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

[cianfa72]

TL;DR

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.

 

[ergospherical]

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*}

 

[cianfa72]

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 $\otimes ^ k (M)$.

 

[ergospherical]

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

 

[cianfa72]

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.

 

[ergospherical]

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 $p \in M$.

 

[cianfa72]

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..

 

[ergospherical]

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.