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\begin{document}
\title{\bf An Introduction To\\
Algebraic Deformation Theory}
\author{Thomas F. Fox\thanks{This work was partially supported by grants from the FCAR of Qu\'ebec and the NSERC of Canada.}\\
McGill University\\
Montr\'eal, Canada }\maketitle
Our purpose in writing these notes is to give a short elementary introduction to the theory of deformations of algebraic structures, with an eye towards elucidating the categorical point of view. Presently the best overview of the subject is the long paper by Gerstenhaber and Schack [18], while Gerstenhaber's original papers remain eminently readable [14-17]. We intend to cast these notes at a more rudimentary level, in the hope that they will make the general theory more accessible to the non-expert.
Algebraic deformation theory was introduced for associative algebras by Gerstenhaber [14], and was extended to Lie algebras by Nijenhuis and Richardson [23,24]. Their work closely parallels the theory of deformations of complex analytic structures, initiated by Kodaira and Spencer [21]. The fundamental results connect deformation theory with the appropriate cohomology groups, and it was assumed that these results would extend to any equationally-defined class of algebras. What was missing was a suitably general cohomology theory. Early attempts to create such a theory were unconvincing [15,26], and it wasn't until the discovery of triple cohomology that the program could be completed. This paper is divided into two chapters. In the first, we give simple examples of deformations of commutative algebras, especially curves, and then present the basic definitions and theorems of the classical theory. In the second chapter, we will show how the classical theory may be extended by using triple cohomology.
All of the results contained in this paper have appeared elsewhere, particularly in [13-18] and [6-11]. We assume the reader is familiar with the basic ideas of homological algebra [22]. If $A$ is an associative $k$-algebra, where $k$ is a commutative ring, recall that a Hochschild $n$-cochain is just a $k$-linear map from $A^{\ten n}$ to $A$ (we will only consider coefficients in $A$ itself). The group of all $n$-cochains is denoted $C^n$, and the boundary map $\d :C^{n-1}\to C^n$ is defined by $$\d f(a_1,...,a_n) = a_1f(a_2,...,a_n) - f(a_1a_2,a_3,...,a_n) +$$ $$f(a_1,a_2a_3,...,a_n) - ...+(-1)^{n+1}f(a_1,...,a_{n-1})a_n$$ The kernel of $\d$ in $C^n$ is the group of $n$-cocycles, and is denoted $Z^n$. The image of $\d$ in $C^n$ is the group of $n$-coboundaries, and is denoted $B^n$. The Hochschild cohomology groups $\Hoch^n(A,A)$ (or just $H^n(A,A)$) are defined to be $Z^n/B^n$ [20]. We also assume the reader is familiar with the language of categories and functors. In chapter II we give a brief introduction to triple cohomology. For further information on triples, the reader should consult [2] and the introduction to that volume.
\vskip3ex \noindent {\Large\bf I: THE CLASSICAL THEORY}
\section{Definition and examples of deformations}
Let $X$ be an object with structure (algebraic, analytic, topological...). Roughly speaking, a deformation of $X$ is a family ${X_t}$ of objects whose structures are obtained by ``deforming'' the structure on $X$ as $t$ varies over a suitable space of parameters in a smooth way. If $A$ is an algebra over a commutative ring $k$, a one-parameter algebraic deformation of A is a family of algebras $\{A_t\}$ parameterized by $k$ such that $A_0\iso A$ and the multiplicative structure of $A_t$ varies algebraically with $t$. Before giving a formal definition, we will look a a simple example (to which the formal definition must apply).
\sube Let $A = k[x,y]/(y^2-x^3)$ and $A_t = k[x,y,t]/(y^2-x^3-x^2t)$. \vskip3ex\begin{picture}(400,140)\put(100,140){\line(0,-1){120}}\put(300,140){\line(0,-1){120}}\put(30,80){\line(1,0){140}}\put(230,80){\line(1,0){140}}\put(100,0){\makebox(0,0)[b]{$A$}}\put(300,0){\makebox(0,0)[b]{$A_t$}}\put(250,85){$-t$}\end{picture}
\vskip3ex
Geometrically it looks inviting to think of $A_t$ as a ``deformation'' of $A$. The question is, in what sense is this deformation achieved by deforming the algebraic structure of $A$? The key is noticing that $y\. y = x^3$ in $A$, while $y\. y = x^3 +x^2t$ in $A_t$, i.e. the product $y\. y$ varies with $t$. We will consider the case where $A$ is an associative $k$-algebra and the deformation of $A$ is again associative.
