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geometricK
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Let $D=i\frac{d}{dx}$ be the Dirac operator on $\mathbb{R}$, acting on the spinor bundle $S=\mathbb{R}\times\mathbb{C}$$\mathbb{R}\times\mathbb{C}$. The bounded operator $\frac{D}{\sqrt{D^2+1}}$$F=\frac{D}{\sqrt{D^2+1}}$ has a coarse index in the group $K_1(C^*(|\mathbb{R}|))$,

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

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 this fact. 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.

Let $D=i\frac{d}{dx}$ be the Dirac operator on $\mathbb{R}$, acting on the spinor bundle $S=\mathbb{R}\times\mathbb{C}$. The bounded operator $\frac{D}{\sqrt{D^2+1}}$ has a coarse index in the group $K_1(C^*(|\mathbb{R}|))$, 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, it would be nice and instructive to see a detailed calculation showing this fact. 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 $\mathbb{T}^n$.

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

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.

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geometricK
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  • 10
  • 11

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 $S=\mathbb{R}\times\mathbb{C}$. The bounded operator $\frac{D}{\sqrt{D^2+1}}$ has a coarse index in the group $K_1(C^*(|\mathbb{R}|))$, 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, it would be nice and instructive to see a detailed calculation showing this fact. 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 $\mathbb{T}^n$.

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