Broken ray tomography and related inverse problems
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- Leslie Payne
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1 Broken ray tomography and related inverse problems Inverse problems seminar University of Washington Joonas Ilmavirta University of Jyväskylä Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
2 Outline 1 Introduction Broken rays Broken ray transform Applications Counterexamples 2 Approaches 3 Periodic broken ray transform 4 Algebraic tomography problems Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
3 Broken rays Let M be a compact Riemannian manifold with boundary (or the closure of a bounded smooth Euclidean domain). Split its boundary in two disjoint parts: M = E R. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
4 Broken rays Let M be a compact Riemannian manifold with boundary (or the closure of a bounded smooth Euclidean domain). Split its boundary in two disjoint parts: M = E R. Broken rays have endpoints on E and reflect finitely many times on R (like light reflecting from a perfect mirror). Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
5 Broken rays. Let M be a compact Riemannian manifold with boundary (or the closure of a bounded smooth Euclidean domain). Split its boundary in two disjoint parts: M = E R. Broken rays have endpoints on E and reflect finitely many times on R (like light reflecting from a perfect mirror). If E = M and R =, then broken rays are just maximal geodesics in M. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
6 Broken rays. A domain. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
7 Broken rays. A line through the domain. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
8 Broken rays. A broken ray. E is blue and R is red. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
9 Broken rays. Large E. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
10 Broken rays. A broken ray. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
11 Broken rays. Another broken ray. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
12 Broken rays. Small E. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
13 Broken rays. A broken ray. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
14 Broken rays. An annulus. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
15 Broken rays. A broken ray. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
16 Broken rays. Another broken ray. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
17 Broken rays. We assumed that all broken rays reflect only finitely many times. Sometimes there are trapped broken rays which never find their way back to E. We only consider non-trapped broken rays here. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
18 Broken ray transform Broken ray tomography problem: Can we recover a function from its integrals over all broken rays? Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
19 Broken ray transform Broken ray tomography problem: Can we recover a function from its integrals over all broken rays? Let Γ be the set broken rays in M. (This set depends on E and R!) The broken ray transform of a continuous function f : M R is Gf : Γ R, Gf(γ) = fds. γ Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
20 Broken ray transform Broken ray tomography problem: Can we recover a function from its integrals over all broken rays? Let Γ be the set broken rays in M. (This set depends on E and R!) The broken ray transform of a continuous function f : M R is Gf : Γ R, Gf(γ) = fds. If E = M, then G is the X-ray transform. γ Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
21 Broken ray transform. Broken ray tomography problem: Can we recover a function from its integrals over all broken rays? Let Γ be the set broken rays in M. (This set depends on E and R!) The broken ray transform of a continuous function f : M R is Gf : Γ R, Gf(γ) = fds. If E = M, then G is the X-ray transform. Broken ray tomography problem: Is G injective? How does this depend on M, E, and regularity assumptions on f? γ Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
22 Applications Calderón s problem with partial data: Can we find the conductivity at every point inside an object by making electrical measurements on a part of the boundary? (Some of the boundary is completely inaccessible.) Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
23 Applications Calderón s problem with partial data: Can we find the conductivity at every point inside an object by making electrical measurements on a part of the boundary? (Some of the boundary is completely inaccessible.) Kenig and Salo (2013) showed that this partial data problem on a tube-shaped domain can be reduced to the injectivity of the broken ray transform on the cross section of the tube. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
