Inversion and Symmetries of the Star Transform

For a unit vector $\psi\in\mathbb{R}^{2}$ and ${t}\in\mathbb{R}$ define the half plane transform of $f$ as

Geometrically, $F_{\psi}({t})$ is the integral of $f$ over the half plane on one side of the line $l(\psi,{t})$ (see Fig. 2).

Now, let’s look at the derivative of $F$ as a function of ${t}$

Note that $F_{\psi}({t}+h)-F_{\psi}({t})$ is the integral of $f$ over the infinite strip between the lines $l(\psi,{t})$ and $l(\psi,{t}+h)$, which plays a central role in our construction. Using Fubini’s theorem and the fundamental theorem of calculus, we get

where $\mathcal{R}$ is the standard Radon transform. Thus we have proved the following

A key step in our proof is the possibility of expressing the half plane transform $F_{\psi}$ of $f$ in terms of its star transform $\mathcal{S}f$. We show this by first finding a relation between $F_{\psi}$ and $\mathcal{X}_{\gamma}(f)$.

If $\langle\psi,\gamma\rangle\neq 0$ we can integrate $\mathcal{X}_{\gamma}(x)$ along the line $l(\psi,{t})$ to get an expression in terms of $F_{\psi}({t})$. Namely

It is easy to notice that if $\mathcal{Z}_{1}\cup\mathcal{Z}_{2}$ is finite, then the singularities appearing in the right hand side of formula (15) are removable. Namely,

In other words, the full Radon data can be recovered from the star transform.

We finish this section with statements of two special cases of Theorem 1 and discussion of some pertinent relations. In the case of $m=2$, Theorem 1 yields a new and simple inversion for the V-line transform.

A curious special case of Theorem 1 is the following.

The directional derivative $D_{-\gamma}$, given by $D_{\gamma}h\,(x)=\gamma\cdot\nabla h\,(x)$, is the natural inverse of $\mathcal{X_{\gamma}}$. Hence, for any $h\in U_{\gamma}=\left\{\mathcal{X}_{\gamma}(f):\,f\in C_{c}(\mathbb{R}^{2})\right\}$ in the range of $\mathcal{X}_{\gamma}$, one can write the following identity

The identity (19) is not really new. It is equivalent to a known relation

where $e_{i}$’s are standard orthonormal vectors and $P$ is a given polynomial of two variables. One can get (20) by repeatedly applying (19) and canceling $\mathcal{R}^{-1}\mathcal{R}$ in the resulting telescoping identity. The other direction of equivalence is trivial.

A trivial example is when a pair of rays point to opposite directions, i.e. $\gamma_{i}=-\gamma_{j}$ for some indices $i$ and $j$. Then it is easy to see that the direction normal to them is a singular direction of Type 2. A more interesting example is discussed below.

Consider an arbitrary open half-plane $H=\left\{x:\langle x,n\rangle<0\right\}$, where $n$ is the outward normal to its boundary passing through the origin. Choose a set of distinct ray directions $\left\{\gamma_{i}\right\}_{i=1}^{m}$ so that $\gamma_{i}\in H$ for all $i=1,...,m$. Without loss of generality, let us assume that $\gamma_{i}$’s are indexed according to the growth of their polar angle (see Figure 3(b)).

Consider $\psi=\gamma_{2}^{\perp}$ oriented so that $\langle\psi,\gamma_{1}\rangle>0$. Then $\langle\psi,\gamma_{2}\rangle=0$ and $\langle\psi,\gamma_{i}\rangle<0$ for all $i>2$. Estimating the terms of the sum in (21), it is easy to notice that $\prod_{j\neq 2}\langle\psi,\gamma_{j}\rangle<0$, while all of the other terms are 0. Thus, for $\psi=\gamma_{2}^{\perp}$ we get $P_{2}(\psi)<0$.

