Question about tangent vector and tangent space of a manifold

Question

Let $S$ be a manifold of dimension $n$.Let $\phi: S\to\Bbb{R^n}$ be a homeomorphism which is a coordinate system for $S$.That is for $p\in S$ we have $\phi(p)=[\xi^1,...,\xi^n]$ be the coordinate of the point $p\in S$.

Let $f:S\to \mathbb{R}$ be a real-valued function and $\gamma:R\to S$ be a curve then the tangent vector at a point $p=\gamma(a)$ is defined as $$\frac{d \gamma}{dt}=\frac{d(f\circ \gamma(t))}{dt}$$ at $t=a.$

Okay, so my question is suppose we have a manifold $S=\mathbb{S^1}$=$\{(x,y)\in \mathbb{R^2}/x^2+y^2=1\}$ and let I have to calculate the tangent vector at point $p=(0,1)$ then how can I apply the above definition to find it. I mean what will be my $f$ and $\gamma$?

Answer 1

The short answer is that $f$ is arbitary, and I've constructed a family $\gamma_s$ in my answer below.

Let me first answer this question using the standard embedding of $S^1\subset \mathbb{R}^2$. Thus, let me produce for you a set of curves $\gamma_s:\mathbb{R}\to S^1$ such that for any $s\in T_{(0,1)}S^1\cong \mathbb{R}\times 0$, we have $$\frac{d\gamma_s}{d\theta}\bigg\vert_{\theta=\frac{\pi}{2s}}=-s.$$ This negative jumps out because of the insistence to parameterize the circle counterclockwise: it's a purely a matter of cosmetics. I will show the tangent vectors at $p$ is in 1-1 correspondence with the following curves: $\gamma_s(\theta):=(\cos(s\theta),\sin(s\theta))$.

By Pythagoren theorem, the image of these curves lie on $S^1$. By construction, $\gamma_s(\frac{\pi}{2s})=(\cos(s\frac{\pi}{2s}),\sin(s\frac{\pi}{2s}))=(0,1)$. Furthermore, $$\frac{d\gamma_s}{d\theta}\bigg\vert_{\theta=\frac{\pi}{2s}}=(-s\sin(s\frac{\pi}{2s}),s\cos(s\frac{\pi}{2s}))=(-s,0)=-s.$$ The issue is that on general manifolds we don't have a (canonical) embedding into $\mathbb{R}^n$, so we seek to find an alternative characterization to the tangent space. In the above, technically $s$ is a vector at $T_{(1,0)}S^1$ but I conflated this with a scalar using the isomorphism. Let us now not conflate the two.

To understand this alternative characterization, we first need to recall the whole point of wanting to have a notion of a tangent space in the first place. The idea is to capture velocity information as you travel through a point. But how do we practically use this velocity information? The answer is by taking directional derivative with respect to this velocity. Namely given any smooth function $f$, at a point $p$, we use velocity information $v$ by computing at $p$, $$D f\cdot v=df(v)=vf.$$ I want to emphasize again that this is a pointwise calculation even though I'm suppressing the $p$ in the notation above (welcome to differential geometry). For example, if $v=\frac{\partial}{\partial x^i}$ is from a coordinate system $x^1,...,x^n$, then we would be computing the standard partial derivatives, as $$df(\frac{\partial}{\partial x^i})=\frac{\partial f}{\partial x^i}.$$ For concreteness, let us fix a smooth function $f(x,y):S^1\to \mathbb{R}$, say $f(x,y)=3xy$ as example. In this case, we compute $$\frac{d(f\circ \gamma_s)}{d\theta}=\frac{df(\cos(s\theta),\sin(s\theta))}{d\theta}=\frac{d(3\cos(s\theta)\sin(s\theta))}{d\theta}=-3s\sin^2(s\theta)+3s\cos^2(s\theta).$$ Evaluating at $\theta=\frac{\pi}{2s}$, we get $$\frac{d(f\circ \gamma_s)}{d\theta}\bigg\vert_{\theta=\frac{\pi}{2s}}=-3s.$$

Answer 2

From https://en.m.wikipedia.org/wiki/Differentiable_manifold#Tangent_vector_and_the_differential, a tangent vector can be defined as an equivalence class of curves passing through a point $p$ on the manifold while being tangent to each other at $p$.

From https://en.m.wikipedia.org/wiki/Differentiable_manifold#Directional_differentiation, the directional derivative of a function $f$ at a point $p$ in a differentiable manifold, for a given tangent vector, is $\frac{d}{dt} [ (f \circ \gamma) (t) ] \bigg\vert_{t=0} $ with $\gamma(0)=p$ and $\gamma$ a curve in the tangent vector’s equivalence class. So $f$ can be any differentiable function at $p$, meaning that $f \circ \phi^{-1}$ is differentiable, where $\phi$ is a coordinate chart.

As an example, we could arbitrarily choose the following:

\begin{align*} \gamma(t) &=(-\sin t,\cos t), \\ f(p) &= x, \\ \theta = \phi(p) &= - \arctan(x/y),\\ f \circ \phi^{-1}(\theta) &= -\sin \theta. \end{align*}

We see that $f \circ \phi^{-1}$ is differentiable. The tangent vector at $p$ is the equivalence class to which $\gamma(t)$ belongs. Note that the tangent vector is independent of the choices of $f$ or the coordinate chart $\phi$.

The directional derivative at the point $p=(0,1)=\gamma(0)$ is therefore

\begin{align*} \frac{d}{dt} [ (f \circ \gamma) (t) ] \bigg\vert_{t=0} &= \frac{d}{dt}[ - \sin(t )]\bigg\vert_{t=0} \\ &= -1. \end{align*}