Components of Tangent Space Vector on Parametrized Curve

[Shirish]

I'm studying 'A Most Incomprehensible Thing - Notes towards a very gentle introduction to the mathematics of relativity' by Collier, specifically the section 'More detail - contravariant vectors'.

To give some background, I'm aware that basis vectors in tangent space are given by $\big\{\frac{\partial}{\partial x^i}\big\}$. I'm also aware that if we act these operators on the coordinate functions $x^i$, then we get a specific basis $\{\vec e_i\}$ whose elements are tangent to the coordinate curves that that point. This specific basis is commonly used as far as I understand.

Then in the 'More detail - contravariant vectors' section:

We can now state that a contravariant vector is a tangent vector to a parameterised curve. Let's see how this works. If the parameter of the curve is $\lambda$, and using a coordinate system $x^i$ the components of the tangent vector $\vec V$ are given by $$V^i=\frac{dx^i}{d\lambda}$$

I'm a little confused by this paragraph. Down the line there's another related paragraph:

In more advanced texts you may see the vector $\vec V$ written as $$\vec V=V^i\frac{\partial}{\partial x^i}$$ where $V^i$ are the vector's components and the partial derivative operators $\frac{\partial}{\partial x^i}$ are the coordinate basis vectors. In order to make sense of this formulation, consider an infinitesimal displacement $df$ at a point $p$ on the manifold, where $f$ is a function of some coordinate system $x^i$. If the displacement is along a curve parameterised by $\lambda$ we can drop $f$ into the above to get $$\vec V=V^i\frac{\partial f}{\partial x^i}=\frac{dx^i}{d\lambda}\frac{\partial f}{\partial x^i}=\frac{df}{d\lambda}$$ where $\frac{\partial f}{\partial x^i}$ are the coordinate basis vectors at $p$.

Again, I'm not able to make much sense of the whole parameterisation w.r.t. $\lambda$ thing.

First question: Why does he say that a contravariant vector is tangent to a parameterised curve? I can understand that a contravariant vector, being a member of the tangent space at some point $p$, must be tangent to some curve passing through $p$. But what makes us say that it should be parameterised?

Second question: Why does he claim that $\partial f/\partial x^i$ are basis vectors? I don't see a reason why, for an arbitrary function $f(x^i)$, $\partial f/\partial x^i$ should form basis vectors.

Third question: Why does he claim that $V^i=dx^i/d\lambda$ are the components w.r.t. this specific basis? What am I missing here?

Would appreciate some guidance!

 

[haushofer]

Question 1: you need a parameter (here: $\lambda$) to make the chain rule work.

Question 2: I'm a bit confused there. Compare the expression $V^i = \frac{dx^i}{d\lambda}$ with $V = V^i \partial_i$, and you'll see that the partials $\partial_{i}$ serve as basis vectors. Be aware: the index i labels whole vectors, not just components! How I see this, is that there is a one-to-one correspondence between the basis vectors and partial derivatives. A nice comparison is between linear algebra and derivative operators, which can be represented by matrices.

Question 3: because of the chain-rule.

 

[PeterDonis]

I'm aware that basis vectors in tangent space are given by $\big\{\frac{\partial}{\partial x^i}\big\}$. I'm also aware that if we act these operators on the coordinate functions $x^i$, then we get a specific basis $\{\vec e_i\}$ whose elements are tangent to the coordinate curves that that point. This specific basis is commonly used as far as I understand.

Your understanding is not quite correct. The correct statement is that the coordinate basis vectors in tangent space are given by $\big\{\frac{\partial}{\partial x^i}\big\}$, and the notation $\{\vec e_i\}$ is just a different notation for the same basis vectors. But you can also have non-coordinate basis vectors, which are not given by $\big\{\frac{\partial}{\partial x^i}\big\}$.

I'm not able to make much sense of the whole parameterisation w.r.t. $\lambda$ thing.

Parameterization just means you label each point on the curve with a real number $\lambda$ in a continuous way, i.e., you treat the curve as if it were a copy of the real number line. Then you treat each of the coordinates $x^i$ as a function of $\lambda$; if you like, you can imagine picking a coordinate chart and seeing how each of the coordinates $x^i$ change with $\lambda$ as you move along the curve.

