Maps with the same image are actually different curves?

[cianfa72]

TL;DR Summary

Are two maps with the same image actually considered two different curves ?

Hi,

I've a doubt about the definition of curve. A smooth curve in $\mathbb R^2$ is defined by the application $\gamma : I \rightarrow \mathbb R^2$.

Consider two maps $\gamma$ and $\gamma'$ that happen to have the same image (or trace) in $\mathbb R^2$. At a given point on the (common) image the two maps can have different velocities (i.e. different tangent vectors).

The question is: are they actually two different curves ?

Thank you.

 

[Orodruin]

They are technically different curves and their tangents differ by a scalar multiplication at each point. However, they are also reparametrisations of each other and in that sense being "the same". (In other words, there is an equivalence relation in which they belong to the same equivalence class - curves that are reparametrisations of each other.)

 

[cianfa72]

They are technically different curves and their tangents differ by a scalar multiplication at each point. However, they are also reparametrisations of each other and in that sense being "the same".

Consider an injective map $f: \mathbb R \rightarrow \mathbb R^2$ that is not regular at point p (i.e. it is not an immersion since the differential/pushforward is not injective at p).

In this case the tangent vector at p is null (all the reparametrisations within the same equivalence class must have tangent null at point p). So for the given fixed image in $\mathbb R^2$ we have different equivalence classes, namely the family of parametrization that give a non-null tangent vector at p vs. the family that instead give a null tangent vector at p.

Does it make sense ?

 

[Orodruin]

Consider an injective map $f: \mathbb R \rightarrow \mathbb R^2$ that is not regular at point p (i.e. it is not an immersion since the differential/pushforward is not injective at p).

In this case the tangent vector at p is null (all the reparametrisations within the same equivalence class must have tangent null at point p). So for the given fixed image in $\mathbb R^2$ we have different equivalence classes, namely the family of parametrization that give a non-null tangent vector at p vs. the family that instead give a null tangent vector at p.

Does it make sense ?

You are thinking of a different equivalence relation relating to the tangents. Two curves in the equivalent up to reparametrisation will generally not have the same tangent except by coincidence. Their tangent at p will be the tangent of a representative that has non-zero tangent multiplied by a non-negative number.

 

[cianfa72]

Two curves in the equivalent up to reparametrisation will generally not have the same tangent except by coincidence

ok, consider the following Lemniscate curve $\beta(t)= (\sin 2t, \sin t) , \text{ } t \in (-\pi, \pi)$

using the parametrisation above the tangent vector's components at point $(0,0)$ are $(2,1)$ whereas their limit for $t \rightarrow \pm \pi$ are actually $(2,-1)$. That means there is not a smooth vector field $X$ defined on $\mathbb R^2$ such that restricted to the Lemniscate gives the tangent vector along it.

So the question is: is there a reparametrisation of the Leminscate's trace/image such that such a smooth vector field $X$ does exist ? This question is related to this thread.

Thanks.

 

[cianfa72]

So the question is: is there a reparametrisation of the Leminscate's trace/image such that such a smooth vector field $X$ does exist ?

Any idea ? In other words: does it exist a reparametrisation of the Leminscate's trace such that the module of the tangent vector (i.e. the velocity) is null at the origin $(0,0)$ ?

 

[lavinia]

Summary:: Are two maps with the same image actually considered two different curves ?

Hi,

I've a doubt about the definition of curve. A smooth curve in $\mathbb R^2$ is defined by the application $\gamma : I \rightarrow \mathbb R^2$.

Consider two maps $\gamma$ and $\gamma'$ that happen to have the same image (or trace) in $\mathbb R^2$. At a given point on the (common) image the two maps can have different velocities (i.e. different tangent vectors).

The question is: are they actually two different curves ?

Thank you.

I think there are smooth curves with the same image that can not be reparameterized to be the same. If one takes a reparameterization to mean a smooth map ϒ: I →I whose derivative is everywhere greater than zero then traversing the curve in the opposite direction has the same image but can not be reparameterized into the original. This may seem like a trivial example but it is important in defining the fundamental group of a topological space.