Let $A\pst $ be the $k\pst $-module of formal power-series with coefficients in the $k$-module $A$ , i.e. $A\pst = A\ten _kk\pst $ as a module. The algebra $A$ is a submodule of $A\pst $, and we could make $A\pst $ an algebra by bilinearly extending the multiplication of $A$, but we may also impose other multiplications on $A\pst$ that agree with that of $A$ when we specialize to $t=0$. Working backwards, suppose a multiplication $F:A\pst \ten _{k\pst }A\pst \to A\pst $ is given by a formal power-series of the form \beqn F(a,b) = f_0(a,b) + f_1(a,b)t + f_2(a,b)t^2\ldots \eeqn Since we are defining $F$ over $k\pst $, it is enough to consider $a$ and $b$ in $A$, and we further presume that each $f_n$ is a linear map $A\ten A\to A$. Since we want the specialization $t=0$ to give the original multiplication on $A$, we insist that $f_0(a,b) = ab$ (multiplication in $A$). \subd A {\em one-parameter formal deformation} of a $k$-algebra $A$ is a formal power-series $F = \sum_{n=0}^\infty f_nt^n$ with coefficients in $\Hom_k(A\ten A,A)$ such that $f_0:A\ten A\to A$ is multiplication in $A$. The deformation is {\em associative} if $F(F(a,b),c) = F(a,F(b,c))$ for all $a,b,c$ in $A$.
\vskip10pt We often refer to $A\pst $ with the multiplication defined by $F$ as the deformation of $A$, and we may write this $A\pst _F$ or $A_F$. If $F$ is finite, or at least finite for each pair $(a,b)$ in $A\ten A$, the multiplication may be defined on $A[t]$ over $k[t]$. Back to example 1.1, where $A = k[x,y]/(y^2-x^3)$ and $A_t = k[x,y,t]/(y^2-x^3-x^2t)$. Of course $\{ x^n,yx^n\}$ is a $k[t]$-basis for $A[t]$, and we define a multiplication $F$ on $A[t]$ by $$ F(x^m,x^n) = x^{m+n}\quad F(y,y) = x^3 + x^2t$$ $$ F(y,x^n) = F(x^n,y) = x^ny \quad F(yx^m,yx^n) = x^{m+n+3} + x^{m+n+2}t $$ Then $A[t]_F\iso A_t$. Note that $f_1(yx^m,yx^n) = x^{m+n+2}$. A simpler example, though less geometric, is given by deforming $k[x]/(x^2)$ to $k[x,t]/({x^2-t})$. The latter is isomorphic to $k[x]/(x^2)[t]$ with a multiplication given by $F(x,x) = t$, i.e. $f_1(x,x) = 1$. Of course $F(1,x) = F(x,1) = x$. The next example will explain why we defined deformations in terms of power-series, and not just polynomials, as well as showing that not any old power-series will serve to yield an associative multiplication.
\sube Let A = $k[x]$ and define $F:A[t]\ten _{k[t]} A[t]\to A[t]$ by $$ F = f_0 + f_1t \hskip4em f_1(x^m,x^n) = mnx^{m+n}$$ This multiplication is associative $mod\>t^2$, but not $mod\>t^3$, since $F(F(x^2,x),x) = F(x^3+2x^3t,x) = x^4 + 5x^4t + 6x^4t^2$ while $F(x^2,F(x,x)) = x^4 + 5x^4t + 4x^4t^2$. This may be patched up by {\em extending} $F$ to $F_1$ defined by $$ F_1 = f_0 + f_1t + f_2t^2 \hskip4em f_2(x^m,x^n) = \frac{m^2n^2}{2}x^{m+n}$$ Now we find that $F_1(F_1(x^2,x),x) = x^4 + 5x^4t + 25/2x^4t^2 + 15x^4t^3 + 9x^4t^4$ and $F_1(x^2,F_1(x,x)) = x^4 + 5x^4t + 25/2x^4t^2 + 10x^4t^3 + 4x^4t^4$, so $F_1$ is associative $mod\>t^3$ but not $mod\>t^4$. Further extensions may be made to patch this up, leading to $F_*$ defined by $$ F_* = \sum_{r=0}^\infty f_rt^r \hskip4em f_r(x^m,x^n) = \frac{m^rn^r}{r!}x^{m+n} $$ $F_*$ defines an associative multiplication on $A\pst$ which cannot be defined on $A[t]$. If the reader objects to the formal power-series $F_*(a,b)$, we can modify this example so that $F(a,b)$ is finite for each pair of elements of $A$ even though $F$ itself is a formal power-series, as follows: \sube Again let $A = k[x]$, and let $Dx^n = nx^{n-1}$. If we begin with $F_1(a,b) = ab + Da\, Db\, t$, which is not associative, we are led to $F(a,b) = \sum_{r=0}^\infty \frac{D^ra\, D^rb}{r!}t^r$ which is finite for given $a$ and $b$ in $k[x]$, and thus defines a multiplication on $k[x,t]$ (see [16, p.13]).
