Showing that a locus is a sub-manifold

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?

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 $$\mathbb{R}^2.$$
• 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 $\mathbb{R}^n$? – golanor Dec 17 '18 at 5:20
• Also, if $c=0$ and we omit the origin, it becomes a submanifold, correct? – golanor Dec 17 '18 at 6:01
• 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. – positrón0802 Dec 17 '18 at 16:43