In this post, I begin discussing the magnetic boundary rigidity problem, following this paper by Dairbekov, Paternain, Stefanov, and Uhlmann.

Definitions

Simple Magnetic System

This section follows Chapter 2 of Merry and Paternain’s notes and Section 1 of Daiberkov et al.

The ultimate question behind the boundary rigidity program is the following: Given a Riemannian manifold with boundary and two metric \(g_1, g_2\), we can study their boundary distance functions \(d_1, d_2\) which are defined to be the restriction of the usual distance function to the boundary. Is it true that if these are equal, then the metrics are isometric? Taking the question at face value, the answer is no. An easy enough example is to just take the disc in \(\mathbb{R}^n\) and on a small open neighborhood of the origin define two metrics which are very big on this open set. Then every geodesic is going to avoid this open set, so therefore the two metrics will agree on the boundary, however they are clearly two different metrics on the open set so they cannot be isometric.

With a slight modification, it turns out that the problem reaches a doable point. We say that \((M, \partial M, g)\) is a simple manifold if two conditions are satisfied:

1) The exponential map is everywhere a diffeomorphism.

2) \(\partial M\) is strictly convex in the sense that the second fundamental form \(S\) satisfies the following on \(\partial M\):

\[S_x(u,w) \geq 0 \text{ with equality if and only if } u =w.\]

Recall that the second fundamental form is defined as follows. Take the vector field \(\nu : \partial M \rightarrow T_{\partial M}M\) which sends a point to the unit inward pointing normal. That is, it takes it to the unique vector \(v\) with \(v \in T_x^\perp \partial M\) and \(\|v\| = 1.\) Then the second fundamental form is the map

\[S : T \partial M \times T \partial M \rightarrow \mathbb{R}, \ \ S_x(u,w) := - \langle \nabla_u \nu(x), w \rangle.\]

In dimension two, it has been shown by Pestov and Uhlmann that simple manifolds are boundary rigid within the class of simple metrics, meaning that if two simple metrics agree on the boundary, then they are isometric. The goal of the magnetic boundary rigidity problem is to try to translate this problem to be in terms of magnetic flows instead of geodesic flows.

The first step is to try to translate the conditions for a simple manifold to be in terms of a magnetic flow. Recall that a magnetic flow is given in terms of the tuple \((M, g, \sigma)\), where \(\sigma\) is a closed \(2\)-form on \(M\). We’ll fix the tuple \((M, g, \sigma)\) to be our magnetic system unless otherwise stated. The magnetic flow will be denoted by \(\psi^t\). Daiberkov et al define \(\partial M\) to be strictly magnetic convex if we have

\[S_x(v,\nu(x)) > \sigma_x(v, \nu(x)).\]

They also define the magnetic exponential map at \(x\) to be the map satisfying the following:

\[\exp^\mu_x : T_x M \rightarrow M, \ \ \exp_x^\mu(t v) = \pi \circ \psi^t(v), \ t \geq 0, x \in T^1_xM.\]

Thus we define the system \((M, \partial M, g, \sigma)\) to be simple if two conditions are satisfied:

1) \(\exp_x^\mu\) is a diffeomorphism everywhere.

2) \(\partial M\) is strictly magnetic convex.

Remark: Just like in the usual simple manifold case, we have that if the above two conditions are satisfied then \(M\) is diffeomorphic to the unit ball in \(\mathbb{R}^n.\) Thus we have that \(\sigma\) is in fact an exact form. In this case, we have an exact magnetic system with \(d \alpha = \sigma\), and we’ll sometimes write \((M, g, \alpha)\) to denote the system.

Free Time Action

A significant tool in the study of the boundary rigidity problem is the energy functional. Given a curve \(\gamma : [a,b] \rightarrow M\), the energy functional is defined as

\[E(\gamma) := \int_a^b \|\dot{\gamma}(t)\|^2 dt.\]

To see the power behind the energy functional, let’s work with the following simplified version of the boundary rigidity problem. Let \(\{g_s : s \in (-\epsilon, \epsilon)\}\) be a family of simple metrics for \((M, \partial M)\) with negative sectional curvature. Let \(d_s := d_{g_s}\) denote the boundary distance function. Suppose \(d_s = d_0\) for all \(s\). The goal is to show that the family is trivial in the Guillemin-Kazhdan sense – that is, we can find a smooth family of diffeomorphisms \(\{\psi_s\}\) so that \(\psi_0 = \text{Id}\), \(\psi_s \mid_{\partial M} = \text{Id}\), and \(g_s = \psi_s^*(g_0)\). Take \(x,y \in \partial M\) and note that \(T := d_0(x,y) = d_s(x,y)\) for each \(s\). Thus we can take a variation with respect to \(s\). If we let \(E_s\) denote the energy functional with respect to the metric \(g_s\), then using the same argument as in the above post we can calculate

\[0 =\frac{\partial}{\partial s} E_s(\gamma_s) = \int_{\gamma_0} \dot{E}, \ \ \dot{E} = \frac{\partial}{\partial s} E_s \mid_{s=0}.\]

