Signature of $f: \Lambda^2(\mathbb{R}^4) \times \Lambda^2(\mathbb{R}^4) \to \mathbb{R}$, $f(\omega, \omega') = \omega \wedge \omega'$ Define$$f: \Lambda^2(\mathbb{R}^4) \times \Lambda^2(\mathbb{R}^4) \to \Lambda^4(\mathbb{R}^4) \cong \mathbb{R}, \quad f(\omega, \omega') = \omega \wedge \omega'.$$

What is the signature of $f$?

I read some things on Wikipedia about signature and a bit in Wald's General Relativity book, but they were somewhat confusing about explaining something as simple as this....
 A: Indeed, $f$ is a symmetric form, since $\omega$ and $\omega '$ are Grassmann-even: $$(\text dx \wedge \text d y)\wedge (\text d z \wedge \text d t)=(\text d z \wedge \text d t)\wedge(\text dx \wedge \text d y)$$etc.. Now, to calculate the signature, you should find a basis which diagonalizes $\omega$, the dimension of the space is $6$. A basis is given by:$$E_1=\frac{\text dx \wedge \text d y+\text dz \wedge \text d t}{2}$$ $$F_1=\frac{\text dx \wedge \text d y-\text dz \wedge \text d t}{2}$$
and cyclic permutations on $(x,y,z)$. Since $$ E_1 \wedge E_1 =\text dx\wedge \text dy \wedge \text d z \wedge \text d t =-F_1 \wedge F_1$$ and $ \text dx\wedge \text dy \wedge \text d z \wedge \text d t $ is invariant under cyclic permutations of $(x,y,z)$ the entries on the diagonal are $(1,1,1,-1,-1,-1)$. Moreover, it's easy to see that $E_i \wedge F_j =0$ and $E_i \wedge E_j  =F_i \wedge F_j = 0$ for $i\neq j$. So the signature of $f$ is $((-1)^3,1^3)$.

As implied by the OP, here I'm freely using the fact that on an oriented $n$-dimensional vector space $V$ we have a canonical isomorphism beetween $n$-forms and scalars, given by: $$\lambda \text d \nu \leftrightarrow \lambda,$$ where $\text d \nu$ is the volume form of $V$, which I have defined as $\text d x \wedge \text d y \wedge \text d z \wedge \text d t$ for Minkowski's space.
