In [1, eq. $7.8.10$] the author gives the coordinate expression for the torsion of a non-linear connection. Here a non-linear connection, in my understanding, is essentially a splitting of the double tangent bundle $TTM$ in a horizontal and vertical part (although his precise definition [1, def. $7.2.1$] is quite a bit more involved, but I can imagine that in the end it is equivalent). The formula for torsion in natural coordinates $(x,y)$ in the tangent bundle reads

$$ T^k_{ij}(x,y) = \bar\partial_j N^k_i(x,y) - \bar\partial_i N^k_j(x,y),\qquad \bar\partial_i \equiv \frac{\partial}{\partial y^i}, $$

where $N^i_j$ are the connection coefficients. This indeed reduces to the usual expression for the torsion tensor $T^k_{ij} = \Gamma^k_{ij} - \Gamma^k_{ji}$ when the connection is linear.

For linear connections my intuitive understanding of torsion has always been that it measures the degree to which an infinitesimal parallelogram does not close on itself. More precisely, given two vectors $v,w\in T_pM$, if we infinitesimally parallel transport $u$ along $v$ and $v$ along $u$ then the difference $\delta$ between the two outcomes is given precisely by the torsion tensor

$$ \delta^k = T^k_{ij}v^i w^j.$$

However, I noticed that this is not true anymore when the connection is non-linear. Indeed, for homogeneous non-linear connections, i.e. $ N(x,\lambda y) = \lambda N(x,y)$, I find that

$$\delta^k = (\bar\partial_j N^k_i(x,v) - \bar\partial_i N^k_j(x,u))u^iv^j.$$

Although the expression in between brackets looks a lot like the torsion of the non-linear connection, it is not the same, because of the additional dependence on $u$ and $v$ in $N^i_j$.

My standard geometrical interpretation of torsion is apparently not valid anymore in the non-linear realm, so my question comes down to the following:

Is there an alternative geometric interpretation of the torsion of a non-linear connection?

[1] Szilasi - Connections Sprays and Finsler Structures


I believe I finally (kind of) solved it, at least heuristically, so I'll answer my own question. But to be honest I'm not quite satisfied with it, so if anyone else has a more enlightening answer, please don't hesitate to share!

We can simplify the expression

$$\delta^k = (\bar\partial_j N^k_i(x,v) - \bar\partial_i N^k_j(x,u))u^iv^j$$

a bit more by noticing that $v-u$ is infinitesimal as well, so we may write

$$ \bar\partial_j N^k_i(x,v) = \bar\partial_j N^k_i(x,u + (v-u)) = \bar\partial_j N^k_i(x,u) + \bar\partial_\ell\bar\partial_j N^k_i(x,u)(v^\ell-u^\ell)$$ to first order. As this is multiplied by $u^i v^j$ in the expression for $\delta^k$, the term containing $\bar\partial_\ell\bar\partial_j N^k_i(x,u)(v^\ell-u^\ell)$ does not contribute to the first order expression of $\delta^k$. Therefore we conclude that

$$\delta^k = (\bar\partial_j N^k_i(x,u) - \bar\partial_i N^k_j(x,u))u^iv^j = T^k_{ij}(x,u)u^i v^j = T^k_{ij}(x,v)u^i v^j.$$

Thus the torsion tensor does seem to describe the gap in an infinitesimal parallelogram after all.


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