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How can I prove that $M=\{(x,y)\in \mathbb{R}^2\ ; y=|x|\}$ is not an embedded smooth submanifold of $\mathbb{R}^2$?

Stefan Hamcke
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bigli
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    Take the definition of "embedded smooth submanifold", and show that it breaks at $(0,0)$. – Daniel Fischer Aug 08 '13 at 13:14
  • Please explain exactly. – bigli Aug 08 '13 at 13:19
  • Related: http://math.stackexchange.com/questions/462502/is-it-possible-to-make-the-set-m-x-y-in-mathbbr2y-x-into-a-differ?rq=1 and http://math.stackexchange.com/questions/184485/why-m-x-x-x-in-mathbbr-is-not-an-embedded-submanifold?rq=1 – Seirios Aug 08 '13 at 13:21
  • Without knowing with which definition you work, I can't explain exactly. Of course, the definitions are (except for the question whether "smooth" means $C^1$ or $C^\infty$) equivalent, but proving the equivalence would be more than good here. Basically, it always boils down to : show that $x\mapsto \lvert x\rvert$ isn't differentiable in $0$. – Daniel Fischer Aug 08 '13 at 13:22
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    One has to be a little careful here since $x\mapsto x^{1/3}$ isn't differentiable at 0, either, but the graph is a smooth submanifold. – Mikhail Katz Aug 08 '13 at 14:08
  • If it were an embedded submanifold (obviously connected), the dimension would have to be 1. But around $(0, 0)$, it cannot be locally diffeomorphic to $\mathbb{R}$, since all nonzero tangent vectors at a point in $\mathbb{R}$ are scalar multiples of one another, whereas this is clearly not the case for the graph. – user43208 Aug 08 '13 at 14:29
  • Here is a proof. – Georges Elencwajg Aug 09 '13 at 21:50

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Being an embedded smooth submanifold of codimension $k$ implies (either as the definition or a consequence of it) that every point $p\in M$ has a neighborhood $U$ such that $$M\cap U = \{x\in U: g_1(x)=\dots = g_k(x)=0\}$$ where the functions $g_1,\dots,g_k$ are smooth and their gradients $\nabla g_j(p)$ are linearly independent.

If your set $M$ was a smooth embedded submanifold, $\nabla g_1(0,0)$ would be orthogonal to both vectors $(1,1)$ and $(1,-1)$. Indeed, the directional derivatives $\lim_{h\to 0}\frac{1}{h}(g_1(\pm h,h)-g_1(0,0))$ are zero because $g_1$ vanishes on $M$. A nonzero vector in two dimensions can't be orthogonal to two linearly independent vectors.

By the way, $M$ is a $1$-dimensional embedded topological submanifold.