I have a doubt regarding Pag. 119/120 of Spin Geometry by Lawson and Michelson. Namely, I am trying to calculate the Dirac Operator on the Euclidean space $\mathbb{R^2}$.
The Clifford Algebra is $Cl_2=\{1,e_1,e_2,e_1e_2\}$ which we can identify with $\mathbb{H}$ via $1=1, i=e_1, j=e_2,k=e_1e_2$ and to $\mathbb{C}\oplus\mathbb{C}$ via $(1,0)=1,(i,0)=e_1e_2,(0,1)=e_1,(0,i)=e_2$.
Now, the irreducible representation $\rho:Cl_2\to End_\mathbb{R}(\mathbb{H})$ is given by left multiplication.
The (unique) spin structure is trivial andgives the trivialization of the spinor structure as follows:
$$(\mathbb{R}^2\times Spin_2)\times_\rho \mathbb{H}\to\mathbb{R}^2\times\mathbb{H}$$ such that $[(x,y),z]\mapsto(x,\rho(y)(z))$ and inverse $(x,z)\mapsto[(x,1),z]$. Analogously, for the tangent bundle we have: \begin{align*} &(\mathbb{R}^2\times Spin_2)\times_{Ad} \mathbb{R}^2\to\mathbb{R}^2\times\mathbb{R}^2\\ &[(x,y),z]\mapsto(x,Ad(y)(z)). \end{align*} Clifford multiplication should have the form (the first thing in parenthesis represents the tangent bundle, the second the spinor bundle): \begin{align*} (\mathbb{R}^2\times\mathbb{R}^2)\times(\mathbb{R}^2\times\mathbb{H})\to\mathbb{R}^2\times\mathbb{H}\\ ((x,y),(x,z))\mapsto(x,\rho(y_1 e_1+y_2 e_2)z). \end{align*} Hence, the Dirac operator would be: $$ D\sigma=\sum_{i=1}^2e_i\cdot\frac{\partial \sigma}{\partial x_i}=i\frac{\partial \sigma}{\partial x_1}+j\frac{\partial \sigma}{\partial x_2}, $$ where $\sigma$ is assumed to be a map from $\mathbb{R}^2$ to $\mathbb{H}$.
My problem is that this result, after passing to $\mathbb{C}\oplus\mathbb{C}$, does not agree with the usual Cauchy Riemann operator as stated in the reference (there is a sign problem which does not occur if we multiply on the right instead). Any help is much appreciated.
Edit: Apparently, I misread (several times) the isomorphism $\mathbb{H}\cong\mathbb{C}\oplus \mathbb{C}$ in Spin Geometry by Lawson and Michelson. Instead of $(i,0)=e_1e_2$ we should put $(i,0)=e_2e_1$. Now, everything should be fine.