# A detail in the construction of the coarse index of a Dirac operator in “Roe: An Index Theorem on Open Manifold, I”

Hi,

I'm currently wreading "Roe: An Index Theorem on Open Manifolds, I, J. Differential Geometry 27 (1988), p. 87-113" and there is a detail in the construction of the coarse index of a Dirac operator that I do not understand.

In Section 4 he explains the construction of abstract indices: Suppose we are given an algebra A with unit and an ideal $B \subset A$. Let M and N be finite, projective right A-modules and let $P: M \to N$ be A-linear (this is bold, because this is the thing which I do not understand later when we look at the Dirac operator). P is called abstractly elliptic if there is a parametrix $Q: N \to M$ such that $Im(QP-1) \subset M \otimes B$ and $Im(PQ-1) \subset N \otimes B$, i.e. P is an isomorphism after tensoring with A/B. He then goes on and explains how to get an index $\operatorname{Ind}(P) \in K^{alg}(B)$.

In Section 7 he then uses this construction to get a coarse index of a Dirac operator: Let $S \to M$ be a Clifford bundle, equipped with a grading $\eta$ and let D be the corresponding Dirac operator. In Lemma 7.6 he shows then (under the assumption of bounded geometry) that the operator D is abstractly elliptic between the $\mathcal{U}(S)$-modules given by the eigenprojections $(1+\eta)/2$ and $(1-\eta)/2$, i.e. has an index in $K^{alg}(\mathcal{U}_{-\infty}(S))$.

Now the thing that I do not understand is that for $D_+: S_+ \to S_-$ being abstractly elliptic means that $D_+$ is, among other things, $\mathcal{U}(S)$-linear. This algebra $\mathcal{U}(S)$ is the algebra of all uniform operator, i.e. all operator $L: C_c^\infty(S) \to C^\infty(S)$ which have for a fixed k (the order of L) a continuous extension to a operator $H^r(S) \to H^{r-k}(S)$ for each r (this extensions must also be a quasilocal operator, but I do not think that this matters here), where $H^r$ are the Sobolev spaces. $\mathcal{U}_{-\infty}(S)$ are the operator of order $-\infty$.

Why does the Dirac operator, resp. just $D_+$ and $D_-$, commute with all of $\mathcal{U}(S)$? Or do I just not correctly understand the construction?

This is a "left and right" issue. The point is that if $A$ is an algebra, the operation of left multiplication by a fixed $a\in A$ (considered as a map $A\to A$) is linear as a map of right $A$-modules.
• Thanks for your answer. To reformulate it a bit, since I did not understand it first: The Dirac operator is meant to act by left multiplication between the right $\mathcal{U}(S)$-modules of uniform operators $S \to S_+$ and $S \to S_-$. – AlexE Apr 29 '11 at 16:47