Prove that the Lie derivative of a vector field equals the Lie bracket: $\frac{d}{dt} ((\phi_{-t})_* Y)|_{t=0} = [X,Y]$

Question

Let $X$ and $Y$ be vector fields on a smooth manifold $M$, and let $\phi_t$ be the flow of $X$, i.e. $\frac{d}{dt} \phi_t(p) = X_p$. I am trying to prove the following formula:

$\frac{d}{dt} ((\phi_{-t})_* Y)|_{t=0} = [X,Y],$

where $[X,Y]$ is the commutator, defined by $[X,Y] = X\circ Y - Y\circ X$.

This is a question from these online notes: http://www.math.ist.utl.pt/~jnatar/geometria_sem_exercicios.pdf .

Answer 1

Here is a simple proof which I found in the book "Differentiable Manifolds: A Theoretical Phisics Approach" of G. F. T. del Castillo. Precisely it is proposition 2.20.

We denote $(\mathcal{L}_XY)_x=\frac{d}{dt}(\phi_t^*Y)_x|_{t=0},$ where $(\phi^*_tY)_x=(\phi_{t}^{-1})_{*\phi_t(x)}Y_{\phi_t(x)}.$

Recall also that $(Xf)_x=\frac{d}{dt}(\phi_t^*f)_x|_{t=0}$ where $\phi_t^*f=f\circ\phi_t$ and use that $(\phi^*_tYf)_x=(\phi^*_tY)_x(\phi^*_tf).$

We claim that $(\mathcal{L}_XY)_x=[X,Y]_x.$

Proof:$$(X(Yf))_x=\lim_{t\to 0}\frac{(\phi_t^*Yf)_x-(Yf)_x}{t}=\lim_{t\to 0}\frac{(\phi^*_tY)_x(\phi^*_tf)-(Yf)_x}{t}=\star$$

Now we add and subtract $(\phi^*_tY)_xf.$ Hence $$\star=\lim_{t\to 0}\frac{(\phi^*_tY)_x(\phi^*_tf)-(\phi^*_tY)_xf+(\phi^*_tY)_xf-Y_xf}{t}=$$ $$=\lim_{t\to 0}(\phi^*_tY)_x\frac{(\phi^*_tf)-f}{t}+\lim_{t\to 0}\frac{(\phi^*_tY)_x-Y_x}{t}f=Y_xXf+(\mathcal{L}_XY)_xf.$$ So we get that $XY=YX+\mathcal{L}_XY.$

Answer 2

Here is a proof that is more-or-less equivalent to the one given by Fallen Apart, but with details further explicated/clarified.

Let $M$ be a smooth manifold and $X$, $Y$ smooth vector fields. The Lie derivative is defined as $$(\mathcal{L}_{X} Y) (p) = \lim_{t\to 0} \frac{\phi^{-t}_{\star} Y (\phi^{t} (p)) - Y(p)}{t}$$ where $\phi^t$ denotes the flow of $X$. Taking a test function $f \in C^{\infty} (M)$, we apply the Lie derivative to $f$ to obtain $$(\mathcal{L}_{X} Y)(p)f = \lim_{t\to 0} \frac{\phi^{-t}_{\star} Y (\phi^{t} (p))f - Y(p)f}{t} = \lim_{t\to 0} \frac{Y(\phi^t (p))(f\circ \phi^{-t}) - Y(p)}{t}$$ Thus, setting $$H(x,y) = Y(\phi^x (p))(f\circ \phi^{-y})$$ we have that $$(\mathcal{L}_{X} Y)(p) f = \frac{d}{dt} \Big\vert_{t = 0} H(t,t) = \frac{\partial H}{\partial x} (0,0) + \frac{\partial H}{\partial y} (0,0)$$ where the second equality is due the chain rule. So, to complete the calculation we just have to compute the partial derivatives of $H$. We find $$\frac{\partial H}{\partial x} (0,0) = \frac{\partial}{\partial x} \Big\vert_{x= 0} (Yf)(\phi^x (p)) = X(p)(Yf)$$ For the other partial derivative of $H$, we introduce a curve $\alpha: (-\epsilon, \epsilon) \to M$ such that $\alpha(0) = p$ and $\alpha'(0) = Y(p)$. Then we have $$\frac{\partial H}{\partial y} (0,0) = \frac{\partial}{\partial y} \Big\vert_{y = 0} Y(p) (f\circ \phi^{-y}) = \frac{\partial}{\partial y} \Big\vert_{y = 0} \frac{d}{ds} \Big\vert_{s= 0} (f\circ \phi^{-y} \circ \alpha)(s)$$ $$= \frac{d}{ds} \Big\vert_{s= 0} \frac{\partial}{\partial y}\Big\vert_{y = 0} (f\circ \phi^{-y} \circ \alpha)(s) = \frac{d}{ds} \Big\vert_{s= 0} -(Xf)(\alpha(s)) = -Y(p)(Xf)$$ We conclude $$(\mathcal{L}_{X} Y)(p)f = X(p) (Yf) - Y(p) (Xf) \implies \mathcal{L}_{X} Y = XY - YX =: [X, Y]$$

