Consider a metric, for example that of a sphere with fixed radius $R$, i.e. $$ds^2 = R^2 d\theta^2 + R^2\sin^2\theta^2d\varphi^2,$$ and a curve on that sphere $\gamma = (\theta_0, \varphi)$, where $\theta_0 = const.$ and $\varphi\in[0,2\pi)$, together with a vector $X_0=(X^\theta_0,X^\varphi_0)$. I can parallel transport this vector and in this specific example even determine its components at every specific point of the curve (for a derivation see Section 2 of these lecture notes), with the result $$ \begin{aligned} X^{\theta}(\varphi) &=X_{0}^{\theta} \cos \left(\varphi \cos \theta_{0}\right)+X_{0}^{\varphi} \sin \theta_{0} \sin \left(\varphi \cos \theta_{0}\right), \\ X^{\varphi}(\varphi) &=X_{0}^{\varphi} \cos \left(\varphi \cos \theta_{0}\right)-\frac{X_{0}^{\theta}}{\sin \theta_{0}} \sin \left(\varphi \cos \theta_{0}\right). \end{aligned} $$

My Question:

In an exam I was asked to reproduce the above result using the Lie derivative. I think the principle behind this is to use the fact that the Lie derivative can be seen as the infinitesimal generator of the push-forward, since by definition $$(L_YT)_p := \left.\frac{d}{dt}(\varphi_{-t}^*T_{\varphi_t(p)})\right|_{t=0},$$ if $Y$ is the vector field which generates the flow $\varphi_t$. Intuitively I would say that one can now take the above vector $X$ and just "push it forward" using the Lie derivative in the direction of the curve, i.e. in the direction of the vector field $\frac{d\gamma}{dt} = (0, d\varphi/dt)$.

The issue is that I don't really know how one is supposed to do that explicitly. How exactly do I act with the Lie derivative on $X$ and what am I supposed to do with this expression?

Side note: It's been a while since the exam, so I had time to think about the problem. Unfortunately I don't really see how this is supposed to work. I always thought of parallel transport and Lie derivative as two distinct things that each on their own have applications for certain problems.

  • 1
    $\begingroup$ If the question needs further details or is not clear, let me know and I will try to fix it. $\endgroup$
    – Sito
    May 20, 2020 at 15:44
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    $\begingroup$ Do I understand it clearly : the question is to compare, given an initial vector $V$, its parallele transport at a point $\gamma(t)$ of the curve and its sort of "Lie transportation" at the same point ? By "Lie transportation", I mean the result at $t$ of the flow of $\gamma'(t)$ acting on $V$. $\endgroup$
    – Didier
    May 25, 2020 at 9:54
  • $\begingroup$ @DIdier_ Since the comment was slightly changed since my last answer, let me make it clear: yes, that is exactly what I'm asking. $\endgroup$
    – Sito
    May 25, 2020 at 12:20
  • $\begingroup$ The assigned task doesn't make sense to me. For example, $L_Y Y = 0$ for any vector field $Y$, but $\nabla_Y Y = 0$ only when $Y$ is the (constant length) tangent vector to a geodesic. More conceptually, the Lie derivative just sees the differentiable structure and has nothing whatsoever to do with the Riemannian structure or Levi-Civita connection. The best you can do is say that for a torsion-free connection (which the Levi-Civita is), we have $\nabla_X Y - \nabla_Y X = [X,Y] = L_X Y$. Now that I stop to think about it, maybe that is what your prof had in mind. $\endgroup$ May 26, 2020 at 19:05
  • $\begingroup$ @TedShifrin I think here, the Lie derivative has to do with the riemannian structure as the question was to ask about the Lie derivative with respect to the tangent vector field of a geodesic! But yes, I don't really understand what was intended behind that question. $\endgroup$
    – Didier
    May 27, 2020 at 13:00

2 Answers 2


NOTE: I am taking $R=1$ for convenience.

Although it's not clear to me what was intended by the exam question, what I said at the end of my comment does lead to a solution. Let $X=\dfrac{\partial}{\partial\theta}$ and $Y=\dfrac1{\sin\theta}\dfrac{\partial}{\partial\phi}$ be the unit tangent vector fields along the longitudes and lines of latitudes, respectively. Since the longitudes are geodesics, $\nabla_X X = 0$; since $Y$ makes a constant angle with $X$, $\nabla_X Y = 0$ as well (i.e., $Y$ is parallel along the longitudes).

