Differential operator vs one-form (covector field)

Hi, I'd like to ask for clarification about the definition of differential of a smooth scalar function ##f: M \rightarrow \mathbb R## between smooth manifolds ##M## and ##\mathbb R##.

As far as I know, the differential of a scalar function ##f## can be understood as:

1. a linear map ##df()## between tangent spaces defined at each point of domain and target manifolds (##T_{p}M## and ##T_{q}\mathbb R##)
2. a one-form or covector field ##df## defined on the domain manifold ##M##

In both cases ##d()## operator is actually the exterior derivative operator and both definitions should be actually equivalent.

Does it make sense ? Thank you.

cianfa72 said:

TL;DR Summary: About the definition of differential operator of a scalar function as one-form or covector field

Hi, I'd like to ask for clarification about the definition of differential of a smooth scalar function ##f: M \rightarrow \mathbb R## between smooth manifolds ##M## and ##\mathbb R##.

As far as I know, the differential of a scalar function ##f## can be understood as:

1. a linear map ##df()## between tangent spaces defined at each point of domain and target manifolds (##T_{p}M## and ##T_{q}\mathbb R##)
2. a one-form or covector field ##df## defined on the domain manifold ##M##

In both cases ##d()## operator is actually the exterior derivative operator and both definitions should be actually equivalent.

Does it make sense ? Thank you.

Yes. These are two ways to look at derivatives. It all depends on what you consider to be a variable: function, location, manifold, tangent vectors, or combinations of them.

The two definitions above are equivalent if you consider the union over all locations in the first definition, i.e. make it a (co-)vector field.

fresh_42 said:

The two definitions above are equivalent if you consider the union over all locations in the first definition, i.e. make it a (co-)vector field.

You basically mean "extend" the definition of differential operator on all the points on the manifold turning it into a field defined on it, right ?

Your first definition was a specific linear map at ##p## and the second one was a vector field from the beginning. This means you need to define ##\displaystyle{ df\, : \,\bigsqcup_{p\in M} T_pM\rightarrow \bigsqcup_{p\in M} T_{f(p)}\mathbb{R}}## simply to make them the same thing.

fresh_42 said:

Your first definition was a specific linear map at ##p## and the second one was a vector field from the beginning.

You mean a co-vector field, I believe.

cianfa72 said:

You mean a co-vector field, I believe.

This also only depends on the perspective. I admit that I tend to confuse the two, and from a mathematical point of view, they aren't very different. What is a slope ##s##? Is it the linear map "times ##s##" or is it the line with the slope ##s,## or is it the scalar ##s## we attach to the derivative a some point?

Your example consists of linear mappings from vector spaces into vector spaces. However, the codomain vector spaces are all ##T_q\mathbb{R}=\mathbb{R}.## So do you map vectors into the scalar field ##\mathbb{R}## making them covectors, or into the tangent space ##T_q\mathbb{R}## which would be a linear map? Linear maps on the other hand can be seen as ##(1,1)##-tensors, i.e. a linear combination of covectors.

I think the latter is the correct point of view in this case. We have a linear map ##d_pf\, : \,T_pM\longrightarrow T_{f(p)\mathbb{R}}## which can be written as
$$
d_pf(v)=\left(\sum_\rho u_\rho^*\otimes V_\rho\right)(v)=\sum_\rho u_\rho^*(v)\cdot V_\rho
$$
with covectors ##u_\rho^* \in (T_pM)^*\, , \,v\in T_pM,## and ##V_\rho\in T_{f(p)}\mathbb{R}.## I have treated ##\mathbb{R}## as a manifold here, i.e. as if ##f:M\rightarrow N## where a function between manifolds. That's why I have ##V_\rho \in T_{f(p)}\mathbb{R}## although in fact we only have one dimension and the entire vector space is spanned by ##1.## This means ##V_\rho = 1## in the above case. I just found the general case with ##V_\rho## better if we want to see what is happening. ##\ldots \cdot 1## isn't very enlightening.

So you are right, covector field. I have written vector field because it is the more general term. A covector field is always a vector field, too, since covectors are vectors. This way, i.e. by writing vector field, I tried to avoid thinking about what you urged me to write in this post here anyway.

Edit: The ##u_\rho## are multiples of the basis ##dx_\rho## in this case.

fresh_42 said:

This means you need to define ##\displaystyle{ df\, : \,\bigsqcup_{p\in M} T_pM\rightarrow \bigsqcup_{p\in M} T_{f(p)}\mathbb{R}}## simply to make them the same thing.

Here ##\displaystyle { \bigsqcup_{p\in M} T_pM}## and ##\displaystyle { \bigsqcup_{p\in M} T_{f(p)}\mathbb R}## are actually tangent bundles ?

cianfa72 said:

Here ##\displaystyle { \bigsqcup_{p\in M} T_pM}## and ##\displaystyle { \bigsqcup_{p\in M} T_{f(p)}\mathbb R}## are actually tangent bundles ?

Yes.

You can find the whole nine yards here (5 parts):
https://www.physicsforums.com/insights/the-pantheon-of-derivatives-i/

fresh_42 said:

Linear maps on the other hand can be seen as ##(1,1)##-tensors, i.e. a linear combination of covectors.

##(1,1)##-tensors should be actually a linear combination of tensor product of covectors times vectors.

cianfa72 said:

##(1,1)##-tensors should be actually a linear combination of tensor product of covectors times vectors.

Sure, but the covectors can be evaluated. That gives you a linear combination of vectors
$$
d_pf(v)=\left(\sum_\rho u^*_\rho \otimes V_\rho\right)(v)=\sum_\rho u^*_\rho(v) \cdot V_\rho.
$$
Your example is
$$
d_pf=\sum_\rho u^*_\rho \otimes V_\rho =\sum_\rho u^*_\rho \otimes 1 =\sum_\rho u^*_\rho = \sum_\rho a_\rho {d_p}x_\rho
$$
which is a linear combination of the basis covectors ##dx_k## at ##p.##

Everything is a matter of perspective.

fresh_42 said:

A covector field is always a vector field, too, since covectors are vectors.

Covectors are vectors in the sense that every element of a vector space is by definition a "vector" !

Edit: in my example above we have just one vector ##V## in the target 1-dimensional tangent space ##T_p\mathbb R=\mathbb R## at ##p##, hence it should be:

$$d_pf=\sum_\rho u^*_\rho \otimes V =\sum_\rho u^*_\rho \otimes 1 =\sum_\rho u^*_\rho = \sum_\rho a_\rho {d_p}x_\rho$$