24 Applications. Calderón s problem with partial data: Can we find the conductivity at every point inside an object by making electrical measurements on a part of the boundary? (Some of the boundary is completely inaccessible.) Kenig and Salo (2013) showed that this partial data problem on a tube-shaped domain can be reduced to the injectivity of the broken ray transform on the cross section of the tube. Eskin (2004) related the broken ray transform to an inverse boundary value problem for the magnetic Schrödinger equation. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
25 Applications Many inverse problems with full data have been previously reduced to the X-ray transform. It seems that in the case of partial data, the X-ray transform is often replaced by the broken ray transform. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
26 Applications. Many inverse problems with full data have been previously reduced to the X-ray transform. It seems that in the case of partial data, the X-ray transform is often replaced by the broken ray transform. Example: Linearization (w.r.t. the metric) of broken ray scattering relation leads to broken ray transform of 2-tensor fields. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
27 Counterexamples The broken ray transform can only be injective if the X-ray transform is injective. The integral over a broken ray is a sum of integrals over geodesics. (The broken ray transform can be seen as a partial data version of the X-ray transform.) Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
28 Counterexamples. The broken ray transform can only be injective if the X-ray transform is injective. The integral over a broken ray is a sum of integrals over geodesics. (The broken ray transform can be seen as a partial data version of the X-ray transform.) There are counterexample to the injectivity of the broken ray transform in Euclidean domains. (Pictures coming.) Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
29 Counterexamples. Some function supported in the gray area are invisible. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
30 Counterexamples. Some function supported in the gray area are invisible. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
31 Outline 1 Introduction 2 Approaches Four angles of attack Reflection Direct calculation Boundary reconstruction Energy estimate for a PDE 3 Periodic broken ray transform 4 Algebraic tomography problems Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
32 Four angles of attack As far as I know, there are four known ways to prove anything about the broken ray transform: Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
33 Four angles of attack As far as I know, there are four known ways to prove anything about the broken ray transform: reflection arguments Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
34 Four angles of attack As far as I know, there are four known ways to prove anything about the broken ray transform: reflection arguments explicit calculation of the transform Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
35 Four angles of attack As far as I know, there are four known ways to prove anything about the broken ray transform: reflection arguments explicit calculation of the transform boundary reconstruction via boundary geodesics Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
36 Four angles of attack As far as I know, there are four known ways to prove anything about the broken ray transform: reflection arguments explicit calculation of the transform boundary reconstruction via boundary geodesics energy estimates Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
37 Four angles of attack. As far as I know, there are four known ways to prove anything about the broken ray transform: reflection arguments explicit calculation of the transform boundary reconstruction via boundary geodesics energy estimates The main goal is to reduce the problem to a more tractable one, like the X-ray transform, manipulation of Fourier series, or unique solvability of a PDE. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
38 Reflection Reflections at the boundary are what makes the broken ray transform difficult. We want to get rid of them. Can we make the domain reflect but keep the rays straight? Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
39 Reflection. Reflections at the boundary are what makes the broken ray transform difficult. We want to get rid of them. Can we make the domain reflect but keep the rays straight? Yes, at least if R is flat. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
40 Reflection Proposition The broken ray transform is injective for the domain Ω = {x B(0, 1) R 2 ; x 1 > 0} with R = {x Ω; x 1 = 0}. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
