\documentclass[12pt,titlepage]{article} \usepackage{amsmath} \usepackage{mathrsfs} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsthm} \usepackage{mathtools} \usepackage{graphicx} \usepackage{color} \usepackage{ucs} \usepackage[utf8x]{inputenc} \usepackage{xparse} \usepackage{hyperref} %----Macros---------- % % Unresolved issues: % % \righttoleftarrow % \lefttorightarrow % % \color{} with HTML colorspec % \bgcolor % \array with options (without options, it's equivalent to the matrix environment) % Of the standard HTML named colors, white, black, red, green, blue and yellow % are predefined in the color package. Here are the rest. \definecolor{aqua}{rgb}{0, 1.0, 1.0} \definecolor{fuschia}{rgb}{1.0, 0, 1.0} \definecolor{gray}{rgb}{0.502, 0.502, 0.502} \definecolor{lime}{rgb}{0, 1.0, 0} \definecolor{maroon}{rgb}{0.502, 0, 0} \definecolor{navy}{rgb}{0, 0, 0.502} \definecolor{olive}{rgb}{0.502, 0.502, 0} \definecolor{purple}{rgb}{0.502, 0, 0.502} \definecolor{silver}{rgb}{0.753, 0.753, 0.753} \definecolor{teal}{rgb}{0, 0.502, 0.502} % Because of conflicts, \space and \mathop are converted to % \itexspace and \operatorname during preprocessing. % itex: \space{ht}{dp}{wd} % % Height and baseline depth measurements are in units of tenths of an ex while % the width is measured in tenths of an em. \makeatletter \newdimen\itex@wd% \newdimen\itex@dp% \newdimen\itex@thd% \def\itexspace#1#2#3{\itex@wd=#3em% \itex@wd=0.1\itex@wd% \itex@dp=#2ex% \itex@dp=0.1\itex@dp% \itex@thd=#1ex% \itex@thd=0.1\itex@thd% \advance\itex@thd\the\itex@dp% \makebox[\the\itex@wd]{\rule[-\the\itex@dp]{0cm}{\the\itex@thd}}} \makeatother % \tensor and \multiscript \makeatletter \newif\if@sup \newtoks\@sups \def\append@sup#1{\edef\act{\noexpand\@sups={\the\@sups #1}}\act}% \def\reset@sup{\@supfalse\@sups={}}% \def\mk@scripts#1#2{\if #2/ \if@sup ^{\the\@sups}\fi \else% \ifx #1_ \if@sup ^{\the\@sups}\reset@sup \fi {}_{#2}% \else \append@sup#2 \@suptrue \fi% \expandafter\mk@scripts\fi} \def\tensor#1#2{\reset@sup#1\mk@scripts#2_/} \def\multiscripts#1#2#3{\reset@sup{}\mk@scripts#1_/#2% \reset@sup\mk@scripts#3_/} \makeatother % \slash \makeatletter \newbox\slashbox \setbox\slashbox=\hbox{$/$} \def\itex@pslash#1{\setbox\@tempboxa=\hbox{$#1$} \@tempdima=0.5\wd\slashbox \advance\@tempdima 0.5\wd\@tempboxa \copy\slashbox \kern-\@tempdima \box\@tempboxa} \def\slash{\protect\itex@pslash} \makeatother % math-mode versions of \rlap, etc % from Alexander Perlis, "A complement to \smash, \llap, and lap" % http://math.arizona.edu/~aprl/publications/mathclap/ \def\clap#1{\hbox to 0pt{\hss#1\hss}} \def\mathllap{\mathpalette\mathllapinternal} \def\mathrlap{\mathpalette\mathrlapinternal} \def\mathclap{\mathpalette\mathclapinternal} \def\mathllapinternal#1#2{\llap{$\mathsurround=0pt#1{#2}$}} \def\mathrlapinternal#1#2{\rlap{$\mathsurround=0pt#1{#2}$}} \def\mathclapinternal#1#2{\clap{$\mathsurround=0pt#1{#2}$}} % Renames \sqrt as \oldsqrt and redefine root to result in \sqrt[#1]{#2} \let\oldroot\root \def\root#1#2{\oldroot #1 \of{#2}} \renewcommand{\sqrt}[2][]{\oldroot #1 \of{#2}} % Manually declare the txfonts symbolsC font \DeclareSymbolFont{symbolsC}{U}{txsyc}{m}{n} \SetSymbolFont{symbolsC}{bold}{U}{txsyc}{bx}{n} \DeclareFontSubstitution{U}{txsyc}{m}{n} % Manually declare the stmaryrd font \DeclareSymbolFont{stmry}{U}{stmry}{m}{n} \SetSymbolFont{stmry}{bold}{U}{stmry}{b}{n} % Manually declare the MnSymbolE font \DeclareFontFamily{OMX}{MnSymbolE}{} \DeclareSymbolFont{mnomx}{OMX}{MnSymbolE}{m}{n} \SetSymbolFont{mnomx}{bold}{OMX}{MnSymbolE}{b}{n} \DeclareFontShape{OMX}{MnSymbolE}{m}{n}{ <-6> MnSymbolE5 <6-7> MnSymbolE6 <7-8> MnSymbolE7 <8-9> MnSymbolE8 <9-10> MnSymbolE9 <10-12> MnSymbolE10 <12-> MnSymbolE12}{} % Declare specific arrows from txfonts without loading the full package \makeatletter \def\re@DeclareMathSymbol#1#2#3#4{% \let#1=\undefined \DeclareMathSymbol{#1}{#2}{#3}{#4}} \re@DeclareMathSymbol{\neArrow}{\mathrel}{symbolsC}{116} \re@DeclareMathSymbol{\neArr}{\mathrel}{symbolsC}{116} \re@DeclareMathSymbol{\seArrow}{\mathrel}{symbolsC}{117} \re@DeclareMathSymbol{\seArr}{\mathrel}{symbolsC}{117} \re@DeclareMathSymbol{\nwArrow}{\mathrel}{symbolsC}{118} \re@DeclareMathSymbol{\nwArr}{\mathrel}{symbolsC}{118} \re@DeclareMathSymbol{\swArrow}{\mathrel}{symbolsC}{119} \re@DeclareMathSymbol{\swArr}{\mathrel}{symbolsC}{119} \re@DeclareMathSymbol{\nequiv}{\mathrel}{symbolsC}{46} \re@DeclareMathSymbol{\Perp}{\mathrel}{symbolsC}{121} \re@DeclareMathSymbol{\Vbar}{\mathrel}{symbolsC}{121} \re@DeclareMathSymbol{\sslash}{\mathrel}{stmry}{12} \re@DeclareMathSymbol{\bigsqcap}{\mathop}{stmry}{"64} \re@DeclareMathSymbol{\biginterleave}{\mathop}{stmry}{"6} \re@DeclareMathSymbol{\invamp}{\mathrel}{symbolsC}{77} \re@DeclareMathSymbol{\parr}{\mathrel}{symbolsC}{77} \makeatother % \llangle, \rrangle, \lmoustache and \rmoustache from MnSymbolE \makeatletter \def\Decl@Mn@Delim#1#2#3#4{% \if\relax\noexpand#1% \let#1\undefined \fi \DeclareMathDelimiter{#1}{#2}{#3}{#4}{#3}{#4}} \def\Decl@Mn@Open#1#2#3{\Decl@Mn@Delim{#1}{\mathopen}{#2}{#3}} \def\Decl@Mn@Close#1#2#3{\Decl@Mn@Delim{#1}{\mathclose}{#2}{#3}} \Decl@Mn@Open{\llangle}{mnomx}{'164} \Decl@Mn@Close{\rrangle}{mnomx}{'171} \Decl@Mn@Open{\lmoustache}{mnomx}{'245} \Decl@Mn@Close{\rmoustache}{mnomx}{'244} \makeatother % Widecheck \makeatletter \DeclareRobustCommand\widecheck[1]{{\mathpalette\@widecheck{#1}}} \def\@widecheck#1#2{% \setbox\z@\hbox{\m@th$#1#2$}% \setbox\tw@\hbox{\m@th$#1% \widehat{% \vrule\@width\z@\@height\ht\z@ \vrule\@height\z@\@width\wd\z@}$}% \dp\tw@-\ht\z@ \@tempdima\ht\z@ \advance\@tempdima2\ht\tw@ \divide\@tempdima\thr@@ \setbox\tw@\hbox{% \raise\@tempdima\hbox{\scalebox{1}[-1]{\lower\@tempdima\box \tw@}}}% {\ooalign{\box\tw@ \cr \box\z@}}} \makeatother % \mathraisebox{voffset}[height][depth]{something} \makeatletter \NewDocumentCommand\mathraisebox{moom}{% \IfNoValueTF{#2}{\def\@temp##1##2{\raisebox{#1}{$\m@th##1##2$}}}{% \IfNoValueTF{#3}{\def\@temp##1##2{\raisebox{#1}[#2]{$\m@th##1##2$}}% }{\def\@temp##1##2{\raisebox{#1}[#2][#3]{$\m@th##1##2$}}}}% \mathpalette\@temp{#4}} \makeatletter % udots (taken from yhmath) \makeatletter \def\udots{\mathinner{\mkern2mu\raise\p@\hbox{.} \mkern2mu\raise4\p@\hbox{.}\mkern1mu \raise7\p@\vbox{\kern7\p@\hbox{.