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Morita equivalence for (commutative) unital algebras


\begin{defn} If $A$, $B$ are \emph{unital} $\bC$-algebras, we say that they ... ...modules. We say that such an $\sE$ is an \lq\lq equivalence bimodule''. \end{defn}

In general, we may choose $\sF \isom \sE^\3 := \Hom_A(\sE, A)$ to be the ``dual'' right $A$-module with a specified action of $B$. We can then identify $\End_A\sE \isom B$.


\begin{fact} $\End_A\sE \isom B$ whenever $\sE$ is an equivalence $B$-$A$-bimodule. \end{fact}


\begin{fact} Since $A$, $B$ are unital, each $\sE$ is finitely generated and projective (and full). \end{fact}


\begin{remark} There is a C*-version, due to Rieffel, whereby all bimodules ar... ...ry in the unital case: remember that $M$ is taken to be compact. \end{remark}


\begin{notn} For isomorphism classes $[\sE]$ of bimodules, we form the set \b... ...\Pic(A) := \Mrt(A,A)$; this is called the \lq\lq Picard group'' of $A$. \end{notn}

We call an $A$-bimodule $\sE$ symmetric if the left and right actions are the same: $a \lt x = x \rt a$ for $x \in \sE$ and $a \in A$. When $A$ is commutative, a symmetric $A$-$A$-bimodule can be called, more simply, an ``$A$-module'' --as we have already been doing. Even when $A$ is commutative, an $A$-$A$-equivalence bimodule $\sE$ need not be symmetric. Indeed, suppose $\phi,\psi \in \Aut(A)$. Then we define ${}_\phi \sE_\psi$ to be the same vector space $\sE$, but with the bi-action of $A$ on $\sE$ twisted as follows:

\begin{displaymath} a_1 \lt a_0 \rt a_2 := \phi(a_1) a_0 \psi(a_2) \end{displaymath}

For $\phi = \psi = \id$, this is the original $A$-$A$-bimodule (when either $\phi = \id$ or $\psi = \id$, we shall not write that subscript). In particular, we can apply this twisting to $\sE = A$ itself.


\begin{lem} If $A$ is a unital algebra, there exists an $A$-$A$-bimodule isomorphism $\th \: A \to {}_\phi A$ if and only if $\phi$ is inner. \end{lem}


\begin{proof} If $\th\: A \to {}_\phi A$ is an $A$-bimodule isomorphism, then ... ... so that $\phi(a) = u a u^{-1}$, where $u = \th(1)$ is invertible. \end{proof}

Thus the ``outer automorphism group'' $\Out(A) := \Aut(A)/\Inn(A)$ classifies the asymmetric $A$-bimodules. When $A$ is commutative, so that $\Inn(A)$ is trivial, this is just $\Aut(A)$.

Recall that

\begin{displaymath} \Aut(C(M)) \isom \Homeo(M), \qquad \Aut(C^\infty(M)) \isom \Diff(M), \end{displaymath}

where $\phi(f) \: x \mapsto f(\phi^{-1}x)$ for $f \in C(M)$. We shall write $\Pic_A(A)$, following [BW], to denote the isomorphism classes of symmetric $A$-bimodules. (This repairs an oversight in [GVF, Chap. 9], which did not distinguish between $\Pic(A)$ and $\Pic_A(A)$, as was pointed out to me by Henrique Bursztyn.)


\begin{fact} $\Pic(A) \isom \Pic_A(A) \rtimes \Aut(A)$ as a semidirect product... ...t ([\sF], \psi) = ([{}_\psi\sE_\psi \ox_A \sF], \phi \circ \phi)$. \end{fact}

The proof is not difficult, but we refer to the paper [BW].


\begin{lem} For $A = C(M)$ or $C^\infty(M)$, $\Pic_A(A) \isom \rH^2(M;\bZ)$. \end{lem}


\begin{proof} Since invertible $A$-modules $\sL$ are given by $\sL = \Ga(M, L)... ... L_2)$, it is again a module of sections for a $\bC$-line bundle. \end{proof}

next up previous contents
Next: Classification of spinor modules Up: Spinor modules over compact Previous: The existence of structures   Contents
Pawel Witkowski 2006-03-14