# How does the Torsion of two vector fields act on their corresponding flows?

Let $X$ and $Y$ be vector fields defined on an open neighborhhod of a smooth manifold $M$ endowed with an (arbitrary) affine connection $\nabla$ (i'm not assuming anything apart from it being a connection on $TM$).

I'm trying to understand the torsion $T(X,Y)=\nabla_X Y - \nabla_Y X - [X,Y]$ of the connection in terms of flows. Here's what i have so far:

Denoting the local flows of $X$ and $Y$ by $\varphi^X_t$ and $\varphi^Y_t$ resp. and their commutator by $\alpha(t)= \varphi^Y_{-t} \varphi^X_{-t}\varphi^Y_t\varphi^X_t$. We have the following relation between the lie bracket and the $\alpha$:

$$[X,Y] = \frac{1}{2} \alpha ''(0)$$

So what i'm left with is finding a way to express $\nabla_X Y - \nabla_Y X$ in terms of flows. Obviously there must be some input from the connection. I tried to compute the flow by exponentiating from the lie algebra of vector fields but i didn't get very far...

This problem made me realize i have no idea how integral curves and parallel are related. A word about how the they relate to each other would in any case be very helpful.

Ideally I'd like to have an expression for $\nabla_X Y - \nabla_Y X$ in terms of parallel transport and the flows of $X$ and $Y$. Is there such a charactrization?

• Perhaps, It would help If you wrote the covariant derivatives in terms of the lie derivative. This page does the opposite (lie derivative in terms of covariant) but in a 'physicist' notation physicspages.com/2013/02/17/… – MonkSphere Sep 23 '15 at 18:18
• I don't think it's possible to write the covariant derivative in terms of lie derivatives. After all I could just as well add an $End(TM)$-valued 1-form to the connection and the lie derivatives wouldn't notice. – Saal Hardali Sep 23 '15 at 18:35
• Take a look at this paper. Apparently it's done in the fifth reference calpoly.edu/~kmorriso/Research/LieBracket.pdf – MonkSphere Sep 23 '15 at 18:56
• In Kobayashi & Nomizu, Foundations of Differential Geometry the torsion tensor comes to torsion form $\Theta$, since the riemann connection is obtained from connection on the Frames bundle L(M), the connection $\nabla_X Y$ is expressed in terms of horizontal lift of integral curves, maybe be usefull. – Donyarley Sep 24 '15 at 17:09
• @Saal Hardali: Could you post what you find. The answer seems like it should be pretty interesting. – MonkSphere Sep 24 '15 at 23:55

Back in Élie Cartan's papers (but I don't remember specifically which) there is a discussion of the torsion as the translational part of the holonomy of an affine connection, whereas curvature is the rotational part (in the Riemannian case). You should also read this discussion on mathoverflow.

So from Samelson Lie Bracket and Curvature, its shown that

$$(\nabla_Xs)^v =[X^h,s^v]$$ where $X \in \Gamma(TM)$ and $s \in \Gamma(E)$ and $\nabla$ is the covariant derivative on $E$. The $s^v$ is the vertical extension and is defined as below Like wise the $X^h$ is the horizontal lift of the vector field and is defined as So $$\nabla_XY-\nabla_YX = \pi([X^h,Y^v] - [Y^h,X^v])$$

Where $\pi$ is the vertical projection.Therefore

$$\nabla_XY - \nabla_YX = \pi(\frac12(\beta^{''}(0)-\gamma^{''}(0)))$$

Where $$\beta = \varphi_{-t}^{Y^v}\varphi_{-t}^{X^h}\varphi_{t}^{Y^v}\varphi_{t}^{X^h}$$ and $$\gamma = \varphi_{-t}^{X^v}\varphi_{-t}^{Y^h}\varphi_{t}^{X^v}\varphi_{t}^{Y^h}$$

where $\varphi_{t}^{Y^v}$,$\varphi_{t}^{X^h}$,$\varphi_{t}^{X^v}$ and $\varphi_{t}^{Y^h}$ are the flows of $Y^v$, $X^h$, $X^v$ and $Y^h$ respectively.

I don't know if this makes any sense but I think by finding a way to horizontally lift and vertically extend the flows you would have a characterization of the covariant derivative in terms of the flows of $X$ and $Y$.