41 Reflection Proposition The broken ray transform is injective for the domain Ω = {x B(0, 1) R 2 ; x 1 > 0} with R = {x Ω; x 1 = 0}. Proof Suppose f C( Ω) integrates to zero over all broken rays. Let Ω = B(0, 1) and let π : Ω Ω be the folding map π(x1, x 2 ) = ( x 1, x 2 ). Define f : Ω R by f(x) = f(π(x)). Now f integrates to zero over all lines in Ω. Because the X-ray transform is injective, f = 0. Thus also f = 0. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
42 Reflection. Proposition The broken ray transform is injective for the domain Ω = {x B(0, 1) R 2 ; x 1 > 0} with R = {x Ω; x 1 = 0}. Proof Suppose f C( Ω) integrates to zero over all broken rays. Let Ω = B(0, 1) and let π : Ω Ω be the folding map π(x1, x 2 ) = ( x 1, x 2 ). Define f : Ω R by f(x) = f(π(x)). Now f integrates to zero over all lines in Ω. Because the X-ray transform is injective, f = 0. Thus also f = 0. The idea of reducing the broken ray transform on Ω to the X-ray transform of a reflected domain Ω works in great generality. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
43 Reflection Lemma (Helgason s support theorem) Let f : R n R be a compactly supported continuous function and K R n a compact, convex set. If the integral of f is zero over every line that doesn t meet K, then f is zero outside K. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
44 Reflection Lemma (Helgason s support theorem) Let f : R n R be a compactly supported continuous function and K R n a compact, convex set. If the integral of f is zero over every line that doesn t meet K, then f is zero outside K. Theorem (I. 2013) Let C R 2 be a cone. (In polar coordinates, r > 0 and 0 < θ < θ 0 for some θ 0.) Let Ω C be a bounded domain. Then the broken ray transform in Ω with R = C Ω is injective. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
45 Reflection. Lemma (Helgason s support theorem) Let f : R n R be a compactly supported continuous function and K R n a compact, convex set. If the integral of f is zero over every line that doesn t meet K, then f is zero outside K. Theorem (I. 2013) Let C R 2 be a cone. (In polar coordinates, r > 0 and 0 < θ < θ 0 for some θ 0.) Let Ω C be a bounded domain. Then the broken ray transform in Ω with R = C Ω is injective. Proof will be given as pictures. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
46 Reflection. A domain with opening angle θ 0 = π/m for an integer m. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
47 Reflection. Glue together m copies of Ω (every other one reflected). The broken ray folds out to a straight line. Use injectivity of X-ray transform. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
48 Reflection. A domain with general opening angle θ 0. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
49 . Reflection Glue together enough copies of Ω so that the shaded sector becomes convex. Apply Helgason s support theorem. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
50 Direct calculation If M is a disc in the plane, we can try to calculate the broken ray transform explicitly. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
51 Direct calculation If M is a disc in the plane, we can try to calculate the broken ray transform explicitly. In polar coordinates (r, θ), we can write a function as a Fourier series: f(r, θ) = k Z e ikθ a k (r). Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
52 Direct calculation. If M is a disc in the plane, we can try to calculate the broken ray transform explicitly. In polar coordinates (r, θ), we can write a function as a Fourier series: f(r, θ) = k Z e ikθ a k (r). A line (a chord) in the disc can be parametrized by the polar coordinates of the closest point to the center. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
53 Direct calculation The integral of f over a line given by (r, θ) is If(r, θ) = k Z e ikθ A k a k (r) where A k is a generalized Abel transform. (A 0 is the usual Abel transform.) Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
54 Direct calculation The integral of f over a line given by (r, θ) is If(r, θ) = k Z e ikθ A k a k (r) where A k is a generalized Abel transform. (A 0 is the usual Abel transform.) These integral transforms A k are injective, and this procedure can be used to invert the X-ray transform: If = 0 all Fourier components of If are zero A k a k = 0 for all k a k = 0 for all k f = 0. (Cormack used this method for inversion.) Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
55 Direct calculation. The integral of f over a line given by (r, θ) is If(r, θ) = k Z e ikθ A k a k (r) where A k is a generalized Abel transform. (A 0 is the usual Abel transform.) These integral transforms A k are injective, and this procedure can be used to invert the X-ray transform: If = 0 all Fourier components of If are zero A k a k = 0 for all k a k = 0 for all k f = 0. (Cormack used this method for inversion.) A broken ray consists of finitely many line segments which are rotations of each other. The integral of f over a broken ray can be expressed explicitly, but it is more complicated than the above sum for the X-ray transform. Careful analysis of this expression shows that the broken ray transform is injective with some regularity assumptions. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