The argument diverges from Guillemin-Kazhdan here, since we can no longer use the Livschitz theorem – we only know the vanishing condition on the boundary, not on all periodic orbits. The goal now is to fix this gap.

We define the X-Ray transform to be a map \(I\) which takes symmetric two tensor \(\beta\) and gives the following:

\[I[\beta] : \partial M \times \partial M \rightarrow \mathbb{R}, \ \ I[\beta](x,y) = \int_0^T (\beta)_{\gamma(t)}(\dot{\gamma}(t), \dot{\gamma}(t)) dt,\]

where \(\gamma\) is the unique unit speed geodesic connecting \(x\) to \(y\). If we let \(\beta = \frac{\partial g_s}{\partial s} \mid_{s=0}\), we see that we have a symmetric \(2\)-tensor, and moreover we actually have that \(\beta \in \ker(I).\) The goal is to build a function \(u : SM \rightarrow \mathbb{R}\) so that \(X(u) = \beta.\) If we can do so, then we have that \(\beta\) is a trivial \(2\)-tensor, and the result essentially follows from the same argument as in the GK post.

Consider the function \(\tau^+, \tau^- : SM \rightarrow \mathbb{R}\) so that if \((x,v) \in SM\) and \(\gamma\) is the unique geodesic (viewed in terms of a function on \(SM\)) associated with \((x,v)\), then \(\gamma(\tau^-(x,v)), \gamma(\tau^+(x,v)) \in \partial(SM).\) The difference between these two vectors is that we have

\[\langle \gamma(\tau^+(x,v)), \nu(\pi(\gamma(\tau^+(x,v)))) \rangle \geq 0, \ \ \langle \gamma(\tau^-(x,v)), \nu(\pi(\gamma(\tau^-(x,v)))) \rangle \leq 0.\]

It is non-trivial to show that the functions \(\tau^+, \tau^-\) are smooth off of \(T^1(\partial M)\), however it follows by the simplicity condition. We define then

\[u(x,v) := \int_{\tau^-(x,v)}^0 \beta_{\gamma(t)}(\dot{\gamma}(t), \dot{\gamma}(t)) dt.\]

The first observation is that \(u(x,v) = 0\) for all \((x,v) \in \partial(SM).\) Suppose \((x,v)\) is such that \(\langle v, \nu(x) \rangle \leq 0,\) then clearly we have that \(\tau^-(x,v) = 0\) so that the integral is \(0\). If \(\langle v, \nu(x) \rangle \geq 0\), then we can use the fact that \(\beta\) is in the kernel to get that \(u(x,v) = 0\). The second observation is that \(u\) is smooth where \(\tau^-\) is smooth, so it is smooth on \(SM \setminus T^1(\partial M)\) by simplicity.

The next observation is that we have \(X(u) = \beta\) where \(u\) is smooth. This is just a matter of slightly varying \(x\) and \(v\) and then differentiating the result. Let \((x,v)\) be where \(u\) is smooth, let \(\gamma : [\tau^-(x,v), \tau^+(x,v)] \rightarrow M\) be a curve defined so that \(\gamma(0) = x\), \(\dot{\gamma}(0) = v.\) We can vary it with sufficiently small \(s\) so that \(x_s = \gamma(s), v_s = \dot{\gamma}(s)\). Letting \(\gamma_s\) be the unique curve defined by \(\gamma_s(0) = x_s,\) \(\dot{\gamma}_s(0) = v_s,\) we have that

\[\gamma_s(t) = \gamma(t+s), \ \ \tau^-(x_s, v_s) = \tau^-(x,v) - s.\]

Now under this new notation, we have

\[u(x_s, v_s) = \int_{\tau^-(x,v)}^s \beta_{\gamma(t)}(\dot{\gamma}(t), \dot{\gamma}(t))dt.\]

Differentiate this and set it equal to \(0\) to get the desired result.

The goal is to integrate \(u\) over all of \(SM.\) Since we have some singularities, we need to worry about convergence of the integral. However, convergence is not too hard to check once one has the Gauss-Ostrogradskii formulas; see Sharafutdinov’s book, Section 3.6. With these formulas, it’s a matter of choosing the right coordinates and transforming the integral correctly in order to apply the Lebesgue dominated convergence theorem. I’ll omit the calculation.

