\begin{theorem}[The Riemannian Cauchy--Crofton formula]
In the case where the scalar product in each point depends on a
metric tensor $M(x)$ varying continuously over our space, the
- Cauchy--Crofton formula becomes
+ Cauchy--Crofton formula becomes \fixme{assuming some stuff about
+ stuff}
\begin{equation}
- \abs{C}_M = \int_\mathcal{L} \sum_{p \in l_{\nu, \rho} \cap C}
- \, \frac{\det M(x)}{2 \left( \nu^T \cdot M(x) \cdot \nu \right)}
+ \abs{C}_M = \int_\mathcal{L} \sum_{x \in l_{\nu, \rho} \cap C}
+ \, \frac{\det M(x)}{2 \left( \nu^T \cdot M(x) \cdot \nu
+ \right)^{\sfrac{3}{2}}}
\, d\mathcal{L}(l_{\nu, \rho}).
\end{equation}
\end{theorem}
C \cap U_i) \, w_i(\nu) \, d\mathcal{L}(l_{\nu, \rho})
\label{eq:mpi_approx}
\end{equation}
- where $w_i$ is the weight-function used in the set $U_i$ using the
- constant tensor $M_\pi(x_i)$ and can be written
+ where $w_i$ is the weight-function used in the set $U_i$ which can
+ be written
\begin{equation}
w_i(\nu) = \frac{\det M(x_i)}{\left(\nu^T M(x_i) \nu
\right)^{\sfrac{3}{2}}}.
\begin{equation}
w_\pi(\nu, x) = \frac{\det M_\pi(x)}{\left(\nu^T M_\pi(x) \nu
\right)^{\sfrac{3}{2}}}.
+ \label{eq:wpi_def}
\end{equation}
Using this weight in \eqref{eq:mpi_approx} we can get rid of the sum
over the partition $i$ and form a sum of all intersection point of
Recall from \fixme{ref} that the left hand side is calculated as
\begin{equation}
- \abs{C}_{M_\pi} = \int_C \sqrt{ \dot{C}^T M_\pi \dot{C} } \, dt.
+ \abs{C}_{M_\pi}
+ = \int_C \abs{\dot{C}}_{M_\pi} \, dt
+ = \int_C \sqrt{ \dot{C}(t)^T M_\pi\big( C(t) \big) \dot{C}(t) }
+ \, dt.
\end{equation}
- We know that $M_\pi(x)$ converges pointwise to $M(x)$ and using the
- fact that the largest eigenvalue of $M$ is equal to 1, we know that
+ We know that $M_\pi(x)$ converges pointwise to $M(x)$, and thus
+ $\abs{\dot{C}}_{M_\pi}$ converges pointwise to $\abs{\dot{C}}_M$.
+ The largest eigenvalue of $M$ is equal to 1 and we can parametrize
+ $C$ by its arclength parameter such that $\abs{\dot{C}} = 1$. Thus
the integrand is bounded and we can apply Lebesgue's dominated
convergence theorem to see that $\abs{C}_{M_\pi} \to
- \abs{C}_{M(x)}$.
+ \abs{C}_{M}$.
We apply the same theorem to show that the right hand side of
- \fixme{ref} converges, but showing that $w_\pi(\nu, x)$ is bounded
- is a bit more involved.
-
- If we manage to bound the singular values and keep them away from
- $0$, that would be great. The singular values \fixme{or eigenvalues}
- of our tensor is
+ \fixme{ref} converges, but showing that $\sum_x w_\pi(\nu, x)$ is
+ bounded is a bit more involved. Recall the definition of $w_\pi$ in
+ \eqref{eq:wpi_def}. The numerator is equal to $\sigma_1^2
+ \sigma_2^2$ and thus from the construction in \fixme{ref} bounded
+ from above by $1$.
