# Coarse index of Dirac operator on $\mathbb{R}$

Let $$D=i\frac{d}{dx}$$ be the Dirac operator on $$\mathbb{R}$$, acting on the spinor bundle $$\mathbb{R}\times\mathbb{C}$$. The bounded operator $$F=\frac{D}{\sqrt{D^2+1}}$$ has a coarse index

$$\text{Ind}(F)\in K_1(C^*(|\mathbb{R}|)),$$

where $$C^*(|\mathbb{R}|)$$ is the Roe algebra of the metric space (as opposed to the group) $$\mathbb{R}$$.

It has been stated in various places (for example, Higson & Roe's book "Analytic $$K$$-Homology" chapter 12) that this coarse index in fact generates the group $$K_1(C^*(|\mathbb{R}|))\cong\mathbb{Z}$$.

However, I haven't seen this explicitly computed anywhere, and it would be nice and instructive to see a detailed calculation showing. For example, the analogous fact that the Dirac operator on $$\mathbb{R}^n$$ has non-trivial coarse index can be used to show that there exists no Riemannian metric of positive scalar curvature on $$\mathbb{T}^n$$.

To be specific, my question is just about the one-dimensional case, from which I would assume higher-dimensional computations follow:

Question: Show that $$\text{Ind}(F)$$ generates $$K_1(C^*(|\mathbb{R}|))\cong\mathbb{Z}$$.

So if there are experts in the area reading this who wouldn't mind sharing their thoughts, that would be very helpful.

Decompose $$\mathbb{R}$$ as the union of the rays $$\mathbb{R}^+ = [0, \infty)$$ and $$\mathbb{R}^- = (-\infty, 0]$$, intersecting at a point. This decomposition admits Mayer-Vietoris sequences in both K-homology and and coarse K-theory, and there is a coarse index map from the former sequence to the latter. This gives a commuting square:
$$\require{AMScd} \begin{CD} K_1(\mathbb{R}) @>Index>> K_1(C^*(\mathbb{R})) \\ @VVM{V}V @VVM{V}V \\ K_0(\text{point}) @>Index>> K_0(C^*(\text{point})) \end{CD}$$
Now, the Mayer-Vietoris boundary maps are isomorphisms because the rays $$\mathbb{R}^\pm$$ have trivial K-homology and trivial coarse K-theory - this follows from homotopy invariance of K-homology and an Eilenberg swindle argument for coarse K-theory. The group $$K_1(\mathbb{R})$$ is generated by the K-homology class of the Dirac operator, and a quick calculation shows that the Mayer-Vietoris boundary map in this case is just the ordinary boundary map induced by the inclusion of a point into $$\mathbb{R}$$. A difficult but standard K-homology calculation shows that this boundary map sends the Dirac class to a generator of $$K_0(\text{point})$$, i.e. a Fredholm operator of index 1.
That completes the calculation, up to the construction of the Mayer-Vietoris sequences and and the index maps between them. I worked all that out in my PhD thesis, and you can see a preprint here. By inductively chopping $$\mathbb{R}^n$$ into half-spaces you get the Gromov-Lawson theorem about the torus for free, and the argument has a nice and geometric flavor to it. That said, it's hard to turn the argument into explicit formulas - calculating the boundary of the Dirac class is hard, and I doubt it's any cleaner to go the other way around the diagram.
• Thanks - do you have a reference for the difficult but standard calculation that you refer to, showing that the $K$-homology class of the Dirac operator on $\mathbb{R}$ maps to an operator of index $1$ under $MV$? – ougoah Jun 15 '19 at 15:31
• The assertion that the ordinary K-homology boundary map sends the Dirac class to $1$ is basically proposition 9.6.6 in Higson and Roe's Analytic K-homology, though the proof with all details uses most of the machinery built in chapters 8 and 9. Passing from this to a calculation with Mayer-Vietoris boundary maps is sort of folk wisdom in operator K-theory; I honestly don't know a place where it's all written out other than my PhD thesis, available here: pwsiegel.github.io/resources/paul-siegel-phd-thesis.pdf. This is not a claim of priority; someone else might know another reference. – Paul Siegel Jun 15 '19 at 17:29