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Let's say I have a set of tangent vectors given at different points ($\vec{v}_i \in T_{p_i}(M)$) on a riemannian manifold with metric and compatible levi civita connection and I like to calculate e.g. the "mean" of all vectors of this set. So I need to add vectors in some sense.

Now I have learned that without connection addition of vectors does not make sense since they are all defined in different tangent spaces.

How does the concept of parallel transport fix ($\vec{v}_1,p_1$) + ($\vec{v}_2,p_2$) ?

Do I understand this correctly that I define a vector addition at one point e.g. $p_1$ ($\vec{v}_1,p_1$) + ($\vec{v}_2,p_2$) as:

Find the geodesic between $p_1$ and $p_2$ with tangent vector $\vec{v}_2$ at $p_2$ (solve this set of ODEs) and calculate the tangent vector $\vec{\hat{v}}_2$ of this curve at $p_1$ so that at $p_1$ ($\vec{v}_1,p_1$) + ($\vec{v}_2,p_2$) := ($\vec{v}_1,p_1$) + ($\vec{\hat{v}}_2,p_1$) = ($\vec{v}_1+\vec{\hat{v}}_2,p_1$)

Is this the correct conceptual approach or do I miss a property of the connection which makes it obvious how this should work now?

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  • $\begingroup$ I think that you are correct. As far as I understand, this is more or less what you do when you compute the covariant derivative of a vector field. $\endgroup$ – Giuseppe Negro May 25 '16 at 9:48
  • $\begingroup$ I'm not sure if I get your construction, but note that if you fix a point $p$ and a vector $v_p$ then the geodesic starting at that point in the direction $v_p$ is uniquely determined. There is no way to additionaly force it to connect $p$ to some given point $q$. A connection allows you to parallely transport vectors from one tangent space to another one along curves (which need not be geodesics), but you have to aware of the fact that the transport may be path dependent. In integrals you usually integrate the action of forms on vectors (which is then a function). $\endgroup$ – Thomas May 25 '16 at 9:57
  • $\begingroup$ @Thomas, ah I see, for the geodesic I only need $v_p$ and $p$. But I also need the curve to pass through $q$. I thinks this confuses me. So if I choose one curve which connects $p$ and $q$, does "transport may be path dependent" mean that the vector $v_p$ is no longer well defined in $T_q$. Since I end up with different vectors $v_p$ in $T_q(M)$ if I choose two different paths? If this is the case, how would one define such an addition properly? Does this question still make sense, mathematically speaking? $\endgroup$ – Bort May 25 '16 at 11:35
  • $\begingroup$ See levap's answer. The vector $v_p$ is mapped to is not the same as the tangent vector to the curve. Rather you map the whole tangent space. $\endgroup$ – Thomas May 25 '16 at 13:53
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The parallel transport mechanism allows you to identify tangent planes at different points $p, q \in M$ only when you specify in addition a curve $\gamma \colon [0,1] \rightarrow M$ connecting $p$ and $q$ (so that $\gamma(0) = p, \gamma(1) = q$). Then you get a linear isomorphism $P = P_{\gamma,0,1} \colon T_pM \rightarrow T_q M$ and given $v \in T_pM, w \in T_qM$, you can use $P$ to define the addition of $v$ and $w$ in $T_pM$ as $v + P^{-1}(w)$. In general, the map $P$ strongly depends on the path $\gamma$ and choosing different paths will result in different linear isomorphisms.

However, usually one uses this identification only for "infinitesimally near" points and in this case, it does not depend on the path one uses to identify the tangent planes. For example, if you will define the derivative of a vector field $X$ with respect to a direcion $v \in T_pM$ at a point $p$ by

$$ ((\nabla_{v})(X))(p) = "\lim_{t \to 0} \frac{X(\gamma(t)) - X(p)}{t}" = \lim_{t \to 0} \frac{P_{\gamma,t,0}^{-1}(X(\gamma(t)) - X(p)}{t} = \frac{d}{dt} P_{\gamma,t,0}^{-1}(\gamma(t))|_{t=0} $$

where $\gamma \colon (-\varepsilon, \varepsilon) \rightarrow M$ is a path with $\gamma(0) = p, \dot{\gamma}(0) = v$ then the result won't dependent on the choice of the path $\gamma$ and will result in the notion of a covariant derivative of a vector field. The expression $P_{\gamma,t,0}^{-1}(\gamma(t)) - \gamma(0)$ will depend on the choice of the path $\gamma$ but its derivative at $t = 0$ will not.

If you are working with a Riemannian metric, then given $p, q \in M$ with $q$ close enough to $p$, there is a unique geodesic $\gamma$ connecting $p$ to $q$ and you can use this "more canonical" choice of path in order to identify $T_pM$ and $T_qM$ but this only works with points $p,q$ that are close to each other.

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  • $\begingroup$ I will think about it a bit more. Since vector addition of different tangent spaces works if $M=\mathbb R^n$ because all isomorphism are trivial, I thought that parallel transport enables this on M uniquely as well. Obviously, I am mistaken. Thank you. $\endgroup$ – Bort May 25 '16 at 16:13
  • $\begingroup$ The fact that parallel transport is usually path dependent and not absolute reflects the fact that the connection/space has curvature. You can see a nice demonstration of it in the wikipedia article (en.wikipedia.org/wiki/Parallel_transport). When the identification between different tangent spaces is path independent, we say that the connection has trivial holonomy. This can happen only if the connection is flat which is the case for $\mathbb{R}^n$ (but this is not enough). $\endgroup$ – levap May 25 '16 at 19:56

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