 

[Shirish]

Parameterization just means you label each point on the curve with a real number $\lambda$ in a continuous way, i.e., you treat the curve as if it were a copy of the real number line. Then you treat each of the coordinates $x^i$ as a function of $\lambda$; if you like, you can imagine picking a coordinate chart and seeing how each of the coordinates $x^i$ change with $\lambda$ as you move along the curve.

Thank you! That perfectly answers the first question.

Your understanding is not quite correct. The correct statement is that the coordinate basis vectors in tangent space are given by $\big\{\frac{\partial}{\partial x^i}\big\}$, and the notation $\{\vec e_i\}$ is just a different notation for the same basis vectors. But you can also have non-coordinate basis vectors, which are not given by $\big\{\frac{\partial}{\partial x^i}\big\}$.

Okay, but $\{\vec e_i\}$ has been defined (in the Luscombe book explicitly and Collier book implicitly) as $$\vec e_i=\frac{\partial \mathbf{r}}{\partial x^i}$$ where I'm guessing $\mathbf{r}$ (assuming 3d manifold) is a triple of the coordinate functions $(x^1, x^2, x^3)$. So what I'm struggling to understand is - how is $\partial \mathbf{r}/\partial x^i$ equivalent to $\partial/\partial x^i$?

 

[Shirish]

Question 3: because of the chain-rule.

Hmm but the chain rule is used in the last equality in $$\vec V=V^i\frac{\partial f}{\partial x^i}=\frac{dx^i}{d\lambda}\frac{\partial f}{\partial x^i}=\frac{df}{d\lambda}$$ So unless we use the fact that $\vec V=\partial f/\partial \lambda$, I'm not seeing where the chain rule comes into play.

 

[PeterDonis]

how is $\partial \mathbf{r}/\partial x^i$ equivalent to $\partial/\partial x^i$?

The apparent difference here is just a confusing (and sloppy) choice of notation.

First, as @haushofer has already pointed out, the index $i$ in $\partial/\partial x^i$ labels vectors, not components. But it's easy to forget that since the same index is often used to label components of a single vector. And it's also easy to confuse or conflate the two usages, as will be illustrated below.

Second, the equivalence is an equivalence between tangent vectors and directional derivatives. So the notation $\partial/\partial x^i$ describes a directional derivative in the direction in which only the coordinate $x^i$ is changing.

Now look at the notation $\vec{e}_i = \partial \mathbf{r}/\partial x^i$. The notation $\mathbf{r}$ is supposed to denote a vector from the origin to the point $x$, and the notation $\partial \mathbf{r}/\partial x^i$ is supposed to denote a vector describing how the vector $\mathbf{r}$ changes as we change the point $x$. But this interpretation is based on viewing vectors as little arrows from one point to another, and that viewpoint does not work in a curved space. That fact is why the preferred viewpoint in differential geometry, which has to deal with curved spaces, is to view vectors as directional derivatives, based on the equivalence described above.

If we view vectors as directional derivatives, then the notation $\vec{e}_i = \partial \mathbf{r}/\partial x^i$ is just a confusing (and sloppy) way of saying that the directional derivative of the n-tuple of coordinates, with respect to one coordinate $x^i$ (note that here the index $i$ is labeling the coordinate), in the direction in which only that coordinate is changing, is simply $1$. Or, to put it another way, a directional derivative has to be applied to a function. If we apply it to the function $x^i$, i.e., the function given by the coordinate that is changing in the direction of our directional derivative, then the result is just $1$. And presto! we now have a unit vector in that direction. Do this $n$ times (for an $n$ dimensional space) and you have a coordinate basis.

 

[Shirish]

The apparent difference here is just a confusing (and sloppy) choice of notation.

First, as @haushofer has already pointed out, the index $i$ in $\partial/\partial x^i$ labels vectors, not components. But it's easy to forget that since the same index is often used to label components of a single vector. And it's also easy to confuse or conflate the two usages, as will be illustrated below.

Second, the equivalence is an equivalence between tangent vectors and directional derivatives. So the notation $\partial/\partial x^i$ describes a directional derivative in the direction in which only the coordinate $x^i$ is changing.

Now look at the notation $\vec{e}_i = \partial \mathbf{r}/\partial x^i$. The notation $\mathbf{r}$ is supposed to denote a vector from the origin to the point $x$, and the notation $\partial \mathbf{r}/\partial x^i$ is supposed to denote a vector describing how the vector $\mathbf{r}$ changes as we change the point $x$. But this interpretation is based on viewing vectors as little arrows from one point to another, and that viewpoint does not work in a curved space. That fact is why the preferred viewpoint in differential geometry, which has to deal with curved spaces, is to view vectors as directional derivatives, based on the equivalence described above.