If one allows derivatives of the curve to be zero then I imagine there are many peculiar examples. For instance suppose that the image of a curve is two circles in the plane that are tangent at a point p.

Starting at p first traverse one circle then afterwards traverse the second. I think such a curve can be made infinitely differentiable by forcing all of its derivatives to be zero at p, not sure. (Example of such a parameterization?)

Next traverse the second circle first and the first circle second. Keep the velocity vectors the same.

There is no reparameterization that makes these two curves the same.

 

[jbergman]

Summary:: Are two maps with the same image actually considered two different curves ?

Hi,

I've a doubt about the definition of curve. A smooth curve in $\mathbb R^2$ is defined by the application $\gamma : I \rightarrow \mathbb R^2$.

Consider two maps $\gamma$ and $\gamma'$ that happen to have the same image (or trace) in $\mathbb R^2$. At a given point on the (common) image the two maps can have different velocities (i.e. different tangent vectors).

The question is: are they actually two different curves ?

Thank you.

What is the context? If you think of them as integral curves, then, yes they are different because the tangent vectors at each point will have different magnitudes.

 

[cianfa72]

Sorry @lavinia, I see some typos in your post #7.

 

[jbergman]

ok, consider the following Lemniscate curve $\beta(t)= (\sin 2t, \sin t) , \text{ } t \in (-\pi, \pi)$
View attachment 298511
using the parametrisation above the tangent vector's components at point $(0,0)$ are $(2,1)$ whereas their limit for $t \rightarrow \pm \pi$ are actually $(2,-1)$. That means there is not a smooth vector field $X$ defined on $\mathbb R^2$ such that restricted to the Lemniscate gives the tangent vector along it.

So the question is: is there a reparametrisation of the Leminscate's trace/image such that such a smooth vector field $X$ does exist ? This question is related to this thread.

Thanks.

I think the answer is no. A curve with a velocity of 0 is a fixed point, meaning, $\forall t \in [-\infty, \infty] : c(t)=p$. The orbits of a vector field are either periodic, fixed points, or non-periodic. If the velocity is 0 at a point then it is a fixed point and the integral curve is a single point.

 

[cianfa72]

I think the answer is no. A curve with a velocity of 0 is a fixed point, meaning, $\forall t \in [-\infty, \infty] : c(t)=p$. The orbits of a vector field are either periodic, fixed points, or non-periodic. If the velocity is 0 at a point then it is a fixed point and the integral curve is a single point.

Sorry, not sure to got it. For sure there exist non-regular parameterization of a curve.
Or are you simply saying that the integral curve of a null vector field at a point is the single point itself ?

 

[lavinia]

Sorry @lavinia, I see some typos in your post #7.

I fixed it. I went to breakfast with some friends and the cats walked on the computer while I was out.
They also posted it before I was done.

 

[jbergman]

What is the context? If you think of them as integral curves, then, yes they are different because the tangent vectors at each point will have different magnitudes.

Ahh, right. You are correct. Ignore my reply.

 

[jbergman]

Any idea ? In other words: does it exist a reparametrisation of the Leminscate's trace such that the module of the tangent vector (i.e. the velocity) is null at the origin $(0,0)$ ?

Note, that the vector field has to be smooth so it would require that the speed on the curve approaches 0 as it approaches the origin. So you are asking if there is a parametrization of the lemniscate such that the velocity is 0 at the origin and approaches 0 from the four directions bordering the origin.

My instinct is that his impossible but I have to think about it more to try and offer a proof.

 

[cianfa72]

So you are asking if there is a parametrization of the lemniscate such that the velocity is 0 at the origin and approaches 0 from the four directions bordering the origin.

Yes, exactly. That's my question.

 

[cianfa72]

They also posted it before I was done.

:-)

Starting at p first traverse one circle then afterwards traverse the second. I think such a curve can be made infinitely differentiable by forcing all of its derivatives to be zero at p.