With \(u\) defined directly, we now want to show that \(\beta\) is potential. In order to do that we need to show that \(u\) is a \(1\)-form (i.e. a function on \(SM\)). This amounts to a clever application of Pestov’s identity, which I’ll omit. This can be seen in Theorem 12.4 of Merry and Paternain’s notes. The result follows nicely at this point.

The moral of the story is that at this point, we are able to deduce the result once we know that \(I[\beta] \equiv 0.\) The goal, then, is to find some other functional which, as long as it remains constant along smooth perturbations of our dynamical system, tells us that \(I[\beta] \equiv 0.\) This leads us to the concept of the free time action functional. Given a magnetic system \((M, \partial M, g, \alpha)\) (where \(\alpha\) is a \(1\)-form) and a curve \(\gamma : [0,T] \rightarrow M\) we have

\[A(\gamma) := \int_0^T \| \dot{\gamma}(t)\|^2 dt - \int_\gamma \alpha.\]

The claim now is that unit speed magnetic geodesics minimize the time free action. That is, if I take a smooth variation of a unit speed magnetic geodesic, say \(\{\gamma_s\}\), then we should have

\[\frac{\partial A}{\partial s}(\gamma_s) \mid_{s=0} = 0,\]

in the same way that unit speed geodesics minimize the energy action. This actually follows from a calculation similar to that in the Guillemin-Kazhdan post. If we denote the smooth variation as \(\Gamma\) and let

\[S(t) := \frac{\partial \Gamma}{\partial s}(0,t), \ \ T := \frac{\partial \Gamma}{\partial t}(0,t),\]

then we can calculate

\[\frac{\partial}{\partial s} A(\gamma_s) \mid_{s=0} = \frac{\partial}{\partial s} \int_0^T \langle T(s,t), T(s,t) \rangle dt \mid_{s=0}- \frac{\partial}{\partial s}\int_0^T \alpha(\gamma_s(t), \dot{\gamma}_s(t))dt \mid_{s=0}.\]

Supposing that \(\gamma\) is a closed unit speed magnetic geodesic, an easy calculation shows

\[\frac{\partial}{\partial s} \int_0^T \langle T(s,t), T(s,t) \rangle dt \mid_{s=0} = - \int_0^T \langle S, \nabla_T T \rangle dt.\]

Recall that we have

\[\sigma(T,S) = \langle \nabla_T T, S \rangle,\]

so we have

\[- \int_0^T \langle S, \nabla_T T \rangle dt = \int_0^T \sigma(S,T) dt.\]

It takes a little bit of work, but it’s not too hard to see that

\[\frac{\partial}{\partial s}\int_0^T \alpha(\gamma_s(t), \dot{\gamma}_s(t))dt \mid_{s=0} = \int_0^T \sigma(S,T)dt.\]

The idea is to note that we can create a simplex \(\Sigma_s\) so that

\[\int_{\gamma_s} \alpha - \int_{\gamma_0} \alpha = \int_{\Sigma_s} \sigma.\]

Then taking the derivative with respect to \(s\), we get our desired result. We can then conclude that

\[\frac{\partial}{\partial s} A(\gamma_s) \mid_{s=0} = 0,\]

as desired. If we’re a little more careful with our set up, we can show that this holds for \(\gamma\) a unit speed magnetic geodesic connecting points on the boundary as well. With this, if we define

\[A_s(\gamma) = \int_0^T \|\dot{\gamma}(t)\|_s^2 dt - \int_{\gamma} \alpha_s,\]

then we have

\[\frac{\partial }{\partial s} A_s(\gamma_s) \mid_{s=0} = \int_0^T [\beta_{\gamma(t)}(\dot{\gamma}(t), \dot{\gamma}(t)) - \dot{\alpha}(\gamma(t), \dot{\gamma}(t))]dt.\]

Thus we have \(I[\beta - \dot{\alpha}] \equiv 0\), and we’re in a similar set up to before. The issue here is that we don’t have a similar lemma to before – i.e. we don’t have a way of just deducing that \(\beta\) or \(\dot{\alpha}\) are trivial. Thus we will need to do a little more work before we’re done. For now, we’ll leave this result where it’s at.

A quick remark is that this type of Guillemin-Kazhdan argument actually works in the general magnetic case, as seen in Theorem C in this paper by Daiberkov and Paternain, although a little more effort is needed to actually define the action functional in this case.