Existence of the Lie Derivative for Smooth Vector Fields

Question

The Lie derivative is a fundamental concept in differential and symplectic geometry. For two smooth vector fields $ X $ and $ Y$ on a smooth manifold $ M $, the Lie derivative of $Y $ with respect to $ X $ is defined as: $ (\mathcal{L}_X Y)_p = \lim_{t \to 0} \frac{(\varphi_{-t})_*(Y_{\varphi_t(p)}) - Y_p}{t} = \frac{d}{dt} \bigg|_{t=0} (\varphi_{-t})_*(Y_{\varphi_t(p)}), $ where $ \varphi_t $ is the flow of $ X $. This limit is taken in the tangent space $ T_p M $, which is a finite-dimensional vector space. (Although, I am unsure about the specific topology being used here, as no additional inner product or norm structure seems to be assumed.)

My question concerns the existence of this Lie derivative in the context of smooth vector fields. Specifically, I want to verify that $ (\mathcal{L}_X Y)_p $ always exists, which boils down to showing that $ (\varphi_{-t})_*(Y_{\varphi_t(p)}) $ is smoothly dependent on $ t $.

Outline of My Calculation

1. Local Coordinates and Flow:

Consider a point $p \in M $ with a local chart $ U $ around $ p $, having coordinates $ (x_1, \dots, x_n) $. Let $ (\varphi_t)^i(p) = \Phi_i(t, p) = (x_i \circ \Phi)(t, p) $, where $ \Phi(t, p) $ denotes the local expression of the flow of $ X $.

2. Pushforward of Vector Fields:

In local coordinates, the pushforward $ (\varphi_t)_* $ acts on the basis vector fields as follows: $ (\varphi_t)_* \left( \frac{\partial}{\partial x_j} \bigg|_p \right) = \sum_i \frac{\partial \Phi_i}{\partial x_j}(t, p) \frac{\partial}{\partial x_i} \bigg|_{\varphi_t(p)}. $ If $ Y = \sum_j b_j \frac{\partial}{\partial x_j} $, then we have: $ Y_{\varphi_t(p)} = \sum_j b_j(\varphi_t(p)) \frac{\partial}{\partial x_j} \bigg|_{\varphi_t(p)}, $ and pushing this forward by $ \varphi_{-t} $, we get: $ (\varphi_{-t})_*(Y_{\varphi_t(p)}) = \sum_j b_j(\varphi_t(p)) (\varphi_{-t})_* \left( \frac{\partial}{\partial x_j} \bigg|_{\varphi_t(p)} \right). $

3. Expression for Pushforward:

Using the Jacobian of $ \varphi_{-t} $, we evaluate the pushforward: $ (\varphi_{-t})_* \left( \frac{\partial}{\partial x_j} \bigg|_{\varphi_t(p)} \right) = \sum_i \frac{\partial \Phi_i}{\partial x_j}(-t, \varphi_t(p)) \frac{\partial}{\partial x_i} \bigg|_p. $

4. Smoothness in $ t $:

Substituting this into the previous expression, we obtain: $ (\varphi_{-t})_*(Y_{\varphi_t(p)}) = \sum_{i,j} b_j(\Phi(t, p)) \frac{\partial \Phi_i}{\partial x_j}(-t, \Phi(t, p)) \frac{\partial}{\partial x_i} \bigg|_p. $ Since $ X $ and $ Y $ are smooth, the functions $ b_j $ and $ \Phi_i $ are smooth, ensuring that the right-hand side depends smoothly on $ t $.

Thus, $ (\varphi_{-t})_*(Y_{\varphi_t(p)}) $ is smooth in $ t $, implying that the derivative $ \frac{d}{dt} \bigg|_{t=0} (\varphi_{-t})_*(Y_{\varphi_t(p)}) $ exists. Consequently, $ (\mathcal{L}_X Y)_p $ exists.

Request for Verification

Can someone please verify if the calculations and reasoning are correct? Is there anything that I may have missed, particularly regarding the topology or smoothness considerations? I appreciate any feedback or corrections!

Answer 1

Seems correct to me. You can also find the proof “Introduction to smooth manifolds” by J.M. Lee. It is Lemma 9.36 and he does exactly what you are doing.

Answer 2

One needs to be careful about the domain of the charts involved in the computation. Given the coordinate chart $(U, \psi)\equiv(U, (x^i))$ with $p\in U$, let $\varepsilon > 0$ and $U_0$ be a neighborhood of $p$ be such that $\varphi((-\varepsilon, \varepsilon)\times U_0)\subseteq U$. For a fixed $t\in (-\varepsilon/2, \varepsilon/2)$, we compute the coordinate representation of $\varphi_{-t}$ with respect to the charts $(\varphi_{t}(U_0), \varphi)$ and $(U, \varphi)$ (the reason for choosing $\varepsilon/2$ shall soon be apparent).

Next, we compute the coordinate representation of $\varphi$ with respect to the charts $((-\varepsilon/2, \varepsilon/2)\times\varphi((-\varepsilon/2, \varepsilon/2)\times U_0), \text{id}\times\varphi)$ and $(U, \varphi)$. Note that $$\varphi((-\varepsilon/2, \varepsilon/2)\times U_0) = \bigcup_{s\in (-\varepsilon/2, \varepsilon/2)}\varphi_{s}(U_{0})$$ and is thus an open subset of $U$. Let $\mathcal{D}$ denote the flow domain of the (global) flow $\varphi$. If $|t_{0}|, |t_{1}| < \varepsilon/2$ and $q\in U_{0}$, then $(t_{0}, \varphi_{t_{1}}(q))\in\mathcal{D}$ since $$t_{0}\in \mathcal{D}^{(\theta_{t_{1}}(q))} = \mathcal{D}^{(q)}-t_{1}.$$ Here we have used the facts that $|t_{0}+t_{1}| < \varepsilon$ and $(-\varepsilon, \varepsilon)\times U_{0}\subseteq\mathcal{D}$. Moreover, by the group law we have $$\varphi(t_{0}, \varphi_{t_{1}}(q)) = \varphi(t_{0}+t_{1}, q)\in U,$$ so this choice of charts makes sense. We have the following equality of coordinate representations: $$\widehat{\varphi}_{-t}(x) = \widehat{\varphi}(-t, x)\text{ for all }x\in\psi(\varphi_{t}(U_0)),$$ and thus $$\frac{\partial \varphi^{i}_{-t}}{\partial x^{j}}(q) = \frac{\partial\varphi^i}{\partial x^{j}}(-t, q)\text{ for all }q\in \varphi_{t}(U_{0}).$$