Differentiable manifolds in $\Bbb{R}^n$ Let $S$ be a surface in $\Bbb{R}^4$ which consists of the points which are the solutions of the following equations 
 $x+y+z+t =0$ and $x^2+y^2+z^2+t^2=12$. How can we show that it's a two dimensional submanifold  ($C^1$ surface)? I have found the corresponding Jacobian matrix which has rank 2. What does this tell about the surface? Any hints of how we say that its manifold of dimension 2 and what is its relation with the the rank of the Jacobian? Any help. A small discussion type answers if possible. 
 A: As a consequence of the constant rank theorem; if $f: M \to N$ is a smooth function and $Df(p)$ has constant rank on $M$, then for any $q \in f(M)$ we have $f^{-1}(q) \subset M$ is a regular sub-manifold. To tackle this problem, you must first show that both of these objects are regular and then you can show that they are transverse. Since the intersection of two transverse sub-manifolds, is transverse then you will be done. If you are not familiar with transversality, in your case it just the fact that;
$$T_p M + T_p N = \mathbb{R}^4, \forall p \in M \cap N$$
Here I am taking $M,N$ to be the "submanifolds" given by your set of points. The first equation is a hypersurface i.e the analogue of a plane in 4-dimensions, so you are fine. You can express the second as;
$$f(x,y,z,t) = x^2+y^2+z^2+t^2 - 12$$
Then you have that $f^{-1}(0) = M$, choosing to call this set of points $M$. Now if you can show that $0$ is a regular value for this smooth function you are done. Here $f: \mathbb{R}^4 \to \mathbb{R}$ i.e you just need to show that the differential has rank $1$. That follows immediately since;
$$Df = \langle 2x, 2y, 2z, 2t \rangle = \textbf{0} \iff x = y = z = t = 0 \not \in M$$
Therefore, $M$ is a regular submanifold of $\mathbb{R}^4$. For the transversality step, it is enough to show that the normal vectors for the corresponding tangent planes aren't parallel. Well is,
$$\langle 1,1,1,1 \rangle = \lambda \langle 2x, 2y, 2z, 2t \rangle \  \ ??$$
The above will imply $x = y = z = t = \frac{1}{2\lambda}$. Since $(x,y,z,t) \in M \cap N$, just plug this into the equation $x+y+z+t = 0$ and it follows that the sum is $\frac{2}{\lambda} \not = 0$ and so the tangent spaces $T_pM, T_pN$ space $\mathbb{R}^4$ i.e $M,N$ are transverse and so $M \cap N = S$ is regular. 
A: It's enough to show that $S^n$ is a manifold. Use the stereographic projection with two poles. Please consider this picture.

