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I am using the definition in "Introduction to Smooth Manifolds" by Lee

Defn: Suppose $M$ is a smooth manifold. An embedded smooth manifold of $M$ is a subset $S\subset M$ which is a manifold in the subspace topology, endowed with a smooth structure with respect to which the inclusion map $i:S\to M$ is a smooth embedding.

My question has to do with showing that {$(x,y):x=|y|=\sqrt{y^2} $} is not a smooth submanifold of $R^2$, but more specifically to do with the inclusion map. I suppose a second part to this question is if I am thinking about this the correct way, or if am I completely wrong.

I want to show that the inclusion map isn't a smooth embedding (hopefully this is the correct approach).

I am wondering if I can define $i:S\to M$ by $$i(x,y)=(x,y)=(|y|,y)$$ so that my derivative is $$(1,\frac{y}{\sqrt{y^2}});(0,1)$$

My conclusion is that this doesn't exist at $(0,0)$ so the inclusion map isn't a smooth embedding.

(I don't know how to make a matrix on here, so first set of parenthesis is first row of Jacobian, second set is second row )

I understand that this question is similar to some from the past, by I haven't found any regarding how to define the inclusion map to break the definition.

Thanks!

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  • $\begingroup$ For the inclusion map to be a smooth embedding you have to first decide what smooth structure you are going to put on $S$, for otherwise being a smooth embedding simply does not mean anything. $\endgroup$ – Mariano Suárez-Álvarez Feb 12 '15 at 20:13
  • $\begingroup$ on the other hand, you cannot define $i$ to be anything: when you decided that $i$ is the inclusion mapping, then that completely determined it and there is nothing else to be defined. $\endgroup$ – Mariano Suárez-Álvarez Feb 12 '15 at 20:30
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So one thing to start with here regarding the 2x2 Jacobian you have, if $S$ is a submanifold, it has to be a 1-manifold. You can see that by noting that the sets $S^+ = \{(y,y) | y > 0\}$ and $S^- = \{(-y,y) | y < 0\}$ (the two rays that make up the V-shape of S) individually are embedded submanifolds of $\mathbb{R}^2$ and they're each embedded 1-submanifolds of $\mathbb{R}^2$.

If $\iota$ was a smooth embedding, its restriction to $S^+$ would be a smooth embedding. We can show an example of $S^+$ being a smooth embedded 1-submanifold, and the smooth structure that makes $\left.\iota\right|_{S^+}$ a smooth embedding is unique, so that means any smooth structure we hook up on $S$ has to be that of a 1-manifold when restricted to $S^+$, so $S$ itself has to be a 1-manifold.

That means we can pick a chart around the point $(0,0)$, $\phi : U \rightarrow \mathbb{R}$. Shrinking $U$ if necessary, and translating, we can say $\phi(U) = (-\epsilon,\epsilon)$ for some $\epsilon > 0$ and $\phi(0) = (0,0)$ (shrinking to a coordinate ball centered at $(0,0)$). $\phi$ is a diffeomorphism and $\iota$ is a smooth immersion, so $g = \iota \circ \phi^{-1} : (-\epsilon,\epsilon) \rightarrow \mathbb{R}^2$ is a smooth immersion of $(-\epsilon,\epsilon)$ into $\mathbb{R}^2$. We can write it in coordinates as $g(t) = (g_x(t),g_y(t))$, swapping $g(t)$ with $g(-t)$ if necessary, and noting the shape of $S$, we really have:

$$ g(t) = \left\{\begin{aligned} &(g_y(t),g_y(t)) && t < 0\\ &(-g_y(t),g_y(t)) && t > 0 \\ & (0,0) && t = 0 \end{aligned} \right.$$

We can also evaluate the tangent vector of this away from $t=0$: $$ g'(t) = \left\{\begin{aligned} &g_y'(t)(1,1) && t < 0\\ &g_y'(t)(-1,1) && t > 0 \\ \end{aligned} \right.$$

Now the only way that can be smooth is if $g_y'(t)$ smoothly goes to zero at $t = 0$, and takes the value $g_y'(0) = 0$.

If it does that, since $|g_x'(t)| = |g_y'(t)|$ near $t=0$, the whole tangent vector to the curve, $g'(t)$ must smoothly go to zero. That's not possible though, because $g$ is a smooth immersion, so it has an injective pushforward. $g'(t)$, the tangent vector, is related to the pushforward of $g$ by $g_*(\left.\frac{d}{dt}\right|_0) = g'(0) = 0$, but $\frac{d}{dt}$ isn't zero so that's not injective. That means $\iota$ can't be a smooth embedding.

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  • $\begingroup$ @Mariano Suárez-Alvarez I don't have enough points to put this right below what Mariano was saying :). I think if we want to show S is not a submanifold we need to show that i is not a smooth embedding for any smooth structure on S. The definition he's using says the smooth structure is part of the data that makes S a submanifold (It turns out it's unique if it exists, but that's a theorem in Lee). $\endgroup$ – mmcfa Feb 12 '15 at 21:56

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