Problem 1-6 (Smooth manifolds have uncountably many distinct smooth structures)

Question

Let $M$ be a nonempty topological manifold of dimension $n \geq 1$. If $M$ has a smooth structure, show that it has uncountably many distinct ones.
ewline

[Hint: first show that for any $s>0, F_{S}(x)=|x|^{s-1} x$ defines a homeomorphism from $\mathbb{B}^{n}$ to itself, which is a diffeomorphism if and only if $s=1$.]

Elu Thingol · Accepted Answer

Let $M$ be a manifold and let $\mathcal{A}$ be a smooth structure on $M$. The idea is to modify the smooth structure $\mathcal{A}$ at a single point $p \in M$ in a very nice way so that the resulting new family of smooth structures $(\mathcal{A}_s^{\prime\prime\prime})_{s>0}$ remains smooth but becomes smoothly incompatible with other members of the family. We defer the proof of the hint to the end.

\begin{proof}
Fix a $p \in M$ and let $(U,\varphi)$ be a chart in $\mathcal{A}$ that covers $p$. Denote $\widehat{p}:=\varphi(p)$ and $\widehat{U} := \varphi(U)$. We can make $(U,\varphi)$ the only chart that covers $p$ by subtracting $\{p\}$ from all other chart domains\footnote{Subtracting a point from an open set, we obtain an open set since singletons are closed.}
\begin{equation*}
    \mathcal{A}^{\prime} := \left\{ \left(V\setminus \{p\}, \psiight) \mid (V,\psi) \in \mathcal{A} ight\} \cup \{ (U, \varphi) \}.
\end{equation*}

Obviously, $\mathcal{A}^{\prime}$ remains a smooth chart. Now we carefully modify the chart $(U,\varphi)$ containing $p$. Since $\widehat{U} \subseteq \mathbb{R}^n$ is open, we can find a radius small enough such that $B(\widehat{p},r) \subseteq \widehat{U}$. Then $W:=\varphi^{-1}(B(\widehat{p},r)) \subseteq U$. Thus, we can alternatively look at a slightly smaller chart around $p$:
\begin{equation*}
    \mathcal{A}^{\prime\prime} := \left\{ \left(V\setminus \{p\}, \psi ight) \mid (V,\psi) \in \mathcal{A} ight\} \cup \{ (W, \varphi) \}.
\end{equation*}

Now, we can modify $\varphi$. Define $\gamma := \dfrac{\varphi - \widehat{p}}{r}$. Since we simply translated and scaled $\varphi$, the resulting function $\gamma$ is also a homeomorphism but its range is the unit ball $\mathbb{B}^n := B(\mathbf{0},1)$. Also, notice that $\varphi(p) = \mathbf{0}$ now stands at the center of the ball. At this point, we can use the hint from the exercise. Since $F_s$ are homeomorphisms for all $s>0$ by the exercise hint, the composition $F_s \circ \gamma$ is a homeomorphism to the unit ball as well. Thus, we define a family of atlases: 
\begin{equation*}
    \forall s > 1: \quad \mathcal{A}_{s}^{\prime\prime\prime} := \left\{ \left(V\setminus \{p\}, \psi ight) \mid (V,\psi) \in \mathcal{A} ight\} \cup \{ (W, F_s \circ \gamma) \}.
\end{equation*}

We now argue that $\mathcal{A}_s^{\prime\prime\prime}$ are smooth yet not smoothly compatible with any other $\mathcal{A}_t^{\prime\prime\prime}$ for $t 
eq s$. 
\begin{itemize}
    \item Fix an $s>0$. To see why $\mathcal{A}_s^{\prime\prime\prime}$ is smooth, notice that every pair of charts $(V\setminus\{p\},\psi)$ and $(V^{\prime}\setminus\{p\},\psi^{\prime})$ not counting $(W, F_s \circ \gamma)$ are inherently smoothly compatible with each other. To show that $(V\setminus\{p\},\psi)$ is also compatible with $(W, F_s \circ \gamma)$, consider the transition function
    $$(F_s \circ \gamma) \circ \psi^{-1}: \psi(W \cap V \setminus \{p\}) 	o \mathbb{B}^n$$
    We have already argued that $\gamma$ is smooth and that $F_s$ is smooth everywhere on $\mathcal{B}^n$ except at the origin (by the hint in the exercise). Since the origin is excluded by the intersection with $V \setminus \{p\}$ anyway, $F_s \circ \gamma$ is smooth everywhere on the given domain, as is the transition function.

