I am trying to see how one can think of defining lie derivative of(along) a vector field on a smooth manifold.

Let $M$ be a smooth manifold and $X$ be a vector field on $M$. We wish to define lie derivative of $X:M\rightarrow TM$ as another vector field $L_X:M\rightarrow TM$ imitating the definition of derivative of a real valued function on real line.

Let $p\in M$. As a first attempt we define $$(L_X)_p=\lim_{t\rightarrow 0}\frac{X(p+t)-X(p)}{t}.$$ As $M$ does not have an extra structure of addition with real numbers in general, it is not clear what is $p+t$ for $t\in \mathbb{R}$. Suppose $M$ is an open subset of $\mathbb{R}$, then $p+t$ is defined and $p+t\rightarrow p$ as $t\rightarrow 0$. So, we are evaluating $X$ on a collection of elements of $M$ varying along a curve (i.e., in $1$ variable) converging to $p$.

Surprisingly, the statement evaluating $X$ on a curve containing $p$ makes perfect sense in any arbitrary manifold and do not need any extra structure. So, as a second attempt, we define $(L_X)_p$ as $$(L_X)_p=\lim_{t\rightarrow 0}\frac{X(\gamma_p(t))-X(p)}{t}$$ where $\gamma_p:(-\epsilon,\epsilon)\rightarrow M$ is a smooth curve in $M$ with $\gamma_p(0)=p$.

Next trouble is, given $t\in (-\epsilon, \epsilon)$, we have $X(\gamma_p(t))\in T_{\gamma_p(t)}M$ and $X(p)\in T_pM$ and it is not clear how do we subtract vectors from different vector spaces.

One obvious way to come across this is to push forward the vector $X(\gamma_p(t))$ to $T_pM$ and then subtract the vector there. For this, we need a smooth map $M\rightarrow M$ sending $\gamma_p(t)$ to $p$. This we need for each $t\in (\epsilon,\epsilon)$. We assume there are such maps. So, we have a collection of smooth maps $\{\eta_t:M\rightarrow M\}$ indexed by elements of $(-\epsilon,\epsilon)$. We can think of this collection as a single map $$F:(-\epsilon,\epsilon)\times M\rightarrow M$$ such that, $F(t,-):M\rightarrow M$ is just $\eta_t$. We also want $F$ to be compatible with $\gamma_p$ i.e., $F(-,p):(-\epsilon,\epsilon)\rightarrow M$ is $\gamma_p$. So, we want a smooth map $F:(-\epsilon,\epsilon)\times M\rightarrow M$ such that

  1. $F(-,p):(-\epsilon,\epsilon)\rightarrow M$ is $\gamma_p$.
  2. $F(t,-):M\rightarrow M$ is $\eta_t$ for each $t\in (-\epsilon,\epsilon)$.
  3. $F(t,\gamma_p(t))=p$ for all $t\in (-\epsilon,\epsilon)$. In particular, $F(0,p)=F(0,\gamma_p(0))=p$.

Now the question is, do we have such map given a vector field $X$. We do have some thing which is similar to this.

Given $p\in M$ we have a local flow $\varphi:(-\epsilon,\epsilon)\times U\rightarrow M$ such that, if we write $\varphi(t,q)=\varphi_t(q)$ then, $\varphi_t:U\rightarrow M$ is such that

  1. $\varphi(0,p)=p$. More generally, $\varphi(0,q)=q$ for all $q\in U$.
  2. $\varphi\circ \varphi_{-t}=1=\varphi_{-t}\circ \varphi_t$.
  3. $\dfrac{\partial}{\partial t}\varphi_t(q)=X(\varphi_t(q))$.

