Compute the pushforward of the vector field $\frac{\partial}{\partial x^1}$ via a given function $\phi$

We define $\phi:=(\phi^1,\phi^2):\Omega\subset\mathbb{R}^2\to\phi(\Omega)$ with $\Omega$ such that $\phi$ is a diffeomorphism by

$$x^1:=\phi^1(r,\theta)=r\cos\theta\qquad\text{and}\qquad x^2:=\phi^2(r,\theta)=r\sin\theta$$
for $r,\theta\in\mathbb{R}$.

Compute the pushforward $(\phi^{-1})_*V$ of the vector field $V(x^1,x^2) = \frac{\partial }{\partial x^1}$

Solution 1:

The OP clearly does not know where to start, so I will try to give a full answer.

Forget all of the fancy notation. Your function $\phi$ changes from polar coordinates to Cartesian coordinates. If you give it an $r$ and a $\theta$ then $\phi$ will give you an $x$ and a $y$. We have:

$$\phi : (r,\theta) \longmapsto (r\cos\theta,r\sin\theta).$$

The Jacobian matrix of $\phi$ is the matrix of partial derivatives: $$J_{\phi} = \left[\begin{array}{cc} \partial\phi_1/\partial r & \partial\phi_1/\partial \theta \\ \partial\phi_2/\partial r & \partial\phi_2/\partial \theta\end{array}\right] = \left[\begin{array}{cc} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{array}\right]$$

Notice that $\det(J_{\phi}) = r\cos^2\theta + r\sin^2\theta \equiv r$, and so $\phi$ is a diffeomorphism if and only if $r \neq 0$.

Given a vector field in polar form, say $v=a(r,\theta)\partial_r + b(r,\theta)\partial_{\theta}$, we can find $(\phi_*)(v)$:

$$(\phi_*)(v) \sim \left[\begin{array}{cc} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{array}\right]\left[\begin{array}{c} a \\ b \end{array}\right]=\left[\begin{array}{c} a\cos\theta-rb\sin\theta \\ a\sin\theta + br\cos\theta \end{array}\right]$$ Hence $(\phi_*)(v) = (a\cos\theta-rb\sin\theta)\partial_x+(a\sin\theta+br\cos\theta)\partial_y$.

Similarily, given the vector field $\partial_x$, we might want to know which vector field in polar form $v$, gets sent to $\partial_x$. For this you want $(\phi_*)(v)=\partial_x$. In other words, $v=(\phi^{-1}_*)(\partial_x)$. Let $v=a\partial_r+b\partial_\theta$, then $(\phi_*)(a\partial_r+b\partial_{\theta}) = \partial_x$ becomes the matrix equation

$$\left[\begin{array}{cc} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{array}\right]\left[\begin{array}{c} a \\ b \end{array}\right]=\left[\begin{array}{c} 1 \\ 0 \end{array}\right]$$

The matrix of $\phi$ is $J_{\phi}$ and the matrix of $\phi^{-1}$ is $(J_{\phi})^{-1}$. If we multiply on the left by the inverse of $J_{\phi}$ we get $a = \cos\theta$ and $b=-\frac{1}{r}\sin\theta$. Hence:

$$(\phi_*)(\cos\theta\partial_r-\tfrac{1}{r}\sin\theta\partial_{\theta}) = \partial_x$$

Equivalently, this may be written as

$$(\phi^{-1}_*)(\partial_x) = \cos\theta\partial_r-\tfrac{1}{r}\sin\theta\partial_{\theta}$$

However, we need to write $r$ and $\theta$ in terms of $x$ and $y$. Notice that since $x=r\cos\theta$ and $y=r\sin\theta$ we have $r=\sqrt{x^2+y^2}$ and so:

$$(\phi^{-1}_*)(\partial_x) = \frac{x}{\sqrt{x^2+y^2}}\partial_r-\frac{y}{\sqrt{x^2+y^2}}\partial_{\theta}$$

The vector $\partial_x$ based at $(x,y)$ gets sent the what is written above.

Solution 2:

HINT:

If you have fixed coordinates then the Jacobian matrix is the matrix of the push-forward.

Calculate the Jacobian and then multiply it by the appropriate vector.