}}\mkern1mu}} \makeatother %% Fix array \newcommand{\itexarray}[1]{\begin{matrix}#1\end{matrix}} %% \itexnum is a noop \newcommand{\itexnum}[1]{#1} %% Renaming existing commands \newcommand{\underoverset}[3]{\underset{#1}{\overset{#2}{#3}}} \newcommand{\widevec}{\overrightarrow} \newcommand{\darr}{\downarrow} \newcommand{\nearr}{\nearrow} \newcommand{\nwarr}{\nwarrow} \newcommand{\searr}{\searrow} \newcommand{\swarr}{\swarrow} \newcommand{\curvearrowbotright}{\curvearrowright} \newcommand{\uparr}{\uparrow} \newcommand{\downuparrow}{\updownarrow} \newcommand{\duparr}{\updownarrow} \newcommand{\updarr}{\updownarrow} \newcommand{\gt}{>} \newcommand{\lt}{<} \newcommand{\map}{\mapsto} \newcommand{\embedsin}{\hookrightarrow} \newcommand{\Alpha}{A} \newcommand{\Beta}{B} \newcommand{\Zeta}{Z} \newcommand{\Eta}{H} \newcommand{\Iota}{I} \newcommand{\Kappa}{K} \newcommand{\Mu}{M} \newcommand{\Nu}{N} \newcommand{\Rho}{P} \newcommand{\Tau}{T} \newcommand{\Upsi}{\Upsilon} \newcommand{\omicron}{o} \newcommand{\lang}{\langle} \newcommand{\rang}{\rangle} \newcommand{\Union}{\bigcup} \newcommand{\Intersection}{\bigcap} \newcommand{\Oplus}{\bigoplus} \newcommand{\Otimes}{\bigotimes} \newcommand{\Wedge}{\bigwedge} \newcommand{\Vee}{\bigvee} \newcommand{\coproduct}{\coprod} \newcommand{\product}{\prod} \newcommand{\closure}{\overline} \newcommand{\integral}{\int} \newcommand{\doubleintegral}{\iint} \newcommand{\tripleintegral}{\iiint} \newcommand{\quadrupleintegral}{\iiiint} \newcommand{\conint}{\oint} \newcommand{\contourintegral}{\oint} \newcommand{\infinity}{\infty} \newcommand{\bottom}{\bot} \newcommand{\minusb}{\boxminus} \newcommand{\plusb}{\boxplus} \newcommand{\timesb}{\boxtimes} \newcommand{\intersection}{\cap} \newcommand{\union}{\cup} \newcommand{\Del}{\nabla} \newcommand{\odash}{\circleddash} \newcommand{\negspace}{\!} \newcommand{\widebar}{\overline} \newcommand{\textsize}{\normalsize} \renewcommand{\scriptsize}{\scriptstyle} \newcommand{\scriptscriptsize}{\scriptscriptstyle} \newcommand{\mathfr}{\mathfrak} \newcommand{\statusline}[2]{#2} \newcommand{\tooltip}[2]{#2} \newcommand{\toggle}[2]{#2} % Theorem Environments \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{lemma}{Lemma} \newtheorem{prop}{Proposition} \newtheorem{cor}{Corollary} \newtheorem*{utheorem}{Theorem} \newtheorem*{ulemma}{Lemma} \newtheorem*{uprop}{Proposition} \newtheorem*{ucor}{Corollary} \theoremstyle{definition} \newtheorem{defn}{Definition} \newtheorem{example}{Example} \newtheorem*{udefn}{Definition} \newtheorem*{uexample}{Example} \theoremstyle{remark} \newtheorem{remark}{Remark} \newtheorem{note}{Note} \newtheorem*{uremark}{Remark} \newtheorem*{unote}{Note} %------------------------------------------------------------------- \begin{document} %------------------------------------------------------------------- \section*{cofree coalgebra} \hypertarget{context}{}\subsubsection*{{Context}}\label{context} \hypertarget{algebra}{}\paragraph*{{Algebra}}\label{algebra} [[!include higher algebra - contents]] \hypertarget{contents}{}\section*{{Contents}}\label{contents} \noindent\hyperlink{definition}{Definition}\dotfill \pageref*{definition} \linebreak \noindent\hyperlink{examples}{Examples}\dotfill \pageref*{examples} \linebreak \noindent\hyperlink{constructions}{Constructions}\dotfill \pageref*{constructions} \linebreak \noindent\hyperlink{modules_over_a_commutative_ring}{Modules over a commutative ring}\dotfill \pageref*{modules_over_a_commutative_ring} \linebreak \noindent\hyperlink{cofree_coalgebra_over_a_vector_space}{Cofree coalgebra over a vector space}\dotfill \pageref*{cofree_coalgebra_over_a_vector_space} \linebreak \noindent\hyperlink{the_cofree_coalgebra_over_a_1dimensional_space}{The cofree coalgebra over a 1-dimensional space}\dotfill \pageref*{the_cofree_coalgebra_over_a_1dimensional_space} \linebreak \noindent\hyperlink{related_concepts}{Related concepts}\dotfill \pageref*{related_concepts} \linebreak \noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak \hypertarget{definition}{}\subsection*{{Definition}}\label{definition} Given a [[monoidal category]] $M$, let $Coalg(M)$ denote the category of coalgebras (i.e., [[comonoid|comonoids]]) with respect to the [[tensor product]] of $M$. Let $U \colon Coalg(M) \to M$ be the [[forgetful functor]]. The \textbf{cofree coalgebra} refers to a functor \emph{[[right adjoint|right]]} [[adjoint functor|adjoint]] to $U$, if it exists. \hypertarget{examples}{}\subsection*{{Examples}}\label{examples} For example, a cofree coalgebra over a [[vector space]] $V$ consists of a coalgebra $C(V)$ and a [[linear map]] $\varepsilon \colon C(V) \to V$ which is [[universal construction|universal]] in the sense that given a coalgebra $C$ and linear map $f \colon C \to V$, there exists a unique coalgebra map $g \colon C \to C(V)$ such that $f = \varepsilon \circ g$. \hypertarget{constructions}{}\subsection*{{Constructions}}\label{constructions} The [[forgetful functor]] $U \colon Coalg(M) \to M$ [[preserved colimit|preserves]] and [[reflected colimit|reflects]] ([[created limit|creates]]) any [[colimits]] which happen to exist, analogous to the fact that the forgetful functor $Alg(M) \to M$ 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]]. \hypertarget{modules_over_a_commutative_ring}{}\subsubsection*{{Modules over a commutative ring}}\label{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 \begin{displaymath} K\text{-Coalg} \to K\text{-Mod} \end{displaymath} has a right adjoint. An argument is given by Michael Barr in \begin{itemize}% \item \href{http://www.math.mcgill.ca/barr/ftp/pdffiles/coalgebra.pdf}{Coalgebras over a Commutative Ring}; see theorem 4.1. \end{itemize} (This is somewhat at odds with an assertion made by Michiel Hazewinkel in \href{http://arxiv.org/ftp/arxiv/papers/0804/0804.3888.pdf}{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.) \hypertarget{cofree_coalgebra_over_a_vector_space}{}\subsubsection*{{Cofree coalgebra over a vector space}}\label{cofree_coalgebra_over_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. \begin{theorem} \label{FundamentalTheorem}\hypertarget{FundamentalTheorem}{} \textbf{(Fundamental theorem of coalgebras)} Every [[coalgebra]] is the [[filtered colimit]] over the [[diagram]] of its [[finite number|finite]]-[[dimension|dimensional]] subcoalgebras and inclusions between them. \end{theorem} This seems to be due to (\hyperlink{Sweedler69}{Sweedler 69}). For a proof of this result, see for example (\hyperlink{WM}{Michaelis 03}). For more see at \emph{\href{coalgebra#AsFilteredColimits}{coalgebra -- As filtered colimits}}. 