Answer 3

Let $X$ and $Y$ two vector fields then the Lie derivative $L_{X}Y$ is the commutator $[X, Y]$.

the proof:

we have

$$L_{X}Y=\lim_{t\to 0}\frac{d\phi_{-t}Y-Y}{t}(f)$$ $$=\lim_{t\to 0}d\phi_{-1}\frac{Y-d\phi_{t}Y}{t}(f)$$ $$=\lim_{t\to 0}\frac{Y(f)-d\phi_{t}Y(f)}{t}$$ $$=\lim_{t\to 0}\frac{Y(f)-Y(f\circ\phi_{t})\circ\phi_{t}^{-1}}{t}$$

we put $\phi_{t}(x)=\phi(t,x)$ and we apply the Taylor formula with integral remains, then there exists $h(t,x)$ such that:

$$f(\phi(t,x))=f(x)+th(t,x)$$ where $h(0,x)=\frac{\partial}{\partial t}f(\phi(t,x))(0,x)$

by defintion of tangent vector: $X(f)=\frac{\partial}{\partial t}f\circ\phi_{t}(x)(0,x)$

then we have $h(o,x)=X(f)(x)$ so:

$$L_{X}Y(f)=\lim_{t\to 0}\left(\frac{Y(f)-Y(f)\circ \phi_{t}^{-1}}{t}-Y(h(t,x))\circ \phi_{t}^{-1}\right)$$ $$=\lim_{t\to 0}\left(\frac{(Y(f)\circ\phi_{t}-Y(f))\circ\phi_{t}^{-1}}{t}-Y(h(t,x))\circ\phi_{t}^{-1}\right)$$

we have $\lim_{t\to 0}\phi_{t}^{-1}=\phi_{0}^{-1}=id.$

then we conclude that

$$L_{X}Y(f)=\lim_{t\to 0}\left(\frac{Y(f)\circ\phi_{t}-Y(f)}{t}-Y(h(0,x))\right)$$ $$= \frac{\partial}{\partial t}Y(f)\circ\phi_{t}(x)-Y(h(0,x))$$ $$= X(Y(f)) -Y(X(f))= [X,Y]$$

This completes the proof.

Answer 4

Consider two vector fields $X$ and $Y$. Note that starting from the point $p$, the vector field $X$ generates the curve $\phi(p,t)$ via the following equation \begin{align*} \frac{d}{dt}(f(\phi(p,t))|_t = Xf|_t \end{align*} with $\phi(p,0)=p$. Also, note that for small $t$, $f(\phi(p,t))=f(p)+Xf|_pt$. We can define the Lie derivative of $Y$ with respect to $X$ as \begin{align*} (\mathcal{L}_XY)_pf = &\lim_{t \to 0} \frac{(Yf)(\phi(p,t))-(\phi_{*t}Y_p)f}{t} \\ = &\lim_{t \to 0} \frac{(Yf)(p)+X(Yf)|_pt-Y_p(f+X_pft)}{t}\\ =&\lim_{t \to 0} \frac{(Yf)(p)+(XY)_pft-(Yf)_p-(YX)_pft}{t}\\ =&(XY-YX)_pf. \end{align*} In the first line of the above equation $\phi_{*t}Y_p$ denotes the push forward of $Y_p$ along the curve $\phi(p,t)$.

