# Vector fields, flows, and the Lie derivative

Originally published at 狗和留美者不得入内. You can comment here or there.

Let $M$ be a smooth real manifold. A smooth vector field $V$ on $M$ can be considered as a function from $C^{\infty}(M)$ to $C^{\infty}(M)$. Every function $f : M \to \mathbb{R}$ at every point $p \in M$ is by a vector field (which implicitly associates a tangent vector at every point) taken to some real value, which one can think of as the directional derivative of $f$ along the tangent vector. Moreover, this varies smoothly with $p$.

Along any vector field, if we start at any point, we can trace a path along the vector field. Imagine a vector field in water based on the velocity that does not change with time. Take a point particle at a point at any time and we can deterministically predict its path both forward in time and backward in time. We call this an integral curve and it is easy to see that integral curves are equivalence classes.

On a manifold $M$, at a point with chart $(U, \varphi)$, under vector field $V$, we would have $\frac{\mathrm{d}x^{\mu}(t)}{\mathrm{d}t} = V^{\mu}(x(t)), \qquad (1)$

where $x^{\mu}(t)$ is the $\mu$th component of $\varphi(x(t))$ and $V = V^{\mu}\partial / \partial x^{\mu}$. This is an ODE which is guaranteed to have a unique solution at least locally, and we assume for now that the parameter $t$ can be maximally extended.

If we attach the initial condition that at $t = 0$, the integral curve is at $x_0$, and denote the coordinate by $\sigma^{\mu}(t, x_0)$(1) becomes $\frac{\mathrm{d}\sigma^{\mu}(t, x_0)}{\mathrm{d}t} = V^{\mu}(\sigma(t, x_0))$,

Here, $\sigma : \mathbb{R} \times M \to M$ is called a flow generated by $V$, which necessarily satisfies $\sigma(t, \sigma(s, x_0)) = \sigma(t+s, x_0)$

for any $s, t \in \mathbb{R}$.

Within this is the structure of a one-parameter family where

(i) $\sigma_{s+t} = \sigma_s \circ \sigma_t$ or $\sigma_{s+t}(x_0) = \sigma_s(\sigma_t(x_0))$.
(ii) $\sigma_0$ is the identity map.
(iii) $\sigma_{-t} = (\sigma_t)^{-1}$.

We now ask the question how a smooth vector field $W$ changes along a smooth vector field $V$. If our manifold were simply $\mathbb{R}^n$ (with a single identity chart, globally) we would at any point $p$ some direction along $V$ and on an infinitesimal change along that, $W$ would change as well. In this case, it is easy to represent tangent vectors with indexed coordinates. Naively, we could take the displacement in $W$, divide by the amount of displacement along $V$ and take the limit. However, we have not defined addition of tangent vectors on different tangent spaces. To do so, we would need some meaningful correspondence between values on different tangent spaces. Why can we not simply do vector addition? Recall that tangent space elements are defined in terms of how they act on smooth functions from $M$ to $\mathbb{R}$ instead of directly. It is only because they are linear in themselves with respect to any given such function that we can using vectors to represent them.

We resolve this in a more general fashion by defining the induced map on tangent spaces $T_pM$ and $T_{f(p)}N$ for smooth $f : M \to N$ between manifolds. Recall that an element of a tangent space is a map $D : C^{\infty}(M) \to \mathbb{R}$ (that also satisfies the Leibniz property: $D(fg) = Df \cdot g + f \cdot Dg$). If $g \in C^{\infty}(N)$, then $g \circ f \in C^{\infty}(M)$. We define the induced map $\Phi_{f, p} : T_p M \to T_{f(p)} N$

in the following manner. If $D \in T_p(M)$, then $\Phi_{f, p}(D) = D'$, where $D'[g] = D[g \circ f]$.

We notice how we can apply this on $\sigma_t : M \to M$ in our construction of the Lie derivative $\mathcal{L}_V W$ of a vector field $W$ with respect to vector field $V$. Since the flow is along $V$, $\sigma_{-t}^{\mu}(p) = x^{\mu}(p) - tV^{\mu}(p) + O(t^2). \qquad (2)$

We define as the induced map of $\sigma_t(p)$ $\Phi_{\sigma_{-t}, \sigma_t(p)} : T_{\sigma_t(p)} M \to T_p M$.

