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Consider the objects $$A_\mu, ~~F_{\mu\nu}:=\partial_\mu A_\nu-\partial_\nu A_\mu,$$ and the objects $$A:=A_\mu dx^\mu,~~F:=\frac{1}{2!}F_{\mu\nu} dx^\mu\wedge dx^\nu.$$ While reading it from Zee's book on Quantum Field Theory in a Nutshell, I acquired the idea that the entities $A$ and $F$ defined above are examples of a 1-form and a 2-form respectively.

However, now going through Sean Carroll's book Spacetime and Geometry Section 2.9, I find that a $p$-form is defined as $(0,p)$ tensor that is completely antisymmetric. But if this definition is correct, then $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ is a 2-form not $F$. It also says that $\epsilon_{\mu\nu\sigma\rho}$ is a 4-form.

Please explain between $F_{\mu\nu}$ and $F$, what stands for a two-form.

Qmechanic
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Solidification
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    The first F is not a tensor, it is an arrangement/display of components of a 2nd rank tensor in a coordinate basis. Your 2nd F is a tensor, a 2-form – DanielC Oct 15 '21 at 05:11

3 Answers3

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In the physics literature it is quite common to casually refer to the components of a tensor as the tensor. The same practice is used for differential forms.

Qmechanic
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  • Thanks @Qmechanic There are $(p,0)$ tensors, $(0,p)$ tensors and $(m,n)$ tensors. $p$-forms are same as $(0,p)$ tensors, and sometimes the components of $(0,p)$ tensors are referred to as $p$-forms. Am I right? – Solidification Oct 20 '21 at 16:14
  • $\uparrow$ Yes. – Qmechanic Oct 20 '21 at 17:35
  • @Qmechanic is that correct? I've often seen the $(0,1)$ and $(1,0)$ referred to as 2-forms rather than 1-forms, including on the Wiki article on Lorentz group representations. – tomdodd4598 Jan 04 '23 at 11:34
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It's important to distinguish between a tensor and the components of a tensor. A (r,s) tensor is a multilinear map that takes r co vectors and s vectors, and maps them into the reals. In that case $dx^\mu$ takes a vector V and maps into the the reals, $dx^\mu(V) = V(x^\mu) =V^\mu$. Now here comes the tricky bit, $V^\mu$ is not a vector, but a component of a vector, i.e. a real number. Similarly, in your example, $F_{\mu\nu}$ and $A_\mu$ are real numbers, while $F$ and $A$ are a one and two form respectively.

Filipe Miguel
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$F$ is a (0,2) tensor, i.e. a bilinear map $F : V\times V \to K$ , where V is a vector field defined over field K. If $\{\partial_{\mu}\}$ be the basis vector for V and $\{dx^{\mu}\}$ be the corresponding dual basis vector satisfying $dx^a(\partial_b)=\delta^a_b$, then $F_{\mu\nu}=F(\partial_{\mu}, \partial_{\nu})$. If we are considering U(1) gauge theory, then $F\equiv dA$. So, $F_{\mu\nu}=dA(\partial_{\mu},\partial_{\nu})=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$ where we have used the definition of the exterior derivative of a 1-form $\omega$ : $d\omega (X,Y):=X(\omega(Y))-Y(\omega(X))-\omega([X,Y])$

KP99
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