I've read some of the answers to related questions to this, but this is an idea I've been grappling with for a while and still can't fully get my head around.
$\mathbb{S}^1$ is a smooth manifold, and so we can induce a natural smooth structure on the boundary of the unit square, $\partial I$ by some homeomorphism (specifically I am thinking of $\frac{|| \cdot ||_\infty}{|| \cdot ||_2}$).
But then intuitively, the unit square should not be a smoothly embedded submanifold of $\mathbb{R}^2$ – so where exactly does this procedure fail in providing a suitable smooth structure on $\partial I$, which makes it as such?
By definition we need to provide an injective immersion which is also a topological embedding. But isn't the topology induced by the homoemorphism from $\mathbb{S}^1$ precisely the same as the subspace topology induced from $\mathbb{R}^2$? So the inclusion map is a topological embedding. It is obviously injective – so I surmise the immersion criteria must fail, but I cannot see (rigorously) why this is so! Thanks in advance for any help.
Best Answer
The map from $S^1$ to the boundary of the square fails to be differentiable at the four points $(\pi/4, 3\pi/4, 5\pi/4, 7\pi/4)$ that map to the corners of the square. You can check this: write down the $xy$-coordinates as a function of $\theta$, and try to take the derivative.
That doesn't mean that the square's not a smooth submanifold of $R^2$ -- only that this mapping isn't an immersion of the circle.
By "slowing down" as you approach each corner, you can in fact make a differentiable map from the circle onto the square --- one whose derivative is zero at the four points mentioned above --- and that's about as close to an immersion as you can get: a $c^\infty$ map that happens to have zeroes in its derivative here and there.