If we view vectors as directional derivatives, then the notation $\vec{e}_i = \partial \mathbf{r}/\partial x^i$ is just a confusing (and sloppy) way of saying that the directional derivative of the n-tuple of coordinates, with respect to one coordinate $x^i$ (note that here the index $i$ is labeling the coordinate), in the direction in which only that coordinate is changing, is simply $1$. Or, to put it another way, a directional derivative has to be applied to a function. If we apply it to the function $x^i$, i.e., the function given by the coordinate that is changing in the direction of our directional derivative, then the result is just $1$. And presto! we now have a unit vector in that direction. Do this $n$ times (for an $n$ dimensional space) and you have a coordinate basis.

Thanks for the clear explanation! I'll quote a part of the relevant section so you understand where the confusion was arising from:

Although, in the context of curved spaces, we can't meaningfully talk about directed line segments stretching from one point to another, we can define an infinitesimal displacement vector. Figure 5.2 shows a two-dimensional space described by the arbitrary coordinates $u$ and $w$. The index $i$ therefore represents either $u$ or $w$. The infinitesimal displacement from point $P$ to $Q$ is represented by the vector $d\vec x$.

• Define a basis vector $\vec e_u$ that is tangent to the $u$ equals constant curve and points in the direction of increasing $w$. Similarly for the other basis vector
• Define the lengths of these basis vectors so that the components of $d\vec x$ are the coordinate differentials $du$ and $dw$, i.e. $$d\vec x=du\vec e_u+dw\vec e_w=dx^i\vec e_i$$

The equation is deceptively simple. It is valid because it includes coordinate basis vectors $\vec e_i$ which, by definition, give the coordinate differentials $dx^i$ as the components of $d\vec x$. Other, non-coordinate, basis vectors would not generally give such a straightforward (and useful) result.

So the source of my confusion between $\vec e_i$ and $\partial/\partial x^i$ was the above, because I can write $$d\vec x=\frac{\partial \vec x}{\partial x^i}dx^i$$ which implies that $\vec e_i\equiv\frac{\partial \vec x}{\partial x^i}$, which in turn is different from $\frac{\partial}{\partial x^i}$.

So (as far as I've understood from your answer) does that mean the norm of $\vec e_i=\frac{\partial \vec x}{\partial x^i}=1$?

Also, by non-coordinate bases, I'm guessing you mean that we can use the derivative operator w.r.t. some arbitrary functions $f^i=f^i(p):M\to\mathbb{R}$ other than the coordinate functions themselves (as long as those derivative operators are linearly independent). Is that correct?

 

[PeterDonis]

does that mean the norm of $\vec e_i=\frac{\partial \vec x}{\partial x^i}=1$?

Yes.

 

[Shirish]

Yes.

Great! One more thing if you don't mind. What's your take on the second and third questions? Basically this quote:

In more advanced texts you may see the vector $\vec V$ written as $$\vec V=V^i\frac{\partial}{\partial x^i}$$ where $V^i$ are the vector's components and the partial derivative operators $\frac{\partial}{\partial x^i}$ are the coordinate basis vectors. In order to make sense of this formulation, consider an infinitesimal displacement $df$ at a point $p$ on the manifold, where $f$ is a function of some coordinate system $x^i$. If the displacement is along a curve parameterised by $\lambda$ we can drop $f$ into the above to get $$\vec V=V^i\frac{\partial f}{\partial x^i}=\frac{dx^i}{d\lambda}\frac{\partial f}{\partial x^i}=\frac{df}{d\lambda}$$ where $\frac{\partial f}{\partial x^i}$ are the coordinate basis vectors at $p$.

I don't understand why $\frac{dx^i}{d\lambda}$ should be considered as components of $\vec V$. Or why $\frac{\partial f}{\partial x^i}$ would form a basis in the first place - isn't $\frac{\partial f}{\partial x^i}$ just a scalar for each $i$? (since $f$ is mentioned as a function of some coordinate system $x^i$, i.e. $f:\mathbb{R}^n\to\mathbb{R}$ assuming an $n$-dimensional manifold)

 

[PeterDonis]

I don't understand why $\frac{dx^i}{d\lambda}$ should be considered as components of $\vec V$.

