Vector fields wedge product vs covector field

[cianfa72]

TL;DR

About the difference between the wedge product of $(n-1)$ independent vector fields vs a covector field (1-form)

There are two ways to assign a $(n-1)$-dimensional distribution on the tangent bundle built over a differentiable manifold of dimension $n$. Namely it can be assigned either via the wedge product of $(n-1)$ independent vector fields or via a covector field (1-form).

Which is the difference between them ?

 

[jambaugh]

Under special linear transformations they should behave identically. However, under reflections I believe they behave differently. E.g. under inversion in the x-direction $\hat{x}\wedge \hat{y}\to -\hat{x}\wedge \hat{y}$ while $(\hat{z}•)\to (\hat{z}•)$. (Notation: $(\hat{z}•):\vec{v}\mapsto \hat{z}•\vec{v}$.)

 

[cianfa72]

I took a look at R. Penrose book The Road to Reality section 12.6. He claims that a 1-form should be thought of as a kind of density while the wedge product of $(n-1)$ vectors should not.

 

[martinbn]

To me it isn't clear what you want to know. The reference doesn't help because it seems to be about something different (the exterior derivative).

 

[cianfa72]

To me it isn't clear what you want to know. The reference doesn't help because it seems to be about something different (the exterior derivative).

Sorry, the relevant sections from that source are actually 12.4 (p. 228 in my edition) and 12.5.

 

[martinbn]

Sorry, the relevant sections from that source are actually 12.4 (p. 228 in my edition) and 12.5.

Ok, but what is your question?

 

[cianfa72]

From that section, I think I got the answer: a wedge product of $n-1$ independent vector fields can't be understood as a "density" like a 1-form (covector field).

Btw, the tangent space at each point on a $n$-dimensional differentiable manifold has itself dimension $n$ (i.e. the maximum number of linear independent vectors at point P is $n$). Coming to vector fields, is it possible to have $m > n$ independent vector fields ?

 

[martinbn]

From that section, I think I got the answer: a wedge product of $n-1$ independent vector fields can't be understood as a "density" like a 1-form (covector field).

Yes, but keep in mind that by density he means something that can be integrated.

Btw, the tangent space at each point on a $n$-dimensional differentiable manifold has itself dimension $n$ (i.e. the maximum number of linear independent vectors at point P is $n$). Coming to vector fields, is it possible to have $m > n$ independent vector fields ?

Yes, the algebra of vector fields is infinite dimensional.

 

[cianfa72]

Yes, but keep in mind that by density he means something that can be integrated.

Ok.

Yes, the algebra of vector fields is infinite dimensional.

The set of vector fields (i.e. the set of smooth sections on the tangent bundle over the base manifold) can be assigned a vector space structure. Is this vector space infinite dimensional ?

 

[martinbn]

The set of vector fields (i.e. the set of smooth sections on the tangent bundle over the base manifold) can be assigned a vector space structure. Is this vector space infinite dimensional ?

Yes, that is what I said. It is infinite dimensional over the real numbers. As an example take vector fields on the line. They are of the form $a(x)\frac{\partial}{\partial x}$. Then all of these $\frac{\partial}{\partial x}, x\frac{\partial}{\partial x}, \dots, x^n\frac{\partial}{\partial x},\dots$ are indipendent.

 

[cianfa72]

Btw, maybe a bit off topic, consider the 2-sphere $\mathbb S^2$ and the set of its Killing Vector Fields (KVFs).
Any such KVF generates rotations about a specific axis. They form a subset of the vector space of vector fields $\Gamma(T\mathbb S^2)$ defined on $\mathbb S^2$.

Is it a vector subspace of $\Gamma(T\mathbb S^2)$ with dimension 3 ?

 

[cianfa72]

Yes, that is what I said. It is infinite dimensional over the real numbers. As an example take vector fields on the line. They are of the form $a(x)\frac{\partial}{\partial x}$. Then all of these $\frac{\partial}{\partial x}, x\frac{\partial}{\partial x}, \dots, x^n\frac{\partial}{\partial x},\dots$ are indipendent.

Sorry, I was thinking about the following: $C^{\infty}(\mathbb R)$ isn't a field, hence the set of vector fields on $\mathbb R$ as manifold can't be a vector space over $C^{\infty}(\mathbb R)$. Nevertheless it is infinite dimensional vector space over $\mathbb R$.

Indeed your $a(x)\frac{\partial}{\partial x}, a(x) \in C^{\infty}(\mathbb R)$ are vector fields (i.e. they are smooth sections of $T\mathbb R$) and the unique $\mathbb R$-linear combination of $\frac{\partial}{\partial x}, x\frac{\partial}{\partial x}, \dots, x^n\frac{\partial}{\partial x},\dots$ that gives the null vector field consists of all zeros (i.e. they are $\mathbb R$-linear independent). As an aside, smooth vector fields form a $C^{\infty}(M)$-module -- MSE.

Does it make sense?