Difference between smooth structures being equivalent or diffeomorphic.

I have just begun a course in Differential Topology and am being introduced to topological and smooth manifolds. Smooth manifolds involve having a smooth structure. And there were further concepts of two smooth structures being equivalent or diffeomorphic of distinct and I'm massively confused by these as I don't really understand the definition of when are two smooth structures equivalent and have been unable to find out definitions of them being distinct or diffeomorphic. Any help in understanding this better would be appreciated. I understand that a smooth structure is a maximal smooth atlas, where a smooth atlas means that any two charts in that atlas are smoothly equivalent.

• What resources (textbook, online notes, something else) are you using? What is your definition of equivalent and what is your definition of diffeomorphic? Sep 19 '17 at 15:12
• Not sure what your definition of equivalence is, but I have added an answer for when two manifolds are diffeomorphic. Sep 19 '17 at 15:27

Double check that my definition agrees with yours: a smooth structure $\mathcal{U}$ on a paracompact, second-countable, Hausdorff topological space $M$ is collection of chart neighborhoods $(U,\phi)$ such that:

• the $U$ cover $M$,
• the charts $\phi: U\to\Bbb{R}^n$ is a topological embedding, and
• for every pair $(U,\phi), (V,\psi)$ the transition map $\phi\psi^{-1}$ is a smooth map with everywhere-invertible differential from $\psi(U\cap V)$ to $\phi(U\cap V)$
• $\mathcal{U}$ is maximal with respect to these conditions, so that given another smooth structure $\mathcal{V}$, if $\mathcal{U}\cup\mathcal{V}$ is a smooth structure, then $\mathcal{V}\subset\mathcal{U}$.

I'm going to guess at the precise definitions of equivalent and diffeomorphic provided by your text.

We might say two smooth structures $\mathcal{U}, \mathcal{V}$ are diffeomorphic provided there exists a homeomorphism $h:M\to M$ such that the pullback of $\mathcal{U}$, defined as the structure $$h^*\mathcal{U} = \{(h^{-1}U, h^*\phi)\ |\ (U,\phi)\in\mathcal{U}\},$$ is equal to $\mathcal{V}$. Note that $h$ need not be smooth with respect to $\mathcal{U}$ or $\mathcal{V}$.

We might say two smooth structures are equivalent provided they are diffeomorphic by the identity map.

• Sorry if this is a stupid question, but what is $h^* \phi$ in your answer? Sep 19 '17 at 17:24
• The pullback of $\phi$ by $h$, defined as $h^*\phi(p) = (\phi\circ h)(p)$.
– Neal
Sep 19 '17 at 17:25
• Ah, okay, thanks! Sep 19 '17 at 17:27
• That clears it out, thanks!
– DpS
Sep 20 '17 at 1:38

Two differentiable manifolds are diffeomorphic to each other if there is a map (called a diffeomorphism) between them that is "smooth" and has an inverse which is "smooth".

As for the definition of "smooth"ness above, assuming your manifolds are embedded in Euclidean spaces: a map $f: M \to \mathbb{R}^n$ is smooth if at every point $x$ of (an $m$-manifold) $M$, there exist a relative neighborhood $N_x \cap M$, where $N_x$ is an open set in $\mathbb{R}^m$, and a differentiable function, say $g$, from $N_x$ to an open set in $\mathbb{R}^n$ such that $g\big|_{N_x \cap M} = f\big|_{N_x \cap M}$.

• I understand what it means for two manifolds to be diffeomorphic, I'm asking what it means for two smooth structures on the same manifold to be diffeomorphic.
– DpS
Sep 19 '17 at 15:40

As for the answer by Neil (I am not a frequent user, so "add comment" is not activated), I believe that it is more coherent to say that smooth atlases $$\mathcal U$$ and $$\mathcal V$$ on $$M$$ are (topologically) equivalent (as opposed to diffeomorphic) if there is a homeomorphism $$h:M\to M$$ which pulls $$\mathcal V$$ back into a smooth atlas on $$M$$ which is smoothly compatible with $$\mathcal U$$. Smooth compatibility of $$\mathcal U$$ and $$\mathcal V$$ means of course that their union is a smooth atlas, so this would be what Neil calls equivalence.

With this definition all smooth atlases on $$\mathbb R^n$$ for $$n \neq 4$$ are topologically equivalent but not smoothly compatible. The reason why this does not happen in dimension 4 is pretty involved.