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 $T(V^\ast)$ as the algebra of (non-commutative) polynomial functions on $V$. The dual of $T(V^\ast)$ 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 $T(V^\ast)$ and surjections between them (compare [[profinite completion]]), using the fact that the dual of a finite-dimensional algebra is a finite-dimensional coalgebra. Let $T(V^\ast)^\circ$ be this colimit coalgebra, embedded in $T(V^\ast)^\ast$. The dual of the inclusion $V^\ast \hookrightarrow T(V^\ast)$ gives a map $T(V^\ast)^\ast \to V^{\ast \ast}$, and this restricts to a map $T(V^\ast)^\circ \to V^{\ast \ast}$. For finite-dimensional $V$, there is an isomorphism $V^{\ast \ast} \cong V$, and we have the following result. \begin{lemma} \label{}\hypertarget{}{} If $V$ is finite-dimensional, then the composite $T(V^\ast)^\circ \to V^{\ast \ast} \cong V$ exhibits $T(V^\ast)^\circ$ as the cofree coalgebra over $V$. \end{lemma} \begin{proof} Suppose given a coalgebra $C$ and a linear map $f \colon C \to V$. We want to show this has a unique lift to a coalgebra map $\widehat{f} \colon C \to T(V^\ast)^\circ$. For $x \in C$, there is a minimal finite-dimensional subcoalgebra $C_x$ containing $x$, and if $f_x \colon C_x \to T(V^\ast)^\circ$ is the restriction of $f$, then the value $\widehat{f}(x)$ 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 $f^\ast \colon V^\ast \to C^\ast$ extends uniquely to an algebra map $T(V^\ast) \to C^\ast$, since the tensor algebra is the free algebra on $V$. This algebra map factors as \begin{displaymath} T(V^\ast) \stackrel{onto}{\to} Q \stackrel{into}{\to} C^\ast \end{displaymath} where $Q$ is a finite-dimensional algebra. The dual $Q^\ast$ carries a coalgebra structure that embeds in $T(V^\ast)^\circ$, and by dualizing we obtain coalgebra maps \begin{displaymath} C \to Q^\ast \to T(V^\ast)^\circ \end{displaymath} that provide the desired lift. The uniqueness of the lift is evident from the uniqueness of the extension $T(V^\ast) \to C$ (or equivalently, of $T(V^\ast)_{prof} \to C$, passing to the profinite completion) on the dual algebra side. \end{proof} 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. \begin{theorem} \label{}\hypertarget{}{} The coalgebra colimit \begin{displaymath} K = colim_{V_\alpha \subset V} C(V_\alpha) \end{displaymath} is the cofree coalgebra over $V$. \end{theorem} \begin{remark} \label{}\hypertarget{}{} Happily, this colimit is computed as in $Vect$, and more happily still -- since the underlying functor $Vect \to Set$ 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. \end{remark} Indeed, suppose given a coalgebra $C$ and a linear function $f \colon C \to V$. 