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The problem

As a continuation of this question (where it was shown that $C$ was a closed $1$-dimensional submanifold for $c \neq 1/27$), I'm trying to find out whether or not $$C = \{(x,y) \mid x^3 + xy + y^3 = c \} \subset \mathbb{R}^2$$ is an embedded submanifold of $\mathbb{R}^2$ for $c = 1/27$.

What I have so far

I'm using the definitions from Spivak's A Comprehensive Introduction to Differential Geometry I:

-immersion: a differentiable function $f:M \rightarrow N$ s.t. $\text{rank } f = \dim M$, at all points of $M$,

-immersed submanifold: a subset $M_1$ of $M$ with a differentiable structure s.t. the inclusion map $i: M_1 \hookrightarrow M$ is an immersion,

-embedding: an injective immersion $f$ that is a homeomorphism onto its image,

-submanifold: an immersed submanifold $M_1 \subset M$ s.t. the inclusion map $i: M_1 \hookrightarrow M$ is an embedding.

So for $C$ to be an embedded submanifold, I take it it has to be a submanifold and it has to be an embedding.

My question is just regarding how to proceed in order to prove or disprove the above problem:

Should I consider the inclusion map $i: C \hookrightarrow \mathbb{R}^2$ and try to figure out whether or not this map is an immersion and an embedding?

And if it is, then try to figure out whether or not $C$ is an immersed submanifold (which would mean that it is a submanifold)?

Any help or hints on how to proceed is appreciated!

Maethor
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1 Answers1

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$\newcommand{\RR}{{\mathbb R}}$

If you prove that the inclusion $i:C \to \RR^2$ is

1) an immersion

2) an embedding

then $C$ is by 1) an immersed submanifold and therefore by 2) a submanifold. (The map $f$ here is always $i$).

In fact things are special here: There is a factorization:

$$x^3+x y + y^3 - 1/27 = 1/27\, \left( 3\,x-1+3\,y \right) \left( 9\,{x}^{2}+3\,x-9\,xy+3\,y+1 +9\,{y}^{2} \right)$$

If you call the rightmost factor $g$ you can put it by an affine change of coordinates $x,y$ into the form $x^2+y^2$, so $g=0$ has a single real point, namely $x=-1/3, y= -1/3$.

The affine transformation is given concretely by $$ \begin{align} x & =2/9\,\sqrt {3} \, y_1 - 1/3 \\ y & =1/3\,x_1+1/9\,\sqrt {3} \, y_1-1/3 \end{align} $$

That means, that as a set $C=\{(x,y)\mid 3 x + 3 y - 1 = 0\} \cup \{(-1/3,-1/3)\}$, a line in $\RR^2$ plus a point. So $C$ consists of two distinct components each of which is a submanifold, one of dimension $1$ and the other of dimension $0$. Adhering strictly to your definitions both together can not be called a submanifold, indeed $i: C \to \RR^2$ even fails to be an immersion.

  • Thanks for your answer! I'm with you until (and including) the factorization of $x^3 + xy + y^3 - 1/27$. But I don't see how you change coordinates to get $g$ on the form $x^2 + y^2$. I get that $g$ can be written as $(3x + 1)^2 + (3y + 1)^2 - (3x + 1)(3y + 1)$, but that doesn't help that much... Any hints? – Maethor Apr 29 '15 at 11:20
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    I added the transformation to my answer. In fact you were quite near with your regrouping of $g$: If $u= 3x +1$ and $v = 3y +1$ then you have $g = u^2 + v^2 - u v$ and this goes under $u = u - v/\sqrt{3}$ and $v = u + v/ \sqrt{3}$ into $u^2+v^2$. – Jürgen Böhm Apr 29 '15 at 14:19
  • Thanks! So the inclusion map $i: C \rightarrow \mathbb{R}^2$ fails to be an immersion since immersions are only defined for manifolds that have constant dimension? – Maethor Apr 29 '15 at 15:05
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    Yes, at least this is my understanding of the definition from Spivak. Another definition, namely $f:M \to N$ is immersive at $x \in M$ when $T_x f: T_x M \to T_{f(x)} N$ is injective and $f$ is an immersion if it is everywhere immersive would make $i$ an immersion here. – Jürgen Böhm Apr 29 '15 at 15:13
  • Yes, in Spivak it says that "a function $f: M^n \rightarrow N^m$ is an immersion ...", so using his definitions it seems that $M$ must have constant dimension for $f$ to be an immersion. Thanks for your help! P.S.: Could you have a look at another of my unanswered questions? http://math.stackexchange.com/questions/1237094/vector-fields-that-are-smooth-on-the-open-unit-ball-b-are-smooth-on-mathbb?lq=1 – Maethor Apr 29 '15 at 16:06