Answer 5

Here is another approach in 4 steps (just one of them is hard):

1. Verify that $\mathcal{L}_X(Y+Z)=\mathcal{L}_X(Y)+\mathcal{L}_X(Z)$ for any fields $Y,Z$;

2. Verify that $\mathcal{L}_X(fY)=f\mathcal{L}_X(Y)+X(f)Y$ for any smooth function $f$;

3. Show that $\mathcal{L}_X\left(\frac{\partial}{\partial x_i}\right)=\left[X,\frac{\partial}{\partial x_i}\right]$;

4. Conclude $\mathcal{L}_X(Y)=[X,Y]$.

1. Obvious by $\mathbb{R}$-linearity of $d\phi_{-t}$ and $\frac{d}{dt}$. $_\blacksquare$

2. Just notice that $(d\phi_{-t})(f\,Y)=f(d\phi_{-t})(Y)$ and use Leibnitz's rule. $_\blacksquare$

3. This is the delicate part. Using coordinates, write $X=\sum_ia_i\frac{\partial}{\partial x_i}$ for functions $a_i$ and $\phi(t,x)=(\phi_1(t,x),...,\phi_n(t,x))$ where $x=(x_1,...,x_n)$. Because $\phi(0,x)=x$ we have $\frac{\partial \phi_k}{\partial x_j}(0,x)=\delta_{jk}$ and $\frac{\partial^2 \phi_k}{\partial x_\ell\partial x_j}(0,x)=0$. So: \begin{align*} \mathcal{L}_X\left(\frac{\partial}{\partial x_i}\right)_p&=\left.\frac{d}{dt}\right|_{t=0}(d\phi_{-t})_{\varphi_t(p)}\left(\left.\frac{\partial}{\partial x_i}\right|_{\phi_t(p)}\right)\\ &=\left.\frac{d}{dt}\right|_{t=0}\sum_k\frac{\partial \phi_k}{\partial x_i}(-t,\phi_t(p))\left.\frac{\partial}{\partial x_i}\right|_p\\ &=\sum_k\left(\left.\frac{d}{dt}\right|_{t=0}\frac{\partial \phi_k}{\partial x_i}(-t,\phi_t(p))\right)\left.\frac{\partial}{\partial x_i}\right|_p \end{align*}

We will apply the chain rule to calculate the derivative inside the sum. Since, $\frac{\partial^2\phi_k}{\partial x_j\partial x_i}=0$, we only need to worry about the derivative of $\frac{\partial \phi_k}{\partial x_i}$ with respect to the time coordinate. With that in mind, we see that $\left.\frac{d}{dt}\right|_{t=0}\frac{\partial \phi_k}{\partial x_i}(-t,\phi_t(p))=\left(\left.\frac{d}{dt}\right|_{t=0}\frac{\partial \phi_k}{\partial x_i}(t,p)\right)\cdot\left(\left.\frac{d}{dt}\right|_{t=0}-t\right)=-\left.\frac{d}{dt}\right|_{t=0}\frac{\partial \phi_k}{\partial x_i}(t,p)$. Now:

\begin{align*} \left.\frac{d}{dt}\right|_{t=0}\frac{\partial \phi_k}{\partial x_i}(t,p)&=\left.\frac{d}{dt}\right|_{t=0}\left.\frac{\partial}{\partial x_i}\right|_p\phi_k\\ &=\left.\frac{\partial}{\partial x_i}\right|_p\underbrace{\left.\frac{d}{dt}\right|_{t=0}\phi_k}_{=a_k}\\ &=\frac{\partial a_k}{\partial x_i}(p)\\ \end{align*} Therefore $\mathcal{L}_X\left(\frac{\partial}{\partial x_i}\right)=\sum_k-\frac{\partial a_k}{\partial x_i}\frac{\partial}{\partial x_k}=\sum_k\left[a_k\frac{\partial}{\partial x_k},\frac{\partial}{\partial x_i}\right]=\left[\sum_ka_k\frac{\partial}{\partial x_k},\frac{\partial}{\partial x_i}\right]=\left[X,\frac{\partial}{\partial x_i}\right]_\blacksquare$

