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A similar calculation can be done for the more general case when $\widehat{d^\ast}$ is allowed to be somewhat more general (but still underdetermined elliptic), but I won't go into that unless someone is interested.

A similar calculation can be done for the more general case when $\widehat{d^\ast}$ is allowed to be somewhat more general (but still underdetermined elliptic), but I won't go into that unless someone is interested. I'll just say that, aside from the metrics above, the only metrics that admit a nontrivial solution when $\widehat{d^\ast}$ is an underdetermined elliptic first order differential operator are the metrics of the form $$ g = \bigl( a\ (x^2{+}y^2) + 2b\ x + c\bigr)\ \bigl(dx^2 + dy^2\bigr) $$ where $a$, $b$, and $c$ are constants such that the open set $M$ in the $xy$-plane defined by $a\ (x^2{+}y^2) + 2b\ x + c > 0$ is nonempty. Thus, the metrics that allow this are very restricted.

The first thing to notice is that, since the principal symbol $\sigma^1_\xi$ of $\widehat{\Delta^1}$ is just scalar multiplication by $|\xi|^2$ and the principal symbol of $\widehat{d^\ast}$ is $\alpha\mapsto \xi\cdot\alpha$, it follows from the symbol calculus that the principal symbol of $\widehat{\Delta^0}$ must also be scalar multiplication by $|\xi|^2$, i.e., the differential operator $widehat{\Delta^0} - \Delta^0$ must be of order at most~$1$. However, it must also be self-adjoint, and this implies that it must be of order $0$, i.e., that $widehat{\Delta^0} = \Delta^0 + H$ (where '$H$' means scalar multiplication by a smooth function $H$ on $M$).

The first thing to notice is that, since the principal symbol $\sigma^1_\xi$ of $\widehat{\Delta^1}$ is just scalar multiplication by $|\xi|^2$ and the principal symbol of $\widehat{d^\ast}$ is $\alpha\mapsto \xi\cdot\alpha$, it follows from the symbol calculus that the principal symbol of $\widehat{\Delta^0}$ must also be scalar multiplication by $|\xi|^2$, i.e., the differential operator $\widehat{\Delta^0} - \Delta^0$ must be of order at most $1$. However, it must also be self-adjoint, and this implies that it must be of order $0$, i.e., that $\widehat{\Delta^0} = \Delta^0 + H$ (where '$H$' means scalar multiplication by a smooth function $H$ on $M$).

I thought a little bit about your question, which is phrased a little more generally than I like, but I decided to think about it with the restrictions that $\widehat{\Delta^0}$ be a differential operator and $\widehat{d^\ast}$ be $d^\ast$ plus a lower-order (i.e., zeroth order) differential operator. I also decided to think, not of the most general perturbation of $\widehat{\Delta^1}$, but of perturbations of the form $\widehat{\Delta^1} = \nabla^\ast\nabla +L$, where $L$ means scalar multiplication by a function $L$ on $M$.

Now, $\widehat{d^\ast}\alpha = d^\ast\alpha + \phi\cdot\alpha$ for some $1$-form $\phi$ on $M$. Under the given assumptions, it is not hard to see that the equation $$ \widehat{d^\ast}\widehat{\Delta^1} = \widehat{\Delta^0}\widehat{d^\ast} $$ implies that $\widehat{\Delta^0} = \Delta^0 + H$ for some function $H$ on $M$, and, expanding both sides, one finds that one must also have the identities $$ \nabla \phi = f\ g\qquad\text{and}\qquad df = -K\ \phi $$ for some function $f$ on $M$ while $$ L = K + |\phi|^2 - c\qquad\text{and}\qquad H = |\phi|^2 -2f - c $$ for some constant $c$.

I thought a little bit about your question, which is phrased a little more generally than I like, but I decided to think about it with the restrictions that $\widehat{\Delta^0}$ be a differential operator and $\widehat{d^\ast}$ be $d^\ast$ plus a lower-order (i.e., zeroth order) differential operator. I also decided to think, not of the most general perturbation of $\widehat{\Delta^1}$, but of perturbations of the form $\widehat{\Delta^1} = \nabla^\ast\nabla+L$, where $L$ means scalar multiplication by a function $L$ on $M$.

The first thing to notice is that, since the principal symbol $\sigma^1_\xi$ of $\widehat{\Delta^1}$ is just scalar multiplication by $|\xi|^2$ and the principal symbol of $\widehat{d^\ast}$ is $\alpha\mapsto \xi\cdot\alpha$, it follows from the symbol calculus that the principal symbol of $\widehat{\Delta^0}$ must also be scalar multiplication by $|\xi|^2$, i.e., the differential operator $widehat{\Delta^0} - \Delta^0$ must be of order at most~$1$. However, it must also be self-adjoint, and this implies that it must be of order $0$, i.e., that $widehat{\Delta^0} = \Delta^0 + H$ (where '$H$' means scalar multiplication by a smooth function $H$ on $M$).

Now, $\widehat{d^\ast}\alpha = d^\ast\alpha + \phi\cdot\alpha$ for some $1$-form $\phi$ on $M$. Under the given assumptions, it is not hard to see that the equation $$ \widehat{d^\ast}\widehat{\Delta^1} = \widehat{\Delta^0}\widehat{d^\ast} $$ implies that $\widehat{\Delta^0} = \Delta^0 + H$ for some function $H$ on $M$. (This is what I just explained above.)

Now, the operator $E = \widehat{d^\ast}\widehat{\Delta^1} - \widehat{\Delta^0}\widehat{d^\ast}$, when expanded out, is of first order (not second order, as you might have expected). Looking at the principal symbol of $E$ and setting this equal to zero gives $4$ equations, and these are equivalent to $\nabla\phi = f\ g$, where, for simplicity, I have set $f = \tfrac12(L-K-H)$. Taking the covariant derivative of both sides of $\nabla\phi = f\ g$ and using the definition of $K$, one gets $df = -K\ \phi$. Substituting this back into the expression for $E$, it reduces to a $0$-th order operator which turns out to be $E(\alpha) = (2f\phi +dK - dL)\cdot\alpha$. Setting this equal to zero gives $d(|\phi|^2 + K - L) = 0$, since $d\bigl(|\phi|^2\bigr) = 2f\phi$ (which is a consequence of $\nabla\phi = f\ g$). Thus, $L = K + |\phi|^2 - c$ for some constant $c$.

Thus, one finds that one must have the identities $$ \nabla \phi = f\ g\qquad\text{and}\qquad df = -K\ \phi $$ for some function $f$ on $M$ while $$ L = K + |\phi|^2 - c\qquad\text{and}\qquad H = |\phi|^2 -2f - c $$ for some constant $c$.