{C}[x]_{(x)}$.

First, let us recall the duality between $\mathbb{C}[x]$ and $\mathbb{C}[[x]]$. In one sense, it’s perfectly obvious: the polynomial algebra as a direct sum of copies of $\mathbb{C}$,

$$\mathbb{C}[x] \cong \sum_n \mathbb{C} \cdot x^n,$$

is dual to the direct product

$$\mathbb{C}[[x]] \cong \prod_n \mathbb{C} \cdot x^n,$$

where we view elements of the latter as formal power series

$$\sum_{n = 0}^{\infty} a_n x^n.$$

In this set-up, we think of the linear basis $x^n$ of $\mathbb{C}[x]$ as dual to the TVS basis $x^n$ of $\mathbb{C}[[x]]$:

$$\langle x^n, x^m \rangle = \delta_{m, n},$$

although it may be wiser to interpret formal power series $\sum_n a_n x^n$ as symbols for measures which are functionals on the polynomial algebra of functions. For example, $x^0$ would be the symbol for the Dirac functional

$$\delta_0 = eval_0: \mathbb{C}[x] \to \mathbb{C}$$

and $x^1$ would be the symbol for the functional which assigns to each polynomial $f$ its linear coefficient. This last example can also be interpreted in terms of the derivative of the Dirac functional, where we have

$$\langle f^\prime, \delta_0 \rangle = - \langle f, (\delta_0)^\prime \rangle$$

(the annoying minus sign is an artifact of a “integration by parts” heuristic often used in the theory of distributions).

Following this further, $x^m$ can be interpreted as the symbol for the measure or functional

$$x_m: \mathbb{C}[x] \to \mathbb{C}$$

which returns the coefficient of $x^m$ of a polynomial $f \in \mathbb{C}[x]$; we can also write it in terms of the $m^{th}$ derivative of the Dirac functional:

$$x_m = (-1)^m \frac{\delta_{0}^{(m)}}{m!}.$$

Now let us consider a typical element

$$\xi = \sum_m a_m x_m: \mathbb{C}[x] \to \mathbb{C}$$

of the cofree coalgebra $C$. By our description of $C$, it belongs to the dual of a finite-dimensional algebra quotient $q: \mathbb{C}[x] \to \mathbb{C}[x]/I$; therefore it vanishes on the ideal $I$, generated by a polynomial $g(x) = b_0 + \ldots + b_n x^n$. This ideal $I$ is the span of elements $x^j g(x)$; therefore

$$\langle x^j g(x), \sum_m a_m x_m \rangle = 0$$

which after a short calculation yields linear recurrence relations, one for each $j$:

$$a_j b_0 + a_{j+1} b_1 + \ldots + a_{j+n} b_n = 0.$$

This enables us to recursively solve for all the $a_m$ once we know initial values $a_0, \ldots, a_{n-1}$.

It may be illuminating to work through a simple example. Let $g(x) = 1 - x^2$. The linear recurrence relations say

$$a_j - a_{j+2} = 0$$

for all $j$. Setting $a_0 = 1$ and $a_1 = 1$, we have $1 = a_0 = a_2 = \ldots$ and $0 = a_1 = a_3 = \ldots$, so the corresponding functional becomes in this case

$$x_0 + x_2 + x_4 + \ldots$$

whose symbol is

$$x^0 + x^2 + x^4 + \ldots = \frac{1}{1-x^2}.$$

If we set $a_0 = 0$ and $a_1 = 1$, we are similarly led to the operator

$$x_1 + x_3 + x_5 + \ldots$$

with symbol

$$x^1 + x^3 + x^5 + \ldots = \frac{x}{1-x^2}.$$

Somehow this reminds me a bit of fundamental solutions or Green’s functions for differential operators. The ‘$x$’ is something like the Fourier transform of ‘taking the derivative at 0’, up to some possible finagling with signs, and the expression $\frac{x}{1-x^2}$ is something like the formal Fourier transform of a fundamental solution $u$ to $(1 - D^2)(u) = \delta_{0}^{\prime}$. I’m not claiming too much should be made of this connection – this sort of thing is not exactly my métier in the first place. I just find it sort of suggestive.