If $\Phi_{\sigma_{-t}, \sigma_t(p)}(W) = W'$, then by definition, $W'[f](p) = W[f \circ \sigma_{-t}](\sigma_t(p))$.

That means $\mathcal{L}_V W[f](p) = \left(\displaystyle\lim_{t \to 0}\frac{W'(p) - W(p)}{t}\right)[f] = \displaystyle\lim_{t \to 0}\frac{W'[f](p) - W[f](p)}{t}. \qquad (3)$

Using that by the chain rule, $\frac{\partial}{\partial x^r}(f \circ \sigma_{-t})(\sigma_t(p)) = \frac{\partial \sigma_{-t}^{\rho}}{\partial x^r}(\sigma_t(p)) \frac{\partial f}{\partial x^{\mu}}(p)$,

we arrive at \begin{aligned} W'[f](p) & = W^{\nu}(\sigma_t(p)) \frac{\partial}{\partial x^{\nu}}[f \circ \sigma_{-t}](\sigma_t(p)) \\ & = W^{\nu}(\sigma_t(p)) \frac{\partial \sigma_{-t}^{\mu}}{\partial x^{\nu}}(\sigma_t(p))\frac{\partial f}{\partial x^{\mu}}(p). \qquad (4) \end{aligned}

Using the power series of $\sigma_t(p)$ at $p$, we get $W^{\nu}(\sigma_t(p)) = W^{\nu}(p) + tV^{\rho}(p) \frac{\partial W^{\nu}}{\partial x^{\rho}} + O(t^2). \qquad (5)$

Moreover, by (2), $\frac{\partial}{\partial x^{\nu}} \sigma_{-t}^{\mu}(\sigma_t(p)) = \delta_{\nu}^{\mu} - t \frac{\partial V^{\mu}}{\partial x^{\mu}}(p) + O(t^2). \qquad (6)$

Substituting (5) and (6) into (4) yields \begin{aligned} W'[f](p) & = \left(W^{\nu}(p) + tV^{\rho}(p) \frac{\partial W^{\nu}}{\partial x^{\rho}} + O(t^2)\right)\left(\delta_{\nu}^{\mu} - t \frac{\partial V^{\mu}}{\partial x^{\mu}}(p) + O(t^2)\right)\frac{\partial f}{\partial x^\mu} \\ & = \left(W^{\mu}(p) + t\left(V^{\rho}(p) \frac{\partial W^{\nu}}{\partial x^{\rho}}(p) - W^{\nu}(p) \frac{\partial V^{\mu}}{\partial x^{\nu}}(p)\right) + O(t^2)\right)\frac{\partial f}{\partial x^\mu} \\ & = \left(W^{\mu}(p) + t\left(V^{\nu}(p) \frac{\partial W^{\mu}}{\partial x^{\nu}}(p) - W^{\nu}(p) \frac{\partial V^{\mu}}{\partial x^{\nu}}(p)\right) + O(t^2)\right)\frac{\partial f}{\partial x^\mu}. \qquad (7) \end{aligned}

There is a constant term, a first order term, and an $O(t^2)$. In (3), the constant term is subtracted out, and the $O(t^2)$ contributes nothing to the limit. This means that the Lie derivative is equal to the first order term, with $(\mathcal{L}_V W)^{\mu}(p) = V^{\nu}(p) \frac{\partial W^{\mu}}{\partial x^{\nu}}(p) - W^{\nu}(p) \frac{\partial V^{\mu}}{\partial x^{\nu}}(p). \qquad (8)$

Notice how in (4), there is $\frac{\partial f}{\partial x^{\mu}}$ that we have omitted in (8). This is because we are using $\partial/\partial x^\mu$ as the basis of the tangent vector that is applied onto $f \in C^{\infty}(M)$.

We have in (8) what is the $\mu$th component of the Lie bracket of $[V,W]$ where $[V,W]^{\mu} = V^{\nu} \frac{\partial W^{\mu}}{\partial x^{\nu}} - W^{\nu} \frac{\partial V^{\mu}}{\partial x^{\nu}}. \qquad (9)$

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