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 $f \colon C \to V$ factors through a map $f_\alpha \colon C \to V_\alpha$ where $V_\alpha$ is a finite-dimensional subspace of $V$. This lifts uniquely to a coalgebra map $\widehat{f_\alpha} \colon C(V_\alpha)$, and the desired lift $\widehat{f}$ is the composite \begin{displaymath} C \stackrel{\widehat{f_\alpha}}{\to} C(V_\alpha) \hookrightarrow K \end{displaymath} 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 \href{http://mathoverflow.net/questions/31109/is-there-an-explicit-construction-of-a-free-coalgebra/31126#31126}{here}. See also the nForum discussion \href{http://www.math.ntnu.no/~stacey/Mathforge/nForum/comments.php?DiscussionID=3467&page=1}{here}. \hypertarget{the_cofree_coalgebra_over_a_1dimensional_space}{}\paragraph*{{The cofree coalgebra over a 1-dimensional space}}\label{the_cofree_coalgebra_over_a_1dimensional_space} Specializing still further, we investigate the detailed structure of the cofree coalgebra over 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 \begin{displaymath} (k[x]/I)^\ast \end{displaymath} 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 $(k[x]/I)^\ast \hookrightarrow k[x]^\ast \cong \prod_n k \cdot x^n$ into the space of formal power series in $x$. As a set, $T(k)^\circ$ is simply the union of the subspaces $(k[x]/I)^\ast \hookrightarrow \prod_n k \cdot x^n$. In other words, every $\xi \in T(k)^\circ$ belongs to some $(k[x]/I)^\ast$, 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 $g(x) = b_0 + \ldots + b_n x^n$, where the top coefficient is of course assumed non-zero. We are trying to understand when a formal power series $\xi = \sum_{n \geq 0} a_n x^n \in \prod_n k \cdot x^n$ vanishes on $I$, in the sense that \begin{displaymath} \langle p(x), \sum_n a_n x^n \rangle = 0 \end{displaymath} for all $p(x) \in (g(x))$, where the pairing $\langle , \rangle \colon k[x] \times \prod_n k \cdot x^n \to k$ is the canonical one: \begin{displaymath} \langle x^j, \sum_n a_n x^n \rangle = a_j. \end{displaymath} Taking the case $p(x) = x^j g(x)$, we obtain recurrence relations \begin{displaymath} \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, \end{displaymath} one for every $j$. We may pick $a_0, a_1, \ldots, a_{n-1}$ at random, but from there on the remainder of the sequence $a_n, a_{n+1}, \ldots$ is determined by the recurrence relations. Now let $g^{rev}(x)$ be the polynomial obtained by reversing the order of the coefficients $b_0, \ldots, b_n$ of $g$, \begin{displaymath} g^{rev}(x) = b_n + \ldots + b_0 x^n, \end{displaymath} and consider the problem of multiplicatively inverting $g^{rev}(x)$ (which we may do since the constant coefficient of $g^{rev}$ is non-zero). If $\sum_j a_j x^j$ is the reciprocal, we have \begin{displaymath} 1 = g^{rev}(x)\sum_j a_j x^j = (b_n + \ldots + b_0 x^n)(\sum_j a_j x^j) \end{displaymath} 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 $g^{rev}(x)^{-1}$ yields a $\xi$ which vanishes on $I = (g(x))$, and similarly the power series of \begin{displaymath} \frac{x}{g^{rev}(x)}, \ldots, \frac{x^{n-1}}{g^{rev}(x)} \end{displaymath} also vanish on $I$. They in fact form a basis for the subspace $(k[x]/(g(x)))^\ast \hookrightarrow \prod_n k \cdot x^n$. 