Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 6 down vote accepted

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.

share|improve this answer
2  
welcome to MO!! –  SGP Apr 29 '11 at 1:33
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.