How to show that $T_{(1,0)}\mathbb S^1 \cong \operatorname{span}(\{e_2\})$?

Question

I want to show that $T_{(1,0)}\mathbb S^1 \cong \operatorname{span}(\{e_2\})$ using the stereographic chart and using the definition that $T_xM$ is the set of velocity vectors $v$ where each vector $v$ is the equivalence class of curves that goes through point $x$ and tangent to each other.

I got so far the following:

• Since $\varphi:U\to\mathbb{R}$ is given by $\varphi(x,y)=\frac{x}{1-y}$ and $v=\frac{d}{dt}(\varphi\circ \gamma)(t)\Big|_{t=0}$ for some $\gamma:I\to \mathbb S^1$ with $\gamma(0)=x=(1,0)$, we can compute that \begin{align} v& =\frac{d}{dt}(\varphi\circ \gamma)(t)\Big|_{t=0}\\ &=\frac{d}{dt}\Big(\frac{x(t)}{1-y(t)}\Big)\Big|_{t=0}\\ &=\frac{x^{\prime}(t)(1-y(t))-x(t)(-y^{\prime}(t))}{(1-y(t))^2}\Big|_{t=0}\\ &=\frac{x^{\prime}(0)(1-y(0))+x(0)y^{\prime}(0))}{(1-y(0))^2}\\ &=x^{\prime}(0)+y^{\prime}(0). \end{align} I don't know how to interprete that and how to actually show that $T_{(0,0)}\mathbb S^1$ should be a span of $e_2$.

• I know that if $i:\mathbb S^1\to\mathbb{R}^2$ is an inclusion, then $$di_x:T_x \mathbb S^1\to T_{i(x)}\mathbb{R}^2\text{ is injective}.$$ So, we need somehow show $di_x(v)=\operatorname{span}(\{e_2\})$.

What should I do?

Answer 1

I suggest you take a look at this previous answer of mine, where I explain how to relate the abstract tangent space via various chart-induced isomorphisms to an actual subspace of some $\Bbb{R}^n$, in particular when your manifold is already a submanifold of some cartesian space.

Let $p = (1,0) \in S^1 \subset \Bbb{R}^2$. If you follow that answer (and notation) carefully, you should see that when relating the abstract $T_pS^1$ to an actual subspace of $\Bbb{R}^2$, we're looking at the linear subspace $V= \Phi_{\text{id}_{\Bbb{R}^2}, p}(T_pS^1) \subset \Bbb{R}^2$. And since $\Phi_{\varphi,p}:T_pS^1 \to \Bbb{R}$ is an isomorphism, it follows that the subspace $V$ which we intuitively think of as the tangent space is simply \begin{align} V= \Phi_{\text{id}_{\Bbb{R}^2}, p}(T_pS^1) = [\Phi_{\text{id}_{\Bbb{R}^2}, p}\circ (\Phi_{\varphi,p})^{-1}](\Bbb{R}) \end{align} Now, if you unwind how the maps $\Phi_{\alpha,p}$ (and their inverses) are defined, you should see that this reduces to \begin{align} V&= D(\text{id}_{\Bbb{R}^2}\circ \varphi^{-1})_{\varphi(p)}[\Bbb{R}] \\ &= D(\varphi^{-1})_{\varphi(1,0)}[\Bbb{R}] \\ &= D(\varphi^{-1})_{1}[\Bbb{R}] \end{align} In other words, we just have to look at the inverse map $\varphi^{-1}:\Bbb{R}\to \Bbb{R}^2$ (sure it maps onto a portion of $S^1$, but you can easily view this as a map into $\Bbb{R}^2$), then we calculate it's Frechet-derivative at the base point $1$, $D(\varphi^{-1})_1:\Bbb{R}\to \Bbb{R}^2$, and then calculate its image. This simply amounts to taking the span of the columns of the matrix representation $(\varphi^{-1})'(1)$ (relative to the standard basis). So, since $\varphi^{-1}(s) = \left(\frac{2s}{s^2+1}, 1- \frac{2}{s^2+1}\right)$. A tedious but straightforward computation shows that $(\phi^{-1})'(1) = 2e_2$. It's span is clearly $\text{span}(\{e_2\})$.

