2
$\begingroup$

I am learning a bit about persistence homology and was hoping that someone could point me to a systematic explanation of how to construct a persistence diagram, such as the one pictured below. I understand the basic idea of the level sets and the height function. I also understand that the lifespan of a feature corresponds to the local minima and maxima of the function.

enter image description here

However, I am not clear how this idea of birth and death of a feature corresponds to a point on the persistence diagram. If the diagonal line on the persistence diagram corresponds to the height of the level set, then how is the distance of the point above the diagonal line computed? That Y-coordinate must have some relation to either the length of the feature or something. Hence, I was looking for a good explanation of how the persistence diagram is created.

As a note, I checked the Edelsbrunner and Harer book, but did not see a persistence diagram that matches the shape of the one below.

Any help would be appreciated.

$\endgroup$
2
$\begingroup$

Ahh, after much research I finally have an answer. So the persistence diagram is very similar if not analogous to the barcode. The 45 degree line does track the height of the level set. When a feature is born--at some local minimum of the function--that point is recorded on the 45 degree line. In the provided diagram above, the dashed lines along the x axis represent the height at which some feature was born.

If at a certain height the feature dies--because of a local maximum--then a point is created on the diagram. The x-coordinate of that point is the height at which that point was born. The y-coordinate is the height at which the feature died. So the distance between the 45 degree line and the point is the duration of the persistence for that feature.

In other words, features that were short lived occur close to the 45 degree line, while more persistent features tend to occur at a greater distance from the 45 degree line.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.