+
+ Next we need to bound $\nu^T M_\pi(x) \nu$ away from zero. According
+ to the Rayleigh principle \fixme{sigma squared is a bit meh?}
+ \begin{equation}
+ \sigma_2^2 = \min_{\norm{\xi} = 1} \sqrt{\xi^T M_\pi(x) \xi}
+ \end{equation}
+ and thus $\nu^T M_\pi(x) \nu \geq \sigma_2^2$ where $\sigma_2^2$ is
+ the smallest eigenvalue of $M_\pi(x)$. Recalling the construction in
+ \fixme{ref} we know that
\begin{align}
- \sigma_1 &= 1 \\
- \sigma_2 &= \frac{1}{1 + \frac{(s_1 -
- s_2)^2}{\gamma^2}} \geq \frac{1}{1 + \frac{s_1^2}{\gamma^2}}
- \geq \Kappa > 0
+ \sigma_2^2 &= \left(1 + \frac{(s_1 -
+ s_2)^2}{\gamma^2}\right)^{-1} \geq \left(1 +
+ \frac{s_1^2}{\gamma^2}\right)^{-1},
\end{align}
- and thus we only have to prove that the largest eigenvalue of our
- structure tensor $s_1$ is finite to show that the singular values of
- our metric tensor is bounded.
-
- The structure tensor is made by taking the outer product of the
- gradient of each point in a smoothed version of the noisy image $f$.
- This tensor is then again smoothed component-wise with a Gaussian
- kernel
+ where $s_1$ and $s_2$ are the largest and smallest eigenvalue of our
+ structure tensor respectively. Bounding $s_1$ from above would then
+ imply $\sigma_2^2 \geq K > 0$.
+
+ The structure tensor is constructed as follows:
\begin{equation}
- \nabla f_{\rho_1} \otimes \nabla f_{\rho_1} \mapsto K_{\rho_2} *
- \left( \nabla f_{\rho_1} \otimes \nabla f_{\rho_1} \right).
+ S(x) = \left(K_{\rho} * \left( \nabla \tilde{f}_{\sigma} \otimes \nabla
+ \tilde{f}_{\sigma} \right)\right)(x).
\end{equation}
- This is a smooth map from $\bar{\Omega}$ to $\mathbb{R}^{2\times
- 2}$, and assuming that our original image $f$ is bounded (?), so is
- this tensor.
-
- The singular values are the roots of a polynomial which depend
- continuously on the coefficients of the polynomial which depend
- continuously on the entries of the tensor. Thusly, the singular
- values are also bounded and the eigenvalues of our metric tensor is
- greater than $0$.
+ This is a smooth continuous map from $\bar{\Omega}$ to
+ $\mathbb{R}^{2\times 2}$. As we can see in \fixme{ref} the
+ eigenvalues are the roots of a monic polynomial and thus depend
+ continuously on the coefficients of the polynomial, which in turn
+ are continuous functions of the elements in the structure tensor
+ $S(x)$.
+
+ This proves that our weight function $w_\pi$ is bounded from above,
+ but the sum
+ \begin{equation}
+ \sum_{x \in l_{\nu, \rho} \cap C} w_\pi(\nu, x)
+ \end{equation}
+ might be infinite. However, if this happens for a set of lines with
+ measure greater than zero, the Euclidean Cauchy--Crofton formula in
+ \fixme{ref} implies that the curve is infinitely long.
- According to the Rayleigh principle \fixme{REF}
+ Assuming that our curve has finite length we can therefore bound the
+ integrand of \fixme{ref} and conclude using Lebesgue's dominated
+ convergence theorem \fixme{ref} that
+ \begin{equation}
+ \int_\mathcal{L} \sum_{x \in l_{\nu, \rho} \cap C}
+ w_\pi(\nu, x) \, d\mathcal{L}(l_{\nu, \rho})
+ \to
+ \int_\mathcal{L} \sum_{x \in l_{\nu, \rho} \cap C}
+ w(\nu, x) \, d\mathcal{L}(l_{\nu, \rho})
+ \end{equation}
+ \fixme{introduce $w$ somewhere}
+ which--as both sides of the equality has been shown to
+ converge--leaves us with what we wanted to prove
\begin{equation}
- \sigma_2 = \min_{\norm{x} = 1} \sqrt{x^T A x},
+ \abs{C}_M =
+ \int_\mathcal{L} \sum_{x \in l_{\nu, \rho} \cap C}
+ \frac{\det M(x)}{2 \left( \nu^T \cdot M(x) \cdot \nu
+ \right)^{\sfrac{3}{2}}} \,
+ d\mathcal{L}(l_{\nu, \rho})
\end{equation}
- so the denominator of \fixme{ref} is greater than $0$.
\end{proof}
\chapter{Discrete formulation}
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