Problem related to differential of a map

Question

I dont understand how to solve this problem. Please can you explain the solution clearly? I want to learn how to solve such problems. Thank you

Answer 1

I won't give a full answer because I think that being a simple exercise it's good to get practice with that kind of construction. I'll give you hints on how to do it.

Well, there are two approaches that I think can be useful. The first of then is using directly the definition. I'll write down here the definition of pushforward for maps from $\mathbb{R}^n$ to $\mathbb{R}^m$: Let $\varphi : U\subset\mathbb{R}^n \to \mathbb{R}^m$ then we define the pushforward of $\varphi$ at the point $p \in U$ to be the linear transformation $\varphi_{\ast p} : T_p\mathbb{R}^n \to T_{\varphi (p)}\mathbb{R}^m$ such that given $v \in T_p\mathbb{R}^n$ we have $\varphi_{\ast p} \in T_{\varphi(p)}\mathbb{R}^m$ with the property that for every $f : \mathbb{R}^m \to \mathbb{R}$ differentiable at $\varphi(p)$ we have:

$$\varphi_{\ast p}(v)(f)=v(f\circ \varphi)$$

From this definition it's immediate to show that the matrix of the pushforward with respect to the cannonical basis of the tangent spaces is the jacobian matrix, in other words, if $x \in U$ and $y = \varphi (x) \in \mathbb{R}^m$ then the matrix in those basis is:

$$[\varphi_{\ast p}]=\left [ \dfrac{\strut \partial y^j}{\partial x^i} \right ]$$

Knowing this result I'll give you a hint on how to solve this problem: if you want to do by the definition, take $f$ arbitrary (arbitrary meaning that you leave the letter $f$ there instead of getting some function) and apply the definition directly. You'll have to use the chain rule and in the end you'll say: "well, this holds for all functions $f$ so we conclude that the derivation we've obtained is that".

Using this result however, it's easier. You'll calculate the matrix of the pushforward with respect to the canonical basis (the basis formed by the partials with respect to the coordinates) and so you'll use this matrix to calculate the pushforward and to do everything that's needed.

Also, if you've never found anything showing that this is really the matrix, try to prove by yourself, it'll give you practice working with those things. Try it and ask if anything is needed. Good luck!

EDIT: I'll give a little detail on the specific problem. Let $F : \mathbb{R}^2 \to \mathbb{R}^2$ be defined by:

$$F(x,y) = (x\cos\alpha - y \sin \alpha, x \sin \alpha + y \cos \alpha)$$

Then, the component functions of $F$ are $F_1 : \mathbb{R}^2 \to \mathbb{R}$ and $F_2 : \mathbb{R}^2 \to \mathbb{R}$ given by

$$F_1(x,y) = x\cos\alpha - y \sin \alpha$$

$$F_2(x,y) =x \sin \alpha + y \cos \alpha$$

Hence, if $(u, v) = F(x,y)$ then $u = F_1(x,y)$ and $v = F_2(x,y)$. Now you know how to form the matrix of the transformation because using the relation I gave, the matrix at a point $p \in \mathbb{R}^2$ is:

$$[F_{\ast p}]= \begin{bmatrix} \dfrac{\partial F_1}{\partial x}& \dfrac{\partial F_1}{\partial y} \\ \dfrac{\partial F_2}{\partial x}& \dfrac{\partial F_2}{\partial y} \end{bmatrix}$$

With the partials evaluated at the point $p$. It's straightforward to see that:

$$[F_{\ast p}]= \begin{bmatrix} \cos \alpha& -\sin \alpha \\ \sin \alpha& \cos \alpha \end{bmatrix}$$

Now, what do you know ? You know that given a vector $ v \in T_p\mathbb{R}^2$ such that:

$$v = k_1 \frac{\partial}{\partial x} + k_2 \frac {\partial}{\partial y}$$

We have $F_{\ast p}(v) \in T_{F(p)}\mathbb{R}^2$ and we know that if we use the canonical basis then the matrices of componentes will be related by:

$$\left[F_{\ast p}(v)\right] = \left[F_{\ast p}\right] \left[v\right] $$

And hence it's easy to see that:

$$\left[F_{\ast p}(v)\right] = \begin{bmatrix} \cos \alpha& -\sin \alpha \\ \sin \alpha& \cos \alpha \end{bmatrix} \begin{bmatrix} k_1\\ k_2 \end{bmatrix}$$

So that the components matrix of $F_{\ast p}(v)$ is:

$$\left[F_{\ast p}(v)\right] = \begin{bmatrix} k_1\cos\alpha - k_2\sin\alpha\\ k_1\sin\alpha + k_2\cos\alpha \end{bmatrix}$$

This holds for any vector $ v \in T_p\mathbb{R}^2$. You know that if $X : \mathbb{R}^2 \to T\mathbb{R}^2$ is a vector field then $X(p) = (p, X_p)$ with $X_p \in T_p\mathbb{R}^2$. Can you continue from here ? Hint: what's the derivation $F_{\ast p}(v)$ now that you know it's coordinate matrix with respect to the canonical basis?

EDIT 2: What I mean by cannonical basis is the basis of the partials with respect to the coordinates. For instance, using $(x,y)$ coordinates, the cannonical basis of the tangent space is the set:

$$B=\left\{\frac{\partial}{\partial x}, \frac{\partial}{\partial y}\right\}$$

And using $(u, v)$ coordinates the cannonical bases is:

$$\tilde{B} = \left\{\frac{\partial}{\partial u}, \frac{\partial}{\partial v}\right\}$$

Components with respect to some basis are the scalars that you multiply the basis vectors on the linear combination that gives some vector. The jacobian matrix is the matrix of partials, the case for function from $\mathbb{R}^2$ to $\mathbb{R}^2$ is written in the answer.

EDIT 3: Well, now the answer for the problem. We know that $X(p) = (p, X_p)$ with $X_p \in T_p\mathbb{R}^2$ given by:

$$X_p = -y \frac{\partial}{\partial x} + x \frac{\partial}{\partial y}$$

Hence, the componentes matrix of the pushforward of this vector is just the matrix:

$$\left[F_{\ast p}(X_p)\right] = \begin{bmatrix} -y\cos\alpha - x\sin\alpha\\ -y\sin\alpha + x\cos\alpha \end{bmatrix}$$

And being this the matrix with respect to the canonical bases we have:

$$F_{\ast p}(X_p) = (-y\cos\alpha - x\sin\alpha) \frac{\partial}{\partial u} + (-y\sin\alpha + x\cos\alpha) \frac{\partial}{\partial v}$$

And equating this with the expression with $a$ and $b$ we get:

$$a = -y\cos\alpha - x\sin\alpha$$

$$b = -y\sin\alpha + x\cos\alpha$$