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Suppose $A=A_1dx_1+A_2dx_2$ is a 1-form connection in $\mathbb{R}^2$ and $D_A \phi=d\phi-iA\phi$ is the gauge covariant derivative with $\phi=\phi_1+i\phi_2$ is a complex scalar field. May I ask what the adjoint $D^*_A$ of the gauge covariant derivative? Thank you so much.

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For $D=d+A$,with respect to the usual inner product on $\mathbb{R}^2$ and the ones induced by it on differential forms, one has $D^{*}_{A}=-*D_{A} *$ where $*$ stands for the hodge star operator. For example, $$D^{*}_A (f_1dx_1+f_2dx_2)=-*D_{A} *(f_1dx_1+f_2dx_2)=-*D_{A} (f_1dx_2-f_2dx_1)=-*(\frac{\partial f_1}{\partial x_1}+\frac{\partial f_2}{\partial x_2}+A_1f_1+A_2f_2)dx_1\wedge dx_2=-(\frac{\partial f_1}{\partial x_1}+\frac{\partial f_2}{\partial x_2}+A_1f_1+A_2f_2).$$

To check, one considers turning off $A$, and making the test one form exact, i.e. $f_i=\frac{\partial \phi}{\partial x_i}$, and the above gives $d^{*}d$ to be the usual Laplacian operator on $\mathbb{R}^2$ but with a minus sign, which is indeed what one expects from general computations.

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