User Tools

Site Tools


2020-02-28, Friday

It does not do much justice to give Lie derivative just one day, but it is nice to first meet it then slowly get famliar with it.

$$\gdef\cD{\mathcal D} \gdef\cL{\mathcal L}$$

Let $M$ be a smooth manifold, $X$ be a vector field, $\Phi: \cD \to M$ is the flow, where $\cD \subset M \times \R$ is the open subset of the flow domain. For simplicity, assume $M$ is compact, and then $\cD = M \times \R$.

Give a smooth function $f$, we have seen how to take directional derivatives of $f$ along $X$. How do we take derivatives of a vector field $Y$ along $X$?

Recall that $\Phi_t$ is a diffeomorphism of $M$ for $t \in \R$, hence one can carry everything on $M$ along $\Phi_t$, such as functions, vector fields, one forms. Hence, we define the derivative of $Y$ along $X$ as $$ \cL_X Y: = \lim_{t \to 0} \frac{(\Phi_{-t})_* Y - Y}{t}.$$ To be more concrete, suppose we are standing at the point $p$. Denote $p(t) = \Phi_t(p)$. What is $(\cL_X Y)_p$? For a small time $t$, we ask, what is the value of the vector field $Y$ at $p(t)$. Our neighbor at $p(t)$ tells us the result, but it is a vector at $T_{p(t)} M$, not here $T_p M$. Alas! Remember, we cannot compare tangent vectors at different points. (Until a bit later, when we learn about parallel transport and connection.) However, we can use diffeomorphism $\Phi_{-t}$ to up-root the entire neighborhood of $p(t)$, and put it right on top of $p$. Thus, the tangent vector $Y|_{p(t)}$ becomes $(D \Phi_{-t} |_{p(t)})(Y|_{p(t)}) \in T_p M$, then we can compare with $Y|_p$, we get $$ \cL_X Y|_p = \lim_{t \to 0} \frac{(D \Phi_{-t} |_{p(t)})(Y|_{p(t)}) - Y|_p}{t}.$$ This is the meaning of the above expression.

So far, this is only a limit, who knows if it exists or not? Well, we can do local computation, and realize that it is indeed well-defined. And, in fact, it is easy to compute! We have $$ \cL_X Y = [X, Y]. $$ Note that, judging from the RHS, the role of $X$ and $Y$ somehow become symmetric.

One can also define Lie derivatives of one-forms $\lambda$ $$ \cL_X \lambda: = \lim_{t \to 0} \frac{\Phi_{t}^*\lambda - \lambda}{t}.$$ We could have written $(\Phi_{-t})_*\lambda$, pushing forward a differential form along a diffeomorphism. However, it is customary to 'pull-back' differential form, hence we write $\Phi_{t}^*$. We will tell you a nice formula for Lie derivative on one-form a bit later.

More generally, if one is given a section $T$ of a bundle $(TM)^{\otimes k} \otimes (TM)^{\otimes k}$ (such a section is a so-called $(k,l)$-type tensor), we can define $$ \cL_X T: = \lim_{t \to 0} \frac{(\Phi_{-t})_* T - T}{t}.$$ Note that the definition is exactly the same.

The Lie derivatives satisfies the Leibniz rules. If $Y$ is a vector field, $f$ is a function, we have $$ \cL_X (f Y) = \cL_X (f) Y + f \cL_X (Y) $$ where we note that $$ \cL_X(f) = X(f)$$

Can we deduce how Lie derivative acts on 1-form $\lambda$? We try to guess using Leibniz rule $$ \cL_X( \lambda (Y)) = \cL_X(\lambda) (Y) + \lambda (\cL_X Y) $$ Hence, we get $$ \cL_X(\lambda) (Y) = \cL_X( \lambda (Y)) - \lambda ([X, Y]) $$ So, if in local chart we have $\lambda = \lambda_i dx^i$ and $X = X^i \d_i$, we have $$ (\cL_X \lambda)(\d_j) = X(\lambda_j) - \lambda([X^i \d_i, \d_j]) = X^i \d_i(\lambda_j) - \lambda_k dx^k( - \d_j(X^i) \d_i) = X^i \d_i(\lambda_j) + \lambda_i \d_j X^i. $$

math214/02-28.txt · Last modified: 2020/02/28 00:46 by pzhou