Show that the intersection of two objects is a manifold with boundary

Question

The question is to find $a$ for which the intersection of the solid hyperboloid $x^2+y^2-z^2\leq a$ with $x^2+y^2+z^2 = 1$ is a manifold with boundary.

My attempt: Let $I$ be the intersection. Boundary of $I$, $\partial I$, is $x^2+y^2=\frac{a+1}{2}, z^2=\frac{1-a}{2}$. Can we claim that if we can prove $\partial I$ and $I^{\circ}$ are manifolds where $dim(\partial I)=1$ and that $\partial I$ is closed, we are done?

$\partial I$ is clearly closed and since $x^2+y^2-z^2 = a$ and $x^2+y^2+z^2=1$ are manifolds, $\partial I$ is a manifold too which is a disjoint union of two manifolds corresponding to $x^2+y^2=\frac{a+1}{2}$ and $z^2=\frac{1-a}{2}$. Thus $a \in [-1,1]$, am I correct? But I'm not sure how to show $I^{\circ} = \{(x,y,z)|x^2+y^2-z^2< a\}$ is a manifold? Thanks and appreciate a hint!

Answer 1

But I'm not sure how to show $I^\circ=\{(x,y,z) | x^2+y^2-z^2 < a\}$ is a manifold?

It's an open set in $\mathbb{R}^3$. Any open set in $\mathbb{R}^n$ is an $n$-manifold.
But then, that's not the intersection - we need to restrict the points to be on the sphere if we want the intersection.

The regions of interest here:

• If $a<-1$, the intersection is empty.
• If $a=-1$, the intersection is a pair of points $(0,0,\pm 1)$. We could call that a $0$-manifold, if we felt like it.
• If $-1<a<1$, the intersection is a pair of closed spherical caps, or equivalently a sphere with an open equatorial band removed. That's a $2$-manifold with boundary.
• If $a\ge 1$, the intersection is a full sphere - a $2$-manifold.

Does the full sphere count as a manifold with boundary? That depends on whether you allow an empty boundary in the definitions.