# Lie vs. covariant derivative: Visual motivation

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.

• I found this but it's far from a self contained small example math.stackexchange.com/questions/209241/… Feb 15, 2017 at 14:31
• 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)) Feb 15, 2017 at 14:48
• [removed response to a comment that's edited] Feb 15, 2017 at 14:50
• 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 Feb 15, 2017 at 14:51
• 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. Feb 15, 2017 at 14:53

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

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 $X$.

• Awesome!! This is close to the example I tried to work out below (I designed $V(x,y)$ to compress toward $y=0$. 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.