Before closing this section, we will give a collection of examples which are particularly satisfying from a geometric point of view.
\sube Throughout this example let $A = k[x,y]/(y^2)$. We will define several deformations of $A$ over $k[t]$ by giving the multiplication on $(y,y)$ only (we leave the reader to define the multiplication on terms of the form $(yx^m,yx^n))$. Define $F$,$G$,$H$,$J$, and $K$ by $F(y,y) = t^2$, $G(y,y) = xt$, $H(y,y) = x^2t^2$, $J(y,y) = x^3t$, and $K(y,y) = (x^2 + x^3)t$ respectively. Then $A_F\iso k[x,y,t]/(y^2-t^2)$, $A_G\iso k[x,y,t]/(y^2-xt)$, $A_H\iso k[x,y,t]/(y^2-x^2t^2)$, $A_J\iso k[x,y,t]/(y^2-x^3t)$, and $A_K\iso k[x,y,t]/(y^2-x^2t-x^3t)$. Now we may visualize $A$ itself as a double line (i.e. consider the closed points of $Spec A$), and our deformations of $A$ may be visualized as below:
\vskip3ex \graphs{A}{A_F} \vskip3ex \graphs{A_G}{A_H} \vskip3ex \graphs{A_J}{A_K} \vskip3ex These graphs represent the deformation for general values of $t$. The deformation is actually a family of such curves parameterized by $t$; it may be visualized as a surface in $k^3$ (see [19, p. 90]).
\section{Infinitesimals and obstructions} As is shown by example 1.4, not all formal deformations of $A$ will be associative. This condition restricts our choice of coefficient maps $f_n:A\ten A\to A$. If $F$ is associative we have $F(F(a,b),c) = F(a,F(b,c))$. Expanding both sides of this equation and collecting coefficients of $t^n$ yields
\beqn \sum_{i=0}^n f_i(f_{n-i}(a,b),c) = \sum_{i=0}^n f_i(a,f_{n-i}(b,c))
\eeqn
which gives necessary and sufficient conditions for the associativity of $F$. Let $f_n$ be the first non-zero coefficient after $f_0$ in the expansion $F = \sum f_nt^n$. This $f_n$ is the {\em infinitesimal} of $F$, and 2.1 reads $f_0(f_n(a,b),c) + f_n(f_0(a,b),c) = f_0(a,f_n(b,c)) + f_n(a,f_0(b,c))$, or
\beqn af_n(b,c) - f_n(ab,c) + f_n(a,bc) - f_n(a,b)c = 0 \eeqn
The left side of 2.2 is the Hochschild coboundary of $f_n$, so 2.2 may be written $\d f_n = 0$, yielding the first connection between deformation theory and cohomology theory.
\thm If $F$ is an associative deformation of $A$ then the infinitesimal of $F$ is a Hochschild $2$-cocycle. \eth For arbitrary $n$, equation 2.1 above may be written
\beqn \d f_n(a,b,c) = \sum_{i=1}^{n-1} f_i(f_{n-i}(a,b),c) - f_i(a,f_{n-i}(b,c)) \eeqn
If $f_n$ satisfying 2.4 have been given for $0 < n < m$, the right side of 2.4 with $m=n$ is the obstruction to finding $f_n$ extending the deformation. The most important theorem in deformation theory is the following. \thm The obstruction is a Hochschild $3$-cocycle. \eth \thc If $\Hoch^3(A,A) = 0$ then every $2$-cocycle of $A$ may be extended to an associative deformation of $A$. \eth We will give a proof of theorem 2.5 below (9.4). The pattern above will be repeated several times in what follows. We will be looking for power-series whose coefficients are $n$-cochains of $A$, and it will turn out that the infinitesimal of the series is a cocycle, while the obstructions to extending the infinitesimal to a full power-series will lie in $H^{n+1}(A,A)$.
