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Dear community,

i have a question regarding differential operators acting on vector valued functions and how to "diagonalize" them.

To explain my question i will use an example: Let $V^k$ be the space of twice differentiable functions $U:[0,2\pi] \to \mathbb{R}^k$ with periodic boundary conditions.

Consider the differential operator $L:V^2\to V^2$ defined via

$L :=\begin{pmatrix} -\partial^2 & 0 \\ 0 & -\partial^2 \end{pmatrix}$

That means it acts on $U\in V$, $U(t)=(u(t),v(t))^T$, by mapping it to $LU=(-u''(t),-v''(t))^T$.

$L$ is self adjoined with respect to the scalar product $(U,V) := \int_0^{2\pi} U^T V \ dt$. So it has real eigenvalues. It commutes with the operator $J=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$.

In fact, the spectrum of $L$ consists of all $\lambda=k^2$ with $k \in \mathbb{Z}$ with multiplicity 4. The spectrum of $-\partial^2:V^1 \to V^1$ is the same, except that the multiplicities of the eigenvalues are halved. This is evident from the diagonal form of $L$.

Now, what happens for other self-adjoined operators on $V^2$ that commute with $J$? For example consider the operator $M:V^2 \to V^2$ defined by

$M :=\begin{pmatrix} -\partial^2 & -\partial \\ \partial & -\partial^2 \end{pmatrix}$

It is also self-adjoined with respect to the above mentioned inner product and it also commutes with $J$. Is it possible to "diagonalize" this operator into a form $\begin{pmatrix} m & 0 \\ 0 & m \end{pmatrix}$ with a scalar differential operator $m: V^1 \to V^1$ having the same spectrum as $M$ except for the multiplicities?

Any references would be appreciated. Thanks.

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2 Answers 2

up vote 3 down vote accepted

Another approach, for this particular example is to try to solve the equation $AMA^{-1} - m I = 0$, where $A$ is an invertible $2$-by-$2$ matrix of functions and $m$ is a scalar differential operator. There are a number of solutions to this. For example, $$ A = \begin{pmatrix}\cos(\tfrac12x) & \sin(\tfrac12x)\cr -\sin(\tfrac12x) & \cos(\tfrac12x) \end{pmatrix} $$ and $m = -\partial^2 - \tfrac14$. Unfortunately, $A$ is $4\pi$-periodic, not $2\pi$-periodic. In fact, there are no $2\pi$-periodic solutions.

You can interpret this in two ways. One is that you shouldn't have imposed the $2\pi$-periodicity in the first place, since the conjugacy question for differential operators is really a local one. The other is that $A$ actually represents an isomorphism between two rank $2$-vector bundles over the circle, one with trivial transition, and the other with a twist (so that the sections of the second bundle are represented by functions $U:\mathbb{R}\to \mathbb{R}^2$ such that $U(t+2\pi) = -U(t)$).

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1  
This is a nice solution (minor correction: should be $A^{-1} M A$ since $AMA^{-1}=-\partial^2 + 2J \partial -\frac{3}{4}$). Indeed, your $m$ has the correct eigenvalues if antiperiodic boundary conditions are applied. So by going to the twisted bundle, the problem simplifies as we can reduce the rank from $2$ to $1$. Still this is not quite the solution I was hoping for, because I cannot see how to generalize it. My original motivation was my other question where I was wondering if there exists a "rank 1" version of the Bochner Laplacian (defined on the rank 2 bundle $T^*M$) on a surface. –  Alexander Vais Sep 1 '11 at 14:40
    
Thanks for the correction. Actually, my mistake was the typo of putting the minus sign in the wrong place in the formula for the explicit solution $A$. I've fixed this now. By the way, I should say that the point is that your operator is conjugate to the double of a scalar operator on the nontrivial line bundle over the circle. What is this `other question' that you refer to? –  Robert Bryant Sep 1 '11 at 15:00
    
Here is the other question. It can be summarized as: Is the 1-form Bochner Laplacian conjugate to the double of some scalar operator? The De Rham Laplacian is. Unfortunately I could not find any reference discussing this issue. –  Alexander Vais Sep 2 '11 at 8:24

Your matrix is a matrix with coefficients in the ring $\mathbb C[\partial]$ of differential operators with constant coefficients, which happens to be a principal ideal domain. Therefore we can take your matrix — let's call it $A$ — to its Smith normal form. In this case, the normal form is $$S=\left( \begin{array}{cc} \partial & 0 \\\\ 0 & \partial ^3+\partial \end{array} \right)$$ Indeed, if we let $$P=\left( \begin{array}{cc} -1 & 0 \\\\ -\partial & 1 \end{array} \right)\qquad\text{and}\qquad Q=\left( \begin{array}{cc} 0 & 1 \\\\ 1 & -\partial \end{array} \right),$$ which are invertible, we have $$PAQ=S.$$ This means that you can change coordinates to get a diagonal matrix.

(Of course, in your situation you want to consider $A$ as defining an endomorphism, so you probably want to restrict changes of coordinates which coincide in the domain and the codomain of the map...)

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