# What is wrong with this proof of Cartan's second equation?

Let $$(M,g)$$ be a Riemannian manifold and $$\nabla$$ a metric compatible connection. Let $$\{e_a\}$$ be a local orthonormal basis of vector fields on some open set $$U\subset M$$. We define the connection $$1$$-forms $$\omega^a_b$$ by

$$\nabla_X e_b = \omega^a_b(X)e_a.$$

Let $$\operatorname{Rm}$$ be the Riemann tensor defined by $$\operatorname{Rm}(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z.$$

One defines the curvature $$2$$-forms $$\Omega^a_b$$ by

$$\frac{1}{2}\operatorname{Rm}(X,Y)e_b=\Omega^a_b(X,Y)e_a.$$

I want to prove Cartan's second equation $$\Omega^a_b=d\omega^a_b-\omega_b^c\wedge \omega_c^a.$$

I just computed $$\operatorname{Rm}(X,Y)e_b$$ using the connecton $$1$$-forms:

$$\operatorname{Rm}(X,Y)e_b = \nabla_X\nabla_Y e_b - \nabla_Y \nabla_X e_b - \nabla_{[X,Y]}e_b\\ =\nabla_X\left[\omega^a_b(Y)e_a\right]-\nabla_Y\left[\omega^a_b(X)e_a\right]-\omega^a_b([X,Y])e_a\\ =X(\omega^a_b(Y))e_a+\omega^a_b(Y)\nabla_X e_a-Y(\omega^a_b(X))e_a-\omega^a_b(X)\nabla_Y e_a-\omega^a_b([X,Y])e_a.$$

Now we know that $$d\omega^a_b(X,Y)=X(\omega^a_b(Y))-Y(\omega^a_b(X))-\omega^a_b([X,Y]),$$

hence the above evaluates to

$$\operatorname{Rm}(X,Y)e_b = d\omega^a_b(X,Y)e_a + \omega^a_b(Y)\omega^c_a(X)e_c-\omega^a_b(X)\omega^c_a(Y)e_c\\ =d\omega^a_b(X,Y)e_a-2 \omega^a_b\wedge \omega_a^c(X,Y) e_c.$$

Now this last equation gives, upon using the definition of the curvature $$2$$-forms

$$\Omega^a_b = \dfrac{1}{2}d\omega^a_b - \omega_b^c\wedge \omega_c^a.$$

So there is this $$1/2$$ factor wrong in front of $$d\omega^a_b$$.

I have already read my computations again a few times but did not spot what I'm doing wrong.

So what is wrong with my approach? Why I'm getting this $$1/2$$ in front of $$d\omega^a_b$$?

The mistake is in your formula for $$d\omega^a_b$$. You seem to be using the following convention for the wedge product (which I call the Alt convention in my book Introduction to Smooth Manifolds): $$\alpha\wedge\beta(X,Y) = \frac1 2 (\alpha(X)\beta(Y) - \alpha(Y)\beta(X)).$$ Using that convention, the formula for $$d$$ of a $$1$$-form should be $$d\alpha (X,Y) = \frac 1 2 (X(\alpha(Y)) - Y(\alpha(X)) - \alpha([X,Y]).$$ Your formula for $$d\omega^a_b$$ is missing the factor $$\frac 1 2$$.
You can also do this computation using the other wedge product convention (which I call the determinant convention): $$\alpha\wedge\beta(X,Y) = \alpha(X)\beta(Y) - \alpha(Y)\beta(X).$$ With this convention, the formula for $$d\alpha$$ doesn't have the factor of $$1/2$$, and you also have to define the curvature form without the factor $$1/2$$: $$\text{Rm}(X,Y)e_b = \Omega_b^a(X,Y)e_a.$$ In both cases, the structure equation comes out the way you wrote it: $$\Omega_b^a = d\omega_b^a - \omega^c_b \wedge \omega^a_c.$$