\section{Equivalent and trivial deformations} The next problem to consider is when two deformation are significantly different from one another. Given associative deformations $A_F$ and $A_G$ of $A$ we ask when there is an isomorphism $A_F\to A_G$ which keeps $A$ fixed. By a {\em formal isomorphism} $\Psi :A_F\to A_G$ we mean a $k\pst$-linear map $A\pst_F\to A\pst_G$ that may be written in the form \beqn \Psi(a) = a + \psi_1(a)t + \psi_2(a)t^2 + \psi_3(a)t^3 + \ldots \eeqn Since $\Psi$ is defined over $k\pst$, it is enough to consider $a$ in $A$, and we further presume that each $\psi_n$ is a $k$-linear map $A\to A$. We called such a map an ``isomorphism'', and not just a ``homomorphism'', because it has an inverse of the same form (its leading coefficient, the identify, is invertible). If $\Psi$ is multiplication-preserving, we say it is an {\em algebraic} isomorphism. The condition that $\Psi$ preserve multiplication means that $G(\Psi(a),\Psi(b)) = \Psi(F(a,b))$ for all $a$ and $b$ in $A$. Expanding both sides of this equation and collecting the coefficients of $t^n$ yields \beqn \sum_{i+j+k=n} g_i(\psi_j(a),\psi_k(b)) = \sum_{i+j=n} \psi_i(f_j(a,b)) \eeqn
We say that $F$ and $G$ are {\em equivalent} if such a $\Psi$ exists, and we write $A_F\iso A_G$ (it is easy to check that this is an equivalence relation on the set of deformations of $A$). In that case 3.2 yields $\d\psi_1 = f_1 - g_1$, i.e. the $2$-cocycles $f_1$ and $g_1$ are in the same cohomology class.
Given $\psi_1$, we may ask when it may be extended to an isomorphism from $A_F$ to $A_G$. For general $n$, equation 3.2 may be written \beqn \d\psi_n(a,b) = \sum_{{i+j=n}\atop {i\neq n}} \psi_i(f_j(a,b)) - \sum_{{i+j+k=n}\atop {j,k\neq n}} g_i(\psi_j(a),\psi_k(b)) \eeqn Given a truncated algebraic isomorphism $\Psi = \sum \psi_it^i$, $ i < n$, from $A_F$ to $A_G$, the right side of 3.3 defines the obstruction to finding $\psi_n$ extending the isomorphism. This obstruction is a Hochschild $2$-cocycle, and if its class in $\Hoch^2(A,A)$ is zero then $\psi_n$ exists. If all such obstructions vanish, then $A_F\iso A_G$.
\thm If $\Hoch^2(A,A) = 0$, then all deformations of $A$ are isomorphic. \eth The most trivial deformation of $A$ is the $A$-algebra $A\pst $, i.e. $A_K$ where $K(a,b) = ab$ for all $a$ and $b$ in $A$. A deformation $F$ is {\em trivial} if $A\pst_F\iso A\pst$. One consequence of theorem 3.4 is that $\Hoch^2(A,A) = 0$ implies all deformations of $A$ are trivial, i.e. $A$ is {\em rigid}. Suppose $F$ is non-trivial, and suppose that its infinitesimal $f_n$ is a coboundary, say $\d \psi = f_n$. Let $\Psi$ be the formal automorphism of $A\pst$ defined by $\Psi = \psi_0 + \psi t^n$ and define $F'(a,b) = \Psi^{-1}F(\Psi(a),\Psi(b))$. Then $F'\iso F$ by construction, and $\Psi^{-1} = \psi_0 - \psi t^n + \psi_2t^{2n} - \ldots$ , so $$ f'_n(a,b) = \sum_n \psi_h^{-1}f_i(\psi_j(a),\psi_k(b)) = -\psi(ab) + f_n(a,b) + a\psi(b) + \psi(a)b = 0 $$ Clearly $f'_m = 0$ for $m < n$, so the index of the infinitesimal of $F'$ must be greater than $n$. We may repeat this process of killing off an infinitesimal that is a coboundary, and the process must stop because $F$ is non-trivial.
