# Riemannian-connection equations with one-forms (basic)

I am learning the basics of reformulating vector field equations in terms of differential forms. In particular, I am trying to express the equations for the Riemannian connection using differential forms. The Riemannian connection equations for an affine connection $\nabla$ on a Riemannian manifold $(M,g)$ are

$$X (g(Y,Z)) - g(\nabla_X Y, Z) - g(Y,\nabla_XZ) = 0, \ \forall X,Y,Z \in \Gamma(TM),$$ $$\nabla_XY - \nabla_YX - [X,Y] = 0, \ \forall X,Y \in \Gamma(TM).$$

However, I'm finding it difficult to convert each to one-form equations and would appreciate any pointers or references on this.

## Work (resolved)

We work in an orthonormal framing $\{ E_i \}$ on the tangent bundle, with dual basis $\{ \omega^i \}$. We define the one-form $\omega^j_i$ by $\omega^j_i (X) := \langle \nabla_X E_i, E_j \rangle$. Then, working on the first equation, and letting $\langle X,Y \rangle := g(X,Y),$ we get

\begin{align} \langle \nabla_XY, Z \rangle &= \ldots = Z^j(X^iE_i(Y^j) + Y^i\omega^{j}_{i}(X)), \\ \langle Y,\nabla_X Z \rangle &= \ldots = Y^j(X^iE_i(Z^j) + Z^i \omega^j_i (X)), \\ X \langle Y,Z \rangle &= \ldots = X^i Y^j E_i (Z^j) + X^i Z^j E_i(Y^j), \\ \implies 0 &= (Z^jY^i\omega^j_i + Y^jZ^i\omega^j_i)(X), \ \forall X,Y,Z \in \Gamma(TM) \\ \implies 0 &= (Z^jY^i)(\omega^j_i + \omega^i_j)(X), \ \forall X,Y,Z \in \Gamma(TM) \\ \implies 0 &= \omega^j_i + \omega^i_j. \end{align}

This part is fine. Now, on setting $[E_i,E_j]=c_{ij}^k E_k$ and slotting in the coordinates to the torsion condition, I get $$\Gamma^k_{ij} - \Gamma^k_{ji} - c^k_{ij} =0,$$ where $\Gamma^i_{jk}E_i:=\nabla_{E_j}E_k$. Now clearly $\Gamma^{k}_{ij} = \omega^k_j(E_i)$ and $-c^{k}_{ij}=d\omega^k(E_i,E_j)$, so we have \begin{align} \omega^k_j(E_i) - \omega^k_i (E_j) + d\omega^k(E_i,E_j) &=0, \ \forall i,j. \end{align} However, $$(\omega^p \wedge \omega^{k}_p)(E_i,E_j) = \begin{vmatrix} \omega^p(E_i) & \omega^p(E_j) \\ \omega^k_p(E_i) & \omega^k_p(E_j) \end{vmatrix} = \ldots = \omega^k_i(E_j) - \omega^k_j(E_i).$$ Putting this all together, we get \begin{align} d\omega^k(E_i,E_j) &= \omega^p \wedge \omega^k_p (E_i,E_j), \ \forall i,j \\ \implies d\omega^k &= \omega^p \wedge \omega^k_p. \end{align}

So our corresponding differential system is

\begin{align} \omega^j_i + \omega^i_j &= 0, \\ d\omega^i - \omega^j \wedge \omega^i_j &= 0.\end{align}

• Yes, what you have is right. The metric-compatibility of $\nabla$ is equivalent to the skew-symmetry $\omega^i_j = -\omega^j_i$, so that's it. The torsion-free condition on $\nabla$ will ultimately be a formula describing $d\omega^i$. – Jesse Madnick Jun 20 '17 at 10:23
• Thanks for that, Jesse. Any pointers on how to start with the torsion-free condition? I have attempted to apply the metric to both sides with an arbitrary vector field, but this doesn't seem to be fruitful. – Andrew Whelan Jun 20 '17 at 10:38
• I haven't done it, but... I'd probably start by substituting $X = X^iE_i$ and $Y = Y^jE_j$ into the equation, just as you did with the other. Use the Leibniz rule for $\nabla$ at some point. Maybe write $[E_i, E_j] = c^k_{ij}E_k$ for some $c^k_{ij}$. Recall that $[E_i, E_j] = c^k_{ij} E_k$ is equivalent to $d\omega^k = -c^k_{ij}\omega^i \wedge \omega^j$ (I think... double-check that). Maybe use the invariant formula for exterior derivatives: $d\omega(X,Y) = X(\omega(Y)) - Y(\omega(X)) - \omega([X,Y])$ (and double-check that formula, too). Just some thoughts. – Jesse Madnick Jun 20 '17 at 10:44
• Thanks! I'll try this and see if I get anywhere. – Andrew Whelan Jun 20 '17 at 11:05
• Hi Jesse, thanks a million for those hints - I only had to use the first identity out of the two, but that was very helpful - I've posted my solution above. If you like you could post your hint as an answer and I will accept it. – Andrew Whelan Jun 20 '17 at 13:00