A little further thought shows this is no coincidence: the linear recurrence relations required to compute the dual of $\mathbb{C}[x]/(g(x))$ as embedded in $\mathbb{C}[[x]]$ are the same as the linear recurrence realtions required to compute $g(x)^{-1}$. Thus, the coalgebra dual to $\mathbb{C}[x]/(x^j g(x))$, where $g(x)$ is an $n^{th}$ degree polynomial with nonzero constant coefficient, is the coalgebra of symbols spanned by formal power series expressions which represent

$$\frac{1}{g(x)}, \frac{x}{g(x)}, \ldots, \frac{x^{n+j-1}}{g(x)}.$$

Since every $\xi$ belonging to the cofree coalgebra $C$ is contained in the dual of some finite-dimensional algebra of the form $\mathbb{C}[x]/(x^j g(x))$ where $g(0) \neq 0$, it follows that its symbol belongs to $\mathbb{C}[x] \cdot \frac{1}{g(x)}$. Hence the space of symbols for the cofree coalgebra is the localization of $\mathbb{R}[x]$ obtained by inverting all polynomials $g(x)$ with nonzero constant coefficient. That is, we may identify $C$ with the localization $\mathbb{C}[x]_{(x)}$.

Finally, a few words on the coalgebra structure on $\mathbb{C}[x]_{(x)}$. The comultiplication $\Delta$ is adjoint to multiplication on the algebra $\mathbb{C}[x]$, and so we have

$$\langle x^m \cdot x^n, \xi \rangle = \langle x^m \otimes x^n, \Delta(\xi) \rangle.$$

For $\xi = \sum_n a_n x_n$, this forces the familiar rule

$$\Delta(\xi) = \sum_n a_n (\sum_{i+j = n} x_i \otimes x_j)$$

(although this doesn’t give a finite linear combination, i.e., an element of $\mathbb{C}[[x]] \otimes \mathbb{C}[[x]]$, unless the $a_n$ satisfy linear recurrence relations coming from the fact that $\xi$ belongs to $\mathbb{C}[x]_{(x)}$).

The only reasonable explicit formulas I have so far for $\Delta$ exploit the theory of partial fraction decompositions. If $g(x)$ is a nonconstant polynomial with nonzero constant coefficient, then (working over $\mathbb{C}$) there is a unique finite decomposition of the form

$$\frac{1}{g(x)} = \sum_i \frac{A_i}{(1 - r_i x)^{n_i}}$$

where the $A_i$ are constants and the pairs $(r_i, n_i)$ are distinct. So it’s enough to determine $\Delta((1 - r x)^{-n})$. For $n = 1$, this is easy: a short calculation yields

$$\Delta(\frac{1}{1 - r x}) = \frac{1}{1 - r x} \otimes \frac{1}{1 - r x}$$

so that $(1 - r x)^{-1}$ is a grouplike element. For $n$ > $1$, the formula is more complicated; I seem to get

$$\Delta(\frac{1}{(1 - r x)^{n+1}}) = \sum_{i + j = n} \frac{1}{(1 - r x)^{i+1}} \otimes \frac{1}{(1 - r x)^{j+1}} - \sum_{i + j = n-1} \frac{1}{(1 - r x)^{i+1}} \otimes \frac{1}{(1 - r x)^{j+1}},$$

although I don’t guarantee a mistake hasn’t slipped in somewhere!

I’d like to continue with my monologue on cofree cocommutative coalgebras, on the off-chance that it might actually turn out to be useful one of these days, and also because these days I’m enjoying contemplating them.

One reason I’m enjoying them is that now they don’t seem nearly so scary as this description (essentially what I read in Getzler and Goerss) might suggest:

The cofree cocommutative coalgebra $Cof(V)$ cogenerated by a finite-dimensional vector space $V$ is the continuous linear dual of the pro-completion of the free commutative algebra on the linear dual $V^\prime$, where the pro-completion of an algebra is formed by taking the inverse limit of its finite-dimensional quotient algebras (with their standard topologies).

Fine, as abstract descriptions go, unless you want to do calculations with the darned things – then it becomes a headache! The point of this comment is to make the cofree construction much more transparent and amenable to calculation.

The cofree cocommutative coalgebra on one cogenerator.

In the case where $V$ is 1-dimensional, I indicated last time how $Cof(V)$ could be described as a certain coalgebra structure on the localization of $\mathbb{C}[x]$ with respect to the prime ideal $(x)$. Let me now make this a little more explicit. In the first place, by “cofreeness” we mean there is a universal arrow

$$\pi: Cof(V) \to V$$

which is universal among linear maps $A \to V$ out of coalgebras $A$. In the 1-dimensional case $V = \mathbb{C}$, it is the composite

$$\mathbb{C}[x]_{(x)} \stackrel{i}{\hookrightarrow} \prod_{n} \mathbb{C} \cdot x^n \stackrel{\pi}{\to} \mathbb{C} \cdot x^1 = \mathbb{C}$$

where the first arrow interprets rational functions $\frac{p(x)}{g(x)}$ in $\mathbb{C}[x]_{(x)}$ as formal power series, and the second projects power series onto their linear coefficients.

