Rank of Homology Groups?

asked 2024-05-29 19:38:56 +0200

I'm trying to compute Magnitude Homology, but when compiling and running the code I keep getting the error: 'HomologyGroup_class_with_category' object has no attribute 'rank.'

I've tried other methods for trying to compute the rank of the Homology Groups, but it's the same thing. Not all the typical functions used for Abelian Groups carries over to Homology in sagemath. The only functions I've found that work are .ngens(), .invariants(), .gens(), and .order().

Can someone please explain to me how to find the rank of the Homology groups I'm trying to work with? I'm looking for torsion, so any explanation on how to work with these will be a big help!

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Please provide an actual code illustrating the issue.

Max Alekseyev ( 2024-05-29 22:17:22 +0200 )edit

You could define any simplicial complex "C" and then ask for C.rank() and this will get you the error. Other functions like C.ngens(), C.invariants(), and C.gens() will work but an error always shows up for C.rank().

WinterStorm7 ( 2024-05-29 22:40:11 +0200 )edit

FrédéricC ( 2024-05-30 07:43:25 +0200 )edit

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answered 2024-05-29 22:20:32 +0200

John Palmieri

8771 ●19 ●71 ●189 http://www.math.washin...

You can use the invariants method:

sage: K = simplicial_complexes.KleinBottle()
sage: H = K.homology(1)
sage: H.invariants()
(2, 0)
sage: T3 = simplicial_complexes.SurfaceOfGenus(3)
sage: T3.homology(1)
Z^6
sage: T3.homology(1).invariants()
(0, 0, 0, 0, 0, 0)

If you want the rank, count how many zeroes are in this list. If you want the torsion, look at the nonzero entries.

sage: T3.homology(1).invariants().count(0)
6

sage: H
Z x C2
sage: [x for x in H.invariants() if x > 0]
[2]

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Ah, thank you this helps a lot!

WinterStorm7 ( 2024-05-29 22:38:37 +0200 )edit

also one can work over QQ

sage: C=simplicial_complexes.Sphere(4)
sage: h = C.homology(base_ring=QQ)
sage: h[4].dimension()
1

FrédéricC ( 2024-05-30 21:12:02 +0200 )edit

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