\thm A non-trivial deformation is equivalent to a deformation whose infinitesimal is not a coboundary. \eth If $F$ is a trivial deformation, it is not true that its infinitesimal $f_n$ need be a coboundary (unless the infinitesimal is $f_1$ ) since 3.2 does not give $\d\psi_n = f_n $ (see 4.5 below). \section{Automorphisms of the deformed algebra} An {\em algebraic automorphism} of $A_F$ is just an algebraic isomorphism $\Psi :A_F\to A_F$. If $ \psi_n$ is the infinitesimal of $\Psi$ (the first non-zero coefficient after $0$), then 3.2 gives us $\d\psi_n = 0$, or $$ \psi_n(ab) = \psi_n(a)b + a\psi_n(b) $$ Hence, the infinitesimal of an algebraic automorphism is a derivation of $A$, and we may ask when a derivation of $A$ may be extended to an automorphism of $A_F$. If a derivation $\psi_1$ has been extended to a truncated automorphism $\Psi = \sum \psi_it^i$, $it^2$, since $t = \Xi(x^2) \equiv (x +\frac{1}{2}xt)^2 mod\>t^2$. But if we want to make $\Xi$ an algebraic automorphism $mod\>t^3$ by extending it to $\Xi_1(a) = a + \psi(a)t + \xi(a)t^2$, we find ourselves trying to solve $1 + 2x\xi(x) = 0$,which is again impossible. Thus the truncated automorphism $\Xi$ is obstructed. It is by no means a coincidence that the $2$-cocycle $g$ is the infinitesimal of the deformation, as we will see in the followingsection.
\section{Jump deformations}
As explained in section 1, an algebraic deformation $F$ of $A$ gives rise to a family of algebras parameterized by the elements of the ring $k$. The algebra $A_F$ is but a generic element of this family. As $t$ varies over $k$, we may have non-isomorphic specializations of $A_F$. A jump deformation of $A$ is one such that these specializations are all isomorphic except perhaps, thespecialization to $t=0$, which must be $A$ itself. Of course we ask thatthe isomorphisms between the specializations of $F$ arise in a genericalgebraic manner. If $F =\sum f_nt^n$ is a deformation of $A$, let $F^\nu$ be the deformation of $A_F$ defined by
\beqn F^\nu(a,b) = \sum f_n(a,b)(t(1+u))^n \eeqn
$F^\nu$ should be thought of as a generic element for the family of specializations of $F$. If $F^\nu$ is a trivial deformation of $F$, then those specializations are isomorphic to $F$ itself. Of course, $F^\nu$ may be written as a power-series in $u$ with coefficients in $Hom_{k\pst}(A_F\ten A_F,A_F)$, which yields
\beqn
F^\nu = \sum_{n=0}^\infty f^\nu_nu^n\quad f^\nu_n = \sum_{m=n}^{\infty} \choose({m}{n}) f_m t^m
\eeqn
\subd If $F^\nu$ is a trivial deformation of $A_F$, then $F$ is a {\em jump} deformation of $A$ .
\vskip10pt From 6.2 above we see that $f^\nu_0 = F$, while $f^\nu = f_1t + 2f_2t^2 + 3f_3t^3 + \ldots$
Now suppose that $k$ is a field of characteristic zero (or an algebra over the rationals). Then $f^\nu_1$ is the infinitesimal of $F^\nu$ . If $F$ is a jump deformation, then $f^\nu_1$ must be a 2-coboundary of $A_F$, as seen in section 3. On the other hand, if $F$ is a non-trivial deformation of $A$, we may assume that its infinitesimal is not a 2-coboundary of $A$ (theorem 3.5). But the infinitesimal of $F$ lifts to $f^\nu_1$, yielding the following result due to J.P. Coffee (see [17]):
\thm If $F$ is a non-trivial jump deformation of $A$ in characteristic zero, then there is a non-trivial $2$-cocycle of $A$ that lifts to atrivial $2$-cocycle of $A_F$, namely the infinitesimal of $F$. \eth
\thc If $F$ is a non-trivial jump deformation of $A$ in characteristic zero, then $dim_{k\pst}H^2(A_F,A_F) < dim_kH^2(A_F,A_F)$ and \\$dim_{k\pst}H^1(A_F,A_F) < dim_kH^1(A_F,A_F)$. \eth
This is exactly what happened in example 5.17 above. Theorems 3.4 and 3.5 show that $A$ can be non-trivially deformed only if $H^2(A,A)$ is not zero. In particular, $A$ cannot be regular (if $A$ is the local coordinate ring of a curve, there must be a singularity). In as much as the cohomology of $A$ measures the deficiencies of $A$, such as singularity or incomplete intersection, corollary 6.5 says that a non-trivial jump deformation must improve $A$. The reader may look again at the examples given in 1.6, all of which are jump deformations.