56 Boundary reconstruction Suppose M is strictly convex and E M is open. If a broken ray starts very close to the boundary and almost tangential to it, it will remain this way for some time. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
57 Boundary reconstruction Suppose M is strictly convex and E M is open. If a broken ray starts very close to the boundary and almost tangential to it, it will remain this way for some time. We can control how well it stays near the boundary. This is similar how an ideally bouncing ball stays near the surface in slowly varying gravitation. (Now gravitation is replaced by the second fundamental form of the boundary.) Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
58 Boundary reconstruction. Suppose M is strictly convex and E M is open. If a broken ray starts very close to the boundary and almost tangential to it, it will remain this way for some time. We can control how well it stays near the boundary. This is similar how an ideally bouncing ball stays near the surface in slowly varying gravitation. (Now gravitation is replaced by the second fundamental form of the boundary.) If we take a sequence of broken rays that stay closer and closer to the boundary, the limit curve is a curve on the boundary M. This curve is actually a geodesic on M. Boundary geodesics as limits of broken rays (= billiard trajectories) are known by many names: glancing billiards, creeping rays, whispering gallery trajectories, gliding rays. These boundary geodesics can be assumed to have endpoints on E. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
59 Boundary reconstruction If we know the broken ray transform of f, we can find its integrals over boundary geodesics. If the X-ray transform on M (or more properly R) is injective, we can recover f at M from its broken ray transform. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
60 Boundary reconstruction If we know the broken ray transform of f, we can find its integrals over boundary geodesics. If the X-ray transform on M (or more properly R) is injective, we can recover f at M from its broken ray transform. If certain weighted X-ray transforms on R are injective, then we can also recover the normal derivatives (of all orders) of f at the boundary. The weight depends on curvature. Intuitively, this is because a broken ray is on average further away from the boundary where the curvature is largest. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
61 Boundary reconstruction. If we know the broken ray transform of f, we can find its integrals over boundary geodesics. If the X-ray transform on M (or more properly R) is injective, we can recover f at M from its broken ray transform. If certain weighted X-ray transforms on R are injective, then we can also recover the normal derivatives (of all orders) of f at the boundary. The weight depends on curvature. Intuitively, this is because a broken ray is on average further away from the boundary where the curvature is largest. When recovering derivatives of order k, the weight is the second fundamental form to the power k/3. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
62 Boundary reconstruction. A broken ray in an ellipse. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
63 Energy estimate for a PDE Suppose dim M = 2. Partition as before: M = E R. Let SM be the sphere bundle of M. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
64 Energy estimate for a PDE Suppose dim M = 2. Partition as before: M = E R. Let SM be the sphere bundle of M. For (x, v) SM, let γ x,v be the unique broken ray starting at x in direction v and ending at E. We assume that every broken ray reaches E in finite time. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
65 Energy estimate for a PDE. Suppose dim M = 2. Partition as before: M = E R. Let SM be the sphere bundle of M. For (x, v) SM, let γ x,v be the unique broken ray starting at x in direction v and ending at E. We assume that every broken ray reaches E in finite time. For x M, let ρ : S x M S x M be the reflection map v v 2(v ν x )ν x where ν x is the (inner or outer) unit normal vector at x. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
66 Energy estimate for a PDE For a C 2 function f : M R, let u f : SM R be the integral of f along this broken ray: u f (x, v) = fds. γ x,v Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
