##C^{\infty}##-module of smooth vector fields can lack a basis

[cianfa72]

TL;DR

About the fact that the $C^{\infty}$-module of smooth vector fields is not guaranteed to have a basis (not even infinite dimensional)

In this lecture, the lecturer claims that the $C^{\infty}$-module of smooth vector fields defined on a smooth manifold can lack to admit a basis (not even infinite dimensional).

Indeed the set of smooth vector fields can be given an (infinite dimensional) vector space structure over the field $\mathbb R$ (i.e. defining it as a $\mathbb R$-vector space). However the set of $C^{\infty}$ functions on a smooth manifold can't be given a field structure but just a (commutative) ring structure.

Take now the span of a collection of smooth vector fields defined on a smooth manifold. By definition of span, they define a vector subspace of the $\mathbb R$-vector space of smooth vector fields.

Viewed as subspace of $C^{\infty}$-module of smooth vector fields, I believe it has a basis though. Nevertheless I'm concerned about the uniqueness of representation when considering as components of vector fields in the span the smooth functions from the ring of $C^{\infty}$ functions and not the real numbers from the field $\mathbb R$.

 

[jbergman]

TL;DR Summary: About the fact that the C∞-module of smooth vector fields is not guaranteed to have a basis (not even infinite dimensional)

In this lecture, the lecturer claims that the C∞-module of smooth vector fields defined on a smooth manifold can lack to admit a basis (not even infinite dimensional).

Indeed the set of smooth vector fields can be given an (infinite dimensional) vector space structure over the field R (i.e. defining it as a R-vector space). However the set of C∞ functions on a smooth manifold can't be given a field structure but just a (commutative) ring structure.

Take now the span of a collection of smooth vector fields defined on a smooth manifold. By definition of span, they define a vector subspace of the R-vector space of smooth vector fields.

Viewed as subspace of C∞-module of smooth vector fields, I believe it has a basis though. Nevertheless I'm concerned about the uniqueness of representation when considering as components of vector fields in the span the smooth functions from the ring of C∞ functions and not the real numbers from the field R.

Nitpick, but a subspace for a module is called a submodule. Also, if you take a subspace of a vector space, that set will not be a sub-module and so you would need to take the span of that set as a module. But in general modules are tricky to work with so it isn't obvious to me that it even has a basis.

 

[cianfa72]

Nitpick, but a subspace for a module is called a submodule. Also, if you take a subspace of a vector space, that set will not be a sub-module and so you would need to take the span of that set as a module. But in general modules are tricky to work with so it isn't obvious to me that it even has a basis.

Ah ok. You mean that a vector subspace of a vector space doesn't even form a submodule, therefore it makes no sense to talk about a basis for it (since it isn't even a submodule).

 

[martinbn]

His comment is general, it is just algebra, nothing to do with manifolds. A vector space always has a basis. A module not necessarily.

 

[cianfa72]

A vector space always has a basis. A module not necessarily.

Ok. In some special cases, however, a module may have a basis.

 

[cianfa72]

Ok. In some special cases, however, a module may have a basis.

Btw, can you give an example of module that admit a basis ?

 

[martinbn]

Btw, can you give an example of module that admit a basis ?

Of course, any free module, if $A$ is a ring, then $A$ as an $A$-module has a basis, namely the identity of the ring. Or $A^n$ has a basis.

 

[cianfa72]

Of course, any free module, if $A$ is a ring, then $A$ as an $A$-module has a basis, namely the identity of the ring. Or $A^n$ has a basis.

Ah ok, you mean the set of $A$ elements viewed/understood as an $A$-module over itself (as ring).

Your example is similar to, say, the set $\mathbb R$ understood as affine space over $\mathbb R$ as translation vector space.

 

[mathwonk]

a module has a basis if and only if it is a free module. the module of smooth vector fields on a smooth manifold is only "locally free", not necessarily free, i.e. at least its restriction to each coordinate neighborhood is free. A basis would be given by a collection of vector fields whose vectors form a basis at each point, such as occurs for a 2- torus. The fact that the module of smooth vector fields on a 2-sphere is not free is implied by the fact that every smooth vector field on a sphere has a zero.

"Locally free" modules are also (at least if finitely generated over a noetherian ring) equivalent to "projective" modules algebraically, i.e. equivalent to being a direct summand of a free module. (Notice the module of vector fields tangent to a 2-sphere is a submodule of the restriction to the sphere of the free module of vector fields on all of R^3. I.e. we get a free module if we remove the requirement that the vectors should be tangent to the 2-sphere. It seems to me that this free module is the direct sum of the tangent vector fields and the normal vector fields, i.e. we can project each vector onto its normal and tangential components .) Thus they have some but not all properties of a free module; e.g. every surjective morphism to a projective module has a right inverse.

https://en.wikipedia.org/wiki/Projective_module

 

[cianfa72]

the module of smooth vector fields on a smooth manifold is only "locally free", not necessarily free, i.e. at least its restriction to each coordinate neighborhood is free. A basis would be given by a collection of vector fields whose vectors form a basis at each point, such as occurs for a 2- torus. The fact that the module of smooth vector fields on a 2-sphere is not free is implied by the fact that every smooth vector field on a sphere has a zero.

I believe the problem on the (smooth) 2-sphere is that if one insists to include a smooth vector field $X$ into a basis then, at the point it vanishes (say P), one is forced to include two other vector fields say $Y,Z$ which evaluated at P must give two linearly independent vectors $Y(p)$ and $Z(p)$.

By continuity in any open neighborhood of P (say $U$) any smooth vector field $W$ evaluated at a point $q \in U$ wouldn't have an unique representation by using $C^{\infty}$ functions evaluated at that point.