They're the components of $\vec{V}$ if $\vec{V}$ is the tangent vector to the curve we are displacing along, whose curve parameter is $\lambda$. In other words, the components of $\vec{V}$ are telling you what proportion of the displacement along the curve is accounted for by displacement in each coordinate basis direction.

Or why $\frac{\partial f}{\partial x^i}$ would form a basis in the first place - isn't $\frac{\partial f}{\partial x^i}$ just a scalar for each $i$?

I think this is the book being sloppy again. The basis vectors are $\frac{\partial}{\partial x^i}$ (no $f$). Since those basis vectors are (equivalent to) directional derivative operators, you can apply them to any function $f$ to get four scalars, as you say. But those four scalars are not "basis vectors". In fact what should be done with them is to add them; the equation with $f$ included should really be

$$
\vec{V} \cdot \text{d} f = \Sigma_i V^i \frac{\partial f}{\partial x^i} = \Sigma_i \frac{dx^i}{d\lambda} \frac{\partial f}{\partial x^i} = \frac{df}{d\lambda}
$$

Here $\text{d} f$ is the gradient 1-form of the function $f$, and $\vec{V} \cdot \text{d} f$ is the inner product of the vector $\vec{V}$ with that gradient. In other words, the equation is saying that if we treat $\vec{V}$ as a directional derivative, its inner product with the gradient 1-form of a function $f$ gives us the derivative of $f$ along a curve for which $\vec{V}$ is the tangent vector.

 

[Shirish]

They're the components of $\vec{V}$ if $\vec{V}$ is the tangent vector to the curve we are displacing along, whose curve parameter is $\lambda$. In other words, the components of $\vec{V}$ are telling you what proportion of the displacement along the curve is accounted for by displacement in each coordinate basis direction.
I think this is the book being sloppy again. The basis vectors are $\frac{\partial}{\partial x^i}$ (no $f$). Since those basis vectors are (equivalent to) directional derivative operators, you can apply them to any function $f$ to get four scalars, as you say. But those four scalars are not "basis vectors". In fact what should be done with them is to add them; the equation with $f$ included should really be

$$
\vec{V} \cdot \text{d} f = \Sigma_i V^i \frac{\partial f}{\partial x^i} = \Sigma_i \frac{dx^i}{d\lambda} \frac{\partial f}{\partial x^i} = \frac{df}{d\lambda}
$$

Here $\text{d} f$ is the gradient 1-form of the function $f$, and $\vec{V} \cdot \text{d} f$ is the inner product of the vector $\vec{V}$ with that gradient. In other words, the equation is saying that if we treat $\vec{V}$ as a directional derivative, its inner product with the gradient 1-form of a function $f$ gives us the derivative of $f$ along a curve for which $\vec{V}$ is the tangent vector.

Ah, now I see. So the part about $\vec V$ being tangent to the parametrized curve was the crucial part. With that the components make complete sense.

As for the correct equation you wrote, I suppose the LHS can be interpreted as the vector "acting" on the differential (or dual vector) $\text{d}f$, which I guess would be equivalent to the dual vector $\text{d}f$ with $\vec V$ as input - $\text{d}f(\vec V)$. It's a bit hard to wrap my head around the concept of operators as vectors sometimes, in time I'll get used to it.

Thanks again!

 

[PeterDonis]

I suppose the LHS can be interpreted as the vector "acting" on the differential (or dual vector)

Yes, that's one way of interpreting it, treating the vector as an operator that acts on a 1-form to produce a scalar.

which I guess would be equivalent to the dual vector $\text{d} f$ with $\vec{V}$ as input

Yes, the usual definition of a 1-form is a linear map (or operator) from vectors to scalars. Treating the vector as the operator instead of the 1-form is just the dual form of this definition.

 

[Shirish]

They're the components of $\vec{V}$ if $\vec{V}$ is the tangent vector to the curve we are displacing along, whose curve parameter is $\lambda$. In other words, the components of $\vec{V}$ are telling you what proportion of the displacement along the curve is accounted for by displacement in each coordinate basis direction.

Could you give one more clarification on this (I'll use a concrete example)? I want to know if my understanding below is correct:

Suppose I have a 3D manifold $M$ whose points I'm representing through the coordinate system $u^1,u^2,u^3$. This basically means that $u^1,u^2,u^3$ can be identified with three families of coordinate curves.

