# How to construct a scalar differential operator having the same spectrum as a non-scalar differential operator exploiting symmetries?

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.

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What kind of restrictions are you willing to assume on your operators? For example, do you want them to be differential operators? Do you want $\widehat{\Delta^0}$ to be self-adjoint? Do you want $\widehat{d^\ast}$ to be first order, or, even more restrictive, to be $d^\ast$ plus a lower-order term? There are certainly special cases in which such formulae hold. For example, if $K$ is constant, then you can just take $\widehat{d^\ast} = d^\ast$ and $\widehat{\Delta^0} = \Delta^0 - K$. This trivial case is not the only one that works, but there aren't many more with these restrictions. –  Robert Bryant Sep 4 '11 at 0:24
@Alexander: I'll try to answer your questions, but for ease of reference (and to avoid running up against the character limit on comments), I will do it by editing my answer below, adding some sentences that may help. –  Robert Bryant Sep 6 '11 at 1:31

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 $c$.

Obviously, there is always the trivial solution $(\phi,f) = (0,0)$, which gives the well-known intertwining of the Hodge Laplacian on $1$-forms and $0$-forms. More interesting is the case when there are nontrivial solutions $(\phi,f)$ to the above equations. This puts severe restrictions on the metric $g$, but these can be understood.

Let's assume that $M$ is connected and complete. Then the pair of equations $\nabla\phi = f\ g$ and $df = -K\ \phi$ forms a linear total differential system, so if the pair $(\phi,f)$ vanishes anywhere, it vanishes identically. Let's assume that it does not. The equation $\nabla\phi = f\ g$ implies that $d\phi = 0$ and that the vector field $F$ that is $g$-dual to $\ast\phi$ is a Killing field. Thus, $(M,g)$ is, at least locally, a surface of revolution. Any fixed points of $F$ are isolated elliptic points. Write $\phi = du$ for some function $u$ (at the moment locally defined). Note that the critical points of $u$ are nondegenerate and of index 0 or 2. It is then not hard to show that $|\phi|^2 = a(u)$ for some function $a$ and that, in the region where $a>0$, the metric $g$ takes the form $$g = \frac{du^2}{a(u)} + a(u)\ d\psi^2$$ for some (locally defined) function $\psi$. In these coordinates, one has $\phi = du$, $f = \tfrac12a'(u)$, and $K = -\tfrac12a''(u)$. One also has $L = a(u) -\tfrac12a''(u)- c$ and $H = a(u) - a'(u) - c$.

Conversely, if one starts with a function $a$ positive on some domain on the $u$-line, then the above formulae give a solution to the intertwining equation. If $a$ is positive and periodic on the $u$-line, then one obtains a solution on the torus.

One can also obtain solutions on the $2$-sphere: If $a$ is smooth and satisfies $a(u_0) = a(u_1) = 0$ (where $u_0 < u_1$) while $a$ is positive between $u_0$ and $u_1$ and satisfies $a'(u_0) = -a'(u_1) = b >0$, then the metric $$g = \frac{du^2}{a(u)} + \frac{4a(u)}{b^2}\ d\theta^2$$ defines a smooth metric on the $2$-sphere with $(u,\theta)$ as 'polar coordinates' (with $\theta$ being $2\pi$-periodic and $u_0\le u\le u_1$, with $u=u_i$ corresponding to the 'poles' of rotation), and this gives a family of solutions to the intertwining equation.

To get $\widehat{\Delta^1}$ to be the Bochner Laplacian, one must have $L = 0$, which is equivalent to $a''(u) = 2\bigl(a(u)-c\bigr)$. This has 'spherical solutions', such as $$a(u) = c - e\ \cosh\bigl(\sqrt{2}\ u\bigr),$$ where the constants $c$ and $e$ satisfy $c > e > 0$. Thus, there is a nontrivial $2$-parameter family of metrics on the $2$-sphere such that the Bochner Laplacian has the desired intertwining property.

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.

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Thanks a lot for your answer . I'm still trying to follow your arguments. Therefore I updated my question. You are right, I should have put restrictions on the operators. I would be happy if $\widehat{\Delta^0}$ could be constructed as a self-adjoined scalar second order differential operator. It wouldn't matter if this requires $\widehat{d^*}$ to be something more complicated than some simple modification of $d^*$, but of course it is natural to try simple modifications first. In fact, I had also been trying to obtain a result along the lines you suggested, but I didn't get very far. –  Alexander Vais Sep 5 '11 at 15:24
First, i want to thank you for the additional remarks and the effort you put into the answer. Now at (at least for me) some questions remain: First: Why is $df = -K\phi$ necessary, rather than $2f\phi - d(L-K) + df + K \phi = 0$? (I can't see why this is implied by the definition of $K$). Second and probably more important in the reasoning: Why does $\nabla \phi = f g$ imply that $\psi=*\phi$ has to be dual to a Killing vector field? I guess there is an easier way to see this, rather than checking $\nabla_i \psi_j+\nabla_j\psi_i = 0$ in local coordinates? –  Alexander Vais Sep 21 '11 at 10:25
You are welcome. It's a somewhat interesting problem, so I didn't mind thinking about it. I don't have time to answer all your questions since I'm about to board a flight back to the US, but let me point out that $\nabla \phi = f g$ implies $df = - K\phi$ because, when you write out $\nabla^2\phi = \nabla (f g)$ and use the fact that $\nabla^2$ is essentially the curvature of $g$, i.e., $K$, you'll see why it is true. (There is some index juggling necessary here, but you should sort this out yourself.) –  Robert Bryant Sep 24 '11 at 3:18