Exterior derivative

Question

Let $$\omega = \frac{1}{2} \sum_{i,j} \omega_{i,j} dx_i \wedge dx_j$$ be an antisymmetric form, so i.e. $\omega_{i,j} = - \omega_{j,i}.$

Now, in some lecture notes I found that

$$d \omega (X_F,X_G,X_H) = \sum_{a,b,c} \left( \frac{\partial \omega_{b,c}}{\partial x_a} +\frac{\partial \omega_{c,a}}{\partial x_b}+\frac{\partial \omega_{a,b}}{\partial x_c} \right)(X_F)_a (X_G)_b (X_H)_c$$ was stated as a result that should be more or less obvious to the reader.

The thing is that I just don't see how you get this out of the exterior derivative. Does anybody see it and could describe what has happened here?

Answer 1

To calculate $d\omega$ use $$d\omega_{i,j}=\sum_s\frac{\partial \omega_{i,j}}{\partial x_s}dx^s,$$ for the components and then ensamble $$d\omega = \frac{1}{2} \sum_{i,j} d\omega_{i,j}\wedge dx_i \wedge dx_j.$$

Answer 2

$d\omega$ is $1/2 \sum \partial \omega_{i,j}/\partial x_k dx_k\wedge dx_i\wedge dx_j$.

Answer 3

Following Spivak Notation:

So we know by definition that $\omega:=\frac{1}{2}\sum_{i_1<i_2}\omega_{i_1,i_2}\space dx^{i_1} \wedge dx^{i_2}$ so $\omega$ is a 2-form.

Thus if we compute $d\omega$ we attain $$d\omega = d\left[ \frac{1}{2}\sum_{i_1<i_2}\omega_{i_1,i_2}\space dx^{i_1} \wedge dx^{i_2} \right]$$ $$\qquad=\frac{1}{2}\sum_{i_1<i_2}d(\omega_{i_1,i_2})\wedge dx^{i_1} \wedge dx^{i_2} $$ $$\qquad \qquad=\frac{1}{2}\sum_{i_1<i_2}\sum_{\alpha=1}^{n} D_\alpha (\omega_{i_1,i_2} ) \cdot \space dx^\alpha \wedge dx^{i_1} \wedge dx^{i_2}$$ $$\equiv \frac{1}{2}\sum_{i_1<i_2}\sum_{\alpha=1}^{n}\frac{\partial \omega_{i_1,i_2}}{\partial x_\alpha} \cdot \space dx^\alpha \wedge dx^{i_1} \wedge dx^{i_2} $$

Note that if $\alpha=i_1$ or $\alpha=i_2$ (whatever $i_1,i_2$ equal!) then by the rules of multivariable calculus such 3-form terms created must be zero (since we repeated indices automatically cancel)