Is this a sub-manifold? $M=\{(x,y,z)\in\mathbb{R} \,\,|\,\, xy-z^2=1, \, x+z=2\}$

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?!

• what happens if $x=0$? Commented Aug 3, 2021 at 17:04
• 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 $z=2$. Commented Aug 3, 2021 at 17:19
• 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. Commented Aug 4, 2021 at 0:19

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.