Differential k-form vs (0,k) tensor field

• Nov 22, 2021

• #1

cianfa72

1,692

192

TL;DR Summary

Clarification about differential k-form vs (0,k) tensor field

Hi,

I would like to ask for a clarification about the difference between a differential k-form and a generic (0,k) tensor field.

Take for instance a (non simple) differential 2-form defined on a 2D differential manifold with coordinates ##\{x^{\mu}\}##. It can be assigned as linear combination of terms ##dx^{\mu} \wedge dx^{\nu}## and it is basically a multi-linear application from ##V \times V## to ##\mathbb R## (##V## is the tangent vector space at each point of the 2D manifold). So I think a 2-form is actually just a particular (0,2) tensor field defined on the 2D manifold.

Is that right ? Thank you.

 

Last edited: Nov 22, 2021

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• Nov 22, 2021

• #2

ergospherical

943

1,261

The key aspect is that forms are totally antisymmetric. That's also why we have the wedge product ##\wedge : \Lambda^k(M) \times \Lambda^l(M) \rightarrow \Lambda^{k+l}(M)##, because the tensor product \begin{align*}
R(\mathbf{e}_1, \mathbf{e}_2) \equiv (S \otimes T)(\mathbf{e}_1, \mathbf{e}_2) = S(\mathbf{e}_1) T(\mathbf{e}_2)
\end{align*}of two forms isn't a form. But the antisymmetrised version is:
\begin{align*}
R'(\mathbf{e}_1, \mathbf{e}_2) \equiv (S \wedge T)(\mathbf{e}_1, \mathbf{e}_2) = S(\mathbf{e}_1) T(\mathbf{e}_2) - T(\mathbf{e}_1) S(\mathbf{e}_2)
\end{align*}The generalisation to forms of higher degree is straightforward, viz\begin{align*}
(U \wedge V)(\mathbf{e}_1, \dots, \mathbf{e}_n) = \dfrac{1}{k! l!} \sum_{\sigma} \mathrm{sgn}(\sigma) U(\mathbf{e}_{\sigma(1)}, \dots) V(\mathbf{e}_{\sigma(k+1)}, \dots)
\end{align*}

 

Last edited: Nov 22, 2021

• Nov 22, 2021

• #3

cianfa72

1,692

192

ergospherical said:

The key aspect is that forms are totally antisymmetric. That's also why we have the wedge product ##\wedge : \Lambda^p(M) \times \Lambda^q(M) \rightarrow \Lambda^{p+q}(M)##, because the tensor product \begin{align*}
R(\mathbf{e}_1, \mathbf{e}_2) \equiv (S \otimes T)(\mathbf{e}_1, \mathbf{e}_2) = S(\mathbf{e}_1) T(\mathbf{e}_2)
\end{align*}of two forms isn't a form. But the antisymmetrised version is:
\begin{align*}
R'(\mathbf{e}_1, \mathbf{e}_2) \equiv (S \wedge T)(\mathbf{e}_1, \mathbf{e}_2) = S(\mathbf{e}_1) T(\mathbf{e}_2) - T(\mathbf{e}_1) S(\mathbf{e}_2)
\end{align*}

I take it as if ##S## and ##T## are two different one-forms -- i.e. two (0,1) tensors defined on manifold-- then their tensor product is not totally antisymmetric.

Instead their antisymmetric version ##S \wedge T## is.

On the other hand the set of k-forms ##\Lambda^k(M)## on the tangent vector space ##T_p(M)## is actually a vector subspace of the tensor product vector space ##\otimes ^ k (M)##.

 

Last edited: Nov 22, 2021

• Nov 22, 2021

• #4

ergospherical

943

1,261

cianfa72 said:

On the other hand the set of k-forms ##\Lambda^k(M)## on the tangent vector space ##T_p(M)## is actually a vector subspace of the tensor product vector space ##\otimes ^ k (M)##.

Careful,

• ##\Lambda^k M## is the set of ##k##-forms on ##M##
• ##\Lambda^k_p M \equiv \wedge^k T_p^* M## is the set of ##k##-forms at the point ##p \in M##, and is contained within ##\otimes^k T_p^* M##

And ##\otimes^k M## doesn't mean anything

 

• Nov 22, 2021

• #5

cianfa72

1,692

192

ergospherical said:

Careful,

• ##\Lambda^k M## is the set of ##k##-forms on ##M##
• ##\Lambda^k_p M \equiv \wedge^k T_p^* M## is the set of ##k##-forms at the point ##p \in M##, and is contained within ##\otimes^k T_p^* M##

ok, so ##\Lambda^k M## is really the set of ##k##-forms defined at each point of the manifold ##M##, right ?

