nLab
cofree coalgebra

Definition

Given a monoidal category $M$, let $\mathrm{Coalg}(M)$ denote the category of coalgebras (i.e., comonoids) with respect to the tensor product of $M$. Let $U:\mathrm{Coalg}(M)\to M$ be the forgetful functor. The cofree coalgebra refers to a functor right adjoint to $U$, if it exists.

Examples

For example, a cofree coalgebra on a vector space $V$ consists of a coalgebra $C(V)$ and a linear map $\epsilon :C(V)\to V$ which is universal in the sense that given a coalgebra $C$ and linear map $f:C\to V$, there exists a unique coalgebra map $g:C\to C(V)$ such that $f=\epsilon \circ g$.

Constructions

The forgetful functor $U:\mathrm{Coalg}(M)\to M$ preserves and reflects (creates) any colimits which happen to exist, analogous to the fact that the forgetful functor $\mathrm{Alg}(M)\to M$ preserves and reflects limits. Therefore, under hypotheses of an adjoint functor theorem (which almost always obtain in cases of practical interest), $U$ will indeed have a right adjoint.

Modules over a commutative ring

The special adjoint functor theorem (SAFT) may be applied to show that for any commutative ring $K$, the forgetful functor

$$K\text{-Coalg}\to K\text{-Mod}$$K\text{-Coalg} \to K\text{-Mod}

has a right adjoint. An argument is given by Michael Barr in

• Coalgebras over a Commutative Ring; see theorem 4.1.

(This is somewhat at odds with an assertion made by Michiel Hazewinkel in Witt Vectors, Part I, section 12.11 (page 58): “Whether the cofree coalgebra over an Abelian group always exists is unknown.” But it seems to me (Todd Trimble) that there is nothing wrong with Barr’s argument.)

Cofree coalgebra on a vector space

In what follows, we give an explicit construction of the cofree coalgebra cogenerated by a vector space.

As a first step, recall the following result.

Fundamental theorem of coalgebras

Every coalgebra is the filtered colimit over the diagram of its finite-dimensional subcoalgebras and inclusions between them.

For a proof of this result, see for example Michaelis. We repeat that colimits of coalgebras are created by taking colimits of their underlying vector spaces.

Let for a vector space $V$, let $T(V)$ be the tensor algebra (or free algebra) generated by $V$. We may consider $T(V^{*})$ as the algebra of (non-commutative) polynomial functions on $V$. The dual of $T(V^{*})$ is not a coalgebra, but in accordance with the fundamental theorem above, we may consider instead the colimit over a system of finite-dimensional coalgebras obtained by dualizing the system of finite-dimensional algebra quotients of $T(V^{*})$ and surjections between them (compare profinite completion), using the fact that the dual of a finite-dimensional algebra is a finite-dimensional coalgebra. Let $T(V^{*}{)}^{\circ }$ be this colimit coalgebra, embedded in $T(V^{*}{)}^{*}$. The dual of the inclusion $V^{*}\hookrightarrow T(V^{*})$ gives a map $T(V^{*}{)}^{*}\to V^{**}$, and this restricts to a map $T(V^{*}{)}^{\circ }\to V^{**}$.

For finite-dimensional $V$, there is an isomorphism $V^{**}\cong V$, and we have the following result.

Lemma

If $V$ is finite-dimensional, then the composite $T(V^{*}{)}^{\circ }\to V^{**}\cong V$ exhibits $T(V^{*}{)}^{\circ }$ as the cofree coalgebra on $V$.

Proof

Suppose given a coalgebra $C$ and a linear map $f:C\to V$. We want to show this has a unique lift to a coalgebra map $\hat{f}:C\to T(V^{*}{)}^{\circ }$. For $x\in C$, there is a minimal finite-dimensional subcoalgebra $C_x$ containing $x$, and if $f_x:C_x\to T(V^{*}{)}^{\circ }$ is the restriction of $f$, then the value $\hat{f}(x)$ must match the value $\widehat{f_x}(x)$. In other words, in view of the fundamental theorem of coalgebras, there is no loss of generality in taking $C$ to be finite-dimensional.

In this case, the map $f^{*}:V^{*}\to C^{*}$ extends uniquely to an algebra map $T(V^{*})\to C^{*}$, since the tensor algebra is the free algebra on $V$. This algebra map factors as

$$T(V^{*})\stackrel{\mathrm{onto}}{\to }Q\stackrel{\mathrm{into}}{\to }{C}^{*}$$T(V^\ast) \stackrel{onto}{\to} Q \stackrel{into}{\to} C^\ast

where $Q$ is a finite-dimensional algebra. The dual $Q^{*}$ carries a coalgebra structure that embeds in $T(V^{*}{)}^{\circ }$, and by dualizing we obtain coalgebra maps

$$C\to Q^{*}\to T(V^{*}{)}^{\circ }$$C \to Q^\ast \to T(V^\ast)^\circ

that provide the desired lift. The uniqueness of the lift is evident from the uniqueness of the extension $T(V^{*})\to C$ (or equivalently, of $T(V^{*}{)}_{\mathrm{prof}}\to C$, passing to the profinite completion) on the dual algebra side.

