How to intersect subspaces of a combinatorial free module?

asked 2022-06-08 04:32:59 +0100

peter.xu
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updated 2024-05-04 21:20:34 +0100

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Hi, I've defined a CombinatorialFreeModule over ZZ with a pretty large named basis, along with a family of subspaces using .submodule(). In the documentation, I see that the class FreeModule_generic_field has a method .intersection() to intersect these subspaces - is there a way to do this for the combinatorial free modules as well? I can think of some very ugly messy solutions by abandoning the CombinatorialFreeModule class, but I was hoping there'd be a better way.

Here's a toy example:

Z = CombinatorialFreeModule(ZZ, ['a','b','c'], prefix="z") 
z = Z.basis()
Z1 = Z.submodule([z['a'],z['b']])
Z2 = Z.submodule([z['b'],z['c']])

I'd like to be able to write something like Z1.intersection(Z2) to get a submodule containing just z['b']. Of course, the real example has a very large number of submodules with much more complicated generators.

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def subm_intersect(Z1,Z2): Z = Z1.basis()[0].lift().parent() assert Z == Z2.basis()[0].lift().parent() # Z1 and Z2 shold have the same parent module z = Z.basis() V = VectorSpace(QQ, len(z)) V1 = V.subspace( [[b.lift().coefficient(k) for k in z.keys()] for b in Z1.basis()] ) V2 = V.subspace( [[b.lift().coefficient(k) for k in z.keys()] for b in Z2.basis()] ) U = V1.intersection(V2) return Z.submodule( [Z.linear_combination(zip(z,u)) for u in U.basis()] )

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