On the other hand, if $\psi=n$, then all terms of the sum in (21) are positive and $P_{2}(\psi)>0$. By continuity of $P_{2}(\psi)$, there has to exist a value of $\psi$, for which $P_{2}(\psi)=0$. $\hfill\blacksquare$

In this section we show that any regular star transform with an odd number of rays (see Figure 3(a)) has a stable inversion. More specifically,

Using the parameterization $\psi(\alpha)=(\cos\alpha,\sin\alpha)$, we denote

We will show that $F(\alpha)\neq 0$ for any $\alpha\in[0,2\pi]$. Our argument is based on the following statement, proved after the proof of Theorem 4.

Due to the symmetry of $P_{2}(\psi)$ with respect to $\gamma_{i}$’s and the symmetric distributions of $\gamma_{i}$’s on $\mathbb{S}^{1}$, it follows that $F(\alpha)$ is periodic with period $2\pi/m$. Therefore all of its Fourier coefficients $F_{k}$ are zero unless $k$ is divisible by $m$. Since $F$ is a trigonometric polynomial, it is equal to its Fourier expansion, and since it is of degree at most $(m-1)$, the only index in its Fourier expansion divisible by $m$ is $k=0$, thus $F$ is constant. By Theorem 2 that constant can not be zero, which completes the proof of Theorem 4. $\hfill\blacksquare$

It easy to notice, that after appropriate cancellations inside the brackets using (8), the last expression does not contain any odd degree of $s$. Substituting $s^{2}=1-r^{2}$ we obtain the desired result. $\hfill\blacksquare$

In the case of $m=3$, there is a simpler proof of Theorem 4. That proof also shows that the transform $\mathcal{S}$ corresponding to the regular star has (in some sense) the most stable inversion. It is based on the following result from [7]:

The regular stars with an odd number of rays are not the only configurations of the transform $\mathcal{S}$ that do not have a singular direction of Type 2. Due to continuous dependence of $P_{2}(\psi)$ on $\alpha_{i}$’s, small perturbations of $\gamma_{i}$’s from symmetry positions will not introduce a singular direction of Type 2. Similarly, small perturbations of uniform weights $c_{1}=\ldots=c_{m}=1$ to non-uniform values will not introduce singular directions of Type 2. Moreover,

For example, modifying the regular star transform of 3 rays (defined by formula (27)) through replacement of $\gamma_{2}$ by $-\gamma_{2}$ and changing the sign of $c_{2}=1$, we get a stable configuration with weights $c_{1}=1$, $c_{2}=-1$, $c_{3}=1$ and aperture vectors:

In general, one can get stable setups when the absolute values of weights are close to each other.

The star transforms with weights of mixed algebraic signs satisfying the condition

play an important role in simultaneous reconstruction of scattering and absorption coefficients in single scattering tomography [9, 10, 15, 26]. Simultaneously satisfying the above condition and keeping the absolute values of weights as close to each other as possible, suggest choosing the weights (up to a proportionality coefficient and re-indexing) as follows:

In the case of $m=3$, this implies choosing the weights proportional to $c_{1}=c_{2}=1$ and $c_{3}=-2$. Such a setup was analyzed in detail in [26].

We finish this section with generalizations of the statements of Section 5.1 to the case of transforms with non-uniform weights. For the stars with an even number of rays, both the result and its proof are almost identical. Namely,

In the case of star transforms with an odd number of rays, it should be clarified in what sense we want to generalize Part 2 of Theorem 3. If one is allowed to choose both the weights $c_{i}$ and the ray directions $\gamma_{i}$ to claim existence of setups with and without Type 2 singularities, then the statement has a trivial proof by choosing uniform weights and referring to Theorem 3. Hence, we consider the following question. Are there configurations (i.e. ray directions $\gamma_{1},\ldots,\gamma_{m}$) with/without Type 2 singularities for a specified set of weights $c_{1},\ldots,c_{m}$?

