I'm beginning my studies of manifolds and am still trying to grasp the basic concepts. The question I want to ask is simple: how do I get the basis of the tangent space from a manifold defined as a level set of a function? I'll give an example:
Imagine the surface $M=\{(x,y) \in \mathbb{R}^2 | x+x^2+y^2=2\}$. This is clearly a circunference centered at the point $(-\frac{1}{2},0)$. We can describe this in two ways:
1)A parametrization. Let $\Psi:\mathbb{R}\rightarrow \mathbb{R}^2;u\mapsto(cos(u)-\frac{1}{2}, sin(u))$. Then the basis of the manifold at a point $p\in M$ is the vector that comes out of the Jacobian matrix, in this case:$$\frac{d}{du}|_p = \begin{pmatrix}
-sin(u) \\
cos(u) \\
\end{pmatrix}$$
2) A level set (I believe this is called a submersion but I'm not sure). If we define $F:\mathbb{R}^2\rightarrow \mathbb{R};(x,y)\mapsto x+x^2+y^2-2$ then $F^{-1}(0)=M$. We know that:$$T_p M=Ker[J(F,p)]$$ where $T_p M$ is the tangent space at $p$ and $J(F,p)$ is the Jacobian Matrix at $p$.
How do I proceed from here to determine the basis of the Tangent Space? How does this generalize for any surface?
Best Answer
There seems to be no general way to approach this problem, but to compute the kernel in each case. For example, in (2), ${\sf D}F(x,y)$ is identified with the gradient $\nabla F(x,y) = (1+2x, 2y)$, and $T_{(x,y)} F^{-1}(0) = \nabla F(x,y)^\perp$. The tangent space in this case is the tangent line, so a direction of the line is given by $(1+2x,-2y)$ (the vector obtained by rotating $\nabla F(x,y)$ by $\pi/2$).
In general, a smooth map $F\colon \Bbb R^n \to \Bbb R^k$ is called a submersion if ${\sf D}F(p)$ is surjective for all $p \in \Bbb R^n$, and in this case $F^{-1}(0)$ is an embedded submanifold of $\Bbb R^n$ of dimension $n-k$, with $T_p F^{-1}(0) = \ker {\sf D}F(p)$. You can represent ${\sf D}F(p)$ by a $k\times n$ matrix and compute its kernel using what you learnt in a first linear algebra course.