The idea is that we have three vectors $X$, $Y$ and $Z$. Starting at a point $p$ in our space, we use our connection to parallel transport $Z$ an infinitesimal amount along a geodesic in the $X$ direction and then along a geodesiccurve in the $Y$ direction. We then parallel transport $Z$ in an infinitesimal amount in the $Y$ direction and then in the $X$ direction. The curvature measures the difference between these two parallel transports. In the formula, the Lie bracket term is there to make sure that everything is nice and tensorial.
In this case, if we parallel transport along a geodesic in the $X$ direction and then along a geodesic in the $Y$ directions versusdirection (see below for how to make this precise), we get a different point from when we parallel transport in the $Y$ direction first then in $X$ directions,direction. When we take the infinitesimal square doesn't close uplogarithm of the differences of these points, and what remainswhat's left is $\epsilon^2 T(X,Y)$ (modulo an error of $\approx \epsilon^3 R(X,Y)(X+Y)$, as Robert Bryant pointed out ). Dividing by $\epsilon^2$ and letting $\epsilon$ to zero, we find the torsionpicture above. Again, in the formula there is a Lie bracket term to make everything nice and tensorial.
There were several commentswas a long discussion about whathow to interpret the pictures are conveying exactly, so I should say a few words about what they mean. Thanks to Robert Bryant and Matt F for their helpful suggestions,
To make the picture slightly more rigorous, we either parallel transport in the direction $X$ by a distance $\epsilon X$ or, (as shown in the picture) we make $X$ a tangent vector whose length is $O(\epsilon)$. We do the same thing with $Y$. On the other hand, we assume that the norm of $Z$ is $O(1)$. To obtain the diagram, we rescale the geometry by $\frac{1}{\epsilon^2}$ and let $\epsilon \to 0$. As Robert Bryant noted, for non-zero epsilon, the $XY$-parallelogram in the first picture does not fully close, but the displacement is essentially $R(X,Y)(X+Y)$, which is $O(\epsilon^3)$. When we rescale and take limits, this error vanishes, which is why the parallelogram closes in the picture. The fact that this picture is infinitesimal in $X$ and $Y$ is also the reason why the geodesics are drawn as straight lines.
If we want to make everything completely rigorous while keeping track of the various tangent spaces and making sure that the final expression lives in in $T_p M $, things get reallymore complicated. I'll write it out for the curvatureHowever, and it's possiblein order to do something similar for the torsion. To doshow that this, we can be done, here's one way to formalize it (using a suggestion by @RobertBryant).
We define the transport functionpoint $Tr(p,X,Z)$$q = \exp_p(\epsilon(X+Y)$ to be the opposite corner of the parallelogram. We parallel transport of $Z\in T_p M$ to$Z$ along the geodesic $T_{\exp_p(X)}Z$$\exp_p(tX)$ for $t$ between $0$ and $\epsilon$ and then parallel transport along the curve $\exp_p(\epsilon X+ t Y)$ until we reach $q$. ThenThis traces out the left path around the parallelogram, in termsbut the second part of the curve is not a geodesic.
We then do the same thing except that we transport map,first in the exponential$Y$ direction and logarithmthen in the $X$ direction. This gives us two vectors at $q$, and we havetake their difference to get a vector. To bring this back to $p$, we can parallel transport the following expressionresult back to our original point using the geodesic from $q$ to $p$ (whose logarithm is $\epsilon(X+Y)$).
\begin{align} R(X,Y)Z = \lim_{\epsilon \to 0} \frac{1}{\epsilon^2} &Tr\big [ q,\log_{q}(p), Tr\big(\exp_p( \epsilon X), Tr(p, \epsilon X, \epsilon Y), Tr(p,\epsilon X,Z)\big)\\ - &Tr\big(r, \log_{r}(q),Tr(\exp_p(\epsilon Y),Tr(p, \epsilon Y, \epsilon X), Tr(p,\epsilon Y,Z) )\big) \big] \end{align} The vector that we obtain by doing this is $$\epsilon^2 R(X,Y)Z+O(\epsilon^3),$$
whereAs such, when we renormalize by $q=\exp_{\exp_p(\epsilon X)}( \epsilon Y)$$\epsilon^2$ and let $r=\exp_{\exp_p(\epsilon Y)}( \epsilon X)$$\epsilon \to 0$, we get the desired expression. I prefer drawing the curvature at $q$, rather than $p$ because it visually shows that I am commuting two covariant derivatives.
I thinkUnfortunately, we can't use this exact idea for the picturessecond picture, because here it really matters that all of the curves are much more meaningfulgeodesics with respect to the connection $\nabla$. Instead, we travel along the geodesic $\exp_p^\nabla(tX)$ until we hit the top left corner. Then we travel along a geodesic in the "direction" $Y$ (more precisely, the parallel translate of $Y$ along the geodesic from $p$ to $\exp_p^\nabla(\epsilon X)$. We then do the same thing except that we first travel in the $Y$ direction and intuitive thanthen the "$X$ direction" (with the same caveat as before). When we do this expression, eventhe resulting "parallelogram" doesn't close up, and if it's a bit harderwe take the logarithm of the differences, what we obtain is $$\epsilon^2 T^\nabla(X,Y)+\epsilon^3 R^\nabla(X,Y)(X+Y) + \epsilon^3 T^\nabla(T^\nabla(X,Y),X+Y)+O(\epsilon^4),$$ after we parallel transport the vector from $q$ back to interpret them rigorously$p$. Normalizing by $\epsilon^2$ and letting $\epsilon \to 0$, we get the torsion exactly.
