I am trying to understand the definition of directional derivative along a vector field in a manifold, but I am having trouble.
Let $M$ be a smooth manifold, $f:M \rightarrow R$ a smooth function and $X:M \rightarrow TM$ a smooth vector field.
The Lie Derivative page on wikipedia says that the directional derivative of $f$ with respect to $X$ at point $p\in M$ is $\lim_{t \rightarrow 0} (f(p+tX(p))-f(p))/t$ but why does this make sens? $tX(p)$ is a vector, why can I sum it with a point on the manifold?


2 Answers 2


It looks as if the Wikipedia page is a little sloppy. A better way to say it would be "Let $u$ be an integral curve of $X$ at $p$, i.e., a function with $$ u(0) = p \\ u'(0) = X(p)\\ u'(s) = X(u(s)) $$ for $-\epsilon < s < \epsilon$; the the derivative is $$ \lim_{t \rightarrow 0} \frac{f(u(p))-f(p)}{t} $$

Now if you look closely, if your manifold is embedded in Euclidean space, and you have a smooth extension $\bar{f}$ of $f$ from the manifold to a neighborhood of the manifold in Euclidean space, you have that

$$ f(u(t)) \approx \bar{f}(p + tX(p)) $$

for small $t$; in particular, for $t = 0$, these are equal, and for $t$ small, they agree to first order, i.e., the derivatives of these two are identical. [All this requires some proof, of course!] So you can see why someone being a bit sloppy might write what's on the Wikipedia page.

  • $\begingroup$ This is much too complicated. $\endgroup$ Feb 14, 2018 at 12:43
  • $\begingroup$ The embedding in Euclidean space is irrelevant, as is the extension of the function from the manifold to such an ambient Euclidean space. $\endgroup$ Feb 14, 2018 at 13:01
  • $\begingroup$ Well, the first part's just a simplified version of the identification of a tangent vector with a germ of curves through the point, which is one of the standard definitions of tangent vector (see, for example, Spivak's Diff'l Geom book). The second addresses one of the situations that routinely arises in physics, where you have some function (e.g., kinetic energy) defined on a (Euclidean) configuration space, but want to differentiate it on some submanifold of that space (e.g., configurations that satisfy some constraints) and you do the differentiation in coordinates on the ambient space. $\endgroup$ Feb 14, 2018 at 13:03
  • $\begingroup$ I see. Another thing I read somewhere is that to calculate the directional derivative of $f$ along a vector field $X$ at point $p$, you can just compute $X_p(f)$. Is that correct? $\endgroup$
    – Orpheus
    Feb 14, 2018 at 16:56
  • $\begingroup$ That is, indeed, correct. And my description with the $u$ function, is one way of doing that. I don't know what your definition of tangent vector actually is (there are about 5 or 6 equivalent ones, nicely described in Spivak's differential geometry, volume 1), so it's a little tough for me to say what $X_p(f)$ might mean to you. :) $\endgroup$ Feb 14, 2018 at 17:36

The long and short of it is that you do everything in coordinates $(u^1,\ldots,u^n)$. This means that you choose a coordinate patch containing your point $p$ and do calculations as if you are in Euclidean space. Here your vector $v$ is replaced by its components $(v_1,\ldots,v_n)$ relative to the standard basis $(\frac{\partial}{\partial u^j})$. Then you show that the result is independent of choosing the patch.

On a more sophisticated level, you can define a vector at a point as a derivation on the space of functions defined on an arbitrarily small neighborhood, and then invariance is obvious.

  • $\begingroup$ It's not invariance that's the question; it's why $f$ is even defined on $p + tX(p)$, which lies in the tangent space at $p$ rather than in the underlying manifold. Answer: it's not. $\endgroup$ Feb 14, 2018 at 12:38
  • $\begingroup$ @JohnHughes, when $p$ is viewed as a point in the coordinate patch (use the notation $\phi^{-1}(p)$ if you prefer), the addition is the ordinary addition in Euclidean space. Details are provided in my lecture notes for my course here. $\endgroup$ Feb 14, 2018 at 12:42
  • $\begingroup$ ...and that would be great if OP were a student in your class. But for most definitions of manifold (e.g., the one use in any of Milnor's books), there's a distinction between a point and its image under a coordinate map, and $p+tX(p)$ is not in the domain of $f$, nd saying "you can identify these, and everything about tangent spaces works out fine" is a pretty big step, justified only by careful use of the chain rule to show that your computation is invariant under choice of coordinate patch (if the point happens to lie in the intersection of two patches). But you clearly know all this. $\endgroup$ Feb 14, 2018 at 12:56
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    $\begingroup$ The truth of the matter is that local calculations are always done by pretending that we are in Euclidean space, and it is very useful for a beginning student to recognize this. As I mentioned, anybody should feel free to replace $p$ by $\phi^{-1}(p)$. $\endgroup$ Feb 14, 2018 at 13:00

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