Now let $V$ be an infinite-dimensional vector space. We may view $V$ as the union or filtered colimit over the system $\{V_{\alpha }\}$ of its finite-dimensional subspaces. Applying the cofree coalgebra construction to these finite-dimensional subspaces, we get a system of coalgebras.

Theorem

The coalgebra colimit

$$K={\mathrm{colim}}_{V_{\alpha }\subset V}C(V_{\alpha })$$K = colim_{V_\alpha \subset V} C(V_\alpha)

is the cofree coalgebra on $V$.

• Remark: Happily, this colimit is computed as in $\mathrm{Vect}$, and more happily still – since the underlying functor $\mathrm{Vect}\to \mathrm{Set}$ also preserves and creates filtered colimits – this colimit is just a set-theoretic union of the $C(V_{\alpha })$ along the canonical inclusions between them.

Indeed, suppose given a coalgebra $C$ and a linear function $f:C\to V$. To construct a coalgebra map $C\to K$ which lifts $f$, we may argue just as we did in the proof of the lemma, and suppose without loss of generality that $C$ is finite-dimensional. Then of course the map $f:C\to V$ factors through a map $f_{\alpha }:C\to V_{\alpha }$ where $V_{\alpha }$ is a finite-dimensional subspace of $V$. This lifts uniquely to a coalgebra map $\widehat{f_{\alpha }}:C(V_{\alpha })$, and the desired lift $\hat{f}$ is the composite

$$C\stackrel{\widehat{f_{\alpha }}}{\to }C(V_{\alpha })\hookrightarrow K$$C \stackrel{\widehat{f_\alpha}}{\to} C(V_\alpha) \hookrightarrow K

Any such lift is obtained in just this way.

It is easy to see that this description is equivalent to that described in a MathOverflow discussion here. See also the nForum discussion here.

The cofree coalgebra on a 1-dimensional space

Specializing still further, we investigate the detailed structure of the cofree coalgebra on a 1-dimensional vector space (over a field) $k$.

Here, the cofree coalgebra $T(k{)}^{\circ }$ is the filtered colimit of finite-dimensional coalgebras of the form

$$(k[x]/I{)}^{*}$$(k[x]/I)^\ast

where $k[x]/I$ is a (finite-dimensional) quotient of the polynomial algebra $k[x]$ modulo a non-zero ideal $I$. The adjoint of the quotient map $k[x]\to k[x]/I$ is an embedding $(k[x]/I{)}^{*}\hookrightarrow k[x{]}^{*}\cong {\prod }_{n}k\cdot x^n$ into the space of formal power series in $x$. As a set, $T(k{)}^{\circ }$ is simply the union of the subspaces $(k[x]/I{)}^{*}\hookrightarrow {\prod }_{n}k\cdot x^n$. In other words, every $\xi \in T(k{)}^{\circ }$ belongs to some $(k[x]/I{)}^{*}$, and it suffices to consider first the structure of a typical such $\xi$ as a formal power series, and then how the comultiplication behaves on it.

Now $I$ is a principal ideal $(g(x))$ for some polynomial $g(x)=b_0+\dots +b_n{x}^n$, where the top coefficient is of course assumed non-zero. We are trying to understand when a formal power series $\xi ={\sum }_{n\ge 0}{a}_n{x}^n\in {\prod }_{n}k\cdot x^n$ vanishes on $I$, in the sense that

$$⟨p(x),\sum _n{a}_n{x}^n⟩=0$$\langle p(x), \sum_n a_n x^n \rangle = 0

for all $p(x)\in (g(x))$, where the pairing $⟨,⟩:k[x]\times {\prod }_{n}k\cdot x^n\to k$ is the canonical one:

$$⟨x^j,\sum _n{a}_n{x}^n⟩=a_j.$$\langle x^j, \sum_n a_n x^n \rangle = a_j.

Taking the case $p(x)=x^{j}g(x)$, we obtain recurrence relations

$$⟨b_0{x}^j+\dots b_n{x}^{n+j},\sum _n{a}_n{x}^n⟩=b_0{a}_j+b_1{a}_{j+1}+\dots +b_n{a}_{n+j}=0,$$\langle b_0 x^j + \ldots b_n x^{n+j}, \sum_n a_n x^n \rangle = b_0 a_j + b_1 a_{j+1} + \ldots + b_n a_{n+j} = 0,

one for every $j$. We may pick $a_0,a_1,\dots ,a_{n-1}$ at random, but from there on the remainder of the sequence $a_n,a_{n+1},\dots$ is determined by the recurrence relations.