4. For $Y=\sum_kb_k\frac{\partial}{\partial x_k}$ use 1), 2), 3) and the fact that $\left[X,b_k\frac{\partial}{\partial x_k}\right]=b_k\left[X,\frac{\partial}{\partial x_k}\right]+X(b_k)\frac{\partial}{\partial x_k}$. $_\blacksquare$

Answer 6

Here is a proof that I find more straight forward. We have $$(\mathcal L_X Y)_p f = \left . \frac{\mathrm d}{\mathrm dt} \right |_{t=0} (\Phi_t^X)^*_pYf = \left . \frac{\mathrm d}{\mathrm dt} \right |_{t=0} \mathrm d_{\Phi_t^X}\Phi_{-t}^X(Y)f$$ where $\Phi^X$ is the flow of $X$. By definition $$\left .\frac{\mathrm d}{\mathrm dt}\right|_{t=0} f \circ \Phi^X_t = Xf$$ We then define the smooth function $g: \mathbb R^2 \to \mathbb R$ $$g(s, u) = \mathrm d_{\Phi^X_s(p)} \Phi^X_u(Y)f$$ We have $$(\mathcal L_X Y)_p f = \left . \frac{\mathrm d}{\mathrm d s}\right |_{s=0}g(s,0) - \left .\frac{\mathrm d}{\mathrm d u} \right |_{u=0} g(0,u)$$ The first term gives by definition $(XY)_pf$. To get the second term is also rather straight forward using the definition of the differential and linearity of differentiation: $$\left . \frac{\mathrm d}{\mathrm d u}\right |_{u=0} \mathrm d_p \Phi_u^X (Y) f = \left .\frac{\mathrm d}{\mathrm d u}\right |_{u=0} Y_p(f \circ \Phi_u^X) = Y_p \left ( \left . \frac{\mathrm d}{\mathrm d u}\right|_{u=0} f \circ \Phi_u^X\right) = (YX)_pf$$ which finally gives $$ \mathcal L_X Y = XY - YX = [X,Y] $$

Answer 7

Why $(\mathcal L_X Y)(f)=\mathcal L_X(Y(f)) - Y(\mathcal L_X f)$ (Leibniz rule for contractions), hence $\mathcal L_X Y=[X,Y]$

Let $M$ be a smooth manifold, and let $X$ and $Y$ be smooth vector fields. The Lie derivative $\mathcal L_X$ along $X$ acts on tensor fields and commutes with all natural tensor operations. In particular, it respects contractions—that is, natural pairings between tensors such as the evaluation of a vector field on a function.

If $\langle T,S\rangle$ denotes such a natural contraction, the defining property is $$ \mathcal L_X\langle T,S\rangle=\langle \mathcal L_X T,S\rangle+\langle T,\mathcal L_X S\rangle. $$ This expresses that the Lie derivative differentiates each argument and sums the results, just as an ordinary derivative acts on a product.

Now take $T=Y$ (a vector field) and $S=f$ (a function). Their contraction is the evaluation $Y(f)$. Applying the contraction rule gives $$ \mathcal L_X\big(Y(f)\big)=(\mathcal L_X Y)(f)+Y\big(\mathcal L_X f\big). $$ Rearranging terms, $$ (\mathcal L_X Y)(f)=\mathcal L_X\big(Y(f)\big)-Y\big(\mathcal L_X f\big). $$

To compute explicitly, recall that the Lie derivative of a function is simply differentiation along $X$: $$ \mathcal L_X f=X(f). $$ Therefore, $$ (\mathcal L_X Y)(f)=X\big(Y(f)\big)-Y\big(X(f)\big)=[X,Y](f), $$ and since this holds for every smooth $f$, we conclude $\mathcal L_X Y=[X,Y]$.

