$M=\{(x,y,z) \in \mathbb{R}^3 : xyz=C\}$ is a manifold $\Rightarrow C \neq 0$

Question

I'm having some troubles on showing that

$M=\{(x,y,z) \in \mathbb{R}^3 : xyz=C\}$ is a manifold $\Leftrightarrow C \neq 0$

I have already proved $\Leftarrow$ but I can't see how to prove $\Rightarrow$.

Suggestions?

Answer 1

Every manifold has at every point $p\in M$ a tangent space $T_pM$. In fact, $T_pM=\text{Im}(d_a\varphi)$ for any parametrization $\varphi:V\to \mathbb R^m$ defined on an open set $V\subset\mathbb R^d$, $d=\dim(M)$, with $a\in V$, $p=\varphi(a)$ and $\varphi(V)$ an open nbhd of $p$ in $M$. In other words $T_pM$ is the linear space generated by the columns of the jacobian matrix of $\varphi$. Equivalently, one shows that $T_pM$ consists of the tangent vectors at $p$ of curves in $M$ through $p$. The dimension of this linear tangent space coincides with the dimension of $M$, because $d_a\varphi$ is injective and the rank of its jacobian is $d$.

In our case, if $M:xyz=0$ were a smooth manifold it would have dimension $<3$, as such an $M$ has empty interior in $\mathbb R^3$, being a union of three planes. Then its tangent space $T_pM$ at the origin $p=(0,0,0)$ would be a linear subspace of $\mathbb R^3$ of dimension $<3$. But looking at curves in $M$ we have: (i) $\gamma_1(t)=(t,0,0)\in M$ has tangent $(1,0,0)\in T$, (ii) $\gamma_2(t)=(0,t,0)\in M$ has tangent $(0,1,0)\in T$, (iii) $\gamma_3(t)=(0,0,t)\in M$ has tangent $(0,0,1)\in T$. Thus $\dim(T)=3$, contradiction.

In fact, $M:xyz=0$ is not a topological manifold, but this requires algebraic topology. See the answer to this: Union of two self-intersecting planes is not a surface.