Now, since the Levi-Civita connection $\nabla$ is torsion-free, we have the "well-known formula" $$\nabla_X Y - \nabla_Y X = [X,Y] = L_X Y.$$ Thus, we have $$\nabla_Y X = L_X Y = -L_Y X.$$ This will tell you the rate at which $X$ is turning as you parallel translate along the $Y$-curve, i.e., the line of latitude, and from this you can compute parallel translation along the line of latitude. (For reference, $L_Y X = -\cot\theta\, Y$. Look familiar?)

  • $\begingroup$ I'm sorry, but you are a bit too fast for me... I'm familiar with the result $\nabla _X Y -\nabla_YX = [X,Y] = L_XY$. This was actually the only thing I wrote down on the exam, but had no clue what to do with it. My biggest problem was that I didn't know what vector field to choose to look at this. So, why are you choosing $X$ and $Y$ the way you do? $\endgroup$
    – Sito
    May 26, 2020 at 20:26
  • $\begingroup$ "This will tell you the rate at which X is turning as you parallel translate along the Y-curve", just to make sure I understand the equation $L_Y X = -\nabla _Y X$. The lhs is supposed to represent the "parallel transport", i.e. pushing $X$ into the direction of the flow generated by $Y$. The rhs is what you call the "rate at which $X$ is turning as you parallel translate along the $Y$ curve". $\endgroup$
    – Sito
    May 26, 2020 at 20:28
  • $\begingroup$ No, the Lie derivative is not anything to do with parallel transport. It's just the way of computing the derivative of the vector field $X$ along the flow of $Y$. The right-hand side has to do with parallel transport (but again a rate of change). It's always natural to work with an orthonormal frame (away from the poles); $X$ and $Y$ are built out of spherical coordinates, of course, and $X$ is particularly convenient because the $X$-curves are geodesics. $\endgroup$ May 26, 2020 at 20:35
  • $\begingroup$ Thanks for the quick response. I'll look into the entire post tomorrow and will maybe come back with more questions. Regardless, +1. $\endgroup$
    – Sito
    May 26, 2020 at 20:53
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    $\begingroup$ Sure thing. It was a thought-provoking question. Keep me posted. $\endgroup$ May 26, 2020 at 21:09

I don't know if this would help but I'll try. In my mind, there is something we can do to compare Lie derivative and parralel transportation (that relies on covariant derivative).

Say $p \in M$ is a point in a riemannian manifold $M$ and that $\exp_p$ is its exponential map. Use polar coordinates for $T_pM$, whre $\mathbb{S}$ is the unit sphere in $T_pM$: this provides a map

\begin{align} E : (0,A) \times \mathbb{S} & \longrightarrow M \\ (t,v) &\longmapsto \exp_p(tv) \end{align} which is a diffeomorphism on its image for small $A$ (because $\mathrm{d}\exp_p(0)$ is invertible). Think of that map as a chart for $M$. Its differential is \begin{align} \mathrm{d}E : T(0,A) \times T\mathbb{S} \to TM. \end{align} Choose $v$ a unitary vector in $\mathbb{S}$. Then $\gamma(t) = E(t,v)$ is a unitary geodesic. In this chart, it is just the function $t \mapsto (t,v)$, and the vector field $(1,0) \in T_t(0;A)\times T_v\mathbb{S}$ is sent by $E$ on $\gamma'(t)\in T_{\gamma(t)}M$. You can, by this chart, identifiate the derivative on the left hand side with respect to the coordinate $t$ with the Lie derivative on the right hand side with respect to $\gamma'(t)$. Moreover, the Gauss lemma says that $\{0\}\times T_v\mathbb{S}$ is identified with $\gamma'(t)^{\perp}$.

We have found a good frame to look at $TM$ : take any orthonormal frame of $v^{\perp} \in T_pM$ and transport it whith constant coordinates in this chart. It gives a frame along $\gamma(t)$ that is "constant" with respect to the Lie derivative $\mathcal{L}_{\gamma'(t)}$.

Thus, what is exactly a vector field along $\gamma$ which is constant in this chart? You can show they are Jacobi fields: knowing a bit of the curvature of $M$ and you can know many things about this chart.

