Let $\mathbb{S}^{2} \subset \mathbb{R}^{3}$ be the $2$-sphere ($\mathbb{S}^{2} = \left\{ (x,y,z) \in \mathbb{R}^3, \; x^2+y^2+z^2 = 1 \right\}$). Let $p \in \mathbb{S}^{2}$ and $\xi \in T_{p}S^{2} = \left\{ p \right\}^{\perp} \simeq \mathbb{R}^2$. When I wanted to compute explicitly the parallel transport of $\xi$ along a geodesic $\gamma$ (such that $\gamma(0)=p$), I used the following "trick" : if $T_{\gamma,0,t}(\xi)$ denotes the parallel transport of $\xi$ from the point $p$ to $\gamma(t)$, along $\gamma$, then :

$$\forall t, \; T_{\gamma,0,t}(\xi) \in T_{\gamma(t)}\mathbb{S}^{2}$$

and an orthogonal basis of $T_{\gamma(t)}\mathbb{S}^{2}$ is given by : $(e_{1}(t),e_{2}(t))=(\gamma'(t), \gamma(t) \wedge \gamma'(t))$. As a consequence, $T_{\gamma,0,t}(\xi)$ writes :

$$ \forall t, \; T_{\gamma,0,t}(\xi) = \alpha(t) e_{1}(t) + \beta(t) e_{2}(t) $$

And, since the parallel transport is an isometry,

$$ \left\langle T_{\gamma,0,t}(\xi),\gamma'(t) \right\rangle = \left\langle \xi,\gamma'(0) \right\rangle = \alpha^{2}(t) \Vert e_{1}(t) \Vert^{2} \tag{1}$$


$$ \Vert T_{\gamma,0,t}(\xi) \Vert^{2} = \Vert \xi \Vert^{2} = \alpha^{2}(t) + \beta^{2}(t) \tag{2}$$

$(1)$ and $(2)$ allow to determine $\alpha$ and $\beta$. But I think the method does not generalize to higher dimension.. Here is my idea : If I want to compute the parallel transport in $\mathbb{S}^{3} \subset \mathbb{R}^{4}$ using the same method, I could determine an orthogonal basis of $T_{\gamma(t)} \mathbb{S}^{3}$, say $(e_{1}(t),e_{2}(t),e_{3}(t))$ and write :

$$ T_{\gamma,0,t}(\xi) = \alpha_{1}(t) e_{1}(t) + \alpha_{2}e_{2}(t) + \alpha_{3}(t) e_{3}(t) $$

but I don't know how I would determine $\alpha_{1},\alpha_{2}$ and $\alpha_{3}$ since the relations $\left\langle T_{\gamma,0,t}(\xi),\gamma'(t)\right\rangle = \left\langle \xi,\gamma'(0) \right\rangle$ and $\Vert T_{\gamma,0,t}(\xi) \Vert^{2} = \Vert \xi \Vert^{2}$ do not give enough information (mainly because two equations are not enough to determine the three $\alpha_{1},\alpha_{2}$ and $\alpha_{3}$)... Is there a way out with this method or shall I compute the parallel transport by solving (when possible) the differential equation ?

  • $\begingroup$ is $S^2$ defined by $x^2+y^2+z^2=1$? $\endgroup$
    – Xipan Xiao
    Jan 21, 2015 at 16:38
  • $\begingroup$ @XipanXiao : Yes, sorry ! You're right, I forgot "$z^2$". I'm correcting it right away. $\endgroup$
    – Odile
    Jan 21, 2015 at 16:41

2 Answers 2


$\newcommand{\Reals}{\mathbf{R}}$Here's a slightly different viewpoint (even for the $2$-sphere). The point $p$ in $S^{n}$ and the normalized initial velocity $v$ of the geodesic $\gamma$ may be viewed as orthogonal unit vectors in $\Reals^{n+1}$, and they span a real $2$-plane $N$ (for "normal space") through the origin.

The tangent space $T_{p} S^{n}$ is identified with the orthogonal complement of $p$, and has an orthogonal decomposition $\operatorname{span}(v) \oplus B$, ($B$ for "binormal space"). An arbitrary vector $\xi$ in $T_{p} S^{n}$ decomposes: let $av = a\gamma'(0)$ denote the tangential component, and $b$ the component of $\xi$ in $B$. (In your notation, $av = \alpha(0) e_{1}(0)$ and $b = \beta(0) e_{2}(0)$. Incidentally, your functions $\alpha$ and $\beta$ are constant.)

The geodesic $\gamma$ has explicit parametrization $$ \gamma(t) = (\cos t) p + (\sin t) v. $$ The tangent space of $S^{n}$ at $q = \gamma(t)$ is spanned by $\gamma'(t)$ and $B$, and under parallel transport the $B$ component of $\xi$ is constant. (This is the crucial point.)

Consequently, the parallel transport of $\xi$ along $\gamma$ from $p$ to $q$ is $a\gamma'(t) + b$. (This expresses the geometric fact that parallel transport along a great circle "depends" only on the subspace of $\Reals^{n+1}$ spanned by $p$, $v$, and $\xi$.)

The preceding isn't all that different from your idea; it merely "agglomerates" the $B$ component of $\xi$ instead of expressing it in terms of an orthonormal basis of $T_{p} S^{n}$.

  • $\begingroup$ Thank you for your kind explanation. The method does not only applies to the sphere, right ? In every manifold, we may write : $T_{p}M=\mathrm{span}(v) \oplus B$. If I'm not mistaken, by following your argument, we show that the parallel transport of a vector along a geodesic is always of the form $a\gamma'(t)+b$, right ? $\endgroup$
    – Odile
    Jan 21, 2015 at 17:26
  • $\begingroup$ @Odile: You're welcome. :) It's crucial to this argument that on a sphere, geodesics lie in a plane. Think of a helix on a circular cylinder in $\Reals^{3}$. $\endgroup$ Jan 21, 2015 at 17:45
  • $\begingroup$ How do you show that $a,b$ must be constant? $\endgroup$
    – koch
    Apr 18, 2017 at 19:12
  • $\begingroup$ Is that because parallel transport is linear? $\endgroup$
    – koch
    Apr 18, 2017 at 19:22
  • $\begingroup$ why is the orthogonal complement parallel? $\endgroup$
    – koch
    Apr 18, 2017 at 19:25

You can always set up the coordinate as $(\sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta)$, and choose the geodesic to be $\phi=0$ and $p=(1, 0, 0)$ be a point at the equatorial. Note that $X=\frac{\partial}{\partial\theta}=(\cos\theta\cos\phi, \cos\theta\sin\phi, -sin\theta)$ and $Y=\frac{1}{\sin\theta}\frac{\partial}{\partial\phi}=(-\sin\phi, \cos\phi, 0)$ are both parallel along the geodesic $\phi=0$ near p.

Now let $\xi\in T_pS^2$, decompose $\xi=aX_p+bY_p$, then the parallel transport of $\xi$ along $\phi=0$ is $aX+bY$.

Since the sphere is homogeneous you can always expect p and the geodesic to be arranged like this.


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