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I'm a novice to analysis but I need to understand the following example. Any help would be greatly appreciated. This might be of interest to some because it gives a way of quantifying changes in eigenspaces as the index of a Fredholm operator.

Let $\alpha: (-\infty,\infty) \to Sym^2(\Bbb{R}^n)^\vee$ be a path to the space of symmetric $n\times n$ real matrices such that $\alpha$ is constant near $-\infty$, and constant near $\infty$. I'll denote these two constant values by $\alpha(\pm \infty)$. Consider the differential operator

$$ d_\alpha(X):= {\partial_s}X + \alpha(s)X$$

which takes a path $X: \Bbb R \to \Bbb R^n$ and outputs another path $d_\alpha X$. I declare the domain to be the space of $W^{1,2}$ maps $X: \Bbb R \to \Bbb R^n$, and naturally the target is the space of $L^2$ maps. (Here $W^{1,2}$ is the Sobolev space of $L^2$ maps whose first derivatives are $L^2$ too.) To calculate the index of $d_\alpha$, I further expand the domain by taking the direct sum with the finite-dimensional vector space spanned by $X$ of the form $$ X(s) = \exp(-s \lambda ) v $$ where $\lambda$ is a negative eigenvalue of $\alpha(\infty)$ and $v$ is an eigenvector for $\lambda$. One easily sees that $d_\alpha X$ indeed lands in $L^2$ so we have an operator $$ \tilde d_\alpha: W^{1,2}(\Bbb R , \Bbb R^n) \oplus \{(e^{-s \lambda} v)\} \to L^2(\Bbb R , \Bbb R^n). $$

Now, I want to prove three things:

(1) that the kernel of this operator is finite-dimensional, and identified with the negative eigenvalue eigenspace of $\alpha(-\infty)$,

(2) that this map is surjective, and

(3) that as a result, $\text{index}(d_\alpha) = \dim \sigma_-(\alpha(-\infty)) -\dim \sigma_-(\alpha(\infty))$, where $\sigma_-$ is the space of vectors with negative eigenvalues.

Here's how I've gone about proving these statements, but I'm stuck in all claims.

(1) To see this, let $\{v_i\}$ be an eigenbasis for $\alpha(-\infty)$ and $\lambda_i$ the corresponding eigenvalues, writing $X(s) = \sum x_i(s) v_i$ near $-\infty$. Then the equation $\tilde d_\alpha = 0$ breaks into first-order ODEs in one variable: $$ \partial_s x_i + \lambda_i x_i = 0. $$ This has the obvious solutions $x_i(s) = \exp(-s \lambda_i)v_i$, but -- to be in $W^{1,2}$ -- $\lambda_i$ must be a negative eigenvalue. (Otherwise the integral $\int_{-\infty}^s |X(s)|^2$ will blow up.)

I now want to say that, since we have identified what a solution $X$ must look like near $-\infty$, by the existence of first-order ODEs, every solution near $-\infty$ extends to a full solution, and by uniqueness of first-order ODEs, this determines all solutions as in one-to-one correspondence with the space of all eigenvectors of $\alpha(-\infty)$ with negative eigenvalue. Here is my question: Does the usual proof of uniqueness and existence of solutions guarantee that my solutions will indeed be in $W^{1,2}$? If you have a reference (or a quick proof) that the $L^2$ norms of both $X$ and its derivative are bounded, I'd be very grateful. Note that $\alpha$ is constant near $\pm \infty$, so its values and derivatives are bounded, but I'm not sure if this is enough.

(2) Let $Y(s)$ be some $L^2$ path. We want to exhibit $X(s)$ such that $\tilde d_\alpha X = Y$. This time, take an eigenbasis for $\alpha(\infty)$, calling it $w_i$, with corresponding eigenvalues $\nu_i$. Again the equation splits into the components of $w_i$ and we are left to solve the first-order ODE $$ \partial_s x_i + \nu_i x_i = y_i. $$ Using the usual tricks from calculus, we see that a solution to this is given by $$ x_i(s) = {\int_{s_0}^s y_i(t) e^{\nu_i t} dt \over e^{\nu_i s} }. $$ One can extend this to a full solution . Again, my question is: How can one guarantee that such a solution is in $W^{1,2}$? And does it require a clever addition of a function of the form $\exp(-\nu_is)w_i$ to exhibit an $L^2$ solution?

(3) To see this, I can just add the indices of each Fredholm operator in the composition $$ W^{1,2}(\Bbb R , \Bbb R^n) \to W^{1,2}(\Bbb R , \Bbb R^n) \oplus \text{span}(e^{-s \lambda} v) \to L^2(\Bbb R , \Bbb R^n). $$

Is there a more clever way of seeing this, or a more natural way?

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1 Answer 1

An interesting question. My answer is far from complete; I hope to find time to return to it tomorrow. To begin with, let's note that $d_\alpha$ folds the direct sum $W^{1,2}\oplus W$ (where $W$ is the set of exponentials defined above) by addition. In other words, the domain of $d_\alpha$ is really the elementwise sum $Z=W^{1,2} + W$, which consists of all functions that can be written as an element of $W$ plus an element of $W^{1,2}$. This decomposition is unique when it exists.

It seems that one of the following should be true: either $\alpha(\infty)$ is invertible, or $W$ also includes the eigenfunctions for eigenvalue $0$. This will come up at the end of next paragraph.

(1) Let's introduce cut-offs $a<b$ such that $\alpha(s)=\alpha(-\infty)$ when $s\le a$ and $\alpha(s)=\alpha(\infty)$ when $s\ge b$. We are looking for functions $X\in Z$ such that $X'(s)=-\alpha(s)X(s)$. As long as $\alpha$ is continuous, the standard existence theorem gives continuously differentiable solutions on $\mathbb R$. Thus, the issue of $X\in W^{1,2}(\mathbb R)$ amounts to its behavior near infinity. You already analyzed the situation near $-\infty$, i.e., in the range $(-\infty,a)$: every solution is a sum of exponentials and in order to be in $W^{1,2}$ we must have decaying exponentials only. In the range $(b,\infty)$ the solution $X$ also decomposes into exponentials. The decaying ones are in $W^{1,2}$ and the rest we included in $W$. Hence, $X\in Z$. I believe this completes the description of the kernel of $d_\alpha$.

(2) I think for surjectivity you also should begin with the interval $(-\infty,a)$. Solve the constant-coefficient equation with initial condition $X(-\infty)=0$; since the solution is given by an explicit integral involving an $L^2$ function, it is in $W^{1,2}(-\infty,a)$. In the positive direction, on $(a,\infty)$, it should be possible to decompose $X$, using its integral representation, into $W$ and $W^{1,2}$ parts.

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