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

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!

• You reference $x^2+y^2-z^2=1$ and $z^2+y^2+z^2=1$ at different times - which is it? – jmerry Feb 28 at 20:28
• Making the edit, sorry. – manifolded Feb 28 at 20:35

## 1 Answer

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 $$-1, the intersection is a pair of closed spherical caps, or equivalently a sphere with an open equatorial band removed. That's a $$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.

• Can you please explain how you got the intersection and the manifolds for $-1<a<1$ and $a\geq 1$? I am looking at $x^2+y^2\leq \frac{a+1}{2}$ and $z^2\geq \frac{1-a}{2}$ for the intersection. – manifolded Feb 28 at 21:18
• The sets I'm looking at are the intersection of the sphere $x^2+y^2+z^2=1$ with the set $x^2+y^2-z^2\le a$. We have those two inequalities you wrote, but then we find the points on the sphere that satisfy them. Requiring $z^2>\frac{1-a}{2}$ means we avoid a certain band around the equator. and the $x^2+y^2$ inequality is the same restriction. – jmerry Feb 28 at 21:23
• Understood, thanks! – manifolded Feb 28 at 21:27