\item To see why $\mathcal{A}_s^{\prime\prime\prime}$ and $\mathcal{A}_t^{\prime\prime\prime}$ are incompatible for $s 
eq t$, consider the transition function
    \begin{equation*}
        (F_s \circ \gamma) \circ (F_t \circ \gamma)^{-1} = F_s \circ \gamma \circ \gamma^{-1} \circ F_t = F_s \circ F_t^{-1} = F_{\frac{s}{t}}
    \end{equation*}
    (see Lemma 1(b)). Since we know this function not to be a diffeomorphism on its domain, the two smooth structures cannot be compatible.
\end{itemize}
\end{proof}

Now that we have seen how helpful the assertion in the exercise hint was, let us prove its validity.
\begin{lemma}[Exercise hint]
    For all $s>0$, define\footnote{The range is well-defined since $\|F_s(x)\| = \|x\|^s \leq 1$ and thus $F_s(\mathbb{B}^n) \subseteq \mathbb{B}^n$.}
    \begin{equation*}
        F_s: \mathbb{B}^n 	o \mathbb{B}^n, %\qquad F_s(x) := \|x\|^{s-1} \cdot x.
        \qquad F_s(x) := \left\{\begin{array}{ll}
             \|x\|^{s-1} \cdot x, &x 
eq \mathbf{0}  \
             \mathbf{0}, & x=\mathbf{0}. 
        \end{array}ight.
    \end{equation*}
    Then the following properties hold:
    \begin{enumerate}[label=(\alph*)]
        \item $F_s$ is a homeomorphism.
        \item $F_s$ is not a diffeomorphism for $s 
eq 1$.
        \item $F_s$ is a diffeomorphism on $\mathbb{B}^n\setminus \{\mathbf{0}\}$.
    \end{enumerate}
\end{lemma}
\begin{proof}
    \begin{enumerate}[label=(\alph*)]
    \item Clearly, $F_s$ is continuous. Furthermore, it is easy to verify the inverse $F_s^{-1}$ of $F_s$ is given by
    \begin{equation*}
        F_s^{-1}(x)=|x|^{\frac{1}{s}-1}x = F_{\frac{1}{s}}
    \end{equation*}
    since, more generally, the following multiplicative identity holds:
    \begin{align*}
        (F_s \circ F_t)(x) &= F_s(F_t^{-1}(x))\ &= \left\|\|x\|^{t-1}xight\|^{s-1} \cdot \left(\|x\|^{t-1}x ight)\
        &= \left\| x ight\|^{{s}{t}-1} \cdot x^{{s}{t}-1} = F_{{s}{t}}(x)
    \end{align*}
    and $F_1$ is the identity function on $\mathbb{B}^n$. Thus, $F_s$ is a bijection and $F_s^{-1}$ is continuous as well.
    
    \item Either for $s < 1$, $F_s$ has a discontinuous partial derivative, or for $s>1$, the inverse $F_s^{-1} = F_{\frac{1}{s}}$ has a discontinuous partial derivative. In the former case,
    \begin{align*}
        \dfrac{\partial F_s}{\partial x_1}(x_1,0,\ldots,0) = (s-1) |x_1|^{s-3} x_1^2  + |x_1|^{s-1}.
    \end{align*}
    Denoting the first standard basis vector by $e_1:=(1,0,\ldots,0)$, we obtain
    \begin{align*}
        \lim_{t 	o 0} \dfrac{\partial F_s}{\partial x_1}(\mathbf{0}+te_1) = \lim_{t 	o 0} (s-1) \cdot |t^{s-3}| \cdot t^2  + t^{s-1} = \lim_{t 	o 0} s |t|^{s-1} = \infty
    \end{align*}

\item $F_s$ is smooth outside of the origin since it is a product and composition of smooth functions.
    \end{enumerate}
\end{proof}

Problem 1-6 (Smooth manifolds have uncountably many distinct smooth structures)

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Problem 1-6 (Smooth manifolds have uncountably many distinct smooth structures)

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