Most general $2$-form in $\mathbb R^3$

Question

I'm currently reading "The Geometry Of Physics An Introduction" by Theodore Frankel.

On page 69 in the chapter introducing the "Grassmann Algebra" he writes down this expression of a "most general $2$-form in $\mathbb{R}^3$":

$$\sum_{i<j} b_{ij} dx^i \land dx^j = b_{12}dx^1 \land dx^2 + b_{13}dx^1 \land dx^3 + b_{23}dx^2 \land dx^3 $$

$$ = b_{23}dx^2 \land dx^3 + b_{31}dx^3 \land dx^1 + b_{12}dx^1 \land dx^2 $$

But as I understood it they're not equal because the tensor in the $b_{31}$ term would be minus the tensor in the $b_{13}$ term due to the antisymmetry? Is this just a definition or have I not understood something ( which is definitely possible since I only started this chapter today and am not very good at math yet...).

Answer 1

The matrix of coefficients $(b_{ij})$ is skew-symmetric, so $$b_{13}dx^1\wedge dx^3=-b_{13}dx^3\wedge dx^1=b_{31}dx^3\wedge dx^1.$$ To see this, note the if we call the form $\omega,$ then $\omega\left(\frac{\partial}{\partial x^1},\frac{\partial}{\partial x^3}\right)=b_{13},$ $\omega\left(\frac{\partial}{\partial x^3},\frac{\partial}{\partial x^1}\right)=b_{31},$ and $$\omega\left(\frac{\partial}{\partial x^1},\frac{\partial}{\partial x^3}\right)=-\omega\left(\frac{\partial}{\partial x^3}\frac{\partial}{\partial x^1}\right).$$