this is a statement that I've seen around, and I thought it's time that I understand it. I know that the LCC is locally given by a matrix $ \omega = (\omega_i^j)$ of 1-forms in a preferred frame $e_i$, so that $$ \nabla f_ie_i = df_i \otimes e_i + f_i \omega_i^j \otimes e_j $$ for any local smooth functions $f_i$. Then, $\omega$ is a matrix representing a linear map on each tangent space.

Now, "$\nabla$ is an $\mathfrak{so}(n)$-valued 1-form" suggests to me that each $(\nabla v)|_p$ is in $\mathfrak{so}(T_pM)$ , but I know that this is only true for $v$ a Killing field.

But perhaps I'm getting confused between $\nabla$ as an object and its representation $\omega$ in a particular frame. So, my next guess is that it means that, in some choices of local frame $e_i$, the matrix $\omega_i^j(v)$ is skew-symmetric for any $v$. Orthormal frame is the probable condition. But this would mean, in particular, that each $\nabla e_i$ is skew-symmetric, since if $v = e_i$, there are no nonconstant components of $v$ to worry about, and '$\nabla = \omega$'. Then again, we'd be at the statement that all the $e_i$ are (local) Killing fields, which is just rubbish - on a generic Riemannian manifold, there are no nontrivial local Killing fields, if I remember right.

So, what does "$\omega$ is $\mathfrak{so}$-valued" mean? Any help would be massively appreciated.

  • $\begingroup$ Well, a one-form is a section of $\Omega^1 (T^\ast X) $, while an $\mathfrak{so}(n)$-valued one-form is a section of $\Omega^1 (T^\ast X) \otimes \underline{\mathfrak{so}(n)}$ where $\underline{\mathfrak{so}(n)}$ denotes the trivial bundle and $\otimes$ is the tensor product of vector bundles $\endgroup$ – Sebastian Schulz Apr 11 '17 at 14:23

The Levi Civita connection is a particular case of an Ehresmann connection defined on the bundle of frames $F$, such a connection is defined by a $1$-form $\omega$ defined on the tanent bundle of $FM$ andwhich is $gl(n,\mathbb{R}$ valued. A Levi Civita connection means that $\omega$ takes its values in $so(n,\mathbb{R})\subset gl(n,\mathbb{R})$. It means olso that for every $x$, $(\omega_i^j(x))$ defines an element of $so(n,\mathbb{R})$.


Generally a Lie-algebra valued $k$-form on $M$ is a smooth section of the bundle $(\mathfrak g\times M)\otimes \bigwedge^k T^*M$. In your case $\mathfrak g=\mathfrak {so}(n)$ and $k=1$.

See here


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