# No diffeomorphism that takes unit circle to unit square

My first time posting in this forum. This is not a homework problem. I am trying to learn my own from John M. Lee Introdcution to Smooth Manifolds.

In Chapter 3, there is the problem 3-4

Let $C \subset \mathbb{R}^2$ be the unit circle, and let $S \subset \mathbb{R}^2$ be the boundary of the square of side 2 centred at orign: $S= \lbrace (x,y) \colon \max(|x|,|y|)=1 \rbrace.$ Show that there is a homeomorphism $F:\mathbb{R}^2 \to \mathbb{R}^2$ such that $F(C)=S$, buth there is no diffeomorhpism with the same property. [Hint: Consider what $F$ does to the tangent vector to a suitable curve in C].

I can construct a homeomorphism (by placing the circle inside the square and then every radial line intersects the square at exactly one point.) But, I dont know how to do the rest of the problem or understand the hint.

I do not know how to write out what tangent space should be for the square. If there were a diffeomorphism than $F_\star$ is isomorphism between any two tangent space. If I show that the tangent space on the corner of square has dimension zero, would it solve problem?

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Have you proved that $S$ is not a smooth submanifold of $\mathbb R^2$ yet? If so, you could use that as a lemma. Argue that if there was such a diffeomorphism $F$, then $S$ would neccessarily be a smooth submanifold of $\mathbb R^2$. – Ryan Budney Jul 27 '11 at 21:24
And if you haven't proven that yet, I think that would be a great way to solve your problem. I like to make this argument using hypothetical charts for $S$ and an application of the Implicit Function Theorem. – Ryan Budney Jul 27 '11 at 21:26
@Ryan Budney -- Thank you for commenting. No, I have not learned submanifolds yet. It is chapter 8 of the book. Does the tangent space at the corner have zero dimensions? I do not know how to prove it. How can I solve this problem with the things I know (defintion of smooth manifolds, maps and tangent space)? Thank you very much for help – Udayana Jul 27 '11 at 21:57
What definition of "tangent space" are you using? – Ryan Budney Jul 27 '11 at 22:43
Show that $K$ and $\mathbb{S}^1$ are isomorphic in $\textbf{Top}$, but $K \not\in ob(\textbf{Diff})$. – user1876508 Apr 4 '14 at 21:01

The original statement above is NOT true. That is a square CAN be given be a differentiable structure!

This follows from the more general theorem

Let M and N be topological manifolds that are homeomorphic with homeomorphism h. If N is also a smooth(differentiable) manifold then a differentiable structure can be defined on M via the pullback defined by h.

And with this differentiable structure on M, h becomes a diffeomorphism from M to N. That M and N are diffeomorphic.

So the homeomorphism given above between the square and the circle can be used to pullback the usual differentiable structure on the circle to the square and with differentiable structure the homeomorphism becomes a diffeomorphism.

Note that this induced differentiable structure on the square is NOT compatible with the usual differentiable structure on R2=R×R. That the inclusion map from the square to R2(identity map restricted to the square) will NOT be differentiable!

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Why is this downvoted? It is a correct statement that the square carries a differentiable structure as constructed in the above comment. I also interpret the title of this post as an incorrect statement. However, if "unit circle" and "unit square" mean being equipped with (topological) embeddings into $\mathbb{R}^2$, then since the square is not smoothly embedded by its set-theoretic identity, the statement is vacuous. So the correct framing of the problem is that as given in the book (and in the body of the poster's question). But there is good content in this response. – John Samples Mar 4 at 6:48

Here's a somewhat rigorous way to see this. Let $\gamma$ be an arc in $C$ such that $F\circ \gamma(0)$ is the corner (1,1). Then (assuming it goes clockwise) there are some functions $x$ and $y$ such that $F \circ \gamma(t) = (1,y(t))$ for $t< 0$ and $F \circ \gamma(t) = (x(t),1)$ for $t > 0$. Thus $F_*\gamma'(t)$ is $(0,y'(t))$ for $t<0$ and $(x'(t),0)$ for $t > 0$. Taking limits, this means that $F_*\gamma'(0) = 0$ contradicting that $\gamma'(0) \ne 0$.

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A diffeomorphism would, among other things, induce an isomorphism between the respective tangent spaces. But look at the corners of the square, i.e., the points {(1,1),(1,0),(0,1),(0,0)} ( in the right coordinate system), and see what happens with the tangent space there. More specifically, the tangent space at the corners is not defined, but it must be the image of the tangent space of some point on the circle where the tangent space is defined. Basically, in the 1-D case, the tangent space ( in one of its versions) is given in terms of the derivative f'(t)dt, but the derivative is not defined at the corners.

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