---

The general message is this: let's say $M$ is an $m$-dimensional submanifold of $\Bbb{R}^l$. Take a point $p\in M$ and a chart $(U,\varphi)$ around $p$. Then, rather than calculating $T_pM$ using one of the abstract definitions, we can calculate what an isomorphic copy (which for the lack of a better name let's call it $\mathcal{T}_pM$) of it looks like. This isomorphic copy is an honest subspace of $\Bbb{R}^l$, and the way it is calculated is as the image of the linear transformation $D(\text{id}_{\Bbb{R}^l}\circ \varphi^{-1})_{\varphi(p)}: \Bbb{R}^m \to \Bbb{R}^l$ (at this stage everything is just plain old multivariable calculus using (Frechet) derivatives); i.e $\mathcal{T}_pM = \text{image } D(\varphi^{-1})_{\varphi(p)} \subset \Bbb{R}^l$.

If you think about this for a moment, this should make sense: $\varphi$ is a chart map, so it takes a portion of the manifold $M$ onto a certain portion of $\Bbb{R}^m$. The inverse map $\varphi^{-1}$ is often called a local parametrization of $M$ about the point $p$ (think of it as mapping the grid lines of $\Bbb{R}^m$ to a bunch of curvy grid lines on the manifold). Then, the derivative $D(\varphi^{-1})_{\varphi(p)}$ is the linear approximation to this map. Which means it maps the $m$-dimensional subspace $\Bbb{R}^m$ bijectively onto a certain $m$-dimensional subspace of $\Bbb{R}^l$, which "approximates" $M$ in some sense (if all you care about is submanifolds of $\Bbb{R}^l$, this is a valid definition of $\mathcal{T}_pM$, in the sense that this subspace is actually independent of the chart).

Answer 2

Since $\mathbb{S}^1$ is an embedded submanifold of $\mathbb{R}^2$, for any $p\in \mathbb{S}^1$, $T_p\mathbb{S}^1\subseteq T_p\mathbb{R}^2$. Identifying $\mathbb{R}^2$ with $T_p\mathbb{R}^2$ in the natural way, we obtain

$T_p\mathbb{S}^1=\{ v=(v_1,v_2)\in \mathbb{R}^2$ $:$ $v_1p_1+v_2p_2=0$ $\}$.

Therefore setting $p=(1,0)$:

$T_p\mathbb{S}^1=\{ (0,b)\in \mathbb{R}^2$ $:$ $b\in \mathbb{R}$ $\}=span(e_2)$

Answer 3

Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ given by $f((x,y))=x^2+y^2$ then $1$ is a regular value of $f$ and hence $f^{-1}(1)$ is a submanifold of $\mathbb{R}^2$ but $S^1=f^{-1}(1)$ and furthermore given $p \in S^1$, $T_p (S^1)=Ker(df_p)$ and calculating we obtain that $df_p=2(p_1,p_2)$ then is easy to see that $$T_p(S^1)=\{x \in \mathbb{R}^2 ; \langle p,x \rangle =0\}=span(p) ^{\perp}$$ But $\mathbb{R}^{2}=span(p) \oplus span(p)^{\perp}$ and $dim(span(p))=1$ then $dim(span(p)^{\perp})=1$ then it is only necessary to find one element of $span(p)^{\perp}$ to characterize it, it is easy to see that if $p=(p_1,p_2)$ then $(-p_2,p_1) \in span(p)^{\perp}$ it follows that $$T_p(S^1)= span((-p_2,p_1))$$

And in particular in your problem taking $p=(1,0)$ then $T_{(1,0)}(S^1)=span((0,1))$