Tangent vector basis and basis of coordinate chart

[AndrewGRQTF]

I am learning the basics of differential geometry and I came across tangent vectors. Let's say we have a manifold M and we consider a point p in M. A tangent vector $X$ at p is an element of $T_pM$ and if $\frac{\partial}{\partial x^ \mu}$ is a basis of $T_pM$, then we can write $$X = X^\mu \frac{\partial}{\partial x^\mu} $$ and if given a curve c on the manifold then we can use that to define the components of the tangent vector such that $X^\mu = \frac{dx^\mu(c(t))}{dt}$

In Nakahara's book, this example is given:

Example 5.11: If X is applied to the coordinate functions $\phi : M \to R^n$ [that come with the manifold] along a curve $c: R \to M$ such that $\phi (c(t))=x^ \mu (t)$, we have $$X (x ^\mu) = (\frac {dx^\mu}{dt}) (\frac{\partial x ^ \mu}{\partial x^ \nu}) = \frac {dx^\mu (t)}{dt}$$

The thing I don't understand is why Nakahara uses the letter $x$ in both the basis of $T_pM$ and for the coordinate functions (by him saying that $\phi (c(t))=x^ \mu (t)$ ). Is it implied that he chose the basis of the tangent space at p as the basis of coordinate chart? The simplification $\delta ^\mu _\nu = \frac{\partial x ^ \mu}{\partial x^ \nu}$ is only possible because of this. And doesn't him saying that $\phi (c(t))=x^ \mu (t)$ defines $x$ to be $$x = \phi \circ c$$ but it is not clear why $$\frac{\partial}{\partial x^\nu} (\phi \circ c) ^\mu= \delta ^\mu _\nu$$.

My point is that $x$ already has a meaning because it is used to write the basis of the tangent space at p. How can we stretch its meaning and say it has anything to do with the coordinate charts or a curve?

 

[fresh_42]

I am learning the basics of differential geometry and I came across tangent vectors. Let's say we have a manifold M and we consider a point p in M. A tangent vector $X$ at p is an element of $T_pM$ and if $\frac{\partial}{\partial x^ \mu}$ is a basis of $T_pM$, then we can write $$X = X^\mu \frac{\partial}{\partial x^\mu} $$ and if given a curve c on the manifold then we can use that to define the components of the tangent vector such that $X^\mu = \frac{dx^\mu(c(t))}{dt}$

In Nakahara's book, this example is given:

Example 5.11: If X is applied to the coordinate functions $\phi : M \to R^n$ [that come with the manifold] along a curve $c: R \to M$ such that $\phi (c(t))=x^ \mu (t)$, we have $$X (x ^\mu) = (\frac {dx^\mu}{dt}) (\frac{\partial x ^ \mu}{\partial x^ \nu}) = \frac {dx^\mu (t)}{dt}$$

The thing I don't understand is why Nakahara uses the letter $x$ in both the basis of $T_pM$ and for the coordinate functions (by him saying that $\phi (c(t))=x^ \mu (t)$ ).

Not exactly. We have $T_p(M) \cong \mathbb{R}^n$ and so they can have the same coordinate system at $p$. But to distinguish them, we write $x^\mu$ for the Cartesian coordinates in $\mathbb{R}^n$ as the chart, and $d x^\mu$ as the basis with coordinates $X^\mu$ in $T_p(M)$. I'm not good with the Einstein notation, so I assume $x^\mu = (x_1,\ldots ,x_n)$ and $X^\mu=(X_1,\ldots ,X_n)=\left(\dfrac{\partial}{\partial x_1},\ldots ,\dfrac{\partial}{\partial x_n}\right)$. Then $X(x^\mu)$ is the total differential $X(x^\mu)=\displaystyle{\sum_{\nu=1}^n} \dfrac{\partial x^\mu}{\partial x^\nu} dx_\nu \,.$

Is it implied that he chose the basis of the tangent space at p as the basis of coordinate chart? The simplification $\delta ^\mu _\nu = \frac{\partial x ^ \mu}{\partial x^ \nu}$ is only possible because of this. And doesn't him saying that $\phi (c(t))=x^ \mu (t)$ defines $x$ to be $$x = \phi \circ c$$ but it is not clear why $$\frac{\partial}{\partial x^\nu} (\phi \circ c) ^\mu= \delta ^\mu _\nu$$.

Yes, $x=(\phi \circ c)$ or in coordinates $(x_1,\ldots ,x_n)=((\phi \circ c)_1,\ldots ,(\phi \circ c)_n)$ resp. as the path it is $(x_1(t),\ldots ,x_n(t))=((\phi \circ c)(t)_1,\ldots ,(\phi \circ c)(t)_n)$ in Cartesian coordinates of $\mathbb{R}^n$.

My point is that $x$ already has a meaning because it is used to write the basis of the tangent space at p. How can we stretch its meaning and say it has anything to do with the coordinate charts or a curve?

