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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!

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

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  • $\begingroup$ 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. $\endgroup$ – manifolded Feb 28 at 21:18
  • $\begingroup$ 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. $\endgroup$ – jmerry Feb 28 at 21:23
  • $\begingroup$ Understood, thanks! $\endgroup$ – manifolded Feb 28 at 21:27

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