Axioms for 2-forms, why should $\omega_x(\Delta x \wedge \Delta x) = 0$

Question

I am having trouble understanding this piece about 2-forms from Terence Tao's "Differential Forms and Integration". I understand the bilinearity requirement in analogy to the one-dimensional case. But I don't understand the last axiom (8). Why should it be zero?

$\omega_x(\Delta x \wedge \Delta x) = 0$

Answer 1

Okay, figured it out. Normally we have two different arguments delta1 x and delta2 x. They are both vectors and represent the sides of a parallelogram extending from our base point. So if delta1 x = delta2 x then the sides are the same and so the parallelogram is "collapsed" and has no area at all. Then it is clear there should be no "flux" through such a parallelogram. This explains why it should be zero.

Answer 2

$\Delta x\wedge \Delta x =0$

$\omega(\Delta x\wedge \Delta x)=\omega(0)=\omega(0 \Delta x)= 0\omega(\Delta x)=0$