Now suppose given a linear functional $f: A \to \mathbb{C}$ on a coalgebra $A$. We are trying to lift this functional to a coalgebra map $\hat{f}: A \to Cof(\mathbb{C})$. In the first place, there is this Very Useful Observation^TM that every coalgebra $A$ is the filtered colimit (union) of its finite-dimensional subcoalgebras, and that the forgetful functor from coalgebras to vector spaces preserves colimits. This implies that once we know how to obtain the lift $\hat{f}$ in the case where $A$ is finite-dimensional, then we can do it in general.

So suppose that $A$ is finite-dimensional. The functional $f: A \to \mathbb{C}$ can be interpreted as an element of the linear dual $A^\prime$, with its induced algebra structure. Consider the algebra map $\mathbb{C}[x] \to A^\prime$ which sends $x$ to $f$. The kernel of this map is an ideal $I$ of finite codimension, and the linear transpose of this map,

$$A \to \mathbb{C}[[x]],$$

factors as the coalgebra map $A \to (\mathbb{C}[x]/I)^\prime$ followed by the mono $j: (\mathbb{C}[x]/I)^\prime \to (\mathbb{C}[x])^\prime \cong \mathbb{C}[[x]]$ which is the transpose of the projection $q: \mathbb{C}[x] \to \mathbb{C}[x]/I$. Now the analysis given last time showed that every such $j$ factors through a mono $(\mathbb{C}[x]/I)^\prime \to \mathbb{C}[x]_{(x)}$ which is a coalgebra map. The composite coalgebra map

$$A \to (\mathbb{C}[x]/I)^\prime \to \mathbb{C}[x]_{(x)}$$

then gives the desired lift $\hat{f}: A \to Cof(\mathbb{C})$ of $f: A \to \mathbb{C}$.

To connect this with the description given by Getzler and Goerss: the coalgebra $\mathbb{C}[x]_{(x)}$ is uniquely determined up to isomorphism as the coalgebra whose linear dual is the algebra obtained by profinite-type completion, i.e.,

$$hom_{Vect}(\mathbb{C}[x]_{(x)}, \mathbb{C}) \cong \lim_I \mathbb{C}[x]/I$$

where the limit is over nonzero ideals $I$. In other words, by taking duals, the localization $\mathbb{C}[x]_{(x)}$ (which, like every ring localization, is a certain filtered colimit) is taken over to the cofiltered limit taken over the system of nonzero ideals (cofiltered, since we can take the intersection of ideals). In some sense which is not yet quite clear to me, this isomorphism feels very closely connected to the theory of partial fraction decompositions.

Aside on partial fraction decompositions.

Partial fraction decomposition is something we teach calculus students as part of their repetoire in computing antiderivatives (in this case of rational functions), and it’s something I’ve always enjoyed thinking about from first principles – there’s some real algebra there! I’m not sure whether that topic is standard in a graduate algebra course; I don’t think it is, even though I never took such a course. So maybe it won’t be taken amiss if I give my own take on this.

Let $O$ be a principal ideal domain, and let $K$ be its field of fractions. Partial fraction decomposition really has to do with the $O$-module structure of $K/O$:

Theorem (partial fraction decomposition): The canonical map

$$\sum_p O[\frac{1}{p}]/O \to K/O,$$

where the sum ranges over generators of prime ideals $p$, is an isomorphism.

The standard example we teach our calculus students is the case where $O = \mathbb{R}[x]$, where $K = \mathbb{R}(x)$ is the field of rational functions. It says that any rational function $f(x)/g(x)$, considered modulo polynomials, has a unique expression as a linear combination of quotients $a_i(x)/p_i(x)^{r_i}$ where the $p_i$ are irreducible factors of $g(x)$. Of course, this is the function field case; it applies equally well to the structure of say $\mathbb{Q}/\mathbb{Z}$ – whenever I have taught calculus, this is the first case I would discuss before embarking on the application to rational functions (on the assumption that students are more comfortable with integers than with polynomials).

