I am interested in eigenvalue problems for differential operators acting on one forms on closed two-dimensional manifolds and how they relate to eigenvalue problems of associated operators acting on scalar functions.

For the discussion I assume $M$ to be a compact two-dimensional Riemannian manifold without boundary. Let $\Delta^k=dd^*+d^*d$ be the De Rham Laplacian on the space $\Omega^k(M)$ of real-valued $k$-Forms. Here $d:\Omega^k(M) \to \Omega^{k+1}(M)$ is the exterior derivative and $d^*: \Omega^{k+1}(M) \to \Omega^{k}$ is its adjoint. For $k=0$ we get the Laplace Beltrami Operator acting on functions: $\Delta^0=d^*d$.

Consider the eigenvalue problem: Find $\lambda \in \mathbb{R}$ and an $1$-Form $\alpha$ such that $\Delta^1\alpha = \lambda \alpha$. Because of the identity $d^* \Delta^1 = \Delta^0 d^* $ any eigenform $\alpha$ of $\Delta^1$ yields an eigenfunction $f:= d^*\alpha$ of $\Delta^0$ for the same eigenvalue, provided that $\alpha$ is not co-closed.

Also, since $\Delta^1$ commutes with the Hodge star we have that with any eigenform $\alpha$, the $1$-form $*\alpha$ (imagine $\alpha$ being rotated pointwise by 90 degrees) is also a linearly independent eigenform for the same eigenvalue. Therefore all eigenspaces of $\Delta^1$ have even dimension and are invariant under the symmetry

$$\alpha \mapsto a \alpha + b (*\alpha) \qquad a,b \in \mathbb{R}$$

To sum up, the spectra of $\Delta^1$ and $\Delta^0$ are closely related: They are essentially the same except for the multiplicities of the eigenvalues. More precisely, any non-zero eigenvalue of the scalar operator $\Delta^0$ with multiplicity $m$ becomes an eigenvalue of the $1$-form operator $\Delta^1$ with multiplicity $2m$.

Now for the **question**: Does a similar relation hold for other differential operators? For example, I consider the Bochner Laplacian $\widehat{\Delta^1} = \nabla^* \nabla$ on $T^*M$. The Weizenböck identity

$$\Delta^1 \alpha = \widehat{\Delta^1} \alpha + K\cdot \alpha$$

shows that the Bochner Laplacian and the De Rham Laplacian on $1$-Forms differ only by a an operator that is a pointwise multiplication with the Gaussian curvature $K$. Moreover, the Bochner Laplacian also commutes with the Hodge star $*$, therefore all eigenspaces of $\widehat{\Delta^1}$ have even dimension and are invariant under the symmetry mentioned above. So far, the situation looks similar. Now, is there an operator $\widehat{\Delta^0}$ acting on $\Omega^0(M)$ and an operator $\widehat{d^*}: \Omega^1(M) \to \Omega^0(M)$ such that the following intertwining relation holds?

$$\widehat{d^*} \ \widehat{\Delta^1} = \widehat{\Delta^0} \ \widehat{ d^* }$$

In that case the the spectrum of the Bochner Laplacian on $1$-Forms would be essentially equal (up to multiplicity as discussed above) to the spectrum of the unknown scalar operator $\widehat{\Delta^0}$. Is this possible? More generally, for what other operators (apart from the De Rham Laplacian) is such a "reduction" possible?

Any references would be appreciated. Thanks.

**Update**: (in reply to the answer of Robert Bryant below)
I have tried to spell out the calculations leading to some of the results in the answer below in some more detail in order to understand them by myself and for future reference. Unfortunately I am not familiar with the principal symbol calculus, so I apologize for my low-level approach. My goal is to calculate the eigenvalues for the Bochner Laplacian numerically. From the implementation point of view it is easier to deal with scalar valued second order equations than with vector (or in this case 1-form valued equations). I think the special case below is instructive. I wonder if it is possible to avoid the need of putting restrictions on the metric. Of course, that would be great. But it would be also interesting to have an argument that says that it is impossible in the general case.

Anyway, In the special case $$\widehat{\Delta^1} \alpha := \nabla^*\nabla \alpha + L \alpha = \Delta^1 \alpha + (L-K) \alpha$$ $$\widehat{\Delta^0f} := \Delta^0 f + H f$$ $$\widehat{d^*}\alpha :=d^* \alpha + \langle \phi,\alpha \rangle $$ we get $$\begin{aligned} E \alpha &:= \widehat{d^*} \widehat{\Delta^1}\alpha - \widehat{\Delta^0} \widehat{d^*} \alpha \\& = \widehat{d^*}(\Delta^1\alpha+(L-K)\alpha)-\widehat{\Delta^0}(d^*\alpha + \langle \phi,\alpha\rangle) \\ & =d^*\Delta^1\alpha+d^*((L-K)\alpha)+\langle \phi,\Delta^1\alpha \rangle + \langle \phi, (L-K) \alpha \rangle \\ & -\Delta^0 d^*\alpha -Hd^*\alpha -\Delta^0\langle \phi,\alpha \rangle -H \langle \phi,\alpha \rangle\end{aligned}$$

Because of the identities $$ \begin{aligned} d^*( (L-K) \alpha ) &= (L-K)d^*\alpha -\langle d(L-K) ,\alpha \rangle \\ \Delta^0 \langle \phi,\alpha \rangle & = \langle \nabla^*\nabla \phi,\alpha \rangle + \langle \phi,\nabla^*\nabla \alpha \rangle - 2 \langle \nabla \phi, \nabla \alpha \rangle \\ \nabla^*\nabla &= \Delta^1 - K\\ d^*\alpha &= -\nabla^i \alpha_i = -g_{ab} g^{ak} g^{bi} \nabla_k\alpha_i = \langle -g, \nabla \alpha \rangle \\ d^*\Delta^1&=\Delta^0 d^* \end{aligned} $$ the third and second order terms in $\alpha$ cancel, yielding the following first-order operator: $$E\alpha = \langle -g(L-K-H) + 2\nabla \phi, \nabla \alpha \rangle + \langle (L-K-H)\phi - d(L-K)- \nabla^*\nabla \phi + K \phi, \alpha \rangle $$

Now I see that in order for the first-order terms to vanish for all $\alpha$ we need $\nabla\phi = f g$ for $f := \frac{1}{2}(L-K-H)$. Taking the covariant derivative of this equation yields $\nabla \nabla \phi = df \otimes g + df \otimes 0$ and taking the trace yields $-\nabla^*\nabla \phi = df$. Therefore, if we set $df = -K\phi$, the zeroth-order part reduces to the operator $$E\alpha = \langle 2f\phi -d(L-K),\alpha \rangle$$

I have two questions at this point: Why is $df =-K\phi$ necessary for $E$ to vanish? More important (since that leads to the restrictions on the metric): Why is $*\phi$ dual to a Killing field? Maybe this is obvious but i don't see it.