In our lectures we had this characterization of sub-manifolds:

$\forall a ∈ M: \,\,\exists R^n = E^k ⊕ E^{n−k}, \,\,\dim E^k = k,$

$\text{and open "surroundings" (I'm not English sorry)}\,\,U' ⊂ E^k,\,\, U'' ⊂ E^{n−k} \text{ such that } a ∈ U' × U'' \text{ and } ϕ ∈ C^α(U', U'') \text{ such that } M ∩ (U'× U'') = \{(x', x'') ∈ U' × U'' \,\,|\,\,x'' = ϕ(x' )\}$

If this is true, then $M$ is a sub-manifold of dimension $k$.

So I have to prove now that $M=\{(x,y,z)\in\mathbb{R} \,\,|\,\, xy-z^2=1, \, x+z=2\}$ is a submanifold of dimension 1.

My solution seems way too easy though:

Just use the whole room as the open surroundings (so $\mathbb{R}=U'$ and $\mathbb{R}^2 = U''$). Then I solve the equation system $xy-z^2=1, \, x+z=2$ this way:

$$z = 2-x, \,\,\,\,\,\,\,\,\,\,\,\,y = \frac{(2-x)^2}{x}$$

And I define $ϕ$ as: $(x,y,z) \mapsto (x,\frac{(2-x)^2}{x}, 2-x)$. Obiously: $ϕ \in C^{oo}$

Also: $M = \{(x', x'') ∈ \mathbb{R} × \mathbb{R}^2 \,\,|\,\,x'' = ϕ(x' )\}$

So this is all we needed, isn't it?!

  • $\begingroup$ what happens if $x=0$? $\endgroup$
    – quarague
    Commented Aug 3, 2021 at 17:04
  • $\begingroup$ I didn't think of this, my bad. But it shouldn't affect the solution since that is not possible: Then we'd have $-z^2=1$, so $z=0$ but also $z=2$. $\endgroup$ Commented Aug 3, 2021 at 17:19
  • $\begingroup$ Better make the domain of $\pi$ to be $\mathbb{R}\backslash\{0\}$. Then the the set is the graph of the function $\phi \colon \mathbb{R}\backslash\{0\} \to \mathbb{R}^2$, so a manifold. Apriori it may be not closed, but then the definition makes it so. It is diffeo to a hyperbola. $\endgroup$
    – orangeskid
    Commented Aug 4, 2021 at 0:19

1 Answer 1


The gradient of the first function is $(y, x, -2 z),$ and of the second $1, 0, 1)$ The only way your space can fail to be a manifold is if these two are linearly dependent. This can happen if

(a) $x=y=z=0$ or

(b) $y=2z, x=0.$

(a) cannot occur, and for (b) $x=0$ implies $z = 2$, implies $y=4,$ but that point is not in your solution set.

So, you are golden.


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