Now if I have another coordinate system $w^1,w^2,w^3$, these can be identified with three different families of coordinate curves. This means that both $\big\{\frac{\partial}{\partial u^i}\big\}$ and $\big\{\frac{\partial}{\partial w^i}\big\}$ are coordinate bases.

One way of looking at $u^i$ or $w^i$ is as coordinate functions from $M\to\mathbb{R}$. I can also interpret these as parameters of the coordinate curves.

I'm thinking this way because if we have a vector $\vec V$ that is tangent to a curve parameterized by $\lambda$, then we've seen that $\vec V[f]=\frac{df}{d\lambda}$, which means we can identify $\vec V$ with the operator $\frac{d}{d\lambda}$. Since, for example, the basis vector $\frac{\partial}{\partial u^1}$ is tangent to the $u^1$ coordinate curve, I can say that this basis vector is tangent to a curve parameterized by $u^1$ (a.k.a. the $u^1$ coordinate curve).

Is the above dual interpretation of $u^i$ and $w^i$ as coordinate functions and parameters reasonable?

 

[PeterDonis]

This means that both $\big\{\frac{\partial}{\partial u^i}\big\}$ and $\big\{\frac{\partial}{\partial w^i}\big\}$ are coordinate bases.

Yes.

One way of looking at $u^i$ or $w^i$ is as coordinate functions from $M\to\mathbb{R}$. I can also interpret these as parameters of the coordinate curves.

Generally speaking, yes. (There is a possible technicality, since the curve parameter $\lambda$ is supposed to be an affine parameter, and I'm not certain that it's always true that a coordinate will be an affine parameter along its corresponding coordinate curve; I think it's possible to choose weird coordinate scalings where that is not true.)

 

[Shirish]

Generally speaking, yes. (There is a possible technicality, since the curve parameter $\lambda$ is supposed to be an affine parameter, and I'm not certain that it's always true that a coordinate will be an affine parameter along its corresponding coordinate curve; I think it's possible to choose weird coordinate scalings where that is not true.)

"Affine parameter" means, crudely speaking, "regularly spaced"?

 

[PeterDonis]

"Affine parameter" means, crudely speaking, "regularly spaced"?

It means a parameter chosen such that the geodesic equation is invariant under affine transformations:

https://en.wikipedia.org/wiki/Geodesic#Affine_geodesics

 

[sunrah]

First question: Why does he say that a contravariant vector is tangent to a parameterised curve? I can understand that a contravariant vector, being a member of the tangent space at some point p, must be tangent to some curve passing through p. But what makes us say that it should be parameterised?

I think it's a bit of a trick at this stage.

When we perform the Lorentz transformation on V^µ, the coordinates x^µ(λ) are transformed but the parameter λ is not. This means we can infer how the components of the tangent vector should changed (as you have shown). The parameterization itself is not so important because the vector V is invariant under Lorentz transformations.

 

[George Jones]

First question: Why does he say that a contravariant vector is tangent to a parameterised curve? I can understand that a contravariant vector, being a member of the tangent space at some point $p$, must be tangent to some curve passing through $p$. But what makes us say that it should be parameterised?

I know that you have received answers to this question, but I want to address another (related) aspect to this, but I am going to up the level of abstraction a bit. In our mind's eye, we often conflate a parameterized curve with the image of a parameterized curve, when, really, they are two different things. A parameterized curve is a (smooth) mapping from an interval in the real line $\mathbb{R}$ (our curve parameters) into a (differentiable) manifold $M$ (our "space"). The image of a parameterized curve is the set of points in our space $M$ that gets mapped onto. It is this image, a (possibly squiggly) "line" in $M$ that we picture as the curve, but this is not the curve! The curve is this abstract thing, the mapping.

Why the distinction? Two quite different parameterized curves can have the same image, but different parameterizations mean that they traverse this common image (the squiggly line) at different rates, i.e., at any point on the common sqiggly line, the two curves can have quite different tangent vectors.

Now for some simple examples where $M$ is the plane $\mathbb{R}^2$ (with standard smooth structure).

Let the curve $\gamma$ be given by
$$\begin{align}
&\gamma : \left[ 0 ,1 \right] \rightarrow \mathbb{R}^2 \nonumber \\
&t \mapsto \left( t , t^2 \right) . \nonumber
\end{align}$$

Here, $\left[ 0 ,1 \right]$ is a closed interval in $\mathbb{R}$, and $\left( t , t^2 \right)$ is a point in $\mathbb{R}^2$.

