What is a compact 2-D submanifold?

Question

Show that the qeuations

$x^3 + y^3 + z^3 + w^3 = 1$

$x^2 +y^2 + z^2 +w^2 =4$

define a compact 2-dimensional submanifold of $\mathbb{R}^4$. Write the equations for its tangent space at a point $(x_0,y_0,z_0,w_0)$.

I'm not sure what a 2-dimensional submanifold is. The second equation looks like it would be a parabaloid of some sort which would be 3 dimensional right?

I'm a little confused as to what this question is asking for. Can anybody help clarify?

Answer 1

Here is one of several equivalent definitions of a submanifold:

Definition. A smooth map $f:M\to N$ between manifolds is an embedding if it is a diffeomorphism onto its range. The range, $f(M)$, of such a map is called a submanifold of $N$.

The dimension of a submanifold is its dimension as a manifold.

If $f:M\to N$ is a smooth map between manifolds, we say $x\in M$ is a regular point if $d_x f$ is surjective. A point $y\in N$ is a regular value if each point of its preimage, $f^{-1}(y)$, is a regular point.

Theorem. If $f:M^m\to N^n$ is a smooth map between manifolds $M$ and $N$ of dimensions $m$ and $n$ respectively, and $y\in f(M)\subseteq N$ is a regular value, then $f^{-1}(y)$ is a submanifold of $M$ of dimension $m-n$.

We will use this theorem to prove that your equations define a $2$-dimensional submanifold of $\mathbb{R}^4$.

Another helpful result is that with the hypotheses of the above theorem, $T_x(f^{-1}(y))=\ker d_xf$.

Define $f:\mathbb{R}^4\to\mathbb{R}^2$ by $$f(x,y,z,w)=(x^3+y^3+z^3+w^3,x^2+y^2+z^2+w^2)\,.$$ Then the Jacobian matrix of $f$ is $$Jf(x,y,z,w)=\left(\begin{array}{cc} 3x^2 & 3y^2 & 3z^2 & 3w^2 \\ 2x & 2y & 2z &2w \\ \end{array}\right)\,.$$ Clearly $Jf(x,y,z,w)=\mathbf{0}$ if and only if $(x,y,z,w)=(0,0,0,0)$. So $(1,4)$ is a regular value of $f$. By the regular value theorem, $f^{-1}(1,4)$ is a regular submanifold of $\mathbb{R}^4$ of dimension $2$.

A manifold is compact if it is a compact topological space.

Answer 2

A set $S\subset{\mathbb R}^4$ is a $2$-dimensional submanifold of ${\mathbb R}^4$ if for any point $p\in S$ there is a $4$-dimensional window $W=U\times V$ with center $p$ such that $S\cap W$ (the part of $S$ lying in $W$) can be written as graph of a differentiable function $(f,g):\>U\to V$ in the form $$(x,y)\mapsto\bigl(x,y,f(x,y),g(x,y)\bigr)\ ,$$ maybe with another choice of the two independent variables.

In the case at hand the set $S$ is the intersection of a cubic hypersurface with a hypersphere of radius $2$. This set $S$ is closed and bounded, hence compact. In order to be sure that $S$ is a $2$-dimensional submanifold of ${\mathbb R}^4$ we have to invoke the implicit function theorem. It says the following: If the Jacobian $$J(x,y,z,w):=\left[\matrix{3x^2&3y^2&3z^2&3w^2 \cr 2x&2y&2z&2w\cr}\right]$$ has rank $2$ in all points $p\in S$ then the two hypersurfaces intersect transversally at all points, and each point $p\in S$ is center of a window as described above.

Therefore we have to check whether it is possible that all six $(2\times2)$-subdeterminants $6xy(y-x)$, $\ \ldots\>$, are zero in a point $p\in S$. This is somewhat tedious, and I leave that to you for the moment.

Answer 3

The jacobian of the map $$(x,y,z,t)\overset{\phi}{\mapsto}( x^2+ y^2 + z^2 + t^2, x^3 + y^3 + y^3 + z^3)$$ is $$\left( \begin{matrix} 2x & 2y & 2z & 2t\\3x^2 & 3y^2& 3z^2 &3t^2 \end{matrix} \right)$$ The jacobian has rank $\le 1$ if and only if its rows are proportional. It is easy to see that is equivalent to : all of the non-zero components of $(x,y,z,t)$ are equal. If for such an $(x_0,y_0,z_0,t_0)$ we had $\phi(x_0,y_0,z_0,t_0)=(4,1)$, then $k a^2 = 4$ and $k a^3= 1$ for some $k\in \{1,2,3,4\}$, and so $a = \frac{1}{4}$, contradiction.

Therefore, every $(x_0,y_0, z_0, t_0)$ in $\phi^{-1}(4,1)$ has two non-zero and non-equal components. Say they are $y_0$, $t_0$. Around this point the set $\phi^{-1}(4,1)$ can be written as the graph of a function $(x,y, f(x,y), g(x,y))$