# Commutator of two vector fields

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?

## Answers and Replies

fresh_42
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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
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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.

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
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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##.

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

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
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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.