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Now i I see that in order for the first-order terms to vanish for all

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.

12 expanded calculations in special case

$\alpha$\alpha \mapsto a \alpha + b (*\alpha) \qquad a,b \in \mathbb{R}$mathbb{R}$$ \Delta^1 \Delta^1 \alpha = \widehat{\Delta^1} \alpha + K\cdot \alphaalpha$$ $\widehat{d^*} $\widehat{d^*} \ \widehat{\Delta^1} = \widehat{\Delta^0} \ \widehat{ d^* }$$ Update: (in reply to the answer of Robert Bryant below):below) I have tried to spell out the calculations leading to some of the results in the answer below in some more detail in order to understand them by myself and for future reference. Unfortunately I am not familiar with the principal symbol calculus, so I apologize for my low-level approach. My goal is to calculate the eigenvalues for the Bochner Laplacian numerically. From the implementation point of view it is easier to deal with scalar valued second order equations than with vector (or in this case 1-form valued equations). I think the special case below is instructive. I wonder if it is possible to avoid the need of putting restrictions on the metric. Of course, that would be great. But it would be also interesting to have an argument that says that it is impossible in the general case. Anyway, In the special case\widehat{\Delta^1} \widehat{\Delta^1} \alpha := \nabla^*\nabla \alpha + L \alpha = \Delta^1 \alpha + (L-K) \alpha,\widehat{\Delta^0f} \widehat{\Delta^0f} := \Delta^0 f + H f, and\widehat{d^*}\alpha \widehat{d^*}\alpha :=d^* \alpha + \langle \phi,\alpha \rangle$$$\widehat{d^*} \begin{aligned} E \alpha &:= \widehat{d^*} \widehat{\Delta^1}\alpha - \widehat{\Delta^0} \widehat{d^*} \alpha \\& = d^*(\Delta^1\alpha + (L-K)\alpha) \widehat{d^*}(\Delta^1\alpha+(L-K)\alpha)-\widehat{\Delta^0}(d^*\alpha + \langle \phi, phi,\alpha\rangle) \Delta^1\alpha \& =d^*\Delta^1\alpha+d^*((L-K)\alpha)+\langle \phi,\Delta^1\alpha \rangle + \langle \phi, (L-K)\alpha 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 \widehat{\Delta^0} \widehat{d^*} begin{aligned}d^*( (L-K) \alpha ) &= (L-K)d^*\alpha -\langle d(L-K) ,\alpha \rangle \\\Delta^0 \langle \phi,\alpha \rangle & = \Delta^0(d^*\alpha +\langle langle \nabla^*\nabla \phi,\alpha \rangle) rangle + H (d^* \langle \phi,\nabla^*\nabla \alpha + \rangle - 2\langle \phi,\alpha nabla \rangle) Since we require \widehat{d^*} phi, \widehat{\Delta^1}\alpha nabla \alpha \rangle \\\nabla^*\nabla &= \Delta^1 - K\\d^*\alpha &= -\nabla^i \alpha_i = -g_{ab} g^{ak} g^{bi} \widehat{\Delta^1} nabla_k\alpha_i = \widehat{d^*} langle -g, \alphanabla \alpha \rangle \\d^*\Delta^1&=\Delta^0 d^* \end{aligned}that the third and second order terms in \alpha$ cancel, yielding thelast two equations implyfollowing first-order operator:$d^*((L-K)\alpha) + $E\alpha = \langle -g(L-K-H) + 2\nabla \phi, \Delta^1+(L-K)\alpha \rangle =\Delta^0\langle nabla \phi,\alpha alpha \rangle +H \langle (d^* L-K-H)\phi - d(L-K)- \alpha nabla^*\nabla \phi + K \langle phi, \phi,\alpha alpha \rangle)$

From here onwards it is not clear to me how to obtain the condition that rangle $K\phi$ has to be exact and deduce the other relationships $Now i see that in order for the answer below. I assume it has first-order terms to follow from some manipulations of this formula involving vanish for all$\alpha$ we need $\nabla\phi = f g$ for $f := \frac{1}{2}(L-K-H)$.Taking the derivatives covariant derivative of the scalar products in it. As an alternative, applying this equation yields $*$\nabla \nabla \phi = df \otimes g + df \otimes 0$ and integrating I arrived at taking the trace yields $\Delta^1 -\nabla^*\nabla \phi + (L-K-H)\phi =df$. Therefore, if we set $df = -dH$. @Robert: Could you please provide some hints in order K\phi$, the zeroth-order part reducesto point my thinking into the right direction?operator $$E\alpha = \langle 2f\phi -d(L-K),\alpha \rangle$$ 11 Expanded question in a special case Update: (in reply to answer of Robert Bryant below): In the special case$\widehat{\Delta^1} \alpha := \nabla^*\nabla \alpha + L \alpha = \Delta^1 \alpha + (L-K) \alpha$,$\widehat{\Delta^0f} := \Delta^0 f + H f$, and$\widehat{d^*}\alpha :=d^* \alpha + \langle \phi,\alpha \rangle $ we get$\widehat{d^*} \widehat{\Delta^1}\alpha = d^*(\Delta^1\alpha + (L-K)\alpha) + \langle \phi, \Delta^1\alpha + (L-K)\alpha \rangle $$\widehat{\Delta^0} \widehat{d^*} \alpha = \Delta^0(d^*\alpha +\langle \phi,\alpha \rangle) + H (d^* \alpha + \langle \phi,\alpha \rangle)$ Since we require $\widehat{d^*} \widehat{\Delta^1}\alpha = \widehat{\Delta^1} \widehat{d^*} \alpha$, the last two equations imply: $d^*((L-K)\alpha) + \langle \phi, \Delta^1+(L-K)\alpha \rangle =\Delta^0\langle \phi,\alpha \rangle + H (d^* \alpha + \langle \phi,\alpha \rangle)$ From here onwards it is not clear to me how to obtain the condition that $K\phi$ has to be exact and deduce the other relationships in the answer below. I assume it has to follow from some manipulations of this formula involving the derivatives of the scalar products in it. As an alternative, applying$*$and integrating I arrived at$\Delta^1 \phi + (L-K-H)\phi = -dH\$. @Robert: Could you please provide some hints in order to point my thinking into the right direction?

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