Compute the index of the Dirac operator on $C_0(R^2)$ to obtain Bott element in $K_0$

Question

I am studying the paper of Baum-Connes-Higson to understand the Connes-Kasparov conjecture. In example 4.23, they discuss the case $G=\mathbb{R}^2$. I have constructed the Dirac operator, but I’m stuck with computing its index. According to the example in the article, we should obtain the Bott element in $K_0(C_r^*G))$, but I cannot see how to get there.

For the computation, we have the following setup (in the notation of [BCH]) $G=\mathbb{R}^2$, $K= \{ e \}$, $\mathfrak{g} = \mathfrak{k} \oplus \mathfrak{p}$ with $\mathfrak{k} =0$ and $\mathfrak{p} = \mathbb{R}^2$.

We take the following representation of the Clifford algebra on the spinor space $\Delta_2 = \mathbb{C}^2$: $$ e_1 \mapsto \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \; \; e_2 \mapsto \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}. $$

The spinor bundle is the trivial bundle $S = \mathbb{R}^2 \times \mathbb{C}^2$, and the sections are given by: $$ C_c^\infty(\mathbb{R}^2,S) =\left\{ \begin{pmatrix} f \\ g \end{pmatrix}: \mathbb{R}^2 \to \mathbb{C}^2 \right\}. $$ The bundle splits as $S = S^+ \oplus S^-$, where $f: \mathbb{R}^2 \to S^+$ and $g: \mathbb{R}^2 \to S^-$. The Dirac operator $D: \Gamma(S) \to \Gamma(S)$ is given by: $$ D= 2 \begin{pmatrix} 0 & \frac{\partial}{\partial \overline{z}} \\ -\frac{\partial}{\partial z} & 0 \end{pmatrix}, $$ where $z= x + i y$. Now as in [BCH], we Fourier transform, so $C_r^*(\mathbb{R}^2)$ becomes $C_0(\mathbb{R}^2)$, and the Dirac operator becomes: $$ D= i \begin{pmatrix} 0 & z \\ - \overline{z} & 0 \end{pmatrix}. $$ To calculate the index we only need $D^+: C_c^\infty(\mathbb{R}^2,S^+) \to C_c^\infty(\mathbb{R}^2,S^-)$. These modules complete to the Hibert modules $C_0(\mathbb{R}^2)$ over $A= C_0(\mathbb{R}^2)$, so we end up calculating the index of the operator: $$ D^+: C_0(\mathbb{R}^2) \to C_0(\mathbb{R}^2), \; f \mapsto i z \cdot f. $$

According to [BCH], the index of $D^+$ should give the Bott element, so: $$ \text{Index}(D^+) = \text{ker}[ iz ] - \text{ker}[ -i\overline{z}] = \left[ \frac{1}{1+z^2} \begin{pmatrix} 1 & \overline{z} \\ z & |z|^2 \end{pmatrix} \right] - \left[ \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\right]. $$

I first tried to compute it by explicitly finding a $T$ such that $1-TD^+$ and $1-D^+T$ are compact, but I couldn't find such $T$. Then I tried to look at the graph projector $G(tD)$, see (def 4.1, Aastrup) and use theorem 4.15. This seems to give the right result, but we need here that 0 is an isolated point in the spectrum of $D^+$, which is not true in our case.

Is there a way to fix this, or another way to calculate the index?

Answer 1

Solved it! (Edit: typo in formula for $\tilde{S}$ and $1-\tilde{T}\tilde{S}$, added i's and signs to obtain exactly Bott element)

Denote by $E=C_0(\mathbb{R}^2)$ our Hilbert module over $A=C_0(\mathbb{R}^2)$. First, we should make $D$ into a bounded operator, we take:

$$ T=\frac{iz}{\sqrt{1+|z|^2}}. $$

Also define: $$ S = \frac{-i\overline{z}}{1+|z|^2}, $$ then $1-TS=1-ST= \frac{1}{1+|z|^2}$, which is a compact operator on $E$. Hence $T$ is $A$-Fredholm, according to (def 3.1, Exel).

Now as $T$ is non-regular, we apply the procedure of (lemma 3.8, Exel) and define: $$ \tilde{T} = \frac{1}{\sqrt{1+|z|^2}}\begin{pmatrix} i z & 0 \\ - i & 0 \end{pmatrix}, \tilde{S} = \frac{1}{\sqrt{1+|z|^2}}\begin{pmatrix} -i \overline{z} & i \\ 0 & 0 \end{pmatrix}. $$

Then the index of $D$ is then given by (see definition 3.10): $$ [\ker \tilde{T} ] - [ \ker \tilde{T^*}] = [1-\tilde{S}\tilde{T}] - [ 1 - \tilde{T} \tilde{S} ] \in K_0(A). $$

We have: $$ 1-\tilde{T} \tilde {S} = 1 - \frac{1}{1+|z|^2} \begin{pmatrix} i z & 0 \\ -i & 0 \end{pmatrix} \begin{pmatrix} -i\overline{z} & i \\ 0 & 0 \end{pmatrix} = \frac{1}{1+|z|^2} \begin{pmatrix} 1 & z \\ \overline{z} & |z|^2 \end{pmatrix} $$ and $$ 1-\tilde{S}\tilde{T} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, $$

so we get precisely the Bott element.