Derivative of this function is injective everywhere

  • #1
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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##.
 
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  • #2
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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?
 
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  • #4
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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.
 
  • #5
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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 :biggrin:
 
  • #6
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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.
 
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  • #7
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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##?
 
  • #8
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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}##.
 
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  • #9
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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?
 
  • #10
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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##.
 
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