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Is the set $\left\lbrace (x,y) \in \mathbb{R}^2 \mid x^2 = y^2 \right\rbrace $ a manifold in $\mathbb{R}^2 $?

I know I could use the level set theorem if I had $\left\lbrace x \in \mathbb{R}^2 \mid x_1^2 = x_2^2=c \right\rbrace $, provided $ c \neq 0$;

but the set I have is in fact the union of all such sets along with the case where $c =0$.

Thanks!

Kerry H
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    Your zero set is the union of two lines in $\mathbb{R}^2$. What appens in a neighborhood of 0? Can this neighborhood be omeomorphic to an open subset of $\mathbb{R}$? – Sabino Di Trani Jan 03 '14 at 12:11

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Your set—let's call it $M$—is not a manifold. Since $x^2 = y^2$ if and only if $|x| = |y|$ if and only if $x = \pm y$, $$ M := \{(x,y) \in \mathbb{R}^2 \mid x^2 = y^2\} = \{(x,y) \in \mathbb{R}^2 \mid y = x\} \cup \{(x,y) \in \mathbb{R}^2 \mid y = -x\}, $$ i.e., your set is the union of the lines $y=x$ and $y=-x$. Away from the origin, $M$ is readily seen to be locally homeomorphic to $\mathbb{R}$, as you would expect for a $1$-manifold, e.g., for $x \neq 0$, $$ \phi_{x,\pm} : \left(-\tfrac{|x|}{2},\tfrac{|x|}{2} \right) \to M, \quad t \mapsto (x+t,\pm(x+t)) $$ defines a would-be coordinate chart at $(x,\pm x)$. However, for any neighbourhood $U$ of $(0,0)$ in $M$, $U$ cannot possibly be homeomorphic to some interval $(-\epsilon,\epsilon)$, for if $\phi : (-\epsilon,\epsilon) \to U$ were such a homemorphism, it would restrict to a homeomorphism $$ (-\epsilon,\epsilon) \setminus \{\phi^{-1}(0,0)\} \to U \setminus \{(0,0)\}, $$ where $(-\epsilon,\epsilon) \setminus \{\phi^{-1}(0,0)\}$ has two connected components, whilst $U \setminus \{(0,0)\}$ has at least four.

In terms of level sets, the key observation is that $M = f^{-1}(0)$ for $f(x,y) = x^2-y^2$, where $$\nabla f(x,y) = (2x,-2y)^T$$ and hence $\nabla f(0,0) = 0$, ruling out the use of the level set theorem. Indeed, this is an excellent illustration of why regularity ($\nabla f(x,y) \neq 0$ everywhere) is essential to get a manifold; things can go horribly wrong precisely where $\nabla f(x,y) = 0$, e.g., at $(x,y) = (0,0)$ in this example.

Branimir Ćaćić
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  • Okay great! What if my set was $\left\lbrace x \in \mathbb{R}^{2n} \mid x_1^2+ \cdots x_n^2= x_{n+1}^2 + \cdots + x_{2n}^2 \right\rbrace $. Will I then have a manifold in $\mathbb{R}^{2n}$? – Kerry H Jan 03 '14 at 12:33
  • Yeah but the level set theorem isn't an if and only if statement, right ? – Kerry H Jan 03 '14 at 13:50
  • Actually, you're absolutely right. For instance, the line $y = x$, which is a perfectly good submanifold of $\mathbb{R}^2$, can also be described as the curve $y^3 = x^3$, i.e., $f^{-1}(0)$ for $f(x,y) = x^3 - y^3$, which has a critical point at $(0,0)$. – Branimir Ćaćić Jan 03 '14 at 14:17
  • So is there a way to show that my set in the first comment is a manifold? – Kerry H Jan 03 '14 at 14:24
  • I still don't think it is. The fact that $f(x) = x_1^2 + \cdots + x_n^2 - x_{n+1}^2 - \cdots - x_{2n}^2$ fails to be regular at $0 \in f^{-1}(0)$ is highly suggestive, if nothing else, that something might go wrong. – Branimir Ćaćić Jan 03 '14 at 15:06