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Suppose we are working in $\mathbb{R}^2$. Suppose I choose some point $p$. The tangent space would be $T_p\mathbb{R}^2$ and cotangent space denoted $T^*_p\mathbb{R}^2$.

I learned today that

  • $\frac{\partial}{\partial x}$, $\frac{\partial}{\partial y}$ is a basis for the tangent space
  • $dx,dy$ is a basis for the cotangent space

I'm just confused what these bases means intuitively.

When I think of vector spaces, I usually think of maps, like Google maps. So in 2D, if I have a vector $(3,5)$ I imagine walking 3m East and 5m North.

So I'm having a lot of difficulty imagining how to walk in a space built of out vectors like $dx,dy$ or $\frac{\partial}{\partial x}$, $\frac{\partial}{\partial y}$, even if I feel comfortable using those vectors in certain contexts (e.g. Differentiation with linear operators).

How do the basis vectors $\frac{\partial}{\partial x},\frac{\partial}{\partial y}$ and $dx,dy$ relate to the more familiar concept of vectors like $(1,0)$ and $(0,1)$ ? What does it mean to "walk" along these vectors in the context of the tangent space and cotangent space? how do I recover intuitive notions of what these concepts mean?

Stan Shunpike
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  • It might be helpful to separate the notions of the basis of a tangent space at a single point, $T_p$, and how to think about "walking" in a direction. The second topic involves tangent spaces at multiple points and how those change along the manifold. If the manifold is $\mathbb{R}^2$ then the tangent planes are all basically the same, but it is more complicated in the more general case. For the first point, here is a good answer – whpowell96 Sep 17 '24 at 01:50

2 Answers2

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$\newcommand\R{\mathbb{R}}$ It's almost easier to explain this for an abstract manifold $M$.

First, recall that $T_pM$ can be viewed as the space of all possible velocity vectors of curves passing through $p$. In particular, given any smooth parameterized curve through $p$, its velocity at $p$ is an element of $T_pM$. Conversely, given any element in $T_pM$, there is a parameterized curve through $p$ whose velocity at $p$ is the given element.

Recall that a coordinate chart (or its inverse) is a smooth map $\Phi: O \rightarrow M$, where $O \subset \R^n$ is open. Denote the coordinates on $O$ by $(x,y)$. Given any $p \in \Phi(O)$, let $$ \Phi(x_0,y_0) = p. $$ Consider the curves obtained by holding one coordinate fixed and using the other as a parameter: \begin{align*} c_1(t) &= \Phi(t,y_0)\\ c_2(t) &= \Phi(x_0,t). \end{align*} Observe that $c_1(x_0) = c_2(y_0) = p$. The definition of the coordinate vector fields at $p$ are \begin{align*} \frac{\partial}{\partial x} &= c_1'(x_0)\\ \frac{\partial}{\partial y} &= c_2'(y_0) \end{align*}

Next, recall that given a smooth scalar function $f: M \rightarrow \R$, its differential at $p\in M$ is an element of the cotangent space at $p$, $$ df(p) \in T^*_pM. $$ Each coordinate defines a scalar function \begin{align*} x: \Phi(O) &\rightarrow \R\\ y: \Phi(O) &\rightarrow \R \end{align*} Therefore, for each $p \in \Phi(O)$, $dx(p), dy(p) \in T_p^*M$.

It can be checked that the definition of coordinate charts imply that $$ \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}\right) $$ is a basis of $T_pM$ and $(dx,dy)$ is a basis of $T_p^*M$. The miracle is that one is the dual basis to the other.

You now can set $O = M = \R^2$ and let $\Phi: \R^2 \rightarrow \R^2$ be the identity map. If you do that, then \begin{align*} \frac{\partial}{\partial x} &= (1,0)\\ \frac{\partial}{\partial y} &= (0,1). \end{align*} Also, $dx$ is the differential of the function \begin{align*} x: \R^2 &\rightarrow \R\\ (a,b) &\mapsto a \end{align*} and $dy$ is the differential of the function \begin{align*} y: \R^2 &\rightarrow \R\\ (a,b) &\mapsto b. \end{align*}

Deane
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  • If I understand correctly, on a smooth manifold, the points on the manifold reside in neighborhoods diffeomorphic to $\mathbb{R}^n$. But why can the tangent vectors live in a vector space if the original points cannot? What allows us to create a vector space from the derivatives of the curve but not vectors for points or on points for the curve itself? – Stan Shunpike Sep 25 '24 at 02:02
  • I like this question better than your original one. There's not enough room to answer it in a comment, and I prefer not to make my answer any longer. Perhaps you could post this as a separate question on this site? If I see it, I'll provide an answer. I'm also pretty sure others will answer it, too. – Deane Sep 25 '24 at 03:51
  • Absolutely! And thanks for taking the time to answer these. I've been trying a while to learn this and finally getting somewhere! – Stan Shunpike Sep 25 '24 at 10:55
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    https://math.stackexchange.com/questions/4976227/why-does-a-manifold-have-a-vector-space-for-its-tangent-vectors-but-not-the-ori – Stan Shunpike Sep 25 '24 at 11:04
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The tangent space represents directions to differentiate in, while the cotangent space represents the possible ways in a which a function can vary to first order around a point. For example the cotangent vector $\mathrm{d} x_p$ and (resp. $\mathrm{d} y_p$) captures the behavior of a function which looks like $x$ (resp. $y$) to first order near $p$.

The tangent vectors $\frac{\partial}{\partial x}|_p$ and $\frac{\partial}{\partial y}|_p$ are just the dual basis of $\mathrm{d}x_p$ and $\mathrm{d}y_p$. Concretely, this means that

\begin{aligned} \left.\frac{\partial}{\partial x}\right|_p \mathrm{d}x_p = 1, \quad \left.\frac{\partial}{\partial x}\right|_p \mathrm{d}y_p = 0,\\ \left.\frac{\partial}{\partial y}\right|_p \mathrm{d}x_p = 0, \quad \left.\frac{\partial}{\partial y}\right|_p \mathrm{d}y_p = 1. \end{aligned}

In other words, we can interpret $\frac{\partial}{\partial x}|_p$ as measuring the rate of change of a function at $p$ along the direction on which $x$ increases at a unit rate and $y$ is constant, and similarly for $\frac{\partial}{\partial y}|_p$.