A Graphing Lie Derivative

Abhishek11235

I am relatively new to differential geometry. I am studying it from Fecko Textbook on differential geometry. As soon as he introduces the concept of lie derivative,he asks to do exercise 4.2.2 in picture. The question is,how do I apply $\phi^*$ to given function $\psi$ . I know that $\phi^*$ transport tensor fields against direction of flow. But how it does,I don't know.

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kent davidge

Consider case (i) first. An integral curve of V is Φ(t, x) = t + x because dΦ/dt = 1 and Φ(0, x) = x. What happens to a point x? Its shifted to t + x. What happens to a function ψ(x)?

Abhishek11235

Consider case (i) first. An integral curve of V is Φ(t, x) = t + x because dΦ/dt = 1 and Φ(0, x) = x. What happens to a point x? Its shifted to t + x. What happens to a function ψ(x)?
It becomes $\phi= e^{-(x+t)^2}$. Is this the answer? Can you give procedure to calculate lie derivative like that for case 2?

Last edited:

kent davidge

It becomes $\phi= e^{-(x+t)^2}$. Is this the answer?
yes, and the problem asks you to draw the graph of $f$. You draw the curve as a function of $t$ for a fixed $x$, which is the initial point. (Don't name the function $\phi$ to avoid confusion with the curve, which the problem already calls $\phi$.)
Can you give procedure to calculate lie derivative like that for case 2?
The Lie Derivative is something else. It's when you take the limit $$\lim_{t \longrightarrow 0} \frac{ \phi^*_t f - f}{t}.$$ What you need to do:

1 - Find the integral curves for the given vector field.
2 - Displace the curve by an infinitesimal amount and take the limit as I gave above.

For case (i) this is $\phi(\epsilon, x) \equiv x' = x + \epsilon V$, but $V = 1$, then $x' = x + \epsilon$. So $f(x') \approx f(x) + \epsilon df/dx$ and $\mathcal L_V f = df / dx$. In our case, $f (x) = \exp (-x^2)$. So $\mathcal L_V \exp (-x^2) = -2x \exp (-x^2)$.

Can you try yourself case (ii)?

• Abhishek11235

"Graphing Lie Derivative"

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