67 Energy estimate for a PDE For a C 2 function f : M R, let u f : SM R be the integral of f along this broken ray: u f (x, v) = fds. γ x,v Let X and V be the geodesic and vertical vector fields on SM. X is the derivative in the direction of the geodesic flow and V is the derivative with respect to direction. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
68 Energy estimate for a PDE. For a C 2 function f : M R, let u f : SM R be the integral of f along this broken ray: u f (x, v) = fds. γ x,v Let X and V be the geodesic and vertical vector fields on SM. X is the derivative in the direction of the geodesic flow and V is the derivative with respect to direction. For x int M this function satisfies Xu f (x, v) = f(x). f does not depend on direction. Thus V Xu f = 0 in int SM. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
69 Energy estimate for a PDE For x R this function satisfies u f (x, v) = u f (x, ρ(v)) because of the way broken ways reflect. If we assume f to have vanishing broken ray transform, then u f (x, v) = 0 whenever x E. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
70 Energy estimate for a PDE For x R this function satisfies u f (x, v) = u f (x, ρ(v)) because of the way broken ways reflect. If we assume f to have vanishing broken ray transform, then u f (x, v) = 0 whenever x E. We have ended up with the PDE V Xu f = 0 with boundary conditions on (SM) = ( M) S 1. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
71 Energy estimate for a PDE. For x R this function satisfies u f (x, v) = u f (x, ρ(v)) because of the way broken ways reflect. If we assume f to have vanishing broken ray transform, then u f (x, v) = 0 whenever x E. We have ended up with the PDE V Xu f = 0 with boundary conditions on (SM) = ( M) S 1. If we can show that this PDE has only the zero solution, then the broken ray transform is injective! Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
72 Energy estimate for a PDE Lemma If u : SM R has sufficient regularity and satisfies the boundary conditions on (SM) mentioned above for u f, then V Xu 2 L 2 (SM) = XV u 2 L 2 (SM) + Xu 2 L 2 (SM) K(x) V u(x, v) 2 SM (SM) where K is the curvature of M and κ is the curvature of M. κ(x) V u(x, v) 2, Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
73 Energy estimate for a PDE. Lemma If u : SM R has sufficient regularity and satisfies the boundary conditions on (SM) mentioned above for u f, then V Xu 2 L 2 (SM) = XV u 2 L 2 (SM) + Xu 2 L 2 (SM) K(x) V u(x, v) 2 SM (SM) where K is the curvature of M and κ is the curvature of M. κ(x) V u(x, v) 2, Suppose K 0 in M and κ 0 in R. If f has vanishing broken ray transform and u f has sufficient regularity, applying the lemma to u f gives so f = Xu f = 0. 0 = Xu f 2 L 2 (SM) + something positive Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
74 Energy estimate for a PDE Theorem (I. Salo 2014) Let ˆM be a nonpositively curved strictly convex Riemannian surface. Let O ˆM be a strictly convex open subset. Let M = ˆM \ O with E = M and R = O. Then the broken ray transform on M is injective on C 2 (M). Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
75 Energy estimate for a PDE. Theorem (I. Salo 2014) Let ˆM be a nonpositively curved strictly convex Riemannian surface. Let O ˆM be a strictly convex open subset. Let M = ˆM \ O with E = M and R = O. Then the broken ray transform on M is injective on C 2 (M). The proof is based on the energy identity (Pestov identity) and establishing sufficient regularity to use the identity. For regularity, we introduce and study Jacobi fields along broken rays. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
76 Outline 1 Introduction 2 Approaches 3 Periodic broken ray transform The problem Counterexamples Reflection 4 Algebraic tomography problems Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
77 The problem For the X-ray transform E = M, but now we take E =. Broken rays are replaced with periodic broken rays. Can a function be recovered from its integrals over all periodic broken rays? Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
78 The problem. For the X-ray transform E = M, but now we take E =. Broken rays are replaced with periodic broken rays. Can a function be recovered from its integrals over all periodic broken rays? There is a similar problem on closed manifolds (without boundary): Can one recover a function from its integrals over all closed geodesics? Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
79 Counterexamples Any counterexample for the usual broken ray transform is a counterexample for the periodic one. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
80 Counterexamples. Any counterexample for the usual broken ray transform is a counterexample for the periodic one. Proposition (I. 2013) There are nontrivial compactly supported smooth functions in the Euclidean disc that integrate to zero over all periodic broken rays. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
