2

Let $E(x, y, z) = (x, y, z)$ be a vector field on $\mathbb{R}^3$.

$α(x, y, z) = x dy ∧ dz − y dx ∧ dz + z dx ∧ dy$ is a 2-form.

Find $\phi^{t*}_E\alpha$.

The flow of X is $\phi^t = (x_0e^t,y_0e^t,z_0e^t)$.

I need to compute $d\phi_t^{*}$ as $\phi_t^{*}\alpha = \alpha_{\phi_t}(d \phi_{t})$.

1) Is the pullbak of a k-form always a k-form?

2) If we consider $\phi_{t}$ as a 1-form, $d\phi_{t}$ is a 2-form, how to compute it?

Thank you for your help!

Conjecture
  • 3,088
  • 1
    How are you arriving at thinking of $\phi_t$ (which is a diffeomorphism of $\Bbb R^3$) as a $1$-form? That's not remotely correct. There's a difference between a function and a vector field (which is naturally dual to a $1$-form). You compute pullback the way you compute pullback by any function. – Ted Shifrin Dec 24 '18 at 21:30
  • Indeed! $\phi_t$ is just a diffeomorphism. However, we can still view $d\phi_t$ as a 1-form right? – Conjecture Dec 24 '18 at 23:28
  • 1
    Because we are living on $\Bbb R^3$ and not a general manifold, yes, it’s a vector-valued $1$-form. – Ted Shifrin Dec 24 '18 at 23:33

1 Answers1

0

I will partially answer my question.

For $2)$ :

Let $v,w$ be two vectors of $\mathbb{R}^3$. Let u be a point in $\mathbb{R}^3$.

The usual expression of the pullback gives:

$\phi_t^* \alpha(u)(v,w) = \alpha(\phi_t(u))(d \phi_t(u).v, d \phi_t(u).w)$

Which gives:

$\phi_t^* \alpha(u)(v,w) = \alpha(e^tu)(ve^t, we^t) = e^{3t}\alpha(u)(v,w)$.

Conjecture
  • 3,088