##\wedge^k T_p^* M## as subset of ##\otimes^k T_p^* M## turns out to be a vector subspace of it.

 

• Nov 22, 2021

• #6

ergospherical

943

1,261

cianfa72 said:

ok, so ##\Lambda^k M## is really the set of ##k##-forms defined at each point of the manifold ##M##, right ?

No. Strictly one should make the distinction between

• ##k##-form fields ##\Lambda^k M \ni \omega : M \rightarrow \wedge^k T_p^* M##
• ##k##-forms at the point ##p##, that is, ##\wedge^k T_p^* M \ni \omega_p : \times^k T_p M \rightarrow \mathbf{R}##.

Which is to say that the ##k##-form field ##\omega## is a smooth assignment of ##k##-forms ##\omega_p## to each point in ##p \in M##.

 

Last edited: Nov 22, 2021

• Nov 22, 2021

• #7

cianfa72

1,692

192

ergospherical said:

No. Strictly one should make the distinction between

• ##k##-form fields ##\Lambda^k M \ni \omega : M \rightarrow \wedge^k T_p^* M##
• ##k##-forms at the point ##p##, that is, ##\wedge^k T_p^* M \ni \omega_p : \otimes^k T_p M \rightarrow \mathbf{R}##.

Which is to say that the ##k##-form field ##\omega## is a smooth assignment of ##k##-forms ##\omega_p## to each point in ##p \in M##.

Sorry, ##\omega_p## should be a multilinear map from ##\times^k T_p M## to ##\mathbf{R}## and not from ##\otimes^k T_p M## to ##\mathbf{R}##, I believe..

 

• Nov 22, 2021

• #8

ergospherical

943

1,261

cianfa72 said:

Sorry, ##\omega_p## should be a multilinear map from ##\times^k T_p M## to ##\mathbf{R}## and not from ##\otimes^k T_p M## to ##\mathbf{R}##, I believe..

Yes, good catch.

 

• Nov 22, 2021

• #9

cianfa72

1,692

192

So basically a ##k##-form ##\omega_p## is just a totally antisymmetric (aka skew-symmetric) ##(0,k)## tensor, right ?

 

Last edited: Nov 22, 2021

1. What is the difference between a differential k-form and a (0,k) tensor field?

A differential k-form is a mathematical object that describes a quantity that varies in different directions and is used in the study of differential geometry. It is a type of differential form that is defined on a smooth manifold. On the other hand, a (0,k) tensor field is a multilinear map that assigns a k-dimensional vector to each point on a manifold. In simpler terms, a differential k-form is a type of tensor field, but not all tensor fields are differential k-forms.

2. How are differential k-forms and (0,k) tensor fields used in physics?

Differential k-forms and (0,k) tensor fields are used in physics to describe physical quantities and their relationships in a geometric way. They are particularly useful in the study of relativity and electromagnetism, where the curvature of space-time and the electromagnetic field can be described using these mathematical objects.

3. Can you give an example of a differential k-form and a (0,k) tensor field?

An example of a differential k-form is the electromagnetic field strength tensor in electromagnetism. It describes the electromagnetic field at each point in space and is a differential 2-form. A (0,k) tensor field example would be the stress-energy tensor in general relativity, which describes the energy and momentum density at each point in space and is a (0,2) tensor field.

4. What are the mathematical operations that can be performed on differential k-forms and (0,k) tensor fields?

Differential k-forms and (0,k) tensor fields can be added, multiplied by a scalar, and multiplied with other differential k-forms or (0,k) tensor fields. They can also be differentiated and integrated over a manifold. These mathematical operations are important in manipulating and solving equations involving these mathematical objects.

5. How do differential k-forms and (0,k) tensor fields relate to each other?

Differential k-forms and (0,k) tensor fields are related in that they both describe quantities that vary in different directions on a manifold. However, they have different mathematical definitions and properties. Differential k-forms are a special type of tensor field that is defined on a smooth manifold, while (0,k) tensor fields are more general and can be defined on any manifold.