---

Remark.
The contraction rule above follows directly from the flow definition of the Lie derivative.
If $\Phi_t$ is the flow of $X$, then for any tensor field $T$, $$ \mathcal L_X T=\left.\frac{d}{dt}\right|_{t=0}(\Phi_t)^*T, $$ where $(\Phi_t)^*$ denotes the pullback by the diffeomorphism $\Phi_t$.
Because pullback by a diffeomorphism preserves all natural tensor operations (tensor product, contraction, evaluation, etc.), we have $$ (\Phi_t)^*\langle T,S\rangle=\langle(\Phi_t)^*T,(\Phi_t)^*S\rangle. $$ Differentiating this identity with respect to $t$ at $t=0$ immediately yields $$ \mathcal L_X\langle T,S\rangle=\langle\mathcal L_X T,S\rangle+\langle T,\mathcal L_X S\rangle, $$ which is the Leibniz rule for contractions used above.

Answer 8

Although there are already several answers here, I decided to present my solution. I believe it offers a clear and intuitive way to see why the "Lie bracket = Lie derivative" relationship is natural.

Let $F(t,x)$ be the flow generated by $X$, and let $G(t,x)$ be the flow generated by $Y$. For convenience, set $M = \mathbb{R}^n$ and $p = 0$. Specifically, we will make use of the fact that in $\mathbb{R}^n$ we can compute the difference between two points. This might look somewhat crude from an abstract manifold perspective, but this approach will soon reveal its usefulness.

Define the following maps in $\mathbb{R}^n$: $$ \Delta_1(t,s) = G(s, F(t, 0)) - F(t, G(s, 0)), $$ $$ \Delta_2(t,s) = F(-t, G(s, F(t, 0))) - G(s, 0). $$ Intuitively, $\Delta_1$’s left-hand side term represents "first flowing along $X$ for time $t$, then along $Y$ for time $s$", while its right-hand side does the reverse. This makes the statement that the "Lie derivative measures the commutativity of two vector fields" explicit.

$\Delta_2$ is performing a similar task, just with a different order of combining flow paths. However, after expanding the definitions, it becomes clear that $ \partial_t \partial_s \Delta_2(0, 0) $ matches the abstract definition of the Lie derivative. If we compute this derivative directly, it will yield the Lie bracket $\partial_X Y - \partial_Y X$, though I aim to connect the concepts in a more intuitive way.

In fact, it’s clear that $$ \Delta_1(t,0) \equiv \Delta_1(0, s) \equiv 0, $$ so performing a second-order Taylor expansion of $\Delta_1$ leaves only one term: $\partial_t \partial_s \Delta_1 \cdot st$. The same holds for $\Delta_2$.

Now, let’s illustrate that $\Delta_1$ indeed corresponds to the concept of the abstract Lie bracket. Given a smooth function $f$, define $$ f_F(t, s) = f(F(t, 0)), \quad f_G(t, s) = f(G(s, 0)), $$ $$ f_{GF}(t, s) = f(G(s, F(t, 0))), \quad f_{FG}(t, s) = f(F(t, G(s, 0))). $$

Then, define $$ \Delta_{1, XY}(t, s) = (f_{GF} - f_F) - (f_G - f_0). $$ $$ \Delta_{1, YX}(t, s) = (f_{FG} - f_G) - (f_F - f_0), $$

According to the definition, $$ \lim_{s, t \to 0} \frac{1}{st} \Delta_{1, XY} = XYf, \quad \lim_{s, t \to 0} \frac{1}{st} \Delta_{1, YX} = YXf. $$

Noting that after omitting higher-order terms, $$ \Delta_{1, XY} - \Delta_{1, YX} = f_{GF} - f_{FG} \approx (\Delta_1(t,s) \cdot \nabla) f, $$ we see that $ \partial_t \partial_s \Delta_1(0, 0) $ is indeed the Lie bracket.