There are some funny little spin-offs of this theorem. In the case $O = \mathbb{Z}$, the summand $\mathbb{Z}[\frac{1}{p}]/\mathbb{Z}$ is an inductive limit (a colimit) along the evident sequence

$$\ldots \hookrightarrow \mathbb{Z}/p^j \hookrightarrow \mathbb{Z}/p^{j+1} \hookrightarrow \ldots$$

of $p$-torsion groups. The dual of this torsion group,

$$hom(\mathbb{Z}[\frac{1}{p}]/\mathbb{Z}, \mathbb{Q}/\mathbb{Z}),$$

is the projective limit of the sequence of quotient maps $\mathbb{Z}/p^{j+1} \to \mathbb{Z}/p^j$ (using the fact that finite abelian groups are self-dual), and this gives a standard presentation of the $p$-adic integers, denoted here as $\hat{\mathbb{Z}}_p$. By partial fraction decomposition, we therefore have

$$hom(\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \cong hom(\sum_p \mathbb{Z}[\frac{1}{p}]/\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \cong \prod_p \hat{\mathbb{Z}}_p.$$

This looks like the maximal compact subring of the (non-archimedean part of the) adeles, and sure enough, multiplication in this ring matches composition of endomorphisms on $\mathbb{Q}/\mathbb{Z}$! Continuing this thought, one can also identify $hom(\mathbb{Q}, \mathbb{Q}/\mathbb{Z})$ with the non-archimedean part of the adeles. I somehow think something interesting could be made of this, but I’m not sure what.

The cofree construction $Cof(V)$ for finite-dimensional $V$.

Now that we have in hand a concrete description of $Cof(V)$ when $V$ is 1-dimensional, we can easily get $Cof(V)$ for any finite-dimensional $V$. This rests on three observations:

• The cofree construction $Cof: Vect \to Coc$ is a right adjoint, and hence preserves products;
• Any finite-dimensional vector space $V$ is a product of 1-dimensional spaces;
• The product of cocommutative coalgebras is given by the tensor product of underlying vector spaces.

So, the cofree cocommutative coalgebra $Cof(\mathbb{C}^n)$ on $n$ cogenerators is an $n$-fold tensor product of copies of $\mathbb{C}[x]_{(x)}$. Since we have

$$\mathbb{C}[x]_{(x)} \cong colim_{g: g(0) \neq 0} \mathbb{C}[x] \cdot \frac{1}{g(x)}$$

where the colimit is along inclusions $\mathbb{C}[x] \cdot \frac{1}{g(x)} \hookrightarrow \mathbb{C}[x] \cdot \frac{1}{h(x)}$ whenever $g$ divides $h$, and since the tensor product of $n$ copies of $\mathbb{C}[x]$ is isomorphic to $\mathbb{C}[x_1, \ldots, x_n]$, we quickly conclude that $Cof(\mathbb{C}^n)$ is the localization of $\mathbb{C}[x_1, \ldots, x_n]$ where we invert products of the form $p_1(x_1)p_2(x_2) \ldots p_n(x_n)$ ($p_i(0) \neq 0$ for all $i$). The coalgebra structure is easily deduced from the way it works in the 1-cogenerator case.

The cofree construction $Cof(V)$ for general $V$.

The Very Useful Observation^TM (that coalgebras are filtered colimits of their finite-dimensional subcoalgebras, preserved and reflected by taking underlying vector spaces) gives a quick description of $Cof(V)$ for general $V$, in terms of $Cof(V^\prime)$ for finite-dimensional $V^\prime$:

$$Cof(V) := colim_{V^\prime \hookrightarrow V} Cof(V^\prime)$$

where the colimit is along inclusions between finite-dimensional subspaces $V^\prime$. For with this definition, we indeed get a universal arrow $Cof(V) \to V$ from this colimit to $V$, induced by the maps

$$Cof(V^\prime) \stackrel{\pi}{\to} V^\prime \hookrightarrow V.$$

Moreover, given a linear map $f: A \to V$ where $A$ is a coalgebra, we get for each finite-dimensional subcoalgebra $A^\prime \hookrightarrow A$ a linear map $A^\prime \to f(A^\prime)$ to a finite-dimensional space, and thence a coalgebra map $A^\prime \to Cof(f(A^\prime))$. Hence we get a coalgebra map

$$A^\prime \to Cof(f(A^\prime)) \hookrightarrow Cof(V)$$

with the definition of $Cof(V)$ above, and then, since $A$ is a colimit of the $A^\prime$, we get an induced coalgebra map $\hat{f}: A \to Cof(V)$ which lifts $f$, as desired.

Next time: I’ll carry the discussion over to the dg world (where actually some things simplify!). Over time, I’d like to work my way toward the explicit calculation of internal homs of dg cocommutative coalgebras.