Commutator of two vector fields

[Pencilvester]

Hello PF, I was reading Carroll’s definition of the commutator of two vector fields in “Spacetime and Geometry”, and I’m having (I think) a simple case of notational confusion.
He says for two vector fields, $X$ and $Y$, their commutator can be defined by its action on a scalar function, $f(x^{\mu})$:$$[X,Y](f) \equiv X(Y(f)) - Y(X(f))$$This is the first time I’m seeing a vector “acting on a function”. What exactly is the nature of this action? I’m very familiar with a vector’s action on a one-form, so my only guess is that this is a shorthand way of saying “a vector acting on the gradient of a function”, so then $X(Y(f))$ would really be $X ( \text{d} ( Y ( \text{d} f ) ) )$. Is that right? Or am I way off?

 

[fresh_42]

Hello PF, I was reading Carroll’s definition of the commutator of two vector fields in “Spacetime and Geometry”, and I’m having (I think) a simple case of notational confusion.
He says for two vector fields, $X$ and $Y$, their commutator can be defined by its action on a scalar function, $f(x^{\mu})$:$$[X,Y](f) \equiv X(Y(f)) - Y(X(f))$$This is the first time I’m seeing a vector “acting on a function”. What exactly is the nature of this action? I’m very familiar with a vector’s action on a one-form, so my only guess is that this is a shorthand way of saying “a vector acting on the gradient of a function”, so then $X(Y(f))$ would really be $X ( \text{d} ( Y ( \text{d} f ) ) )$. Is that right? Or am I way off?

The vector is the gradient in a sense. It is at it's heart the behavior of $f$ along $X$, a directional derivative if you like. And $[X,Y]$ describes how far the endpoints of a rectangle vary if you go along $X$ followed by $Y$ or the other way around. Commuting vector fields mean the two path end at the same point; only that we haven't considered certain points in the above equation where the directional derivatives are evaluated, so the image is a bit sloppy because of the abstraction of the point. Otherwise we would have had $X_p(f) = X|_p(f)= D_p(X)(f)$ or several other notations.

I've tried to be a bit more accurate here: https://www.physicsforums.com/insights/pantheon-derivatives-part-ii/
and you can find an example here: https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/ or in part 4 of the series in the first link.

 

[Orodruin]

I find it hard to believe that Carroll has not mentioned vectors as derivative operators. The typical order of introduction is vectors as derivatives, one-forms, and then gradients and exterior derivatives. Typically you would not say that vectors act on one-forms, rather that one-forms are linear maps from vectors to scalars (and vice versa). The differential $df$ is a one-form such that $Xf = df(X)$. I usually try to avoid the parentheses in $Xf$, but that is just notational.

 

[Pencilvester]

I find it hard to believe that Carroll has not mentioned vectors as derivative operators. The typical order of introduction is vectors as derivatives, one-forms, and then gradients and exterior derivatives.

As far as I can remember, he explained the rationale behind using $\partial _{\mu}$ as the natural coordinate basis for vectors (and how, therefore, vectors can be thought of as directional derivatives), then he had a little bit about the dual basis for one-forms, and then the commutator. I simply did not bring that connection between vectors and directional derivatives over to the commutator.

The differential $df$ is a one-form such that $Xf = df(X)$.

Okay, I think I’m a little shaky on the distinction between the differential $df$ and the gradient $\text {d} f$. What is the difference between $df (X)$ and $\text {d} f (X)$?

 

[fresh_42]

As far as I can remember, he explained the rationale behind using $\partial _{\mu}$ as the natural coordinate basis for vectors (and how, therefore, vectors can be thought of as directional derivatives), then he had a little bit about the dual basis for one-forms, and then the commutator. I simply did not bring that connection between vectors and directional derivatives over to the commutator.

You should read the insights. The first link and especially part 1 of that series could be helpful. Have you seen the notation
$$
d_pf (v) = \langle \nabla f(p) , v \rangle \text{ or } \nabla \times F
$$
These are strong indications, that the gradient $\nabla$ needs to be a vector. At least at some point. Without such a point, the entity of all possible gradients is a vector field. And it is applied to a function.

Okay, I think I’m a little shaky on the distinction between the differential $df$ and the gradient $\text {d} f$. What is the difference between $df (X)$ and $\text {d} f (X)$?

None. There are dozens of ways to write the differential. Usually there is also a point $p$ of evaluation involved: the location of the tangent: $d_pf (X)=X_p(f)=X_pf$ and without that point, i.e. $df(X)=Xf$ we have all possible tangents at $f$.

 

[Pencilvester]

None. There are dozens of ways to write the differential.

Okay, thanks!

Have you seen the notation
$$
d_pf (v) = \langle \nabla f(p) , v \rangle \text{ or } \nabla \times F
$$

I haven’t, but they seem intuitive enough, except for the $\nabla \times F$. I thought that was a notation for curl.

 

[fresh_42]

It is curl, and it uses the gradient as a vector.

 

[Pencilvester]

Oh, gotcha, I was thinking you were meaning $d_pf (v) = \langle \nabla f(p) , v \rangle = \nabla \times F$, but you were asking if I was familiar with the curl notation in addition to the notation for the one-form/vector contraction, yes?

 

[fresh_42]

Yes. The gradient is a vector expressed in the basis $\dfrac{\partial}{\partial x_i}$ and we can evaluate this at a point or apply it to a function. O.k. the other way around.