81 Reflection. Let us consider the periodic broken ray transform in a cube in R n. If we glue together 2n (suitably reflected) copies of the cube and then identify opposite edges, periodic broken rays in the original cube become periodic geodesics on the torus T n. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
82 Outline 1 Introduction 2 Approaches 3 Periodic broken ray transform 4 Algebraic tomography problems The problem Geodesics on tori Notation The Radon transform on tori Lie groups Finite groups Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
83 The problem If M is a closed manifold (compact without boundary), can a function on it be recovered from its integrals over all periodic geodesics? Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
84 The problem If M is a closed manifold (compact without boundary), can a function on it be recovered from its integrals over all periodic geodesics? What if the manifold has algebraic structure? Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
85 The problem. If M is a closed manifold (compact without boundary), can a function on it be recovered from its integrals over all periodic geodesics? What if the manifold has algebraic structure? We assume M to be a Lie group. The torus T n is a Lie group. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
86 Geodesics on tori Let T n = R n /Z n. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
87 Geodesics on tori Let T n = R n /Z n. Every homomorphism T 1 T n is of the form t tv mod Z n for some v Z n. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
88 Geodesics on tori Let T n = R n /Z n. Every homomorphism T 1 T n is of the form t tv mod Z n for some v Z n. For every closed geodesic γ : T 1 T n there are x T n and v Z n \ 0 so that γ(t) = x + tv mod Z n. Geodesics are translates of nontrivial homomorphisms from the group T 1. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
89 Geodesics on tori. Let T n = R n /Z n. Every homomorphism T 1 T n is of the form t tv mod Z n for some v Z n. For every closed geodesic γ : T 1 T n there are x T n and v Z n \ 0 so that γ(t) = x + tv mod Z n. Geodesics are translates of nontrivial homomorphisms from the group T 1. This is true on all compact Lie groups. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
90 Notation We denote f, g = G f(x)g(x)dx for f, g C (T n ). Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
91 Notation We denote f, g = G f(x)g(x)dx for f, g C (T n ). We will invert the Radon transform on the Fourier side. Recall that T n = R n /Z n. For every k define the function e k (x) = e 2πik x. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
92 Notation. We denote f, g = G f(x)g(x)dx for f, g C (T n ). We will invert the Radon transform on the Fourier side. Recall that T n = R n /Z n. For every k define the function e k (x) = e 2πik x. The Fourier transform of a function f : T n C is a function f : Z n C defined by f(k) = f, e k = f(x)e k (x)dx. T n The Fourier transform is injective. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
93 The Radon transform on tori We parametrize the Radon transform of f : T n C as Rf(x, v) = where x T n and v Z n \ f(x + vt)dt, Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
94 The Radon transform on tori We parametrize the Radon transform of f : T n C as Rf(x, v) = where x T n and v Z n \ f(x + vt)dt, Let Rf(k, v) be the Fourier transform of Rf(x, v) in the variable x. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
95 The Radon transform on tori. We parametrize the Radon transform of f : T n C as Rf(x, v) = where x T n and v Z n \ f(x + vt)dt, Let Rf(k, v) be the Fourier transform of Rf(x, v) in the variable x. We need the Radon transform of the basis elements: Re k (x, v) = e k (x)δ 0,k v. For every k Z n we can find v k Z n \ {0} orthogonal to it. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
96 The Radon transform on tori This definition of the Radon transform has a big benefit: the Radon transform is symmetric. The key is to understand the Radon transform as a family of integral transforms indexed by Z n \ 0 instead of a single operator. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
97 The Radon transform on tori This definition of the Radon transform has a big benefit: the Radon transform is symmetric. The key is to understand the Radon transform as a family of integral transforms indexed by Z n \ 0 instead of a single operator. If we fix v, the map f Rf(, v) is continuous C (T n ) C (T n ) (and also from distributions to distributions). Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
98 The Radon transform on tori This definition of the Radon transform has a big benefit: the Radon transform is symmetric. The key is to understand the Radon transform as a family of integral transforms indexed by Z n \ 0 instead of a single operator. If we fix v, the map f Rf(, v) is continuous C (T n ) C (T n ) (and also from distributions to distributions). For any v we have Rf(, v), g = f, Rg(, v). Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
