Derivative of this function is injective everywhere

[kent davidge]

I'm reading a pdf where it's said that the function $f: \mathbb R \longrightarrow \mathbb{R}^2$ given by $f(x) = \langle \sin (2 \pi x), \cos ( 2 \pi x) \rangle$ is not one-to-one, because $f(x+1) = f(x)$. This is pretty obvious to me. What I don't understand is that next they say that the derivative map, which they denote by $Df(x)$, is injective for all $x \in \mathbb R$. How can it be?

Is'nt the derivative given by $2 \pi \langle \cos (2 \pi x), - \sin (2 \pi x) \rangle$? It's easy to see that it's the same for $x$ and $x + 1$.

 

[fresh_42]

Can you upload an image or give the precise wording, the definition of $Df(x)$ included? Injective for all x doesn't make much sense, except if $y \longmapsto (Df(x))(y)$ is the function considered, e.g $(Df(x))(y)= \left. \dfrac{d}{dx}\right|_{x=y} f(x)$. This isn't injective, but e.g. $p \longmapsto (p,Df)$ is. Anyway, injective at a point, what should that be?

 

[kent davidge]

I agree with you in that it's not very precise.

Anyway, injective at a point, what should that be?

I also had this thought

The PDF is here https://www.math.lsu.edu/~lawson/Chapter6.pdf

The considered case is in the example after the first definition (6.1).

 

[Mark44]

I agree with you in that it's not very precise.

I also had this thought

The PDF is here https://www.math.lsu.edu/~lawson/Chapter6.pdf

The considered case is in the example after the first definition (6.1).

Example 6.2 in the PDF has a restricted domain, the open interval $N = (-\pi/4, 3\pi/4)$. On this interval the function of the example is one-to-one.

 

[kent davidge]

Example 6.2 in the PDF has a restricted domain, the open interval $N = (-\pi/4, 3\pi/4)$. On this interval the function of the example is one-to-one.

I'm sorry if I seemed to point to example 6.2. I meant example after definition 6.1 but above example 6.2

 

[fresh_42]

I agree with you in that it's not very precise.

I also had this thought

The PDF is here https://www.math.lsu.edu/~lawson/Chapter6.pdf

The considered case is in the example after the first definition (6.1).

Yes, it is as I suspected. The differential is a mean thing, because it has so many meanings, depending on the perspective, i.e. on the variables regarded. Here we have:
$$
f\, : \,\mathbb{R} \longrightarrow \mathbb{R}^2 \\
D_xf\, : \,T_x\mathbb{R} = \mathbb{R} \longrightarrow T_{f(x)}\mathbb{R}^2=\mathbb{R}^2
$$
where each tangent vector (here a point) is mapped to a pair of tangent vectors. The injective function is
$$
Df(x)=D_xf \, : \, v \longrightarrow \left(\langle Df_1(x),v \rangle,\langle Df_2(x),v \rangle\right) \text{ for all vectors } v\in T_x\mathbb{R}
$$
which are real numbers in this case, so $Df(x)(y)=\left( 2\pi y \cos(2\pi x)\; , \;-2\pi y\sin(2\pi x) \right)$ which is injective in the tangent vector variable $y \in T_x\mathbb{R}=\mathbb{R}$ for any given fixed $x \in \mathbb{R}$, the manifold.

I couldn't resist and once made a list of how a differential can be seen. If you want to have a look:
https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/
It's the list of ten items, quite at the beginning. You're perhaps also interested to read:
https://www.physicsforums.com/insights/the-pantheon-of-derivatives-i/
where I tried to shed some light on the various conceptions, yours included.

 

[kent davidge]

Oh thanks. I will read your linked posts.

$$
Df(x)=D_xf \, : \, v \longrightarrow \left(\langle Df_1(x),v \rangle,\langle Df_2(x),v \rangle\right) \text{ for all vectors } v\in T_x\mathbb{R}
$$

Are those angled brackets inner products? Does that relation holds for any manifolds $\mathcal M$ and $\mathcal N$ or only for this (apparent) special case of $\mathcal M = \mathbb R$ and $\mathcal N = \mathbb{R}^2$?

 

[fresh_42]

Oh thanks. I will read your linked posts.

Are those angled brackets inner products?

Yes, although it's a bit over the top here, as we only have numbers. The differential is generally a linear transformation: "times slope", "gradient", "Jacobi matrix". So it applies to vectors: $(D_xf)(v)$, the direction of the tangent, resp. the differential. In this case the vector is one-dimensional, but nevertheless a vector. In one, resp. each coordinate this can be written as an inner product.
Btw. it's better to write the point at which a derivative is evaluated by an index as $Df(x)=D_xf$ so that it is clearer what is what.

Does that relation holds for any manifolds $\mathcal M$ and $\mathcal N$ or only for this (apparent) special case of $\mathcal M = \mathbb R$ and $\mathcal N = \mathbb{R}^2$?

Yes, it holds in any case. We have $f\, : \,\mathcal{M} \longrightarrow \mathcal{N}$ and $D_pf\, : \,T_p\mathcal{M} \longrightarrow T_{f(p)}\mathcal{N}$.

 

[kent davidge]

Upon thinking about this, I realized that we can bring $x$ to just an interval like $(0, 1]$. In such case it's easy to see that the "differential map matrix" $2 \pi (\cos (2 \pi x), - \sin (2 \pi x))$ is injective everywhere. Wouldn't taking this approach to the case end up being more concise?

 

[fresh_42]

Upon thinking about this, I realized that we can bring $x$ to just an interval like $(0, 1]$. In such case it's easy to see that the "differential map matrix" $2 \pi (\cos (2 \pi x), - \sin (2 \pi x))$ is injective everywhere. Wouldn't taking this approach to the case end up being more concise?

You confuse the functions. Quote: $D_xf$ is injective for all $x$, means the function $v \longmapsto (D_xf)(v)$ is injective. So to say the Jacobi matrix at any point $x$ of evaluation is injective. What you are talking about is $x \longmapsto D_xf$ which is a different function. I haven't checked whether $D_{(.)}f$ depends injective on the point of evaluation. It wasn't stated nor meant. The way the script deals with it, is a fixed point $x$ on the left side and $f(x)$ on the right, but the function is for a certain $x$ mapping tangent vectors $v$.