Abstract description of $T_pS^2$ - space of differentiation - at fixed point $p$ in terms of $\mathbb{R}^3$

Question

I am trying to understand coordinate description of $T_pS^2$, where $$ S^2 = \{(x, y, z) \in \mathbb{R}^3: x^2 + y^2 +z^2 = 1\} $$ at fixed point $p$ in terms of $\mathbb{R}^3$. Geometrically I understand that it's just a $2-$ dimensional tangent plane to $S^2$ at $p$ and moreover i can geometrically find $f(x, y, z) = 0$ - plane's equation. But i am interested how to find such equation abstractly, without geometric intuition, for example with fixed proper $\partial f / \partial x = 1$. What i am looking for, it's just algebraic way of finding it, thinking about $T_pS^2$ as a space of all differentiations at point $p$

Answer 1

The two dimensional sphere is defined as $$ S^2 = \left\{X \in \mathbb{R}^3 \mid \left\langle X,X\right\rangle =1 \right\} $$ where $\left\langle \cdot,\cdot\right\rangle$ is the usual euclidean inner product. In other words, it is the $1$-level set of the map $g \colon \mathbb{R}^3 \to \mathbb{R}$ defined by $g(X) = \langle X,X\rangle$.

First method: submersion. One can show that at each $X \in S^2$, $\mathrm{d}g(X)$ is a surjective linear application. This is another way to say that $g$ is a submersion on a neighbourhood of $S^2$, and hence, $S^2$ is a smooth submanifold of $\mathbb{R}^3$ of codimension $\dim \mathbb{R}=1$. Moreover, its tangent space at $X \in S^2$ is $\ker \mathrm{d}g(X)$. So your question turns out to be equivalent to "how can we find $\ker\mathrm{d}g(X)$?" The answer is pretty simple: just compute the differential and it turns out that $\ker \mathrm{d}g(X) = X^{\perp}$, the orthogonal complement of $X$ in $\mathbb{R}^3$.

Second method: tangent vector to curves. A characterization (or a definition) of the tangent space of a submanifold $M \subset \mathbb{R}^n$ is the following: $$T_pM = \left\{ \gamma'(0) \mid \gamma \colon (-\varepsilon,\varepsilon) \to M \text{ is differentiable and } \gamma(0) = p \right\}.$$ Let $\gamma$ be a differential curve in $S^2$ with $\gamma(0) = X \in S^2$. Then for all $t$ we have $$\langle \gamma(t),\gamma(t) \rangle = 1$$ and differentiating this equality at $0$ yields $$ 2 \langle \gamma'(0),\gamma(0) \rangle = 0 $$ which is equivalent to $\langle\gamma'(0),X \rangle = 0$, that is $ \gamma'(0) \perp X$. Thus, we have the inclusion $T_XS^2 \subset X^{\perp}$. To conclude, here are two ways:

1. use the dimension to conclude that these two linear spaces are equal
2. for $v \in X^{\perp}$, find $\gamma(t)\in S^2$ such that $\gamma(0) = X$, $\gamma'(0) = v$ (for example, the curve $\gamma(t) = X\cos(\|v\|t) + \frac{v}{\|v\|}\sin(\|v\|t)$ if $v \neq 0$.)

Comment: you can check that all these work for higher dimensional spheres, there is nothing special with the two dimensional one.

Edit: I consider tangent spaces to (sub-)manifolds to be linear spaces, while some (like @hm2020, see the other answers) might consider tangent spaces as affine spaces. This is a matter of taste. In the latter case, $T_XS^2 = X + \ker \mathrm{d}g(X)$, etc.

Answer 2

Question: "But i am interested how to find such equation abstractly, without geometric intuition, for example with fixed proper $∂f/∂x=1$. What i am looking for, it's just algebraic way of finding it, thinking about $T_pS^2$ as a space of all differentiations at point $p$"

Answer: There is a general formula for "the embedded tangent space" of any such hypersurface $Z(f) \subseteq \mathbb{R}^n$. In your case let $f(x,y,z):=x^2+y^2+z^2-1$ and let $S^2:=Z(f)$ denote the zero set of the polynomial $f$. An equation for the tangent plane of $S^2$ at the point $p:=(p_1,p_2,p_3)$ is the following: Let $F(x,y,z):= f_x(p)(x-p_1)+f_y(p)(y-p_2)+f_z(p)(z-p_3)$ where $f_x(p)$ is the partial derivative of $f$ at $x$ evalulated at $p$. In your case you get the formula

$$H1.\text{ }2p_1(x-p_1)+2p_2(y-p_2)+2p_3(z-p_3)=0.$$

If $p:=(0,0,1)$ you get the tangent plane $2(z-1)=0$ or equivalently $z=1$.

The polynomial in $H1$ is linear hence its zero set is a plane intersecting $S^2$ in the point $p$. Hence the equation in $H1$ defines the embedded tangent space $T_p(S^2)$ at $p$.

Answer 3

Maybe this answers your (very unclear) question: every smooth manifold have an atlas associated to it. Let a chart $(U,\varphi )$, then by definition $\varphi :U\to V$ is a diffeomorphism, where $V$ is the image of $\varphi $ and it is an open subset of some Euclidean space, say $\mathbb{R}^n$.

Then, for $q\in V$ we have that $T_qV= T_q\mathbb{R}^n\cong \mathbb{R}^n$, so every vector of $\mathbb{R}^n$ can be related uniquely to a vector in $T_pU=T_pM$ (for $p=\varphi ^{-1}(q)$) using the differential of $\varphi^{-1}$ at $q$, that is, $d\varphi^{-1} _q$ establishes an isomorphism between both vector spaces. This bijection can be made explicitly clear knowing that in $\mathbb{R}^n$ derivations defined by vectors are just the partial derivatives in the direction of this vector, that is, for every $w\in \mathbb{R}^n$ we define uniquely a derivation (a vector) $v\in T_pM$ by

$$ vf=(d\varphi_q ^{-1}w)f=w(f\circ \varphi ^{-1})=D_w (f\circ \varphi ^{-1})(q)\tag1 $$

for any smooth functional $f\in C^{\infty }(M)$, where $D_w$ is the well-known directional derivative in direction $w$ of multivariable calculus. As $\varphi =(x^1,\ldots,x^n)$ defines coordinates $\partial/\partial x^1,\ldots,\partial/\partial x^n$ in $T_pM$ then we can write $v=\sum_{k=1}^nv^k \frac{\partial}{\partial x^k}$ for some scalars $v^k\in \mathbb{R}$. Then from (1) we have that

$$ v^k= vx^k=D_w\operatorname{proj}_{k}(q)=(\partial \operatorname{proj}_{k}(q))w=w^k\tag2 $$

where $\operatorname{proj}_{k}$ are the canonical projections on the axis of $\mathbb{R}^n$. Thus from (2) for every $w\in \mathbb{R}^n$ we have a way to explicitly relate it to it corresponding vector in $T_pM$ using the basis defined by the chart.