The situation in characteristic $p$ is complicated by the fact that the infinitesimal of $F^\nu$ may not be $f^\nu_1$ . This situation is discussed at length in [17] and [18]. We have not touched upon many of the important topics in algebraic deformation theory, most notably the classification of deformations of plane curves [1,4], multiparameter deformations and spaces of moduli [14], and perturbation theory for operator algebras [27]. As well, we have barely mentioned the algebraic automorphisms of $A\pst$, which are particularly important to the study of inseparable extensions. These depend on analyzing automorphisms of the form $\exp(\psi t)$, where $\psi$ is a derivation of $A$. For more information about deformation theory, the reader should see [18] and the other papers in the same volume. Our hope is that this chapter has given the reader sufficient understanding of the connection between deformation theory and homological algebra to make the following chapter accessible.
\vskip2em \noindent {\Large\bf II: DEFORMATIONS USING TRIPLES}
\section{Triples and power-series}
If we were to follow the classical approach to algebraic deformation theory, we would now replicate the results of chapter I using commutative algebras and Harrison or Andre cohomology, then we would repeat the exercise for Lie algebras, using Chevalley-Eilenberg cohomology groups, then we would consider Lie triple systems, using Yamaguti-Harris groups, etc. (see [22] for references), or we could wave our hands and say everything works for ``equationally-defined'' classes of algebras. All this is unnecessary, since the categories defined by triples on the category of $k$-modules include all the interesting examples, and triple cohomology unifies the classical theories. Let $\Msc$ denote the category of $k$-modules, and recall that a {\em triple} $(T,\mu,\eta)$ on $\Msc$ is a functor $T:\Msc\to \Msc$equipped with natural transformations $\eta:I\to T$ and $\mu:T^2\to T$satisfying $\mu\. T\mu = \mu\.\mu T$ and $\mu\. T\eta = \mu\.\eta T = I$. A {\em $T$-algebra} $(A,\gka)$ is a $k$-module $A$ equipped with a multiplication map $\gka:TA\to A$satisfying $\gka\.\eta = I$ and $\gka\.\mu = \gka\. T\gka$. A$T$-algebra map $(A,\gka)\to (B,\gkb)$ is a $k$-linear map $\psi:A\to B$ satisfying $\gkb\. T\psi = \psi\.\gka$ . We will use $\Asc$ to denote the category of all $T$-algebras. If we are looking at associative algebras, $TA$ is just the tensor algebra generated by the module $A$, and $\gka$ multiplies everything in sight. The equations $\gka\.\eta = I$ and $\gka\.\mu = \gka\. T\gka$ ensure that $\gka$ is unitary and associative respectively (amongst otherthings [3,10]). If we let $T$ be the symmetric algebra functor, we will getcommutative algebras, while the free Lie algebra functor yields the categoryof Lie algebras, etc. Hence any construction carried out in this generalsetting applies equally well to all these classical special cases.
We will now recast algebraic deformation theory in the setting of triples. Hence, we wish to deform the multiplication $\gka :TA\to A$ on a $T$-algebra to a multiplication $F:T(A\pst)\to A\pst$ on the $k\pst$-module $A\pst$. Of course $T(A\pst)\iso TA\ten_kk\pst$, so it makes sense to consider multiplications given by power-series $F = \sum f_nt^n$ where $f_n:TA\to A$ is a $k$-linear map and $f_0 = \gka$. We now consider what conditions associativity \beqn F\.\mu = F\.TF \eeqn places on the coefficient maps. We will temporarily make the gross assumption that $T$ is additive, even though {\em no triple ofinterest to us is additive} (we will fix this up later). Expanding 7.1 andcollecting coefficients of like power yields
\beqn f_n\.\mu = \sum_{i+j=n} f_i\. Tf_j \eeqn
For $n=1$ this reads $f_1\.\mu = f_1\. T\gka + \gka\. Tf_1$ or
\beqn f_1\. T\gka - f_1\.\mu + \gka\.Tf_1 = 0 \eeqn
which looks suspiciously like a cocycle condition, which it is. Equation 7.3 says that $f_1:TA\to A$ is a $1$-cocycle of $A$, just as in \S 2. However, the complex we use to define triple cohomology groups is not the usual one, as in [2], but the ``non-homogeneous'' complex defined by Beck [3]. Since this does not appear in the standard category theory texts, we will review ithere.
