Does the existence of exotic smooth structure in $\mathbb{R}^4$ imply the existence of an atlas which has a $C^0$ mapping to the Cartesian atlas, but not a $C^k$ mapping (for some finite $k$)? Does the nonexistence of exotic smooth structure in $\mathbb{R}^n$, $n\neq 4$ imply that all atlases therein have smooth mappings to the Cartesian atlas?


Regarding your 1st question, perhaps you meant to ask something else? Any atlas can be composed with a non-smooth homeomorphism to produce an atlas that isn't smooth in the standard sense. For example, $\mathbb R \to \mathbb R$ defined by $t \longmapsto t^{1/3}$ is an atlas on $\mathbb R$ but it's not $C^1$. This answers your 2nd question in the negative.

Alternatively, some exotic smooth $\mathbb R^4$'s are diffeomorphic to open subsets of the standard $\mathbb R^4$, so even for exotic smooth $\mathbb R^4$'s you could potentially have only a one-map atlas, which is smooth in the standard sense.

You might like to read this article: http://en.wikipedia.org/wiki/Exotic_R4

  • $\begingroup$ I'm confused by your $t^{1/3}$ example. This atlas seems non-diffeomorphic to the trivial one. But, I thought the lack of exotic differentiable structure in $\mathbb{R}$ implied that all atlases on $\mathbb{R}$ had to be diffeomorphic to each other. I thought this followed from the fact that a differentiable structure is an equivalence class, under diffeomorphism, of atlases, and only one differentiable structure is allowed for $\mathbb{R}$. What am I missing? $\endgroup$ – Outis Jul 28 '10 at 17:51
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    $\begingroup$ $\mathbb R$ with the standard atlas and the $t \longmapsto t^{1/3}$ atlas are diffeomophic. The non-intuitive thing is that the diffeomorphism isn't the identity map -- it's the $t \longmapsto t^3$ map. $\endgroup$ – Ryan Budney Jul 28 '10 at 17:58
  • $\begingroup$ I guess I need clarification on what it means for two atlases to be diffeomorphic. You're saying that the atlas represented by $t \longmapsto t$ and the atlas represented by $t \longmapsto t^{1/3}$ are diffeomorphic to each other. The domains of each of these maps are trivially diffeomorphic, but the ranges are not. $f: t \mapsto t^{1/3}$ is not smooth at $t=0$, although its inverse is. $\endgroup$ – Outis Jul 29 '10 at 1:43

What do you mean by "smooth mapping" of atlases? I think you maybe using the terminology a bit different from how I do.

A chart or a coordinate chart on a topological manifold $M$ is a pair $(U,\psi)$ where $U\subset M$ is open and $\psi: U \to V\subset \mathbb{R}^n$ is a homeomorphism. Given two charts $(U_1,\psi_1)$ and $(U_2,\psi_2)$ with non-empty intersection $U_1\cap U_2$, we say that the charts are $C^k$ compatible if the transition function $\psi_2\circ\psi_1^{-1} |_{\psi_1(U_1\cap U_2)}$ is $k$-times continuously differentiable. Then a $C^k$ atlas on $M$ is a collection of charts $\{(U_\alpha,\psi_\alpha)\}_{\alpha \in A}$ such that the union $\cup_{\alpha \in A} U_\alpha$ covers $M$ and all the charts are $C^k$-compatible.

Two atlases are said to be $C^k$ compatible if all of their corresponding charts are pair-wise compatible, and hence if $\mathcal{A},\mathcal{B}$ are two compatible atlases, their union also is an atlas. An atlas is said to be maximal if any other compatible atlas must be a subset. It always exists by Zorn's lemma.

A differential structure on the manifold $M$ is a choice of a maximal $C^ \infty$ atlas, we write it as $(M,\mathcal{A})$. Two differentiable manifolds $(M,\mathcal{A})$ and $(N,\mathcal{B})$ are said to be diffeomorphic if there exists a homeomorphism $\Psi: M\to N$ such that for any $(U_{\alpha},\psi_\alpha) \in \mathcal{A}$ and $(V_{\beta}, \phi_{\beta})\in\mathcal{B}$ we have that the function $$ \phi_\beta \circ \Psi \circ \psi_\alpha^{-1} |_{\psi_\alpha(U_\alpha \cap \Psi^{-1}(V_\beta))} $$ is smooth.

Ryan already gave you an answer to your question. But let me elaborate a bit on your question two. Let $\mathbb{R}^2$ be your manifold. Define two charts on it $$ U_1 := \mathbb{R}\times (-1,\infty), U_2 := \mathbb{R}\times (-\infty,1) $$ and define $\psi_1(x) = x$ if $x\in \mathbb{R}\times (-1,2)$ and $\psi_1(x_1,x_2) = (x_1 + x_2 - 2, x_2)$ if $x_2 \geq 2$. Similarly $\psi_2$. Clearly $(U_1,\psi_1), (U_2, \psi_2)$ cover $\mathbb{R}^2$ and the transition function on the strip $\mathbb{R}\times (-1,1)$ is equal to the identity, and hence is smooth, so this gives a smooth atlas. But this smooth atlas is not compatible with the standard Cartesian atlas. What that does not mean that $\mathbb{R}^2$ with this atlas is exotic!

Let $(\mathbb{R}^k,\mathcal{E})$ denote the standard Euclidean space. An exotic smooth structure $(\mathbb{R}^k,\mathcal{A})$ requires that every homeomorphism from $\mathbb{R}^k$ to itself to not extend to a diffeomorphism from $(\mathbb{R}^k,\mathcal{E})\to (\mathbb{R}^k,\mathcal{A})$. What we have here is just that our stupid choice of homeomorphism is non-smooth. It is simple to change our homeomorphism such that the mapping is smooth relative to the fixed atlases.

So the non-existence of exotic smooth structure just say that for any two fixed smooth atlases, there exists some homeomorphism from $\mathbb{R}^k$ to itself such that it extends to a diffeomorphism. If this is what you meant, then the answer to your second question is "yes". But again, I don't understand what you mean by "smooth mapping of atlases".

  • $\begingroup$ Willie, thanks, but let me make sure I understand. If I'm interpreting your construction correctly you've defined the function $\psi_1: (x,y) \mapsto (x,y)$ if $y<2$ and $(x,y) \mapsto (x,y-2)$ if $y\geq 2$. But this function is discontinuous at $y=2$. Doesn't this function then fail to satisfy the definition of a homeomorphism? $\endgroup$ – Outis Jul 28 '10 at 21:44
  • $\begingroup$ Oops, you are right, I forgot to type in the second component of the map. It is not supposed to be a discontinuity, but a "kink", so that $\psi_1$ is not a $C^1$ map in the given coordinates. $\endgroup$ – Willie Wong Jul 28 '10 at 23:17

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