TITLE: Tangent space from a level set
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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.