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

  • #1
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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.
 
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Answers and Replies

  • #2
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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*}
 
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  • #3
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113
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)##.
 
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  • #4
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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
 
  • #5
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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.
 
  • #6
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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##.
 
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  • #7
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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..
 
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  • #8
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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. :smile:
 
  • #9
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So basically a ##k##-form ##\omega_p## is just a totally antisymmetric (aka skew-symmetric) ##(0,k)## tensor, right ?
 
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