\section{Triple Cohomology}
The usual definition of the cotriple cohomology groups $H^n(A,A)$ is as follows: There is a pair of adjoint functors $F:\Msc\to \Asc$ and$U:\Asc\to \Msc$ , the free algebra and underlying module respectively. Moreprecisely, if $\gka:TA\to A$ is in $\Asc$, then $U(A,\gka) = A$, while if$M$ is in $\Msc$, then $FM$ is the $T$-algebra $TM$ with its multiplication$\mu:T^2M\to TM$. These give rise to a cotriple $(G,\gke,\gkd)$ on $\Asc$,where $G = FU$ and the natural transformation $\gke :GA\to A$ is themultiplication on $A$. This yields the following cotriple-generatedsimplicial complex over $A$: \complex{A}{GA}{G^2A}{G^3A}{G^4A}
\noindent where the i-th face $G^{n+1}A\to G^nA$ is $G^{n-i}\gke G^i$, for $0\leq i\leq n$ [2]. In the general theory, one applies a contravariant abelian group-valued functor $E$ to the complex 8.1, yielding a complex of abelian groups\cocomplex{EA}{EGA}{EG^2A}{EG^3A}{} The homology of the associated chain complex defines the cotriple cohomology groups (denoted $H^n(A,E)_G$ in [2]). Note that if $E$ preservesequalizers the term $EA$ is dropped from 8.2, since $0\to EA\to EGA \to EG^2A$ would be exact, yielding trivial low level homology groups. We are only interested in the cohomology of $A$ with coefficients in itself, so the appropriate candidate for $E$ is the functor $\Der_k(-,A)$ [2, 1.3,1.4]. Thus the cohomology groups $H^n(A,A)$ are defined by the ``homogeneous'' chain complex of abelian groups givenbelow:
$$ 0\to \Der_k(GA,A)\to \Der_k(G^2A,A)\to \Der_k(G^3A,A)\ldots $$
\beqn \d^{n-1}f = \sum_{i=0}^n (-1)^if\. G^{n-i}\gke G^i \eeqn
This doesn't look anything like the coboundary formula 7.3. However, since $F$ is the left adjoint of $U$, where $G = FU$ and $T = UF$, we have $$\Der_k(G^{n+1}A,A) = \Der_k(FUG^nA,A)\iso\Hom_k(UG^nA,UA)=\Hom_k(T^nUA,UA)$$
If the switch from $\Der_k$ to $\Hom_k$ seems strange, remember that any linear map lifts to a uniquederivation from the free algebra. Dropping the superflous applications of$U$, we find the complex 8.3 is isomorphic to Beck's ``non-homogeneous''complex $$ 0\to \Hom_k(A,A)\to \Hom_k(TA,A)\to \Hom_k(T^2A,A)\ldots $$ \beqn \d^nf = f\. T^n\gka + \sum_{i=1}^n (-1)^if\. T^{n-i}\mu T^{i-1} + (-1)^{n-1}\gka\. Tf \eeqn In particular $\d^1f =f\.T\gka - f\.\gkm + \gka\.Tf$, which looks like the left side of 7.3, so this is the complex we will work with when doingdeformation theory in the context of triples. Note that the additivity of$T$ is implicit in the definition of the boundary operator. Before returning to deformation theory, we note that there is a product on the cochains of the non-homogeneous complex that makes it a graded ring. If $f\in \Hom_k(T^mB,C)$ and $g\in \Hom_k(T^nA,B)$, then define $f\circ g$ in $\Hom_k(T^{m+n}A,C)$ by \beqn f\circ g = f\. T^mg \eeqn It is quite easy to see that this circle product satisfies the Leibnitz formula \beqn \d(f\circ g) = f\circ\d g + (-1)^n\d f\circ g \eeqn Thus the circle product lifts to the level of homology [9]. This replaces both the cup product and the circle product in the classical theory [13],and will be useful in the obstruction theory to follow. Note that the associativity of $\gka$ may be written $\gka\mu = \gka\circ\gka$. \section{Obstruction theory} It now becomes easy to carry the obstruction theory of chapter I to our general setting, though we will dispense with power-series and only work with sequences of (coefficient) functions. Thus, a one-parameter algebraic deformation $F$ of a $T$-algebra $\gka :TA\to A$ is a sequence of maps $f_n:TA\to A$, $n\geq 0$, such that $f_0 =\gka$ , $F\.