Let the curve $\mu$ be given by
$$\begin{align}
&\gamma : \left[ 0 ,1/2 \right] \rightarrow \mathbb{R}^2 \nonumber \\
&t \mapsto \left( 2t , 4t^2 \right) . \nonumber
\end{align}$$

The curves $\gamma$ and $\mu$ have the same image, a particular parabola that joins the points $\left(0,0\right)$ and $\left(1,1\right)$ in $\mathbb{R}^2$, but they are different parameterized curves (i.e., different mappings).

The point $\left(1/2 , 1/4\right)$ is on this parabola, but the curves $\gamma$ and $\mu$ have different tangent vectors at this point,

Can you think of curve that:

1) has the point $\left(1/2 , 1/4\right)$ in its image (i.e., its squiggly line "goes through" this point);
2) has a different image than $\gamma$;
3) at $\left(1/2 , 1/4\right)$, has the same tangent vector as $\gamma$?

 

[Shirish]

I know that you have received answers to this question, but I want to address another (related) aspect to this, but I am going to up the level of abstraction a bit. In our mind's eye, we often conflate a parameterized curve with the image of a parameterized curve, when, really, they are two different things. A parameterized curve is a (smooth) mapping from an interval in the real line $\mathbb{R}$ (our curve parameters) into a (differentiable) manifold $M$ (our "space"). The image of a parameterized curve is the set of points in our space $M$ that gets mapped onto. It is this image, a (possibly squiggly) "line" in $M$ that we picture as the curve, but this is not the curve! The curve is this abstract thing, the mapping.

Why the distinction? Two quite different parameterized curves can have the same image, but different parameterizations mean that they traverse this common image (the squiggly line) at different rates, i.e., at any point on the common sqiggly line, the two curves can have quite different tangent vectors.

Now for some simple examples where $M$ is the plane $\mathbb{R}^2$ (with standard smooth structure).

Let the curve $\gamma$ be given by
$$\begin{align}
&\gamma : \left[ 0 ,1 \right] \rightarrow \mathbb{R}^2 \nonumber \\
&t \mapsto \left( t , t^2 \right) . \nonumber
\end{align}$$

Here, $\left[ 0 ,1 \right]$ is a closed interval in $\mathbb{R}$, and $\left( t , t^2 \right)$ is a point in $\mathbb{R}^2$.

Let the curve $\mu$ be given by
$$\begin{align}
&\gamma : \left[ 0 ,1/2 \right] \rightarrow \mathbb{R}^2 \nonumber \\
&t \mapsto \left( 2t , 4t^2 \right) . \nonumber
\end{align}$$

The curves $\gamma$ and $\mu$ have the same image, a particular parabola that joins the points $\left(0,0\right)$ and $\left(1,1\right)$ in $\mathbb{R}^2$, but they are different parameterized curves (i.e., different mappings).

The point $\left(1/2 , 1/4\right)$ is on this parabola, but the curves $\gamma$ and $\mu$ have different tangent vectors at this point,

Can you think of curve that:

1) has the point $\left(1/2 , 1/4\right)$ in its image (i.e., its squiggly line "goes through" this point);
2) has a different image than $\gamma$;
3) at $\left(1/2 , 1/4\right)$, has the same tangent vector as $\gamma$?

Thanks so much for the explanation! Coincidentally I was just studying about this today and reading this further reinforced my concepts. I have a major doubt related to formation of bases, so I'll post it in the differential geometry subforum.

 

[kent davidge]

Why the distinction? Two quite different parameterized curves can have the same image, but different parameterizations mean that they traverse this common image (the squiggly line) at different rates, i.e., at any point on the common sqiggly line, the two curves can have quite different tangent vectors.

Now for some simple examples where $M$ is the plane $\mathbb{R}^2$ (with standard smooth structure).

You meant to give an example of a same curve with different images, but you ended up giving an example of two curves with different domains and same image.

 

[PeterDonis]

You meant to give an example of a same curve with different images

No, he didn't. He said explicitly, in the very passage you quoted, that he was talking about two different curves (different parameterizations) with the same image (same set of points in $M$).

you ended up giving an example of two curves with different domains and same image

No, both curves have the same domain; they are both maps from $[0, 1]$ into $\mathbb{R}^2$. They just have different parameterizations.