Given this, we have following :

  1. First thing we have asked for is a curve on $M$ containing $p$. For $p$, we have a smooth map $f_p:=\varphi(-,p):(-\epsilon,\epsilon)\rightarrow M$ such that $f_p(0)=\varphi(0,p)=p$. So, first assumption we have asked for is saisfied. We have got a curve in $M$ containing $p$.
  2. Second thing we have asked for is, given $t\in (\epsilon,\epsilon)$ a smooth map $U\subseteq M\rightarrow M$ that takes $f_p(t)$ to $p$. In this case it should take $\varphi(t,p)=\varphi_t(p)$ to $p$. Second condition on $\varphi$ says that $\varphi_{-t}\circ \varphi_t=1$. In particular, we have $\varphi_{-t}(\varphi_t(p))=p$. So, we also have a smooth map that sends $f_p(t)=\varphi_t(p)$ to $p$. We use this map to pushforward the vector $X(f_p(t))$ to $T_pM$.

So, we define $$(L_X)_p=\lim_{t\rightarrow 0}\frac{\varphi_{-t*}(X(\varphi_t(p)))-X(p)}{t}.$$ I have not yet used the condition that $\frac{\partial}{\partial}\varphi_t(q)=X(\varphi_t(q))$ for all $q\in U$. This notation is really confusing. I am more or less sure that this should mean $X(\varphi_t(p))=\varphi_{t*}(X(p))$. So, we would then have $$\varphi_{-t*}(X(\varphi_t(p)))=\varphi_{-t*}(\varphi_{t*}X(p))=(\varphi_{-t*}\circ \varphi_{t*})(X(p))=(\varphi_{-t}\circ \varphi_t)_* X(p)=1(X(p))=X(p).$$ So, $(L_X)_p$ is then zero for all $p$. So, it is not very interesting to differentiating $X$ using the flow given by $X$ which would turn out to be zero, we had something similar in case of lie brackets, namely $[X,X]=0$. It might be interesting to differentiate a vector field $Y$ along the flow given by $X$. So, given vector fields $X,Y$ we define differentiation of $Y$ along the vector field $X$ to be $$(L_XY)_p=\lim_{t\rightarrow 0}\frac{\varphi_{-t*}(Y(\varphi_t(p)))-Y(p)}{t}.$$

I guess this is how Lie would have thought of defining Lie derivative of a vector field along another vector field.

Any suggestions/comments are welcome.

May be I am missing something silly, but, any suggestion on I have not yet used the condition that $\frac{\partial}{\partial}\varphi_t(q)=X(\varphi_t(q))$ for all $q\in U$. This notation is really confusing. I am more or less sure that this should mean $X(\varphi_t(p))=\varphi_{t*}(X(p))$ is welcome.

EDIT : This $X(\varphi_t(p))=\varphi_{t*}(X(p))$ looks something close to what is called left invariant vector field on a lie group. A vector field $X$ on a lie group is called a left invariant vector field if $(l_g)_{*,e} X(e)=X(g)$ for all $g\in G$ where $l_g:G\rightarrow G$ is given by $h\mapsto gh$. So, we can see the condition of being left invariant as $$(l_g)_{*,e} X(e)=X(l_g(e)).$$ Suppose $(l_g)_{*,e} X(e)=X(l_g(e))$ for all $g\in G$ then, we say $X$ is left invariant vector field. This is exactly what we are asking for flow $\varphi_t$ to satisfy, we want $(\varphi_t)_*(X(p))=X(\varphi_t(p))$. Suppose $Y$ is another vector field and $\varphi_t$ is the flow corresponding to $X$. We would say $Y$ is invariant under the flow if $(\varphi_t)_*(Y(p))=Y(\varphi_t(p))$. It seems reasonable to expect that $X$ is invariant under its own flow. So, it should be certainly true that $$(\varphi_t)_*(X(p))=X(\varphi_t(p)).$$ But, I could not see that from the conditions on the flow.


