How to show that M is a two-dimensional submanifold

smooth-manifoldssubmanifold

I have the following problem:

Let $M=\{(a,b,c,d):ac=b^2, bd=c^3,ad=bc,abcd\neq 0\}$. Show that M is a two dimensional submanifold.

So we had the following definition about submanifolds:

Def: $M\subset \mathbb{R}^n$ is submanifold of dimension k if for all $a\in M$ one of the following three equivalent points holds:

  1. For each $a\in M$ there exists an open neighbourhood $U\subset \mathbb{R}^n$ and a diffeomorphismen $\phi:U\rightarrow V$ where V is open in $\mathbb{R}^n$ such that $$\phi(U\cap M)=V\cap(\mathbb{R}^k\times \{0\})$$
  2. For each $a\in M$ there exists an open Neighbourhood $U\subset \mathbb{R}^n$ and a submersion $F:U\rightarrow \mathbb{r}^{n-k}$ at a such that $$U\cap M=U\cap\{F=0\}$$
  3. For each $a\in M$ there exists an open $O\subset M$ and and open $O'\subset \mathbb{R}^k$ and a homeomorphism $\phi:O'\rightarrow O$ such that $\phi$ is a immersion at $\phi^{-1}(a)$ and $O=\phi^{-1}(O)$

But I somhow don't understand how to work with them, could someone help me please? I am at this exercise for too long.
Thank you very much.

Best Answer

The condition $abcd \neq 0$ implies that $a, b, c, d$ are all non-zero. In particular,

$$ d =\frac{bc}{a}$$ Plug it into $bd = c^3$, we have $$ \frac{b^2 c}{a} = c^3\Rightarrow ac^2 = b^2$$ Since $ac = b^2$, it implies that $c = 1$. Thus $a = b^2, d = 1/b$. Hence

$$M = \{ (b^2, b, 1, b^{-1}) : b\in \mathbb R\setminus \{0\}\}$$

and indeed $M$ is one-dimensional.

With that in mind, one consider

$$ F : U \to \mathbb R^3, F(a, b, c, d) = (a-b^2, c-1, d-b^{-1}),$$ where $U = \{ (a, b, c, d) : abcd\neq 0\}$. It is clear that $F^{-1}(0,0,0) = M$ and

$$DF = \begin{bmatrix} 1 & -2b & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & b^{-2} & 0 & 1\end{bmatrix}$$ clearly has full rank (consider the first, third and fourth column). Thus definition 2, $M$ is a one dimensional submanifold in $\mathbb R^4$.

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