For what values of $(a,b,c)$ is $M$ a manifold?

Question

Let $M$ denote the subset of $\mathbb{R}^3$ defined by the equations:

$x^2 +y^2 + z^2 = 1$

$ax^2 + by^2 + cz^2 = 0$

Where $a, b,$ and $c$ are constants. For what values of $(a,b,c)$ is $M$ a manifold? Is $M$ compact?

Here's my attempt at a solution:

Define $(u,v)$ = $F(x,y,z)$ = $(x^2 +y^2 + z^2 - 1, ax^2 + by^2 + cz^2)$

The Jacobian is computed to be:

$\begin{bmatrix}2x & 2y & 2z\\2ax & 2by & 2cz\end{bmatrix}$

Solving for when the $2 \times 2$ minor matrices are zero we get:

$4bxy - 4axy = 0$

$4cxz - 4axz = 0$

$4cyz - 4byz = 0$

Since the point $(x,y,z) = (0,0,0)$ is not in the solution set to the original given equations, we can divide by $xy, xz, yz$ respectively.

This results in $a = b =c$.

So my answer is, so long as $a \neq b \neq c$, $M$ is a manifold.

Is this reasoning correct? And how can I show that $M$ is compact? Obviously the unit sphere is compact, but what about it's intersection with $ax^2 + by^2 + cz^2 = 0$ ?

Can somebody please help? Thank you kindly.

Answer 1

Since I don't know how to do the "manifold" part yet, here's how to answer the "compact" part. Let $A$ be the unit sphere, $B=\{(x,y,z): ax^2+by^2 + cz^2=0\}.$ Then $M= A\cap B.$ Now $B=f^{-1}(\{0\}),$ where $f(x,y,z) = ax^2+by^2 + cz^2.$ Since $f:\mathbb R^3\to \mathbb R$ is continuous, $B$ is the inverse image of the closed set $\{0\},$ hence is closed. Thus $M$ is the intersection of the compact set $A$ with a closed set. Somewhere you learned that such a thing is compact. So $M,$ whether or not it is a manifold, is compact.

Answer 2

In this case it seems easier to do it directly:

First of all, the solution set is a manifold if

• $a=b=c=0$ (it's the sphere)

• $a=b=0$ and $c\neq 0$ (it's the unit circle in the $x-y$ plane).

When $a=0$ and $b, c\neq 0$. First of all they have to be of different sign, or the solution set is empty. Now assume $b>0$ and $c<0$ and write

$$ by^2 + cz^2 = by^2 - |c| z^2 = (\sqrt b y - \sqrt{|c|} z)(\sqrt b y + \sqrt{|c|} z).$$

Thus the solution set is a union of two circles with intersection. Thus it is not a manifold.

Lastly, assume all three are nonzero. First of all $a, b, c$ cannot be of the same sign, or the solution set is empty.

Assume $a>0$ and $b, c, <0$. Dividing by $a$ if necessary, we assume that $a=1$. Then

$$x^2 = |b|y^2 + |c| z^2.$$

Put into the first equation gives

$$ (1+|b|)y^2 + (1+ |c|)z^2 = 1, x = \pm \sqrt{1-(1+|b|)y^2 + (1+ |c|)z^2}$$

This means that under the projection $(x, y, z) \mapsto (x, y)$ the solution curves projected to the ellipse. Since $b, c$ are nonzero, $x$ is nonzero. Thus the projection, while restricted to $x>0$ (or $x<0$) is a diffeomorphism onto its image. This shows that the solution curves is a manifold and is diffeomorphic to two disjoint ellipse.