I'm self-studying differential geometry using Frankel's ``The Geometry of Physics".

The first problem (1.1(1)) is about determining whether or not the locus $$x^2+y^2-z^2 = c $$ is a submanifold in $\mathbb{R}^3$, for $c\in\{1,0,-1\}$.

My solution was based on an example earlier in the text - that of the unit sphere. We can describe points in each hemisphere as $$ z = \pm F(x,y) =\pm \sqrt{x^2+y^2-c},$$ and we can see that the function is differentiable everywhere for $c<0$, which means (I gather from the aforementioned example).

My question is what happens for $c\geq0$. I would think that this is no longer a submanifold since around the origin, there is a ``hole", and the coordinates are not well defined. Is this correct? Is there a more rigorous approach to generally show whether a locus is a submanifold or not?


You can use this following well know result.

Theorem. Let $X,Y$ be manifolds and $f:X\to Y$ be smooth. Let $y\in Y.$ If for every $x$ such that $f(x)=y$ the differential $df_x:T_x(X)\to T_y(Y)$ is surjective, then the preimage $f^{-1}(y)$ is a submanifold of $X$ with $\dim f^{-1}(y)=\dim X-\dim Y.$

Now, in your case $X=\mathbb{R}^3,$ $Y=\mathbb{R}$ and $f:\mathbb{R}^3\to \mathbb{R}$ is given by $$ f(x,y,z)=x^2+y^2-z^2.$$ If $a=(a_1,a_2,a_3)\in\mathbb{R}^3$ then $df_a:\mathbb{R}^3\to \mathbb{R}$ has matrix $(2a_1,2a_2,-2a_3),$ so $df_a$ is surjective unless $a_1=a_2=a_3=0.$ This shows that the locus is a submanifold of $\mathbb{R}^3$ with dimension 2 if $c\neq 0.$

If $c=0$ then the locus is a cone, with is not a manifold since the origin has no neighborhood homeomorphic to an open subset of $\mathbb{R}^2.$

  • $\begingroup$ Thanks for the answer! A few follow up questions: 1) If $a_1=a_2=a_3=0$ $df_a$ is not surjective because the matrix is singular? 2) The last statement means that each point on a manifold needs to be locally homeomorphic to an open subset of $\mathbb{R}^n$? $\endgroup$ – golanor Dec 17 '18 at 5:20
  • $\begingroup$ Also, if $c=0$ and we omit the origin, it becomes a submanifold, correct? $\endgroup$ – golanor Dec 17 '18 at 6:01
  • $\begingroup$ 1) Since $\dim\mathbb{R}=1$then $df_a$ is surjective iff $df_a\neq 0.$ 2) Yes, by definition every point in a manifold has a neighborhood homeomorphic to an open subset of $\mathbb{R}^n.$ Also note that Frankel's definition of a manifold is more precisely a "smooth manifold" or "differentiable manifold", not to be confused with a "topological" manifold (not necessarily smooth). Hence a (smooth) manifold of dimension 2 in $\mathbb{R}^3$ is just a surface "smooth" in the sense that there is a tangent plane at every point. 3) If you remove the origin it becomes a (smooth) submanifold, yes. $\endgroup$ – positrón0802 Dec 17 '18 at 16:43

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