I'm currently teaching a course that "applies" differential geometry to computational problems but doesn't have time to go through theorems/proofs in detail. We're taking a visual approach to help people see from a high level the differential geometry toolbox. I'd like to cover derivatives of vector fields on surfaces. Both the Lie and covariant derivatives come up in such a lecture.

Is there a clear/concrete example of a pair of vector fields $(X,Y)$ on the plane that illustrates (1) why the Lie derivative $\mathcal L_X Y$ is different from the covariant derivative $\nabla_X Y$ and (2) why both derivatives might be useful in different contexts?

I'm looking for a succinct, plot-able visualization to help explain what's going on.

  • 2
    $\begingroup$ I found this but it's far from a self contained small example math.stackexchange.com/questions/209241/… $\endgroup$ Feb 15, 2017 at 14:31
  • $\begingroup$ The covariant derivative (on a surface in $R^3$) $\nabla_XY$ is at each point the projection on the tangent space at $p$ of the derivative of the field $Y$, viewed as taking values in $R^3$, along the integral curve of $X$ through $p$. This description makes it clear that it is a rather different beast to the Lie derivative, that it depends on the embedding (ie, the metric), and gives it a rather intuitive content. (This interpretation is `half' the content of the so called Gauss formula (the other half is that the normal projection of that derivative is the second fundamental form)) $\endgroup$ Feb 15, 2017 at 14:48
  • $\begingroup$ [removed response to a comment that's edited] $\endgroup$ Feb 15, 2017 at 14:50
  • $\begingroup$ They differ at essentially all non-silly examples. If you pick your favorite surface, two fields and compute both $[X,Y]$ and $\nabla_XY$, chances are the results will be very different.ç $\endgroup$ Feb 15, 2017 at 14:51
  • $\begingroup$ Great! Even better if this is for examples on the plane, for which this projected definition of covariant derivative isn't needed. If you could provide one worked-out, this would be much appreciated --- I'm seeking help identifying a very simple example where they're different and the difference suggests why both might be useful. The latter is where I'm a little stuck. Thanks. $\endgroup$ Feb 15, 2017 at 14:53

2 Answers 2


Here's a qualitative illustration I have lying around, showing the difference between a vector field having zero Lie derivative and zero covariant derivative.

enter image description here

Certainly doesn't capture the whole idea, but might help. I think the main idea I was trying to communicate is that covariant derivatives are all about a single flow line of $X$, while Lie derivatives care about all of $X$.

  • $\begingroup$ Awesome!! This is close to the example I tried to work out below (I designed $V(x,y)$ to compress toward $y=0$. $\endgroup$ Feb 16, 2017 at 4:46

At the encouragement of a commenter, I'll post my attempt at a simple example. My apologies if this is completely incorrect.

Take $V(x,y):=(1,-y)$ and $W(x,y):=(-y,x)$, the circular vector field.

To compute the Lie derivative, note that the field $V$ implies a diffeomorphism of the plane $\psi_t(x,y)=(t+x,ye^{-t})$. This effectively warps the $y$ axis.

The Jacobian of $\psi_t$ is: $$ J_t(x,y)=\left( \begin{array}{cc} 1 & 0\\0 &e^{-t} \end{array} \right) $$

Now, let's compute a Lie derivative: $$ \begin{array}{rl} \mathcal L_V W(x,y)&= \lim_{t\rightarrow0}\frac{1}{t} [ (d\psi_{-t})_{\psi_t(p)}(W_{\psi_t(p)})-W_p ]\\ &= \lim_{t\rightarrow0}\frac{1}{t} \left[ \left( \begin{array}{cc} 1&0\\0&e^{t} \end{array} \right) \begin{pmatrix} -ye^{-t}\\t+x \end{pmatrix} - \begin{pmatrix} -y\\x \end{pmatrix} \right] \\ &=\lim_{t\rightarrow0}\frac{1}{t}\begin{pmatrix} -y(e^{-t}-1)\\ e^{t}(t+x)-x \end{pmatrix} \\ &=\begin{pmatrix} y\\1+x \end{pmatrix} \end{array} $$

Because we're in the plane, the covariant derivative coincides with the directional derivative. Hence: $$ \begin{array}{rl} \nabla_VW(x,y) &= \lim_{t\rightarrow0}\frac{1}{t}(W(x+tV_x(x,y),y+tV_y(x,y))-W(x,y)) \\ &= \lim_{t\rightarrow0}\frac{1}{t}(W(x+t,(1-t)y)-W(x,y)) \\ &= \lim_{t\rightarrow0}\frac{1}{t}\begin{pmatrix} -(y-ty)+y\\ (t+x)-x \end{pmatrix} \\ &=\begin{pmatrix} y\\1 \end{pmatrix} \end{array} $$

Notice the $y$ components of these two derivatives differs.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.