A more interesting example is the case when all $\gamma_{i}$’s belong to the same half-plane. Recall the corresponding setup from the proof of Theorem 3. Without loss of generality, let us assume that the $c_{1}c_{2}>0$, i.e. the weights $c_{1}$ and $c_{2}$ have the same sign. Since we consider the case of $m=2k+1$ for $k\geq 1$, the assumption above is just a matter of re-indexing the weights (if necessary).

Let $\psi_{1}=\gamma_{1}^{\perp}$ and $\psi_{2}=\gamma_{2}^{\perp}$ be vectors obtained by a clockwise rotation by $\pi/2$ from $\gamma_{1}$ and $\gamma_{2}$ correspondingly. Then, it is easy to verify that $P_{2}(\psi_{1})$ and $P_{2}(\psi_{2})$ have different signs. Hence, by continuity of $P_{2}$ there exists some $\psi$ s.t. $P_{2}(\psi)=0$. $\hfill\blacksquare$

Before proceeding to the proof of the second statement of Theorem 6, we define a new class of planes in $\mathbb{R}^{3}$ and discuss their properties.

Not every plane (containing the origin) in $\mathbb{R}^{3}$ is SSAV. The following lemma gives a complete description of admissible normal vectors to an SSAV plane.

The set $A$ of admissible normals to an SSAV plane has an interesting geometric description due to the three triangle inequalities

In the octant $n_{1}>0$, $n_{2}>0$, $n_{3}>0$ it coincides with an infinite polyhedral cone with a vertex at the origin and infinite edges along vectors $(0,1,1)$, $(1,0,1)$ and $(1,1,0)$. We denote the portion of $A$ in this first octant by $A^{+}$. In all other octants $A$ coincides with a similar tetrahedron with the edges along appropriately signed versions of the same vectors.

Formula (26) shows that every plane with a normal vector $n\in A^{+}$ intersects $e_{2}^{-1}(0)$ only at the origin. Moreover, among the SSAV planes, those with a normal vector in $A^{+}$ are the only ones that intersect $e_{2}^{-1}(0)$ only at the origin. Hence, the set $A^{+}$ contains normals to SSAV planes that correspond (through the spanning aperture vectors) to star transforms with uniform weights $c_{1}=c_{2}=c_{3}=1$ and a stable inversion, i.e. without a singular direction of Type 2.

Let

denote the diagonal matrix of weights.

Proof. Let $n^{(1)},n^{(2)}$ denote some vectors normal to the planes $T_{1}$ and $T_{2}$ correspondingly. The proof will follow immediately if we show that for any $c_{1},c_{2},c_{3}>0$ there exist $n^{(1)},n^{(2)}\in A^{+}$ such that $n^{(2)}=Wn^{(1)}$.

We first note that the problem is invariant with respect to multiplication of all weights $c_{1},c_{2},c_{3}>0$ by a positive constant, as well as their permutation. Hence, without loss of generality we will assume that $0<c_{1}\leq c_{2}=1\leq c_{3}$.

A choice of desired pair of vectors $n^{(1)},n^{(2)}$ is given by

which ends the proof. $\hfill\blacksquare$

We will use our knowledge of existence of stable configurations for every star transform with uniform weights (marked by superscript ⁽²⁾ in the proof below) to show existence of stable configurations for star transforms with positive weights (marked by superscript ⁽¹⁾ in the proof below).

By Lemma 6 there exist SSAV planes $T_{1},T_{2}$ such that $T_{2}=W^{-1}T_{1}$ and $T_{i}\cap e_{2}^{-1}(0)=\{0\}$ for $i=1,2$. Let $a^{(1)},b^{(1)}$ and $a^{(2)},b^{(2)}$ be some star aperture vector pairs spanning correspondingly $T_{1}$ and $T_{2}$.

To prove the theorem, it is enough to show that for ray directions $\gamma_{1}^{(1)},\gamma_{2}^{(1)},\gamma_{3}^{(1)}$ corresponding to the aperture vectors $a^{(1)},b^{(1)}$ the following relation holds