Norm of curvature 2-form in coordinates

Question

Set up

Let $M=\mathbb{R}^n$ with Euclidean metric. Suppose $F \in \Omega^2(M)$ is an 2-form (e.g. curvature) and $\phi \in \Omega^1(M)$ be a 1-form. These can be written in coordinates as:

$$F = \sum_{i<j} F_{ij} dx^idx^j \quad \quad \phi = \phi_i dx^i\thinspace .$$

Taking the covariant derivative of $\phi$ we get:

$$F = \sum_{i<j} F_{ij} dx^idx^j \quad \quad \nabla \phi =\sum_{i,j} \nabla_i\phi_j \thinspace dx^i\otimes dx^j\thinspace .$$

Goal: Compute the norm of $F$ and $\nabla \phi$ in coordinates.

Question

Are the below expressions correct (repeated indices to be summed)?

$$|F|^2 = F_{ij}F^{ij} \quad |\nabla \phi|^2 = \nabla_i\phi_j\nabla^i\phi^j\thinspace .$$

By $F_{ij}F^{ij}$ I think we mean:

$$F_{ij}F^{ij} = \sum_{i,j} F_{ij}F^{ij}\thinspace .$$

Confusion

In an article the authors consider the following action:

$$ \mathscr{A} = \int_M d^4x \left(\frac{1}{2} F_{ij}F^{ij} + \nabla_i\phi_j\nabla^i\phi^j \right)\thinspace .$$

So do they mean

1. $ \mathscr{A} = \int_M d^4x \left(\frac{1}{2} |F|^2 + |\nabla \phi|^2 \right)\thinspace$ OR
2. $ \mathscr{A} = \int_M d^4x \left(|F|^2 + |\nabla \phi|^2 \right)\thinspace ?$

Upshot: Confused by a factor of 2, all help is appreciated.