# Uniqueness of connection $1$-forms and curvature $2$-forms

If $(M,\langle\cdot,\cdot\rangle)$ is a pseudo-Riemannian manifold and $\nabla$ denotes the Levi-Civita connection of the metric, we can define the connection $1$-forms and curvature $2$-forms relative to any frame $(E_1,\ldots,E_n)$, not necessarily orthonormal, by $$\nabla_XE_j = \sum_i \omega^i_{\;j}(X)E_i \quad \mbox{and}\quad R(X,Y)E_j = \sum_i \Omega^i_{\;j}(X,Y)E_i,$$ok. Then, if $(\theta^1,\ldots,\theta^n)$ denotes the dual coframe, one proves the structure equations $${\rm d}\theta^i = \sum_{j} \theta^j \wedge \omega^i_{\;j}\quad\mbox{and}\quad\Omega^i_{\;j} = {\rm d}\omega^i_{\;j}+\sum_k \omega^i_{\;k}\wedge \omega^k_{\;j}.$$I recall seeing somewhere that these structure equations actually characterize the connection and curvature forms, but I don't recall where. So I'd like a reference or proof for this result. More precisely:

If $\widetilde{\omega}^i_{\;j}$ and $\widetilde{\Omega}^i_{\;j}$ satisfy$${\rm d}\theta^i = \sum_{j} \theta^j \wedge \widetilde{\omega}^i_{\;j}\quad\mbox{and}\quad\widetilde{\Omega}^i_{\;j} = {\rm d}\widetilde{\omega}^i_{\;j}+\sum_k \widetilde{\omega}^i_{\;k}\wedge \widetilde{\omega}^k_{\;j},$$then $\widetilde{\omega}^i_{\;j} = \omega^i_{\;j}$ and $\widetilde{\Omega}^i_{\;j} = \Omega^i_{\;j}$?

Thanks.

Edit: inspired by the comments... do we get the desired characterization adding the assumption that

$${\rm d}g_{ij}=\sum_k (g_{ik}\omega^k_{\;j}+g_{jk}\omega^k_{\;i})$$

?

I use the conventions

• $R(X,Y)Z = \nabla_X\nabla_YZ - \nabla_Y\nabla_XZ - \nabla_{[X,Y]}Z$;
• $\alpha \wedge \beta = \frac{(k+\ell)!}{k!\ell!}{\rm Alt}(\alpha\otimes\beta)$, for $\alpha \in \Omega^k(M)$ and $\beta \in \Omega^\ell(M)$;
• ${\rm Alt}\gamma= \frac{1}{r!}\sum_{\sigma \in S_r}(-1)^{|\sigma|} \gamma^\sigma$, where $\gamma \in \Omega^r(M)$.
• Have you checked out Kobiyashi and Nomizu's two-volume work? It's full of stuff like this. I think the title is Foundations of Differential Geometry. – Robert Lewis Dec 14 '17 at 19:04
• To uniquely characterize the Levi-Civita connection $1$-forms in the case of an orthonormal frame, you certainly also need the requirement that $\omega$ be $\mathfrak o(n)$-valued. So I'm suspicious, Ivo. – Ted Shifrin Dec 14 '17 at 22:28
• @IvoTerek: My point is that any other torsion-free affine connection $\tilde \nabla$ (in the same frame) would produce forms $\tilde \omega, \tilde \Omega$ satisfying the structure equations, so the best uniqueness theorem you can hope for is that any solution of the structure equations corresponds to some affine-free connection. (This is true at least locally.) If you want to characterize the Levi-Civita connection you of course need to get the metric involved somehow. – Anthony Carapetis Dec 14 '17 at 22:30
• In the case of a non-orthonormal frame, you get something like $dg_{ij} = \sum g_{ik}\omega^k_j + g_{kj}\omega^k_i$ as the requirement ... where $g_{ij} = \langle E_i,E_j\rangle$. – Ted Shifrin Dec 14 '17 at 22:32
• You cannot get uniqueness without it, @Ivo. Sorry to break the news to you. :) That's one of those Cartan lemma symmetry + skew-symmetry implies $0$ arguments for the difference of two connection $1$-forms. – Ted Shifrin Dec 14 '17 at 22:44

Theorem: let $$(M,g)$$ be a pseudo-Riemannian manifold and $$(E_i)$$ a local frame for $$M$$. Then the connection $$1$$-forms $$\omega^i_{~j}$$ of the Levi-Civita connection $$\nabla$$ of $$(M,g)$$ are completely determined by the relations $${\rm d}\theta^i = \theta^j \wedge \omega^i_{~j} \qquad \mbox{and}\qquad \omega_{ij} + \omega_{ji} = {\rm d}g_{ij},$$where $$(\theta^i)$$ is the dual coframe to $$(E_i)$$, $$(g_{ij})$$ are the coefficients of $$g$$ with respect to the frame $$(E_i)$$, and the $$\omega_{ij}$$ are obtained from the $$\omega^i_{~j}$$ by lowering $$i$$ using $$g$$.
Proof: First observe that $$\d{\theta^i}(\vec{E}_j,\vec{E}_k) = -\theta^i([\vec{E}_j,\vec{E}_k])$$. Now use the structure equations to compute $$-\theta^i([\vec{E}_j,\vec{E}_k]) = (\theta^r\wedge \omega^i_{\;r})(\vec{E}_j,\vec{E}_k) = \theta^r(\vec{E}_j)\omega^i_{\;r}(\vec{E}_k) - \theta^r(\vec{E}_k)\omega^i_{\;r}(\vec{E}_j) = \omega^i_{\;j}(\vec{E}_k) - \omega^i_{\;k}(\vec{E}_j).$$Now lower the index $$i$$ and write $$-\pair{\vec{E}_i,[\vec{E}_j,\vec{E}_k]} = \omega_{ij}(\vec{E}_k) - \omega_{ik}(\vec{E}_j)$$. Consider cyclic permutations of $$(ijk)$$: $$\begin{cases} -\pair{\vec{E}_i,[\vec{E}_j,\vec{E}_k]} = \omega_{ij}(\vec{E}_k) - \omega_{ik}(\vec{E}_j) \\ -\pair{\vec{E}_j,[\vec{E}_k,\vec{E}_i]} = \omega_{jk}(\vec{E}_i) - \omega_{ji}(\vec{E}_k) \\ -\pair{\vec{E}_k,[\vec{E}_i,\vec{E}_j]} = \omega_{ki}(\vec{E}_j) - \omega_{kj}(\vec{E}_i) \end{cases}$$ We add the first two equations and subtract the last one. The left side is a combination with Lie brackets that we'll address shortly, while the right side becomes just $$2\omega_{ij}(\vec{E}_k) - \d{g_{ij}}(\vec{E}_k) + \d{g_{jk}}(\vec{E}_i) - \d{g_{ik}}(\vec{E}_j).$$Solve for $$2\omega_{ij}(\vec{E}_k)$$ to recover the Koszul formula: $$2\omega_{ij}(\vec{E}_k) = -\pair{\vec{E}_i,[\vec{E}_j,\vec{E}_k]} - \pair{\vec{E}_j,[\vec{E}_k,\vec{E}_i]} + \pair{\vec{E}_k,[\vec{E}_i,\vec{E}_j]} + \vec{E}_k\pair{\vec{E}_i,\vec{E}_j} - \vec{E}_i\pair{\vec{E}_j,\vec{E}_k} + \vec{E}_j\pair{\vec{E}_i,\vec{E}_k}.$$The explicit expression for $$\omega^i_{\;j}$$ may then be obtained by raising $$i$$ on both sides. We are done. $$\square$$