To prove that the $\partial_s \partial_t$ derivatives of $\Delta_1$ and $\Delta_2$ at $(0,0)$ are equal, we could compute them directly (and it is easy). However, I’ll use a more intuitive method to show their equality. Consider the following auxiliary map: $$ \Delta(t, s, r) = F(-r, G(s, F(t, 0))) - F(-r, F(t, G(s, 0))). $$ Intuitively, it serves as a bridge between $\Delta_1$ and $\Delta_2$: $$ \Delta_1(t, s) = \Delta(t, s, 0), \quad \Delta_2(t, s) = \Delta(t, s, t). $$ Clearly, $\Delta(0, s, r) \equiv \Delta(t, 0, r) \equiv 0$, so its second partial derivatives at $(0,0,0)$ have only $\partial_s \partial_t$ nonzero. Thus, when $t = s = r =0$, $$ \partial_{ts} \Delta_2 = \partial_{ts} \Delta + \partial_{sr} \Delta = \partial_{ts} \Delta = \partial_{ts} \Delta_1, $$ which confirms that the Lie derivative equals the Lie bracket.

As a bonus, we can observe that besides $\Delta_1$ and $\Delta_2$, there is yet another way to combine the flow lines: $$ \Delta_3(t, s) = G(-s, F(-t, G(s, F(t, 0)))), $$ In fact, $\Delta_3(\sqrt{t}, \sqrt{t})$ corresponds to the curve given in this MO answer.

Answer 9

I just learned an elegant proof in Page 226 of "Lectures on Differential Geometry" by Bennett Chow and Yutze Chow.

Take $f\in C^{\infty }(M)$, then compute ($\partial _{t}$ means differentiating at $t=0$) \begin{align*} (\mathcal{L}_{V}W)_{p}f& =\partial _{t}(d\phi _{-t}(W_{\phi _{t}(p)}))f \\ & =\partial _{t}(W_{\phi _{t}(p)}(f\circ \phi _{-t})) \\ & =\partial _{t}(W_{\phi _{t}(p)}(f\circ \phi _{0}))+\partial _{t}(W_{\phi _{0}(p)}(f\circ \phi _{-t})) \\ & =\partial _{t}(W_{\phi _{t}(p)}f)+\partial _{t}(W_{p}(f\circ \phi _{-t})) \\ & =\partial _{t}(\left( Wf\right) (\phi _{t}(p)))+W_{p}(\partial _{t}(f\circ \phi _{-t})) \\ & =V_{p}(Wf)-W_{p}(Vf)=[V,W]_{p}f. \end{align*} Maybe the third equality needs some explanation, which I computed using local coordinate: \begin{equation*} W_{q}=W^{i}(q)\partial _{i}|_{q},\qquad f\circ \phi _{-t}=:f_{t}. \end{equation*} Then $W_{\phi _{t}(p)}(f\circ \phi _{-t})=W^{i}(\phi _{t}(p))\partial _{i}f_{t}(\phi _{t}(p))$, \begin{align*} & \partial _{t}|_{t=0}(W_{\phi _{t}(p)}(f\circ \phi _{-t}))=\partial _{t}|_{t=0}\left( W^{i}(\phi _{t}(p))\partial _{i}f_{t}(\phi _{t}(p))\right) \\ & \qquad=\left( \partial _{t}|_{t=0}W^{i}(\phi _{t}(p))\right) \cdot \left( \partial _{i}f_{0}(\phi _{0}(p))\right) +W^{i}(\phi _{0}(p))\cdot \partial _{t}|_{t=0}\partial _{i}f_{t}(\phi _{t}(p)) \\ & \qquad=\partial _{t}|_{t=0}(W_{\phi _{t}(p)}(f\circ \phi _{0}))+\partial _{t}|_{t=0}(W_{\phi _{0}(p)}(f\circ \phi _{-t}))\text{.} \end{align*}

Answer 10

Here, I will fill in some of the details from the answer by @rmdmc89. Fix $p\in M$ and a coordinate chart $(U, \varphi)\equiv (U, (x^i))$ in a neighborhood of $p$. Denote the flow of $X$ by $\theta$. Let $\varepsilon > 0$ and $U_{0}$ be a neighhorhood of $p$ such that $\theta((-\varepsilon, \varepsilon)\times U_0)\subseteq U$.

