# problem with Exterior Derivative

I have a 3-manifold with Riemannian metric $$g=\omega_1^2+\omega_2^2+\omega_3^2$$, where $$\omega_i$$ are 1-forms (coframe fields).

I am in this situation: $$- (u^{-1}p+\frac{3}{2}u^{-1})du \wedge \omega_2 \wedge \omega_3 + \omega_2 \wedge d\omega_3=d\omega_2 \wedge \omega_3$$, that it should be possible to write like: $$-(u^{-1}p+\frac{3}{2}u^{-1})du \wedge \omega_2 \wedge \omega_3 =d(\omega_2 \wedge \omega_3)$$. where $$p$$ is a constant.

EDITED AFTER "User284193" answer and "jgon" comment:

How I calculate $$\omega_2$$ and $$\omega_3$$?

I don't know how calculate $$\omega_2$$ and $$\omega_3$$

• What do you mean by "get $\omega_2$ and $\omega _3$"? The only meaningful interpretation I can attach to that in this context is the one given in the answer below, which apparently doesn't answer your question. You may want to add more context, for example what would an answer look like? Where is this problem from? What have you tried? – jgon Jun 19 '19 at 17:09
• Sorry for my bad english, I mean: "I don't know how to calculate $\omega_2$ and $\omega_3$" – exxxit8 Jun 19 '19 at 18:38
• You need a lot more information to proceed. – Ted Shifrin Jun 20 '19 at 17:40

If I got your question, you can move $$\omega_2 \wedge d \omega_3$$ on the RHS, so the RHS becomes $$d \omega_2 \wedge \omega_3-\omega_2 \wedge d \omega_3$$.
This is equal to $$d(\omega_2 \wedge \omega_3)$$, since $$d(\omega_2 \wedge \omega_3)=d\omega_2 \wedge \omega_3 + (-1)^1 \omega_2 \wedge d\omega_3=d \omega_2 \wedge \omega_3-\omega_2 \wedge d \omega_3$$ ($$\omega_2$$ is a $$1$$-form)
In general $$d(\alpha \wedge \beta)=d \alpha \wedge \beta + (-1)^p \alpha \wedge d \beta$$ where $$\alpha$$ is a $$p$$-form.
• Yes, my problem is how to find $\omega_2$ and $\omega_3$ – exxxit8 Jun 19 '19 at 8:36