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The homology of a gray scale image

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We will try to understand the meaning of the homology of gray scale images.

Below is an example of a gray scale image and its "frames", i.e., three thresholded versions. These are binary images that we understand as cell complexes (see Cell decomposition of images). Here white is empty.

“Lena”: original “Lena”: T = 50 “Lena”: T = 100 “Lena”: T = 150

For completeness sake one can add a fully white image in the beginning and fully black in the end.

Here is another example:

Coins.jpg Coins 1.png Coins 2.png Coins 3.png Coins 4.png

For complete analysis, let's consider this simple image:

Filtration of rings.png

First the image is "thresholded". The lower level sets of the gray scale function of the image form a filtration: a sequence of three binary images, i.e., cell complexes: $K^{1}\hookrightarrow K^{2}\hookrightarrow K^{3},$ where the arrows represent the inclusions. Suppose $A_{i},B_{i},C_{i}$ are the homology classes that represent the components of $K^{i}$ and $a_{i},b_{i},c_{i}$ are the holes, clockwise starting at the upper left corner. The homology groups of these images also form sequences -- one for each dimension $0$ and $1$.

Filtration of rings and cycles.png

Suppose $F_{1},F_{2}$ are the two homology maps, i.e., homomorphisms of the homology groups generated by the inclusions of the complexes, with $F_{3}=0$ included for convenience. These homomorphisms act on the generators, as follows: \begin{eqnarray*} A_{1} &\rightarrow &A_{2}\rightarrow A_{3}\rightarrow 0,B_{1}\rightarrow B_{2}\rightarrow B_{3}\rightarrow 0, \\ C_{2} &\rightarrow &C_{3}\rightarrow 0,a_{1}\rightarrow a_{2}\rightarrow a_{3}\rightarrow 0, \\ b_{1} &\rightarrow &0,c_{3}\rightarrow 0. \end{eqnarray*}

To avoid double counting, we want to count only the homology classes that don't reappear in the next homology group. As it turns out, a more algebraically convenient way to accomplish this is to count only the homology classes that go to $0$ under these homomorphisms. These classes form the kernels of $F_{1},F_{2},F_{3}$. Now, we choose the homology group of the original, gray scale image to be the direct sum of these kernels: \begin{equation*} H_{0}(\{K^{i}\})=<A_{3},B_{3},C_{3}>,\text{ }H_{1}(\{K^{i}% \})=<b_{1},a_{3},c_{3}>. \end{equation*} Thus the image has three components and three holes, as expected.

As a diagram, the $0$- and $1$-homology groups of the frames and the homomorphisms form two sequences: $$R^2 → R^2 → R^3 → 0,$$ $$R^2 → R^1 → R^2 → 0.$$ Instead of counting objects that do not reappear in the next frame, we use linear algebra to find the cycles in each frame that go to $0$ under $F_{1},F_{2},F_{3}$. These cycles form the kernels $$\ker F_{1}, \ker F_{2}, \ker F_{3}$$ of these linear operators. Then, the homology group of the gray scale image should be $$H_{0}(\{K^{i}\}) = \ker F_{1} \times \ker F_{2} \times \ker F_{3} = 0 \times 0 \times R^3 = R^3,$$ $$H_{1}(\{K^{i}\}) = \ker F_{1} \times \ker F_{2} \times \ker F_{3} = R \times 0 \times R^2 = R^3 .$$ Three components and three holes, again.

This is our conclusion: the homology group of a gray scale image is the product of the kernels of the linear operators generated by the inclusions of its frames.

Exercise: Compute the homology group of the blurred version and the negative of the above image.