Ok, suppose to pick one parametrization that traverse the first circle then afterwards the second. Evaluate the tangent vector (velocity vector) along the complete curve and pick a smooth vector field $X$ that restricts to it along that curve (note that the velocity at point p is zero).

If we write down the differential equations for the $x$ and $y$ coordinates respectively, we get as solution an infinite number of integral curves with starting point p.

 

[jbergman]

:-)Ok, suppose to pick one parametrization that traverse the first circle then afterwards the second. Evaluate the tangent vector (velocity vector) along the complete curve and pick a smooth vector field $X$ that restricts to it along that curve (note that the velocity at point p is null).

If we write down the differential equations for the $x$ and $y$ coordinates respectively, we get as solution an infinite number of integral curves with starting point p.

A basic theorem is that the integral curves for a smooth vector field are unique. Which means that if you have a curve, $c(0)=p$ and $c'(0)=0$ then $c(t)=p$ is a solution and the only solution by the uniqueness theorem.

Conversely, if a smooth integral curve is an immersion then it is regular everywhere along that curve. Therefore, I am having a hard time imagining the reality of any of these examples.

 

[cianfa72]

Conversely, if a smooth integral curve is an immersion then it is regular everywhere along that curve.

ok, so coming back to the lemniscate curve if it is an immersion as smooth integral curve then it must be a regular curve.

 

[jbergman]

ok, so coming back to the lemniscate curve if it is an immersion as smooth integral curve then it must be a regular curve.

Yes. I was thinking about some examples, yesterday.

Imagine the curve $(t^3, 0)$. This curve has zero speed at the origin so it isn't an immersion. If you try and solve for the associated vector field you get,

$$ x'(t) = 3t^2 = f(x(t)) = f(t^3)$$
Which has the solution the vector field, $3x^{2/3} \partial_x$ which is not smooth at the origin. Similarly if you try and parametrize your curve as a flow you find it isn't smooth. $\theta(t,x,y)=((t+x^{1/3})^3, y)$.

 

[cianfa72]

Yes. I was thinking about some examples, yesterday.

Imagine the curve $(t^3, 0)$. This curve has zero speed at the origin so it isn't an immersion. If you try and solve for the associated vector field you get,

$$ x'(t) = 3t^2 = f(x(t)) = f(t^3)$$
Which has the solution the vector field, $3x^{2/3} \partial_x$ which is not smooth at the origin.

ok, that makes sense.

Similarly if you try and parametrize your curve as a flow you find it isn't smooth. $\theta(t,x,y)=((t+x^{1/3})^3, y)$.

Sorry, your $\theta(t,x,y)$ should be actually the flow $\theta_t (x,y)=((t+x^{1/3})^3, y)$. Then working out the derivatives w.r.t. $t$ and evaluate them at $t=0$ we get the vector field $3x^{2/3} \partial_x$.

If I get you correctly, doing the same for the lemniscate curve with zero velocity at the origin, we get a non-smooth vector field.

 

[jbergman]

ok, that makes sense.Sorry, your $\theta(t,x,y)$ should be actually the flow $\theta_t (x,y)=((t+x^{1/3})^3, y)$. Then working out the derivatives w.r.t. $t$ and evaluate them at $t=0$ we get the vector field $3x^{2/3} \partial_x$.

If I get you correctly, doing the same for the lemniscate curve with zero velocity at the origin, we get a non-smooth vector field.

We have to be a bit careful. You can create a vector field that is 0 at the origin and flows around the lemniscate and approaches 0 at the origin. However, this will result in 3 integral curves, the stationary curve at the origin and one for the top and bottom loops.

 

[cianfa72]

You can create a vector field that is 0 at the origin and flows around the lemniscate and approaches 0 at the origin. However, this will result in 3 integral curves, the stationary curve at the origin and one for the top and bottom loops.

Ah ok, since we get 3 solutions (integral curves) from the uniqueness theorem.

Anyway for the reasons pointed out above the lemniscate curve cannot be an integral curve of a smooth vector field defined on $\mathbb R^2$.