Differential operator vs one-form (covector field)

[cianfa72]

TL;DR

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.

 

[fresh_42]

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.

 

[cianfa72]

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 ?

 

[fresh_42]

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.

 

[cianfa72]

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.

 

[fresh_42]

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.

 

[cianfa72]

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 ?

 

[fresh_42]

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/

 

[cianfa72]

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.

 

[fresh_42]

$(1,1)$-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.

 

[cianfa72]

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:

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