# Exterior derivative calculation in Chern and Hamilton paper

Chern and Hamilton in their paper "On Riemannian metrics adapted to three-dimensional contact manifolds" constructed an structure on 3-sphere as follows (Example 3.2 of paper):

Let $$\omega_1=xdy-ydx+zdw-wdz,\quad \omega_2=xdz-zdx+ydw-wdy,\quad \omega_3=xdw-wdx+ydz-zdy,$$ Taking exterior derivative we have: $$d\omega_1=2\omega_2\wedge \omega_3,\quad d\omega_2=2\omega_3\wedge \omega_1,\quad d\omega_3=2\omega_1\wedge \omega_2.$$

But my calculations show that: $$d\omega_1=2(dx\wedge dy+dz\wedge dw)\neq 2\omega_2\wedge \omega_3=2(z^2-w^2)dx\wedge dy+2(x^2-y^2)dz\wedge dw+2(xw-yz)(dx\wedge dz-dy\wedge dw)+2(xz-yw)(dy\wedge dz-dx\wedge dw),$$

Which part of my calculation is wrong?

Is this on the sphere $$x^2+y^2+z^2+w^2=1?$$ If so then $$x\,dx+y\,dy+z\,dz+w\,dw=0$$ there. Try using these relations in addition to your calculations.
• @C.F.G Did you also use $x^2+y^2+z^2=1$? If you really cannot derive these equations from these relations, then they aren't true. – Lord Shark the Unknown Sep 22 '18 at 7:58