More precisely: of $w \in T_v\mathbb{S}$, then $t \mapsto \mathrm{d}E(t,v)tw$ is the normal Jacobi field along $\gamma$ with initial data $Y(0) = 0$ and $\nabla_v Y(0) = w$. Thus, $\mathrm{d}E(t,v) w = \dfrac{Y(t)}{t}$ for the only solution of \begin{align} \nabla_{\gamma'(t)}\nabla_{\gamma'(t)}Y(t) +R\left(\gamma'(t),Y(t)\right)\gamma'(t)=0 \end{align} with the desired initial data, where $R$ stands for the riemannian curvature tensor (careful : the sign depends on the convention. I choose $R(u,v) = \nabla_{[u,v]} - \left[\nabla_u,\nabla_v\right]$).

On the same time, you can create another frame along $\gamma(t)$ wich is just the parallel transport along this curve. What does this frame looks like in the chart $E$?

I think one can understand your question by "can we compare these two frames ?" or "what are the coordinates of the parallel frame in the chart $E$", but I may be completely wrong.

Attempt of application: For a sphere of radius $R_0$, say $\mathbb{S}^{n+1}(R_0)$, choose a point $p$ and look at the exponential map. It provides a chart \begin{align} E : ]0,R_0\pi[ \times \mathbb{S}^{n} & \longrightarrow \mathbb{S}^{n+1} \\ (t, v) & \longmapsto \exp_p(tv) \end{align} (By the way: when you look at the metrics in these coordinates, you have $ E^*g = \mathrm{d}t^2 + {(R_0)}^2\sin^2\left(\dfrac{t}{R_0}\right) g_{\mathbb{S}^n}$ where $g_{\mathbb{S}^n}$ is the standard metrics on the $n$ dimensional sphere. It is pretty much what you wrote on your post!)

Take $w \in T_v\mathbb{S}^n$, and say you continue it by a parallel vector field $w(t)$ along $\gamma(t)=\exp_p(tv)$. Then it is easy to show that the Jacobi field with initial data $Y(0)=0$ and $\nabla_vY(0) = w$ is just $Y(t) = R_0\sin\left(\dfrac{t}{R_0}\right)w(t)$. It is because the curvature here is constant and equals $\frac{1}{{R_0}^2}$, thus the Jacobi equation is \begin{align} \nabla_{\gamma'(t)}\nabla_{\gamma'(t)} Y = -\dfrac{Y}{{R_0}^2} \end{align} and you can solve it in a parallel orthonormal frame along $\gamma$: in this frame, the covariant derivative corresponds exactly to the usual derivative on each component. Thus the "Lie transportation" (in the sense stated above) of $w$ is $W(t) = \dfrac{\sin\left(\dfrac{t}{R_0}\right)}{\left(\dfrac{t}{R_0}\right)}w(t)$.

In that case, the Lie transportation is parallel to the parallel transportation but the curvature affects the norm! Conversly, you can say that the parallel transport differs from the constant transportation by the multiplicative function $\dfrac{\left(\dfrac{t}{R_0}\right)}{\sin\left(\dfrac{t}{R_0}\right)}$.

My personal interpretation of this phenomena is that parallel transportation tries to be an isometry taking account of curvature. Thus the norm is respected. On the contrary, the exponential map tries to flatten your manifold, hence the constant vector fields in this map can look really hideous in the manifold! The real difference is that parallel transportation is a solution of a first order ODE which is $\nabla_{\gamma'}W=0$ while the Lie transportation is related to Jacobi fields, hence a second order linear ODE.

  • $\begingroup$ First of all, thank you for the answer! Your approach is very general, which I see as a positive thing, but I have a hard time translating any of that into the specific example given in my question. Cloud you maybe explicitly show what the "Lie transported" version of $X$ for the given vector field is? $\endgroup$
    – Sito
    May 25, 2020 at 14:49
  • $\begingroup$ @Sito Hi, I added a little paragraph about Jacobi fields to explain a bit what is hidden behind that. Then, I tried to use it to show the difference between the two transportations in the case of an $(n+1)$ dimensionnal sphere of radius $R_0$. I hope it will be useful. There may be some mistakes in the calculations, but I hope it is still understandable. $\endgroup$
    – Didier
    May 25, 2020 at 15:19
  • $\begingroup$ @Sito Apparently this does not answer your question. Sorry! $\endgroup$
    – Didier
    May 26, 2020 at 17:32
  • $\begingroup$ Please don't misunderstand. It's not that this does not answer the question, but I'd like to go through the calculations myself (I'm a physicist just starting to get familiar with DG) before making further comments. $\endgroup$
    – Sito
    May 26, 2020 at 17:52

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