# Pushforward of a vector field

### Can someone help me with that ?

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 $V(x^1,x^2) = \frac{\partial }{\partial x^1}$

• What have you tried? Do you know the formula for computing the pushforward of vector fields in coordinates? – Tyler Holden Sep 25 '13 at 19:53
• I don't know the formula... – Thomas Produit Sep 25 '13 at 20:22

## 3 Answers

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 $(\phi_*)(a\partial_r+\partial_{\theta}) = \partial_x$ 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:

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

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

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.

• What do you mean by vector ? Vector fields apply on function, aren't they ? – Thomas Produit Sep 25 '13 at 20:04
• If you have two manifolds, say $M$ and $N$, and a smooth function between them, say $\operatorname{f} : M \to N$, then the push-forward - also called the differential - is a function between tangent bundles: $\operatorname{f}_* : TM \to TN$. At a given point, say $x \in M$, the push-forward is a linear map between tangent spaces $(\operatorname{f}_*)_x : T_xM \to T_{\operatorname{f}(x)}N$. At each point, it takes a vector in $T_xM$ to a vector in $T_{\operatorname{f}(x)}N$. In general, it takes a vector field on $M$ to a vector field on $N$. – Fly by Night Sep 26 '13 at 15:58
• +1 for the answer, but in the comment a bit of care must be taken. Vectors can always be pushed forward to new vectors, but vector fields may not push forward to vector fields. Specifically, if $f$ fails to be injective with $f(x) = f(y)$, There is no reason $d_x f\, V(x) = d_y f \,V(y)$ should hold, so the image isn't neccesarily a vector field. Think of immersing $S^1$ into $\mathbb{R}^2$ as a figure $8$ for a concrete counter example. – Jason DeVito Sep 26 '13 at 18:00
• @JasonDeVito In general, I agree. But this question deals with a diffeomorphism while your example is an immersion. For a map to be a diffeomorphism, the differential must be bijective at each point. Although, I guess your point is that I should have mentioned that, and in that case: I agree. Thanks for pointing that out. – Fly by Night Sep 26 '13 at 18:26
• @FlybyNight: Agreed about the diffeomorphism vs immersion distinction. I recall in grad school I once talked about push-forwarding a v.f. and the professor harped on me. I guess it stuck ;-). Sorry to bother! – Jason DeVito Sep 26 '13 at 18:35

To elaborate upon Fly by Night's response, if $F: M \to N$ is a smooth map and $p\in M$ and $f: M \to \mathbb R$, then in a coordinate chart $(x^i)$ about $p$ and $(y^i)$ about $F(p)$ we have $$\left(F_*\left. \frac{\partial}{\partial x^i}\right|_p\right)f = \left( \frac{\partial F^j}{\partial x^i}(p) \frac{\partial}{\partial y^j}\right) f$$ So the Jacobian of your transformation describes how to compute pushforwards in coordinates.

• Does it mean that I have to multiply the Jacobian of $\phi$ to $\begin{pmatrix} \frac{\partial}{\partial r}\\ \frac{\partial}{\partial \theta} \end{pmatrix}$ ? – Thomas Produit Sep 25 '13 at 20:20
• @ThomasProduit No, the symbols $\partial/\partial r$ and $\partial/\partial \theta$ represent basis vector fields. Think back to linear algebra. They play the role of $e_1$ and $e_2$. If your vector is $1e_1+2e_2$ then your represent that as $(1,2)^{\top}$. – Fly by Night Sep 26 '13 at 16:03
• I am curious as to why this was downvoted (especially 5 months later). It is a correct statement which answers the question, so...? – Tyler Holden Feb 4 '14 at 21:37
• @TylerHolden: Don't hold your breath, there is not much rime and reason to it; it seems some either enjoy downvoting or believe that hurting you , rather than explaining why you're wrong ( unless the question is of extremely-low quality, which this is not) , is an effective way of learning. Go figure. – gary Jan 18 '18 at 17:25