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Note: below the word embedding is supposed to also be an immersion as in its differential topology definition.

Question:

$$ f : \mathbb{R}^2 \rightarrow \mathbb{R}, f(x,y) = x^3 + xy + y^3 + 1 $$

for which points among $p1 = (0; 0), p2 = (1/3; 1/3), p3 = (-1/3; -1/3)$, is $s_i = f^{-1}(f(p_i))$ an embedding in $\mathbb{R}^2$ (using inclusion) ?

What I have so far:

  • $s_1$: $f(0,0)=1$, hence the set $s_1$ is the subset of $\mathbb{R}^2$ such that $x^3 + xy + y^3 = 0$.
  • $s_2$: $f(\frac{1}{3},\frac{1}{3})=\frac{32}{27}$, hence the set $s_2$ is the subset of $\mathbb{R}^2$ such that $x^3 + xy + y^3 = \frac{5}{27}$.
  • $s_3$: $f(\frac{-1}{3},\frac{-1}{3})=\frac{28}{27}$, hence the set $s_3$ is the subset of $\mathbb{R}^2$ such that $x^3 + xy + y^3 = \frac{1}{27}$.

  • the inclusion map is

    $$ i : s_i \rightarrow \mathbb{R}^2, i(s) = (x,y) $$ where $(x,y)$ are simply the canonical coordinates of $\mathbb{R}^2$

the curves

At this stage, I understand I have to prove something like "$i$ is a homeomorphism with no singularity for $s_i$ to be an embedding". But I'm not sure what this exactly mean in the context of this function $i$, what shall I prove ?

  • It seems to me that a subset of $\mathbb{R}^2$ is always homeomorphic to its image by the inclusion map.
  • How to choose a topology on $s_i$?
user2346536
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1 Answers1

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Partial answer: The first case ($S_1$) is not an embedded submanifold. $s_2$ is an embedded submanifolds. (Only in immersed manifolds you can change the topology).

Proof:

$s_1:$

By definition of embedded submanifold, you do not have freedom to choose the topology. It must be the subspace topology inherited from $\mathbb{R}^2$.

Thus (assuming your pictures are "correct"), then the case of the left picture is not an embedding:

You can see that $s_1$ (with the subspace topology) is not a (topological) manifold at all. Look at the intersection point. Take small enough ball around it. It's intersection with $s_1$ will look like two lines crossing each other. Such a thing cannot be homeomorphic to an open interval of $\mathbb{R}$. This can be seen by considering connected components. If you take out the intersection point, you get $4$ connected components instead of the $2$ you were supposed to get if this was homeomorphic to $\mathbb{R}$.

$s_2:$

Look at the differential of $f$. $df_{(x,y)}=(3x^2+y,3y^2+x)$ so $df_{(x,y)} = 0 \iff x = -3y^2 \, , \, y = -3x^2$.

This implies $x=-27x^4$ which forces $x=0$ or $x^3=-\frac{1}{27} \Rightarrow x = -\frac{1}{3}$.

So $df_{(x,y)} = 0 \iff (x,y)=(0,0) \text{ or } (x,y)=(-\frac{1}{3},-\frac{1}{3})$

So define $M = \mathbb{R}^2 \setminus \{(0,0),(-\frac{1}{3},-\frac{1}{3})\}$. This is an open submanifold of $\mathbb{R}^2$, so in particular a smooth submanifold in it's own right.

Now look at your function $f$ restricted to $M$:

$f:M \to \mathbb{R}$.

By what we have shown $df_p \neq 0$ for all $p \in M$ so $f$. Since the target manifold is one-dimensional, this shows $f$ is a submersion.

Now according to the theorem which says inverse images of submersions are embedded submanifolds*, it follows that $s_2 = f^{-1}(\frac{32}{27})$ is an embedded submanifold, as required.

Note that by restricting the domain to $M$ we didn't lose any points in the inverse image which intersts us, since $f(0,0),f(-\frac{1}{3},-\frac{1}{3}) \neq f(\frac{1}{3},\frac{1}{3})$

About $s_3$ I am not sure.

Asaf Shachar
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