Now let $g^{\mathrm{rev}}(x)$ be the polynomial obtained by reversing the order of the coefficients $b_0,\dots ,b_n$ of $g$,

$$g^{\mathrm{rev}}(x)=b_n+\dots +b_0{x}^n,$$g^{rev}(x) = b_n + \ldots + b_0 x^n,

and consider the problem of multiplicatively inverting $g^{\mathrm{rev}}(x)$ (which we may do since the constant coefficient of $g^{\mathrm{rev}}$ is non-zero). If ${\sum }_j{a}_j{x}^j$ is the reciprocal, we have

$$1=g^{\mathrm{rev}}(x)\sum _j{a}_j{x}^j=(b_n+\dots +b_0{x}^n)(\sum _j{a}_j{x}^j)$$1 = g^{rev}(x)\sum_j a_j x^j = (b_n + \ldots + b_0 x^n)(\sum_j a_j x^j)

which yields a second set of recurrence relations by examining coefficients of powers of $x$. We get the first set of recurrence relations by restricting attention to the coefficients of $x^n$, $x^{n+1}$, etc.

Thus the power series of $g^{\mathrm{rev}}(x{)}^{-1}$ yields a $\xi$ which vanishes on $I=(g(x))$, and similarly the power series of

$$\frac{x}{g^{\mathrm{rev}}(x)},\dots ,\frac{x^{n-1}}{g^{\mathrm{rev}}(x)}$$\frac{x}{g^{rev}(x)}, \ldots, \frac{x^{n-1}}{g^{rev}(x)}

also vanish on $I$. They in fact form a basis for the subspace $(k[x]/(g(x)){)}^{*}\hookrightarrow {\prod }_{n}k\cdot x^n$.

We conclude that an arbitrary element of $T(k{)}^{\circ }$ is of the form $\frac{p(x)}{h(x)}$ for some $h$ with non-zero constant coefficient:

Theorem

As subspaces of the space of formal power series ${\prod }_{n}k\cdot x^n$, there is an identification

$$T(k{)}^{\circ }=k[x{]}_{(x)}$$T(k)^\circ = k[x]_{(x)}

where the right side is the localization of $k[x]$ at the prime ideal $(x)$, whose elements are considered as formal power series.

It remains to identify the coalgebra structure on $k[x{]}_{(x)}$. This extends the standard coalgebra structure on the tensor algebra $k[x]$, where the comultiplication is given by deconcatenation:

$$\delta (x^N)=\sum _{m+n=N}{x}^m\otimes x^n.$$\delta(x^N) = \sum_{m+n = N} x^m \otimes x^n.

We may as well focus on a typical subcoalgebra $(k[x]/(g(x)){)}^{*}$, with basis elements $\frac{x^i}{g^{\mathrm{rev}}(x)}$. By duality, the definition of $\delta (\frac{x^i}{g^{\mathrm{rev}}(x)})$ can be extracted from the multiplication table for residue classes $x^j(modg)(x)$.

The calculations work out cleanest if we assume if $k$ is algebraically closed. For in that case, we can take advantage of partial fraction decompositions of the form

$$\frac{x^i}{g^{\mathrm{rev}}(x)}=\sum _k\frac{A_k}{(1-r_{k}x{)}^{n_k}}$$\frac{x^i}{g^{rev}(x)} = \sum_k \frac{A_k}{(1- r_k x)^{n_k}}

so that it suffices to give $\delta (\frac{1}{(1-rx{)}^n})$, although the algebra works out better with a slightly different basis.

With this in mind, put $e_n(r)=\frac{(rx{)}^n}{(1-rx{)}^{n+1}}$ for $n\ge 0$. The power series expansion is

$$e_n(r)=\sum _{k\ge 0}\left(\genfrac{}{}{0ex}{}{k}{n}\right)r^k{x}^k$$e_n(r) = \sum_{k \geq 0} \binom{k}{n} r^k x^k

