Let $\nabla$ be a linear connection on $M$, let $\{e_i\}$ be a local frame on some open subset $U \subset M$, and let $\{\omega^i\}$ be the dual coframe. We know that there is a uniquely determined matrix of 1-forms $\Theta^i_j$ on $U$, called the connection 1-forms, for this frame, such that $$ \nabla_X e_i = X\lrcorner \Theta^j_i e_j $$ for all $X \in T M$.

There is a well knon formula called Cartan’s first structure equation: $$d\omega^k = \omega^i \wedge \Theta^k_i + \tau^k,$$ where $\{\tau^1 , . . . , \tau^n\}$ are the torsion 2-forms, i.e., $\tau^k$ is the $k$th component of $$ \tau (u, v)= \nabla_u v- \nabla_v u-[u,v] $$

So according to this, if we know the torsion, can we recover the connection (the matrix $\Theta$) by computing the structure functions ($T_{ij}^k$ such that $d\omega^k=\sum_{i<j}T_{ij}^k\omega^i \wedge\omega^j$)?

I know that this is done in some examples, mainly with the Levi-Civita connection of a Riemannian metric and orthonormal frames. In this case there is torsion 0, and moreover the matrix is antisymmetric. But can this always be done?

  • 1
    $\begingroup$ $\Theta^i_j$ is skew-symmetric for the Levi-Civita connection only if the frame is orthonormal $\endgroup$
    – Didier
    Jul 4, 2023 at 16:14
  • 1
    $\begingroup$ certainly not, and you essentially gave the reason why. The Levi-Civita connection is always torsion free, but it’s not necessarily flat. Consider say (an open subset of) the flat Euclidean plane vs hyperbolic plane. $\endgroup$
    – peek-a-boo
    Jul 4, 2023 at 16:21
  • $\begingroup$ @peek-a-boo Ok, I see. I need to know more information about the "relation" of the chosen frame with the connection. Usually it is assumed that the frame is orthonormal, and this is enough data to recover the metric using the torsion (null torsion, for example). But anyway, if you assume orthornormality you can compute the metric, compute Christoffel symbols and then change to the desired frame, not needing Cartan's first structure equations... I guess they are used as a shortcut, but they are not indeed needed... $\endgroup$ Jul 5, 2023 at 7:06

1 Answer 1


No, you can't determine the connection just from the torsion.

The space of linear connections on a given manifold $M$ is an affine space whose associated translation space is the space of type $(1,2)$-tensor fields on $M$. This means that given any two connections $\nabla,\nabla'$ on $M$, then $\nabla'-\nabla$ is a type $(1,2)$-tensor field, and if $\nabla$ is a connection and $A$ is a type $(1,2)$-tensor field, then $\nabla+A$ is a connection.

So, fix a connection $\nabla$ and a such a tensor $A$. Compute:

$$\begin{align}\tau^{\nabla+A}(X,Y) &= (\nabla+A)_XY - (\nabla+A)_YX - [X,Y] \\ &= \nabla_XY+A_XY-\nabla_YX-A_YX-[X,Y] \\ &= \tau^\nabla(X,Y) + A_XY-A_YX.\end{align}$$This means that $\nabla+A$ and $\nabla$ have the same torsion whenever $A$ is symmetric in the sense that $A_XY=A_YX$, for all vector fields $X$ and $Y$.

  • $\begingroup$ Is $A$ what is called the "contorsion"? $\endgroup$ Jul 5, 2023 at 7:17
  • $\begingroup$ No. It is an arbitrary $(1,2)$-tensor field. $\endgroup$
    – Ivo Terek
    Jul 5, 2023 at 15:55

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