1 Answer 1


To answer your question, let me generalize your exposition slightly. The expression defining the directional derivative of a vector field $Y \colon \mathbb{R}^n \rightarrow \mathbb{R}^n$ at a point $p \in \mathbb{R}^n$ in the direction $v \in \mathbb{R}^n$ is given by

$$ \lim_{t \to 0} \frac{Y(p + tv) - Y(p)}{t}. $$

To generalize the expression above so that it will at least make sense on a general manifold, we need to do two things:

  1. Replace the curve $t \mapsto p + tv$ with some curve $\gamma_p(t)$ to get the expression $$ \lim_{t \to 0} \frac{Y(\gamma_p(t)) - Y(p)}{t}. $$
  2. Once we have chosen the curve, provide some mechanism of transporting tangent vectors from $T_{\gamma_p(t)}M$ to $T_pM$ and use it to transport $Y(\gamma_p(t))$ to $T_pM$ so that the subtraction will happen at $T_pM$. Namely, provide a family of maps $\Phi_t \colon T_{\gamma_p(t)} M \rightarrow T_{p} M$ and then consider the expression $$ \lim_{t \to 0} \frac{\Phi_t(Y(\gamma_p(t))) - Y(p)}{t} $$ which now makes sense as a limit of elements in $T_pM$.

Of course, nothing guarantees that the limit above exists or that the result behaves similarly to the familiar notion of a directional derivative but this idea is very general and is behind all the various "derivatives" in differential geometry. For example:

  1. Given a vector field $X$, we can use it to do steps $(1)-(2)$. Namely, take $\gamma_t(p) = \varphi_{t}^{X}(p)$ to be the integral curve of $X$ passing through $p$ at $t = 0$ and $\Phi_t = d\phi_{-t}|_{\gamma_t(p)}$. This leads to the notion of a Lie derivative $\mathcal{L}_X(Y)$ of a vector field.
  2. Given a vector field $X$ and a connection $\nabla$, we can use both to do steps $(1)-(2)$. Namely, take $\gamma_p(t) = \varphi_{t}^{X}(p)$ to be any curve $\gamma_p(t)$ such that $\gamma_p(0) = p$ and $\dot{\gamma}_p(0) = X(p)$ but now take $\Phi_t$ to be the parallel transport induced by $\nabla$ along $\gamma_p(t)$. This leads to the notion of a covariant derivative $\nabla_X Y$ of a vector field.

In particular, in your discussion you don't need to use any properties of $\varphi_t$ other than those that make steps $(1)-(2)$ above work in order to define the expression behind the Lie derivative. Of course, to make sure that the limit exists, compute it and so on you will use the fact that $\varphi_t$ is a flow (so $\frac{d}{dt} \varphi_t(q) = X(\varphi_t(q))$).

  • $\begingroup$ I am asking for lie derivative, you are saying about covariant derivative. I do not think these two are same. Can you please explain your point. $\endgroup$
    – user312648
    Oct 21, 2017 at 9:28
  • $\begingroup$ The point #1 in the second list addresses the Lie derivative. Thanks for the nice explanation @levap, this will help me teach this content! $\endgroup$ Apr 6, 2021 at 21:07
  • $\begingroup$ @levap so the covariant derivative calculates the change in $Y$ along the geodesics in the direction of $X$, and the lie derivative is along the flow of $X$? $\endgroup$ Jul 27, 2021 at 17:24
  • $\begingroup$ @ZiadH.Muhammad: No, the Lie derivative calculates the change of $Y$ along the flow $\phi_t$ of $X$ and uses $d\phi_{-t}$ to transport $Y(\varphi_t(p))$ back to $T_pM$ in order to compare it with $Y(p)$. The covariant derivative calculates the change of $Y$ in the direction of $X$ using any curve $\gamma$ which goes in the direction of $X$ (not necessarily a geodesic) and uses parallel transport to transport $Y(\gamma(t))$ back to $T_pM$ in order to compare it with $Y(p)$. $\endgroup$
    – levap
    Jul 28, 2021 at 10:36
  • 1
    $\begingroup$ @ZiadH.Muhammad: Correct. The Lie derivative uses the flow in order to transport the vectors and does not require a connection. $\endgroup$
    – levap
    Jul 28, 2021 at 13:53

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