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In Tu's manifold book, there is this passage when introducing tangent spaces

enter image description here

I tried to construct an explicit case with $S^2$. Let's say we have a coordinate chart $(U, \varphi)$ where $U=S^2\setminus{\{\text {poles}\}}$, and $\varphi$ is an angular coordinate chart $\theta = [0, 2\pi)$, $\phi = (0, \pi)$ shown below

enter image description here

Let's also say I have some embedding $i:S^2\mapsto \mathbb R^3$ given by $i(\theta,\phi) =(\cos\theta\sin\phi, \sin\theta\sin\phi, \cos\phi)$.

I want to calculate what the basis vectors would look like in the embedded $S^2$ for the tangent space located at $p=(\theta,\phi) = (0, \frac\pi2)$. If I remember correctly, I can do this by applying the "partial derivative" representation of the basis vectors (as seen in the Tu book) to the embedding(?), so something like this $$\left.\frac{\partial}{\partial x^i}\right|_p i= \left.\frac{\partial}{\partial r^i}\right|_{\varphi(p)} i$$

but I am very confused on how to proceed, especially what the difference between $x^i$ and $r^i$ would be in this context.

If I just proceed and assume either one of $x^i,p$ or $r^i,\varphi(p)$ is $\theta,\phi$ and $\left(0, \frac\pi2 \right)$ in the coordinate chart and work out the math, it works

\begin{align*} \vec{v}_\theta & = \left\langle \left.\frac{\partial}{\partial\theta}\left[ \cos(\theta)\sin(\phi) \right]\right|_{\left(0, \frac{\pi}{2}\right)}, \left.\frac{\partial}{\partial\theta}\left[ \sin(\theta)\sin(\phi) \right]\right|_{\left(0, \frac{\pi}{2}\right)}, \left.\frac{\partial}{\partial\theta}\left[ \cos(\phi) \right]\right|_{\left(0, \frac{\pi}{2}\right)} \right\rangle = \langle 0, 1, 0\rangle \\ \vec{v}_\phi & = \left\langle \left.\frac{\partial}{\partial\phi}\left[ \cos(\theta)\sin(\phi) \right]\right|_{\left(0, \frac{\pi}{2}\right)}, \left.\frac{\partial}{\partial\phi}\left[ \sin(\theta)\sin(\phi) \right]\right|_{\left(0, \frac{\pi}{2}\right)}, \left.\frac{\partial}{\partial\phi}\left[ \cos(\phi) \right]\right|_{\left(0, \frac{\pi}{2}\right)} \right\rangle = \langle 0, 0, -1\rangle \end{align*}

but I have no idea why it works.

Also, they both are also interpreted as functions/maps, with $x^i:U\mapsto \mathbb R$ (e.g. $x^2(p) = \frac\pi2$) and $r^i:\mathbb R^n\mapsto\mathbb R$ (e.g. $r^3(a, b, c) = c$), so in a sense we are taking partial derivatives with respect to functions??? What am I misunderstanding.

Max0815
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    We are NOT differentiating with respect to functions. This is a new object with a precise definition of its own, but packaged/notated and intentionally designed to evoke familiarity/comfort with the old notation. See Rate of change of a multivariable function with respect to a change in another multivariable function and this answer for more about coordinates. – peek-a-boo Mar 18 '25 at 06:57
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    Also, see this, this, and this for remarks about how to relate abstract definitions of the tangent space to honest subspaces of the ambient model space. And finally, some computations from Physics examples (not directly related to what you asked, but good to think about nonetheless): here for a GR example and here for thermodynamics. – peek-a-boo Mar 18 '25 at 07:01
  • @peek-a-boo Thank you so much for the links, I'll take a look after I wake up tomorrow trying to learn diffgeo is hard ;-; – Max0815 Mar 18 '25 at 07:07
  • One other thing: if you mean $i:S^2\to\Bbb{R}^3$ is the usual inclusion, then writing $i(\theta,\phi)$ makes no sense. What you’re doing is already fixing a spherical coordinate chart $\sigma:U\subset S^2\to (0,\pi)\times (0,2\pi)$, sending a point $p\in U$ to its polar and azimuthal angles relative to some axis, $\sigma(p)=(\theta(p),\phi(p))$. Then, it is $i\circ\sigma^{-1}: (0,\pi)\times (0,2\pi)\to\Bbb{R}^3$ which can be written as a function defined on $2$-tuples. Again, all of this will hopefully be clearer after reading all the links (and tons of sublinks/related links on the side). – peek-a-boo Mar 18 '25 at 07:10

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