Show that the wedge product $ dX \wedge dX = 0 $ and $dY \wedge dY = 0$

Question

So first I want to give you some background information:

begin of the background information

I'm currently reading an abstact about the Lotka Volterra differential equations:

$$ x^{'} = x -xy $$ $$ y^{'} = -y +xy $$

We know that the most numerical methods give us spiraling solutions instead of cyclic. So I want to try to show that a modified forward Euler Method leads to cyclic solutions.

Here:

$$ \frac{x_{n+1}-x_n}{\Delta t} = x_n -x_ny_n $$ $$\frac{y_{n+1}-y_n}{\Delta t} = -y_n -x_{n+1}y_n $$

There is a proof to show that this modification does not spiral.

end of background information

Now I have some questions:

So let us say to simplify notitation $$ X = \Delta tx +x - \Delta txy$$ and $$ Y = -\Delta ty+ y + \Delta tXy $$ where we set $X:=x_{n+1}, Y:=y_{n+1}$, $x:= x_n$ and $ y := y_n$ solved for the unknown $X$ and $Y$.

Taking derivatives: $$ dX = \Delta tdx +dx - \Delta tdxy - \Delta txdy $$ and $$ dY = -\Delta tdy+ dy + \Delta tdXy + \Delta tXdy $$

Finally we arrived to one point where I stucked.

I need that $ dX \wedge dX = 0 $ and $ dY \wedge dY = 0 $. Can you help me out here?

Answer 1

A defining property of the outer product is its anti-symmetry $$ a∧b = -b∧a, $$ which implies that $$ 2\,a∧a=0. $$