Basis-dependent isomorphism from FiniteRankFreeModule to an object in the category ModulesWithBasis

[mkoeppe]

(This ticket has been narrowed to the topic mainly discussed in the comments. The topic originally in title and ticket description has been moved to #30174)

There are multiple incompatible protocols for dealing with bases of a free module in Sage (#19346).

One such protocol is defined by the category ModulesWithBasis. CombinatorialFreeModule is one of the parent classes in this category.

Another such protocol is defined by the parent class FiniteRankFreeModule, which is in the category Modules.
A FiniteRankFreeModule can be equipped with a finite list of bases (each represented by an instance of FreeModuleBasis), none of which are considered distinguished. FiniteRankFreeModules are unique parents; the operation of equipping a FiniteRankFreeModule does not change the identity of the parent.

We define a method FiniteRankFreeModule. isomorphism_with_fixed_basis that establishes, given a basis, an isomorphism with a CombinatorialFreeModule (and thus a parent in the category ModulesWithBasis.

Depends on #30180

CC: @egourgoulhon @tscrim @mjungmath

Component: categories

Author: Matthias Koeppe, Michael Jung

Branch/Commit: 8a800cc

Reviewer: Michael Jung, Matthias Koeppe, Eric Gourgoulhon, Travis Scrimshaw

Issue created by migration from https://trac.sagemath.org/ticket/30094

[mkoeppe]

Description changed:

--- 
+++ 
@@ -2,6 +2,7 @@
 ```
 sage: Q3 = FiniteRankFreeModule(QQ, 3)
 sage: Q3 in Modules(QQ)
+True
 sage: Modules(QQ)(Q3)
 TypeError: 'FiniteRankFreeModule_with_category' object has no attribute 'vector_space'
 unable to coerce x (=3-dimensional vector space over the Rational Field) into Category of vector spaces over Rational Field

[mkoeppe]

comment:2

Also it seems various maps are missing.

sage: Q3.has_coerce_map_from(QQ^3)
False
sage: Q3((QQ^3).an_element())
ValueError: the None has not been defined on the 3-dimensional vector space over the Rational Field

[tscrim]

comment:4

These are incompatible objects since QQ^3 has a distinguished basis (the standard Euclidean basis) whereas Q3 does not. So there should not be a map between them.

My 2 cents for the error reported in the ticket is to first check that the object is not in the category. The other option is to lift the vector_space method up to the category of modules and then implement a change_ring method for Q3.

[mkoeppe]

comment:5

Replying to @tscrim:

These are incompatible objects since QQ^3 has a distinguished basis (the standard Euclidean basis) whereas Q3 does not. So there should not be a map between them.

Shouldn't there be a map in one direction that forgets the distinguished basis?

[tscrim]

comment:6

Replying to @mkoeppe:

Replying to @tscrim:

These are incompatible objects since QQ^3 has a distinguished basis (the standard Euclidean basis) whereas Q3 does not. So there should not be a map between them.

Shouldn't there be a map in one direction that forgets the distinguished basis?

No, because where would you send the point (1,3,1) to? How would you know how it is expressed in one of the bases of Q3?

[mkoeppe]

comment:7

Ok, I feel we had this discussion before

[mkoeppe]

comment:8

Should FreeModuleBasis get a method that creates the identity map from self.free_module() to a free module with that basis?

[mjungmath]

comment:9

Replying to @mkoeppe:

Should FreeModuleBasis get a method that creates the identity map from self.free_module() to a free module with that basis?

No, I don't think it should. The identity map (like any other automorphism) is a basis independent object.

[tscrim]

comment:10

Replying to @mjungmath:

Replying to @mkoeppe:

Should FreeModuleBasis get a method that creates the identity map from self.free_module() to a free module with that basis?

No, I don't think it should. The identity map (like any other automorphism) is a basis independent object.

I might even go further to say that it would not be the identity map, but an isomorphism of free modules. We have to be careful about distinguishing things up to equality and up to isomorphism. For example, if I have V = span_R{x,y,z}, this is not equal to W = R^3. So there is no identity map from V -> W because I am not looking at End(V), but Mor(V, W). Yes, they are isomorphic, but there is no canonical isomorphism until I choose a distinguished basis for V and W.

[mkoeppe]

comment:11

Replying to @tscrim:

Replying to @mjungmath:

Replying to @mkoeppe:

Should FreeModuleBasis get a method that creates the identity map from self.free_module() to a free module with that basis?

No, I don't think it should. The identity map (like any other automorphism) is a basis independent object.

I might even go further to say that it would not be the identity map, but an isomorphism of free modules.

Of course, that's what I meant.

[mkoeppe]

comment:12

Replying to @tscrim:

Replying to @mkoeppe:

Should FreeModuleBasis get a method that creates the identity map from self.free_module() to a free module with that basis?

We have to be careful about distinguishing things up to equality and up to isomorphism. For example, if I have V = span_R{x,y,z}, this is not equal to W = R^3.
... there is no canonical isomorphism until I choose a distinguished basis for V and W.

In my question above, this would be a method of FreeModuleBasis - so there is a particular basis. And RR^3 also has a distinguished basis.

[tscrim]

comment:13

Replying to @mkoeppe:

Replying to @tscrim:

Replying to @mkoeppe:

Should FreeModuleBasis get a method that creates the identity map from self.free_module() to a free module with that basis?

We have to be careful about distinguishing things up to equality and up to isomorphism. For example, if I have V = span_R{x,y,z}, this is not equal to W = R^3.
... there is no canonical isomorphism until I choose a distinguished basis for V and W.

In my question above, this would be a method of FreeModuleBasis - so there is a particular basis. And RR^3 also has a distinguished basis.

You were asking for a map from self.free_module(), which is the free module without a distinguished basis. If you want to go from a FreeModuleBasis, then that might be a good method as there is a canonical isomorphism of the homspace with the endspace and an identity map. Although I am not sure how much the basis is intended as a basis as a set or the span of the basis elements. I suspect the former, so such a method would not make sense. I would have to see what Eric and Michal say.

[egourgoulhon]

comment:14

Replying to @tscrim:

Replying to @mkoeppe:

Replying to @tscrim:

Replying to @mkoeppe:

Should FreeModuleBasis get a method that creates the identity map from self.free_module() to a free module with that basis?

We have to be careful about distinguishing things up to equality and up to isomorphism. For example, if I have V = span_R{x,y,z}, this is not equal to W = R^3.
... there is no canonical isomorphism until I choose a distinguished basis for V and W.

In my question above, this would be a method of FreeModuleBasis - so there is a particular basis. And RR^3 also has a distinguished basis.

You were asking for a map from self.free_module(), which is the free module without a distinguished basis. If you want to go from a FreeModuleBasis, then that might be a good method as there is a canonical isomorphism of the homspace with the endspace and an identity map. Although I am not sure how much the basis is intended as a basis as a set or the span of the basis elements. I suspect the former, so such a method would not make sense. I would have to see what Eric and Michal say.

I of course agree that one may implement a method in FreeModuleBasis that returns an isomorphism between Q3 and QQ^3, but for which purpose? Such an isomorphism cannot be used for a coercion since a coercion must be unique and cannot depend on a extra parameter such as a
basis-dependent isomorphism.

Coming back to the issue mentioned in the ticket title and description, I would say that the problem arises because "vector spaces" as they are implemented in Sage are not really vector spaces in all their generality but only Cartesian powers F^n, where F is the base field.

[mkoeppe]

comment:15

Replying to @egourgoulhon:

Coming back to the issue mentioned in the ticket title and description, I would say that the problem arises because "vector spaces" as they are implemented in Sage are not really vector spaces in all their generality but only Cartesian powers F^n, where F is the base field.

OK, I think we are getting closer.

When you say, '"vector spaces" as they are implemented in Sage', I assume you mean the ones implemented in sage.modules.free_module and constructed by the function VectorSpace, and then I agree with what you wrote.

What I was asking for in comment 8 is, given a basis, a map from a FiniteRankFreeModule (which is in the category of finite dimensional modules) to a parent that is in the category of finite dimensional modules with basis.

[mjungmath]

comment:16

Replying to @tscrim:

I might even go further to say that it would not be the identity map, but an isomorphism of free modules.

Well, yes. It is a special kind of automorphism. However, I wouldn't see it as an isomorphism between any two free modules of the same rank, in case this is what you are referring to.

Replying to @tscrim:

Although I am not sure how much the basis is intended as a basis as a set or the span of the basis elements. I suspect the former, so such a method would not make sense. I would have to see what Eric and Michal say.

I am not sure what you mean with "the span of the basis elements". This does not make any sense to me because the span is always the free module again. Anyway, I would see an instance of FreeModuleBasis as a representation of a (finite) subset of the corresponding free module which is linearly independent and generates the module (possibly in the sense of formal linear combinations).