Equivalence of the persistence landscape diagram and the barcode?I am studying persistent homology for the...



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Equivalence of the persistence landscape diagram and the barcode?
I am studying persistent homology for the first time. I was reading Peter Bubenik's paper "Statistical Topological Data Analysis using Persistence Landscapes" from 2015 introducing persistent landscapes. I am quite confused on the approach on finding the values of the persistence landscape function using a barcode/persistence diagram. I feel like I have a naive misunderstanding of this topic as I shall attempt to explain.
Suppose X is a finite set of points in Euclidean space. From my understanding, if we consider the (finite length) persistence vector space given by the simplicial complex homology for a fixed dimension l, { H l ( X k ) } k = 1 n with maps { δ k , k }, for 1 k , k n, 1 k , k n, we have
{ H l ( X k ) } k = 1 n i = 1 m I ( b i , d i )
(Theorem 4.10 of this paper) for some multiset { ( b i , d i ) } i = 1 m , where I ( b , d ) gives the persistence vector space of length n,
0 . . . 0 R R R . . . 0 . . . 0
with non-zero vector spaces at values of the specified interval.
This multiset corresponds to the persistence diagram/barcode so that the k-th Betti number can be identified by finding the number of lines of the barcode that intersect the line x=k in R 2
Now Bubenik defines the Betti number of the persistence vector space for an interval [a,b] by β a , b = dim ( im ( δ a , b ) ), and the persistence landscape functions λ k : R [ , ], for k N by
λ k ( t ) = sup { m 0 β t m , t + m k }
Shouldn't β t m , t + m then correspond to the number of lines on the barcode that contain the interval [ t m , t + m ]], so that λ k ( t ) is the largest value of m that has at least k lines of the barcode intersecting [ t m , t + m ]?
I am confused on how the triangle construction is equivalent to the persistence landscape function instead. Any help would be much appreciated!

Answer & Explanation



Beginner2022-06-20Added 21 answers

To read off the value of the landscape function λ k ( t ) from the barcode for fixed k, imagine expanding an interval centered at t. A small interval might be contained in many bars of the barcode, but as the interval grows, fewer and fewer bars of the barcode can contain all of it. Continue expanding the interval until just before it is contained in only k−1 bars. The radius m of that interval is equal to λ k ( t ).
Of course, I prefer to obtain landscapes from persistence diagrams. To do this, drop two lines down and right from each point on the diagram to the diagonal and rotate the entire picture 45 clockwise. Imagine the "diagonal" (which is now horizontal) as the t-axis and the vertical distance as the m-axis. What you see formed from the lines you drew is the union of the graphs of the functions λ k ( t ) for k 1. The topmost ridge formed from the lines is the graph of λ 1 , the second topmost ridge formed is the graph of λ 2 , and so on.

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