99 The Radon transform on tori. This definition of the Radon transform has a big benefit: the Radon transform is symmetric. The key is to understand the Radon transform as a family of integral transforms indexed by Z n \ 0 instead of a single operator. If we fix v, the map f Rf(, v) is continuous C (T n ) C (T n ) (and also from distributions to distributions). For any v we have Rf(, v), g = f, Rg(, v). It is easy to handle distributions and Fourier analysis. There is no need for a normal operator. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
100 The Radon transform on tori If we know Rf(x, v) for all x T n and v Z n \ {0}, then we know Rf(k, v) Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
101 The Radon transform on tori If we know Rf(x, v) for all x T n and v Z n \ {0}, then we know Rf(k, v) = Rf(, v), e k Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
102 The Radon transform on tori If we know Rf(x, v) for all x T n and v Z n \ {0}, then we know Rf(k, v) = Rf(, v), e k = f, Re k (, v) Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
103 The Radon transform on tori If we know Rf(x, v) for all x T n and v Z n \ {0}, then we know Rf(k, v) = Rf(, v), e k = f, Re k (, v) = f, e k δ 0,k v Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
104 The Radon transform on tori If we know Rf(x, v) for all x T n and v Z n \ {0}, then we know Rf(k, v) = Rf(, v), e k = f, Re k (, v) = f, e k δ 0,k v = f(k)δ 0,k v for all k and v. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
105 The Radon transform on tori. If we know Rf(x, v) for all x T n and v Z n \ {0}, then we know Rf(k, v) = Rf(, v), e k = f, Re k (, v) = f, e k δ 0,k v = f(k)δ 0,k v for all k and v. Let us fix any k choose v orthogonal to it. Then f(k) = Rf(k, v). We can thus recover f from Rf! Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
106 Lie groups Theorem (I. 2014, Grinberg 1991) Let G be a compact, connected Lie group with at least two points. The Radon transform is injective on G if and only if G / {S 1, S 3 }. The transforms can be inverted explicitly. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
107 Lie groups Theorem (I. 2014, Grinberg 1991) Let G be a compact, connected Lie group with at least two points. The Radon transform is injective on G if and only if G / {S 1, S 3 }. The transforms can be inverted explicitly. This is true even if the Radon transform acts on distributions instead of smooth functions. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
108 Lie groups Theorem (I. 2014, Grinberg 1991) Let G be a compact, connected Lie group with at least two points. The Radon transform is injective on G if and only if G / {S 1, S 3 }. The transforms can be inverted explicitly. This is true even if the Radon transform acts on distributions instead of smooth functions. The exceptional groups have many names: S 1 = SO(2) = U(1) = T 1 and S 3 = SU(2) = Sp(1). Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
109 Lie groups. Theorem (I. 2014, Grinberg 1991) Let G be a compact, connected Lie group with at least two points. The Radon transform is injective on G if and only if G / {S 1, S 3 }. The transforms can be inverted explicitly. This is true even if the Radon transform acts on distributions instead of smooth functions. The exceptional groups have many names: S 1 = SO(2) = U(1) = T 1 and S 3 = SU(2) = Sp(1). The group SO(3) has been studied by several authors due to applications. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
110 Finite groups We defined geodesics to be translates of homomorphisms S 1 G. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
111 Finite groups We defined geodesics to be translates of homomorphisms S 1 G. If G is a finite group, we can define geodesics as translates of homomorphisms C n G where C n is the cyclic group. We cannot scale all geodesics to have the same length, so we need different ns. Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
112 Finite groups. We defined geodesics to be translates of homomorphisms S 1 G. If G is a finite group, we can define geodesics as translates of homomorphisms C n G where C n is the cyclic group. We cannot scale all geodesics to have the same length, so we need different ns. Theorem (I. 2014) The Radon transform is injective on a finite group G if and only if G is not a Frobenius complement. (In the case of functions from G to a field of characteristic zero.) Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
113 End. Thank you. Questions? Slides and papers available at Joonas Ilmavirta (Jyväskylä) Broken ray tomography /
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