\mu = F\. TF$, and $F\.\eta = I$, i.e. \beqn f_n\.\mu = \sum_{i+j=n} f_i\circ f_j \quad (n\geq 0) \eeqn \beqn f_n\.\eta = 0 \quad (n\geq 1) \eeqn The latter says that the coefficient maps $f_n$ are normalized $1$-cochains and will play no further role here. The associativity formula 9.1 corresponds to 2.1 and may be written \beqn \d f_n = - \sum_{{i+j=n}\atop {i,j\neq n}} f_i\circ f_j \eeqn An immediate consequence of 9.3 is that the infinitesimal of the deformation must be a $1$-cocycle. If $\{f_m\}_{m 0$, and \beqn \Del f_n = \sum_{i+j=n} f_i\ten f_j \eeqn A deformation of a point is an approximation to another point in $C$. If $\{f_n\}_{n>0}$ is an $\infty$-deformation, then the formal sum $F = \sum f_n$ satisfies $\Del F = F\ten F$ and $\gke F = 1$. We will need to apply our generalized functors to Hom sets other than $(A,A)$, so let $M$ and $N$ be $k$-modules, and let $(M,N)$ be the cofree coalgebra over $\Hom_k(M,N)$. The adjunction $(M,N)\to \Hom_k(M,N)$ yields an evaluation map $(M,N)\ten M\to N$. This is just another way of saying that the elements of $(M,N)$ represent linear maps $M\to N$. Now let $T:\Msc\to \Msc$ be the triple defining our category of algebras $A$. Unless $T$ is additive there is no linear map $\Hom_k(M,N)\to \Hom_k(TM,TN)$ defining the action of $T$ on maps, but there is always a coalgebra map $\T:(M,N)\to (TM,TN)$ which agrees with $T$on points [7]. This coalgebra map is the additive enrichment of $T$. We will illustrate this construction using the tensor algebra functor $TM = k + M + M\ten M + M\ten M\ten M + \ldots$ Suppose $f$ in $(M,N)$ is an element such that $f = \sum f_{(1)}\ten f_{(2)}$ (Sweedler's notation [31]). Then $\T f$ in $(TM,TN)$ is defined by $\T f(1) = \gke f$ and \beqn \T f(m_1\ten m_2\ten\ldots\ten m_i) = \sum f_{(1)}m_1\ten f_{(2)}m_2 \ldots\ten f_{(i)}m_i \eeqn Note that if $f$ is a point in $(M,N)$ then $f_{(n)} = f$ for all $n$, so $\T$ acts just like the usual functor $T$. In particular, if $\{f_n\}$ is an $\infty$-deformation in $(M,N)$ then $\T$ and $T$ agree on the formal point $\sum f_n$. More generally, if $F = \sum f_nt^n$ is a formal power-series whose coefficients form a sequence of divided powers in $(M,N)$, then $\T F = TF$, though $\T$ is additive, and $T$ is not. This gets us back to deformation theory. A one-parameter algebraic deformation of a $T$-algebra $\gka :TA\to A$ is a sequence of divided powers $F = \{f_n\}$ over $\gka$ in $(TA,A)$ such that $F\.\mu = F\.\T F$ and $F\.\eta = I$. This leads to obstruction formulas just as in \S 9, except that the complex 8.4 is replaced by the non-homogeneous complex of coalgebras \beqn 0\to (A,A)\to (TA,A)\to (T^2A,A)\ldots \eeqn where $\T$ replaces $T$ in the boundary formula, and the circle product $\circ:(T^mA,A)\ten (T^nA,A)\to (T^{m+n}A,A)$ is defined by $f\circ g = f\.\T^mg$. The question of extending truncated deformations is more complicated in this setting, so we will work out the analogue of theorm 2.4 to illustrate how things work. Suppose $\{f_i\}_{i