No.
Tangents are linear forms and thus noted in coordinates $X^\mu$ to the basis vectors $dx^\mu$.
Points of the manifold have only coordinates via their chart. These are then in $\mathbb{R}^n$ with coordinates $x_\mu$ according to the standard Cartesian basis vectors $(0, \ldots ,1 , \ldots, 0)^\mu$.
$\phi \circ c$ is a path in $\mathbb{R}^n$, the chart at $p$ and therefore has coordinates $x^\mu$, resp. coordinate functions $x^\mu(t)\,.$
$X(x^\mu(t))$ is the total differential of these coordinate functions along $X=X_1 \cdot dx_1 + \ldots + X_n \cdot dx_n \,.$

 

[AndrewGRQTF]

Not exactly. We have $T_p(M) \cong \mathbb{R}^n$ and so they can have the same coordinate system at $p$. But to distinguish them, we write $x^\mu$ for the Cartesian coordinates in $\mathbb{R}^n$ as the chart, and $d x^\mu$ as the basis with coordinates $X^\mu$ in $T_p(M)$. I'm not good with the Einstein notation, so I assume $x^\mu = (x_1,\ldots ,x_n)$ and $X^\mu=(X_1,\ldots ,X_n)=\left(\dfrac{\partial}{\partial x_1},\ldots ,\dfrac{\partial}{\partial x_n}\right)$. Then $X(x^\mu)$ is the total differential $X(x^\mu)=\displaystyle{\sum_{\nu=1}^n} \dfrac{\partial x^\mu}{\partial x^\nu} dx_\nu \,.$

Yes, $x=(\phi \circ c)$ or in coordinates $(x_1,\ldots ,x_n)=((\phi \circ c)_1,\ldots ,(\phi \circ c)_n)$ resp. as the path it is $(x_1(t),\ldots ,x_n(t))=((\phi \circ c)(t)_1,\ldots ,(\phi \circ c)(t)_n)$ in Cartesian coordinates of $\mathbb{R}^n$.

No.
Tangents are linear forms and thus noted in coordinates $X^\mu$ to the basis vectors $dx^\mu$.
Points of the manifold have only coordinates via their chart. These are then in $\mathbb{R}^n$ with coordinates $x_\mu$ according to the standard Cartesian basis vectors $(0, \ldots ,1 , \ldots, 0)^\mu$.
$\phi \circ c$ is a path in $\mathbb{R}^n$, the chart at $p$ and therefore has coordinates $x^\mu$, resp. coordinate functions $x^\mu(t)\,.$
$X(x^\mu(t))$ is the total differential of these coordinate functions along $X=X_1 \cdot dx_1 + \ldots + X_n \cdot dx_n \,.$

The Einstein summation in $X^\mu \frac{\partial}{\partial x^\mu}$ means $\sum \limits _ \mu X^\mu \frac{\partial}{\partial x^\mu}$

On the matter of $X$, I do not know why you say that $\{dx_i\}$ is the basis of $T_pM$. Tangent vectors are vectors, not forms ... what am I misunderstanding? https://en.wikipedia.org/wiki/Tangent_space#Basis_of_the_tangent_space_at_a_point says this. The $X^\mu$ are the components of the tangent vectors so they are real numbers.

Why do you say that tangents are linear forms?

 

[Orodruin]

The basis $\partial_\mu$ is the coordinate basis of the tangent space. You could pick a different basis of course, but here we are looking at the coordinate basis, which naturally is connected to the coordinate functions.

 

[fresh_42]

Why do you say that tangents are linear forms?

Because they transform a direction (vector) along the curve into a slope (scalar) in this direction: $v \longmapsto \langle X_p,v \rangle =X_p(v)$.

But however you look at it - and there are quite a few different perspectives, a linear map like the gradient is just one of them - the main task is to distinguish spaces (manifold $M$, chart $\mathbb{R}^n$, tangent space $T_pM \,\,\cong \mathbb{R}^n$), coordinates and basis vectors to interpret the coordinates. If you do not like the view as a linear form, then you have vectors: tangent $X$ evaluated at $p$, expressed in local Cartesian coordinates, the partial derivatives $\partial_\mu=\dfrac{\partial}{\partial x_\mu}\,.$ In this view, $X_p$ becomes a machinery, which attaches a single tangent vector in some direction $X$ at a certain point $p$ to a curve $c$ in $M$ through $p=c(t_0)$ with the help of the chart $\phi$, since we only have calculus available on $\mathbb{R}^n$:
$$
(X,p,c(t)) \longrightarrow X_p(\phi (c(t))=\left. \dfrac{d}{dt}\right|_{t=t_0} (\phi (c(t)) = \left. \dfrac{d}{dt}\right|_{t=t_0}x^\mu = X_p(x^\mu)
$$
And shouldn't the second $\mu$ in your formula for $X(x^\mu)$ be a $\nu\,,\dfrac{dx^\nu}{dt}$ as the summation index?

 

[Orodruin]

Because they transform a direction (vector) along the curve into a slope (scalar) in this direction: v⟼⟨Xp,v⟩=Xp(v)v⟼⟨Xp,v⟩=Xp(v) v \longmapsto \langle X_p,v \rangle =X_p(v) .

That would be the cotangent space, not the tangent space. $T_pM$ is the set of directional derivatives at $p$.

 

[AndrewGRQTF]

The basis $\partial_\mu$ is the coordinate basis of the tangent space. You could pick a different basis of course, but here we are looking at the coordinate basis, which naturally is connected to the coordinate functions.

Thank you, I now understand.