# Need clarity about the definition and notation of $p$-forms used in physics

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.

• 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 Oct 15, 2021 at 5:11

• 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? Oct 20, 2021 at 16:14
• $\uparrow$ Yes. Oct 20, 2021 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. Jan 4 at 11:34
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.
$$F$$ 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 $$\omega$$ : $$d\omega (X,Y):=X(\omega(Y))-Y(\omega(X))-\omega([X,Y])$$