and we formally calculate

$$\begin{array}{ccc}\delta (e_n(r))& =& \sum _k\left(\genfrac{}{}{0ex}{}{k}{n}\right)r^k\delta (x^k)\\ & =& \sum _k\left(\genfrac{}{}{0ex}{}{k}{n}\right)r^k\sum _{p+q=k}{x}^p\otimes x^q\\ & =& \sum _{p,q\ge 0}\left(\genfrac{}{}{0ex}{}{p+q}{n}\right)r^p{x}^p\otimes r^q{x}^q\\ & =& \sum _{p,q}\sum _{i+j=n}\left(\genfrac{}{}{0ex}{}{p}{i}\right)\left(\genfrac{}{}{0ex}{}{q}{j}\right)r^p{x}^p\otimes r^q{x}^q\\ & =& \sum _{i+j=n}(\sum _p\left(\genfrac{}{}{0ex}{}{p}{i}\right)r^p{x}^p)\otimes (\sum _q\left(\genfrac{}{}{0ex}{}{q}{j}\right)r^q{x}^q)\\ & =& \sum _{i+j=n}{e}_i(r)\otimes e_j(r)\end{array}$$\array{ \delta(e_n(r)) & = & \sum_k \binom{k}{n} r^k \delta(x^k) \\ & = & \sum_k \binom{k}{n} r^k \sum_{p+q=k} x^p \otimes x^q \\ & = & \sum_{p, q \geq 0} \binom{p+q}{n} r^p x^p \otimes r^q x^q \\ & = & \sum_{p, q} \sum_{i+j=n} \binom{p}{i} \binom{q}{j} r^p x^p \otimes r^q x^q \\ & = & \sum_{i+j=n} (\sum_p \binom{p}{i} r^p x^p) \otimes (\sum_q \binom{q}{j} r^q x^q) \\ & = & \sum_{i+j=n} e_i(r) \otimes e_j(r) }

which exhibits comultiplication as deconcatenation, familiar from the usual coalgebra structure on $k[x]$. In particular, we have $\delta (e_0(r))=e_0(r)\otimes e_0(r)$: the elements $e_0(r)$ are grouplike.

A more respectable-looking calculation (but really with the exact same content!) shows that the comultiplication thus defined on $k[x{]}_{(x)}$ is indeed adjoint to multiplication on $k[x]$:

$$\begin{array}{ccc}⟨x^p{x}^q,e_n(r)⟩& =& ⟨x^{p+q},\sum _k\left(\genfrac{}{}{0ex}{}{k}{n}\right)r^k{x}^k⟩\\ & =& \left(\genfrac{}{}{0ex}{}{p+q}{n}\right)r^{p+q}\\ & =& \sum _{i+j=n}\left(\genfrac{}{}{0ex}{}{p}{i}\right)\left(\genfrac{}{}{0ex}{}{q}{j}\right)r^p{r}^q\\ & =& \sum _{i+j=n}⟨x^p,e_i(r)⟩⟨x^q,e_j(r)⟩\\ & =& \sum _{i+j=n}⟨x^p\otimes x^q,e_i(r)\otimes e_j(r)⟩\\ & =& ⟨x^p\otimes x^q,\delta (e_n(r))⟩.\end{array}$$\array{ \langle x^p x^q, e_n(r) \rangle & = & \langle x^{p+q}, \sum_k \binom{k}{n} r^k x^k \rangle \\ & = & \binom{p+q}{n} r^{p+q} \\ & = & \sum_{i+j=n} \binom{p}{i} \binom{q}{j} r^p r^q \\ & = & \sum_{i+j=n} \langle x^p, e_i(r) \rangle \langle x^q, e_j(r) \rangle \\ & = & \sum_{i+j=n} \langle x^p \otimes x^q, e_i(r) \otimes e_j(r) \rangle \\ & = & \langle x^p \otimes x^q, \delta(e_n(r)) \rangle. }

An analogous calculation shows that the counit $\epsilon :k[x{]}_{(x)}\to k$ is defined by $\epsilon (e_n(r))={\delta }_{0n}$ (Kronecker delta).

We summarize the calculations above as follows.

Theorem

Let $k$ be an algebraically closed field. For each nonzero scalar $r\in k$, let $L(r)$ denote the $k$-linear span of the elements $e_n(r)=\frac{(rx{)}^n}{(1-rx{)}^{n+1}}$ in $k[x{]}_{(x)}$, with coalgebra structure given in each case by

$$\delta (e_n(r))=\sum _{i+j=n}{e}_i(r)\otimes e_j(r),\epsilon (e_n(r))={\delta }_{0n}$$\delta(e_n(r)) = \sum_{i+j=n} e_i(r) \otimes e_j(r), \qquad \varepsilon(e_n(r)) = \delta_{0 n}

Let $L(0)=k[x]$ as a subspace of $k[x{]}_{(x)}$. The cofree coalgebra over a 1-dimensional space $k$, identified with $k[x{]}_{(x)}$, decomposes as a direct sum of coalgebras

$$k[x{]}_{(x)}\cong \underset{r\in k}{⨁}L(r)$$k[x]_{(x)} \cong \bigoplus_{r \in k} L(r)

according to the partial fraction decomposition of rational functions. Each of the coalgebras $L(r)$ is isomorphic to $k[x]$ with its standard coalgebra structure.

Related concepts

• The dual concepts are the of free monoid, and tensor algebra,

References

Walter Michaelis, Coassociative Coalgebras, Handbook of Algebra Volume 3, Elsevier (2003).