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: \begin{theorem} \label{}\hypertarget{}{} As subspaces of the space of formal power series $\prod_n k \cdot x^n$, there is an identification \begin{displaymath} T(k)^\circ = k[x]_{(x)} \end{displaymath} where the right side is the localization of $k[x]$ at the prime ideal $(x)$, whose elements are considered as formal power series. \end{theorem} 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: \begin{displaymath} \delta(x^N) = \sum_{m+n = N} x^m \otimes x^n. \end{displaymath} We may as well focus on a typical subcoalgebra $(k[x]/(g(x)))^\ast$, with basis elements $\frac{x^i}{g^{rev}(x)}$. By duality, the definition of $\delta(\frac{x^i}{g^{rev}(x)})$ can be extracted from the multiplication table for residue classes $x^j \pmod g(x)$. 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 \begin{displaymath} \frac{x^i}{g^{rev}(x)} = \sum_k \frac{A_k}{(1- r_k x)^{n_k}} \end{displaymath} so that it suffices to give $\delta(\frac1{(1 - r x)^n})$, although the algebra works out better with a slightly different basis. With this in mind, put $e_n(r) = \frac{(r x)^n}{(1 - r x)^{n+1}}$ for $n \geq 0$. The power series expansion is \begin{displaymath} e_n(r) = \sum_{k \geq 0} \binom{k}{n} r^k x^k \end{displaymath} and we \emph{formally} calculate \begin{displaymath} \itexarray{ \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) } \end{displaymath} 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 \emph{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{displaymath} \itexarray{ \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. } \end{displaymath} An analogous calculation shows that the counit $\varepsilon \colon k[x]_{(x)} \to k$ is defined by $\varepsilon(e_n(r)) = \delta_{0 n}$ (Kronecker delta). We summarize the calculations above as follows. \begin{theorem} \label{}\hypertarget{}{} 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 $e_n(r) = \frac{(r x)^n}{(1 - r x)^{n+1}}$ in $k[x]_{(x)}$, with coalgebra structure given in each case by \begin{displaymath} \delta(e_n(r)) = \sum_{i+j=n} e_i(r) \otimes e_j(r), \qquad \varepsilon(e_n(r)) = \delta_{0 n} \end{displaymath} 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 \begin{displaymath} k[x]_{(x)} \cong \bigoplus_{r \in k} L(r) \end{displaymath} 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. \end{theorem} \hypertarget{related_concepts}{}\subsection*{{Related concepts}}\label{related_concepts} \begin{itemize}% \item The [[duality|dual concepts]] are those of \emph{[[free monoid]]}, and \emph{[[tensor algebra]]}, \end{itemize} \hypertarget{references}{}\subsection*{{References}}\label{references} \begin{itemize}% \item [[Moss Sweedler]], \emph{Hopf algebras}, 1969 \item [[Walter Michaelis]], \emph{Coassociative Coalgebras}, Handbook of Algebra Volume 3, Elsevier (2003). \end{itemize} See also \begin{itemize}% \item Tom Fox, \emph{The construction of cofree coalgebras}. JPAA 84 (1993) 191-198. \href{http://www.math.mcgill.ca/fox/papers.html}{texfile here} \end{itemize} but note that this contains an error which is corrected in \begin{itemize}% \item [[Michiel Hazewinkel ]] (2003), ``Cofree coalgebras and multivariable recursiveness'', \href{http://www.sciencedirect.com/science/article/pii/S0022404903000136}{Journal of Pure and Applied Algebra 183 (1): 61--103} \end{itemize} [[!redirects cofree coalgebras]] [[!redirects free coalgebra]] [[!redirects free coalgebras]] \end{document}