Connections

Arun Ram
Department of Mathematics
University of Wisconsin
Madison, WI 53706 USA
ram@math.wisc.edu

de Rham cohomology

Let $A$ be a commutative algebra. The de Rham cohomology of $A$ is the cohomology of the complex

where the $p$-differential forms of $A$ is

${\Omega }^p\left(A\right)={\Lambda }^p\left({\Omega }^1\right(A\left)\right),{\Omega }^1\left(A\right)=I/I^2,I=\ker (A\otimes A\to A),$

and $d$ is the unique antiderivation of degree 1 which extends

$\begin{array}{cccc}d:& A& ⟶& {\Omega }^1\left(A\right)\\ & x& \mapsto & x\otimes 1–1\otimes x\end{array}\text{and satisfies}{d}^2=0$.

Let $M$ be an $A$-module. A connection on $M$ is an $𝔽$-linear map

$\nabla :M⟶M{\otimes }_A{\Omega }^1\left(A\right)$ such that $\nabla \left(fm\right)=f\nabla \left(m\right)+m\otimes df$,

for $f\in A$, $m\in M$. There is a unique extension of $\nabla$ to

$\cdots ⟶M{\otimes }_A{\Omega }^{i–1}\left(A\right)\stackrel{\nabla }{⟶}M{\otimes }_A{\Omega }^i\left(A\right)\stackrel{\nabla }{⟶}M{\otimes }_A{\Omega }^{i+1}\left(A\right)⟶\cdots$

such that

$\nabla \left(x\omega \right)=\left(\nabla x\right)\omega +(–1{)}^{\mathrm{deg}\left(x\right)}xd\omega \text{,}\text{for }x\in M{\otimes }_A{\Omega }^p(A)\text{, }\omega \in {\Omega }^{\ell }(A).$

The curvature of $\nabla$ is

$R:M⟶M{\otimes }_A{\Omega }^2\left(A\right)\text{given by}R={\nabla }^1\circ {\nabla }^0,$

and $\nabla$ is flat if $R=0$.

If $\nabla$ is a flat connection then (??) is a complex and the de Rham cohomology of $(M,\nabla )$ is the homology of (??).

Then $A$ acts on ${\Omega }^1\left(A\right)$ by

$f(\sum g_i\otimes f_i)=\sum f{g}_i\otimes h_i=\sum g_i\otimes f{h}_i\text{mod}{I}^2$,

for $f\in A$ and $\sum g_i\otimes h_i\in I$. As $A$-modules

$\begin{array}{ccc}{\mathrm{Hom}}_A\left({\Omega }^1\right(A),A)& \stackrel{\sim }{⟶}& \mathrm{Der}\left(A\right)\\ \varphi & \mapsto & \varphi d\end{array}$

and, if ${\Omega }^1\left(A\right)$ is a reflexive $A$-module then

${\Omega }^1\left(A\right)={\mathrm{Hom}}_A\left(\mathrm{Der}\right(A),A)$.

Let $\nabla$ be a connection on $M$ and define

$\begin{array}{ccc}\mathrm{Der}\left(A\right)& ⟶& {\mathrm{End}}_{𝔽}\left(M\right)\\ \partial & \mapsto & {\nabla }_{\partial }\end{array}$ by ${\nabla }_{\partial }=(\mathrm{part}\otimes {\mathrm{id}}_M)\circ \nabla$,

so that ${\nabla }_{\partial }:M\stackrel{\nabla }{\to }{\Omega }^1\left(A\right){\otimes }_{A}M\stackrel{\partial \otimes {\mathrm{id}}_M}{\to }A{\otimes }_{A}M=M$. Then, for $f,f_1,f_2\in A$, $\partial ,{\partial }_1,{\partial }_2\in \mathrm{Der}\left(A\right)$ and $m\in M$,

${\nabla }_{\partial }\left(fm\right)=\partial \left(f\right)m+f{\nabla }_{\partial }\left(m\right),\text{and}{\nabla }_{f_1{\partial }_1+f_2{\partial }_2}\left(m\right)=f_1{\nabla }_{{\partial }_1}\left(m\right)+f_2{\nabla }_{{\partial }_2}\left(m\right).$

If ${\Omega }^1\left(A\right)$ is a reflexive $A$-module then the connection $\nabla$ is determined by the map $\partial \to {\nabla }_{\partial }$ with the properties (a) and (b).

Derivations

Let $A$ be a ring and let $M$ be an $(A,A)$-bimodule. A derivation is an $\mathrm{Fopf}$-linear map $\mathrm{part}?A?M$ such that

$\partial \left(f_1{f}_2\right)=f_1\partial \left(f_2\right)+\partial \left(f_1\right)f_2,\text{for }{f}_1,f_2\in A.$

Let

$I=I=\ker \left(\begin{array}{ccc}A\otimes A& ⟶& A\\ f_1\otimes f_2& \mapsto & f_1{f}_2\end{array}\right)\text{and}\begin{array}{cccc}d:& A& ⟶& I\\ & f& \mapsto & f\otimes 1–1\otimes f\end{array}.$

Then $d:A\to I$ is a derivation and if $\partial :A\to M$ is a derivation then there exists a unique $(A,A)$-module map $\varphi :I\to M$ such that $\partial =\varphi \circ d$.

$\begin{array}{ccc}A& \stackrel{\partial }{⟶}& M\\ d\downarrow & \nearrow \varphi & \\ I& & \end{array}\text{In other words,}\mathrm{Der}(A,M)={\mathrm{Hom}}_{A\otimes A}(I,M)$.

If $A$ is commutative and $M$ is an $A$-module then $M$ is an $(A,A)$-bimodule on which $I$ acts by $0$ (since $(f\otimes 1–1\otimes f)m=\mathrm{fm}–\mathrm{mf}=0$). In fact, since the multiplication map $A\otimes A\to A$ is surjective, $A\otimes A\otimes A)/I$ and $A$-modules are the same thing as $A\otimes A$-modules on which $I$ acts by 0, and (*) becomes

$\mathrm{Der}(A,M)={\mathrm{Hom}}_{A\otimes A}(I,M)={\mathrm{Hom}}_A(I/I^2,M)$.

Thus, $\mathrm{Der}\left(A\right)$ is dual to the space ${\Omega }^1=I/I^2$ of 1-forms for $A$,

$\mathrm{Der}\left(A\right)={\mathrm{Hom}}_A({\Omega }^1,A)$.