As explained in detail in the answer here, for $(t, q)\in (-\varepsilon/2, \varepsilon/2)\times U_{0}$ we have $$\text{d}(\theta_{-t})_{\theta_{t}(q)}(Y_{\theta_{t}(q)}) = \frac{\partial\theta^{i}}{\partial x^{j}}(-t, \theta(t, q))Y^{j}(\theta(t, q))\left.\frac{\partial}{\partial x^{i}}\right|_{q}.$$ Here, the matrix of partial derivatives $\frac{\partial \theta^{i}}{\partial x^{j}}$ is computed relative to the charts $((-\varepsilon/2, \varepsilon/2)\times \theta((-\varepsilon/2, \varepsilon/2)\times U_{0}), \text{id}\times \varphi)$ and $(U, \varphi)$. Thus, \begin{align}(\mathcal{L}_{X}Y)_p =& \left.\frac{\partial}{\partial t}\right|_{t = 0}\left(\frac{\partial\theta^{i}}{\partial x^{j}}(-t, \theta(t, p))Y^{j}(\theta(t, p))\right)\left.\frac{\partial}{\partial x^{i}}\right|_{p}.\end{align} In the following, we denote coordinate representations of the various quantities computed relative to appropriate charts by the symbol $(\widehat{\color{white}{x}})$. Then we have $$(\mathcal{L}_{X}Y)_{p}^{i} = \underbrace{\left.\frac{\partial}{\partial t}\right|_{t = 0}\left(\frac{\partial\widehat{\theta}^i}{\partial x^{j}}(-t, \widehat{\theta^{(p)}}(t))\right)Y^{j}(p)}_{I}+\underbrace{\frac{\partial\theta^{i}}{\partial x^{j}}(0, p)\left.\frac{\partial}{\partial t}\right|_{t = 0}\widehat{Y}^{j}(\widehat{\theta^{(p)}}(t))}_{II}.$$ Applying the chain rule we get \begin{align}I &= \begin{pmatrix}\displaystyle\frac{\partial^{2}\widehat{\theta}^{i}}{\partial t\partial x^{j}}(0, \varphi(p)) & \displaystyle\frac{\partial^{2}\widehat{\theta}^{i}}{\partial x^1\partial x^{j}}(0, \varphi(p)) & \cdots & \displaystyle\frac{\partial^{2}\widehat{\theta}^{i}}{\partial x^n\partial x^{j}}(0, \varphi(p))\end{pmatrix}\begin{pmatrix}-1\\ \widehat{X}^{1}(\varphi(p))\\ \vdots\\ \widehat{X}^{n}(\varphi(p))\end{pmatrix}Y^{j}(p).\end{align} Since $\widehat{\theta}^{i}(0, x) = x^{i}$ for each $x\in \varphi(\theta((-\varepsilon/2, \varepsilon/2)\times U_{0}))$, we have $\displaystyle\frac{\partial^{2}\widehat{\theta}^{i}}{\partial x^k\partial x^{j}}(0, \varphi(p)) = 0$ for all $k$ and thus $$I = -\frac{\partial^{2}\widehat{\theta}^{i}}{\partial x^{j}\partial t}(0, \varphi(p))Y^{j}(p),$$ where we have used the fact that the partial derivatives commute. Moreover, since $\widehat{\theta}^{i}(t, x) = (\widehat{\theta^{(\varphi^{-1}(x))}})^{i}(t)$ for each $(t, x)\in (-\varepsilon/2, \varepsilon/2)\times \varphi(\theta((-\varepsilon/2, \varepsilon/2)\times U_{0}))$, we have $\displaystyle\frac{\partial\widehat{\theta}^{i}}{\partial t}(0, x) = \widehat{X}^{i}(x)$ and thus $$I = -\frac{\partial X^{i}}{\partial x^{j}}(p)Y^{j}(p).$$ Similarly, applying the chain rule to simplify $II$, we get $$II = \delta^{i}_{j}\frac{\partial \widehat{Y}^{j}}{\partial x^{k}}(\varphi(p))\widehat{X}^{k}(\varphi(p)) = \frac{\partial Y^{i}}{\partial x^{k}}(p)X^{k}(p).$$