Why $\mathbb R$ with the chard $t\longmapsto t^3$ is not a submanifold of $\mathbb R^2$ ?

Question

I'm not very used to Manifold. I'm a little bit confuse with something.

1) Let $\phi:t\longmapsto t^3$ with $t\in \mathbb R$. So $(\mathbb R,\phi)$ is a smooth manifold. Now, why this is not a submanifold of $\mathbb R^2$ ? I don't really see the trick here.

2) In the french wikipedia they say that the Lemniscate is not a sub-manifold of $\mathbb R^2$. But can it even be a manifold ? Since no open that contain the intersection point will be euclidien. But if wikipedia specify that it's not a "sub-manifold" of $\mathbb R^2$, may be we can give it a structure of manifold.

This distinction of manifold and submanifold is not very clear for me.

Answer 1

1. I suppose that $(\mathbb R,\phi)$ is not a submanifold of $\mathbb R^2$ because it is not even a subset of $\mathbb R^2$.

2. The lemniscate, with the subspace topology, is not a manifold. With a different topology, it is a manifold. There is a natural continuous bijection $f$ from the $(0,1)$ to the lemniscate; giving the lemniscate the unique topology which makes $f$ a homeomorphism allows the lemniscate to have a manifold structure. The "X" point of the lemniscate now locally looks like a line, because the other two arms of the "X" are no longer "close."
   
   The reason they bring up the lemniscate is to illustrate that the image of an injective immersion is not always a manifold under the subspace topology, so there is a difference between embedded submanifolds and immersed submanifolds. Under the alternate topology in the previous paragraph, the lemniscate is an immersed submanifold, but not an embedded one.

Answer 2

This is more of a complement to Mike Earnest's answer. Considering your comment in his answer, we want to know if the differentiable manifold $\mathbb{R}\times \{0\}$ with differentiable structure generated by the global chart $\phi:\mathbb{R}\times\{0\} \to \mathbb{R}$ given by $\phi(t,0)=t^3$ is a submanifold of $\mathbb{R}^2$.

The answer is still no. To see this, first recall that if $M$ is a submanifold of $N$, then the inclusion of $M$ into $N$ is differentiable. We claim that the inclusion of $\mathbb{R} \times \{0\}$ into $\mathbb{R}^2$ is not differentiable, considering the usual structure on $\mathbb{R}^2$. Indeed, for any point $(t_0,0) \in \mathbb{R}\times \{0\}$, taking $\phi$ as a chart in the domain, and the identity as a chart on the codomain, if we represent the inclusion on these charts we obtain $$ Id \circ \iota \circ \phi^{-1}(t) = (\sqrt[3]{t},0) $$ which is not a differentiable map from $\mathbb{R}$ to $\mathbb{R}^2$ in the usual sense. It follows that the inclusion $\iota$ is not differentiable, so $\mathbb{R} \times \{0\}$ is not a submanifold of $\mathbb{R}^2$ with this exotic differentiable structure.

We may wonder however if there exists an embedding of $\mathbb{R} \times \{0\}$ (with this different structure) into $\mathbb{R}^2$. If my calculations are right, I think that $F(t,0) = (t^3,t^9)$ does the trick. To see this, let us show that the tangent vector $\partial/\partial \phi$ (derivation with respect to the chart $\phi$) gets mapped to a non-zero tangent vector of $\mathbb{R}^2$. Consider any point point $(t_0,0) \in \mathbb{R} \times \{0\}$ and the function $\pi_1 \in C^\infty(\mathbb{R}^2)$ given by projection onto the first factor. By the definition of the differential we have $$ dF_{(t_0,0)}\left(\frac{\partial}{\partial \phi}\Bigg\rvert_{(t_0,0)}\right)(\pi_1) = \frac{\partial(\pi_1 \circ F)}{\partial \phi}\Bigg\rvert_{(t_0,0)}, $$ and by the definition of differentiating with respect to a chart we have $$ \frac{\partial(\pi_1 \circ F)}{\partial \phi}\Bigg\rvert_{(t_0,0)} = \frac{d(\pi_1 \circ F \circ \phi^{-1})}{dt}\Bigg\rvert_{t_0^3}. $$ Notice however, that the function $\pi_1 \circ F \circ \phi^{-1}$ is the identity on $\mathbb{R}$, so the value of the derivative above is $1$. This shows that $dF_{(t_0,0)}$ sends $\partial/\partial\phi\rvert_{(t_0,0)}$ to a non-zero tangent vector of the image, so this differential is injective, and $F$ is then an immersion, therefore an embedding, since it is also an homeomorphism with its image.