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I have this exercise.

Let $(x_0, x_1, x_2, x_3)$ be coordinates in $\mathbb{R}^4$ and let $$ w = dx_0 \wedge dx_1 + dx_2 \wedge dx_3 $$. Let $u: \mathbb{R}^4 \rightarrow \mathbb{R}$ be a smooth function and $X$ be the following vector field: $$ X = -\partial_1(u)\partial_0 +\partial_0(u)\partial_1 -\partial_3(u)\partial_3 +\partial_2(u)\partial_3 $$

Show that if $\phi_t^X$ defines the flow of X and $\phi_1^X$ is defined, then $(\phi_1^X)^* w = w.$

[Hint: Consider $\frac{d}{dt}(\phi_t^X)^* w$]

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  • $\begingroup$ Can you show us how you arrived at the constant solution? $\endgroup$ Commented Nov 7, 2020 at 13:56
  • $\begingroup$ What is $c_t$? ${}{}{}$ $\endgroup$ Commented Nov 7, 2020 at 14:38
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    $\begingroup$ Um.... it's not clear how $\partial_1 u = 1$. You know nothing about the function $u$. Back to your question, you need to calculate $\frac{d}{dt} (\phi^X)^*$, so explicit form of $\phi^X$ is not needed. $\endgroup$ Commented Nov 7, 2020 at 14:52
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    $\begingroup$ It will be just a two line proof if you know Cartan's formula and that $du = \iota_X \omega$. $\endgroup$ Commented Nov 7, 2020 at 14:58
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    $\begingroup$ Perhaps you could fix the title a little bit. Something like "$2$-form invariant under the flow of a vector field"? $\endgroup$ Commented Nov 7, 2020 at 22:50

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Hint Define $\gamma : \mathbb{R} \to \mathbb{R}^4$ to be the solution of $\gamma' = X(\gamma)$ and $\gamma(0) = p \in \mathbb{R}^4$. The flow you want to compute is $\varphi_t^X(p) = \gamma(t)$.

Write $\gamma = (x_0(t),x_1(t),x_2(t),x_4(t))$ in coordinates and find the differential equations verified by the coordinates. For example, for the first and second coordinate: \begin{align} {x_0}'(t) &= -\partial_1u(x_0(t),x_1(t),x_2(t),x_3(t))\\ {x_1}'(t) &= \partial_0 u(x_0(t),x_1(t),x_2(t),x_3(t)) \end{align}

Remark it appears constant are not integral curves of the vector field if $\partial u$ is not zero.

Edit To answer the (new) question there is no need to compute the flow, but just to use its definition. By the very definition of the Lie derivative of a differential form in the direction of a vector field, one has \begin{align} \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}\varphi^*_t\omega = \mathcal{L}_X\omega \end{align} By the Cartan's (magic) formula, \begin{align} \mathcal{L}_X\omega = \mathrm{d}\left(\omega(X,\cdot)\right) + \mathrm{d}\omega\left(X,\cdot\right) \end{align} As $\mathrm{d}\omega = 0$ (easy computation), it is \begin{align} \mathcal{L}_X\omega = \mathrm{d}\left(\omega(X,\cdot)\right) \end{align} But \begin{align} \omega(X,\cdot) &= -\partial_0u\mathrm{d}x^0 -\partial_1u\mathrm{d}x^1-\partial_2u\mathrm{d}x^2-\partial_3u\mathrm{d}x^3 \\ &= -\mathrm{d}u \end{align} and thus \begin{align} \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}\varphi^*_t\omega = \mathrm{d}\left(-\mathrm{d}u\right) = -\mathrm{d}^2u = 0 \end{align} and consequently, $\varphi_t^*\omega$ is constant. As $\varphi_0 = \mathrm{id}$, it is equal to $\omega$.

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    $\begingroup$ @120548 I edited the answer to highlight the dependency of the differential equation with $t$. Without any asumption on $u$ we cannot say much about the general form of $x_i(t)$ I guess. $\endgroup$
    – Didier
    Commented Nov 7, 2020 at 13:53
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    $\begingroup$ @120548 I edited my answer to give a concrete answer to your (new) question. $\endgroup$
    – Didier
    Commented Nov 8, 2020 at 10:06
  • $\begingroup$ Thank you for your answer !!! I have just one doubt, in the exercise I have to show the invariance w.r.t. the flow $\phi_1^X$. Can I still apply your reasoning ? $\endgroup$
    – m120p
    Commented Nov 8, 2020 at 13:55
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    $\begingroup$ @120548 The fact that $\varphi^*_t\omega$ is constant show that it is the same at $t=0$ and $t=1$, and at $t=0$ it is $\omega$, so $\varphi_1^*\omega = \omega$ $\endgroup$
    – Didier
    Commented Nov 8, 2020 at 14:23

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