# Abstract Nonsense

## Smooth Manifolds (Pt. II)

Point of Post: This is a continuation of this post.

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Now, given a smooth atlas we have a notion of what’s differentiable based on the charts in the atlas. That said, there exists, in general, other smooth atlases on our topological manifold, disjoint from the original atlas, which still give us the same set of differentiable functions. In other words, given a topological manifold $M$ it is entirely feasible that we can find two smooth atlases $\mathscr{A}$ and $\mathscr{B}$ for $M$ for a which a function, say, $f:M\to\mathbb{R}$ will be differentiable with respect to $\mathscr{A}$ if and only if it is differentiable with respect to $\mathscr{B}$.Thus, for our purposes, it behooves us to consider $(M,\mathscr{A})$ and $(M,\mathscr{B})$ “the same”. They both admit the same differentiable theory.

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It becomes annoying to have to think about some kind of equivalence relations on topological manifolds with smooth atlases. We would have to say something like “Let $(M,\mathscr{A})\sim(M,\mathscr{B})$ if $\mathscr{A}$ and $\mathscr{B}$ define the same smooth theory on $M$.” It turns out though that this condition (that they define the same theory) will be true if and only if $\mathscr{A}\cup\mathscr{B}$ is also a smooth atlas (i.e. that all the charts in $\mathscr{A}$ are smoothly compatible with all of the charts in $\mathscr{B}$)[Prove this!]. Thus, instead of considering both $\mathscr{A}$ and $\mathscr{B}$ it is easier to just consider their union. Of course, this fixes the problem with the specific smooth atlases $\mathscr{A}$ and $\mathscr{B}$ but still causes issue for any other, not considered, smoothly compatible atlases on $M$.

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It thus makes sense to, in essence, given a smooth atlas $\mathscr{A}$ to instead of considering the smooth theory on $M$ as being determined by $\mathscr{A}$ instead have it be determined by the “union” of all smooth atlases on $M$ which are smoothly compatible with $\mathscr{A}$. Now, how do we formalize this? Well, taking a hint from other areas of mathematics that this comes up we define the poset $(\mathfrak{A}(M),\subseteq)$ of all smooth atlases on $M$ with inclusion. This union then should be a maximal element of $\mathscr{A}$ greater than or equal to $\mathscr{A}$–we call such an atlas maximal on $M$, or a smooth structure on $M$.. Thus, we define a smooth manifold to be a topological manifold $M$ equipped with a maximal smooth atlas $\mathscr{A}$. The dimension of a smooth manifold is the same as its dimension as a topological manifold.

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It is a simple exercise in Zorn’s Lemma to show that every atlas in $\mathfrak{A}(M)$  is contained in a (necessarily unique) maximal atlas. Explicitly, one can construct this maximal atlas by taking a given atlas $\mathscr{A}$ and defining $\mathscr{M}=\mathscr{M}(\mathscr{A})$ to be the set of all continuous charts on $M$ which are smoothly compatible with $\mathscr{A}$. It’s easy to see that $\mathscr{M}$ is, in fact, smooth for if one takes two charts in $\mathscr{M}$ which aren’t already in $\mathscr{A}$, the fact that both are smoothly compatible with a given chart in $\mathscr{A}$ means that you can patch them through to get that they themselves are smoothly compatible. It’s clear that this atlas is maximal since any maximal atlas is contained within it (for all of the charts in the maximal atlas are contained in $\mathscr{M}(\mathscr{A})$.

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With this in mind, it is often convenient for us to only give a smooth atlas $\mathscr{A}$ on a topological manifold $M$ and for us to understand that this to mean that we are considering the smooth manifold $(M,\mathscr{M}(\mathscr{A}))$.

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Consider the two smooth atlases $\{(\mathbb{R}^n, \text{id}_{\mathbb{R}^n})\}$ and $\{(B_1(x),\text{id}_{B_1(x)})\}_{x\in\mathbb{R}^n}$. Clearly since these atlases are smoothly compatible one can see that they induce the same smooth structure on $\mathbb{R}^n$–this is called the standard smooth structure on $\mathbb{R}^n$. It’s also clear that the smooth functions $f:\mathbb{R}^n\to\mathbb{R}$ with the standard structure are precisely the smooth functions in the sense of multivariable calculus. A natural question is, “what exactly does this structure look like?” In other words, can we explicitly describe $\mathscr{M}(\{(\mathbb{R}^n,\text{id}_{\mathbb{R}^n})\}$. Well, suppose that $U,V\subseteq\mathbb{R}^n$ is open and suppose that $f:U\to V$ is a diffeomorphism, then evidently $f$ is smoothly compatible with $(\mathbb{R}^n,\text{id}_{\mathbb{R}^n})$ since $f\circ \text{id}_{\mathbb{R}^n}^{-1}$ and $\text{id}_{\mathbb{R}^n}\circ f^{-1}=f^{-1}$ are both diffeomorphisms. Conversely, if $(U,\varphi)$ is a chart smoothly compatible with the standard structure on $\mathbb{R}^n$ then $\varphi(U)\subseteq\mathbb{R}^n$ is open and $\varphi\circ\text{id}_{\mathbb{R}^n}^{-1}=\varphi$ is a diffeomorphism $U\cap\mathbb{R}^n\to\varphi(U\cap\mathbb{R}^n)$, or from $U\to\varphi(U)$. Thus, we see that

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$\mathscr{M}(\{(\mathbb{R}^n,\text{id}_{\mathbb{R}^n})\})=\left\{f:U\to V:U,V\subseteq\mathbb{R}^n\text{ are open and, }f\text{ is a diffeomorphism}\right\}$

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The smooth atlas given by stereographic projection on $\mathbb{S}^n$ defines a smooth structure on $\mathbb{S}^n$ which is called the standard smooth structure on the sphere $\mathbb{S}^n$.

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The smooth atlas given for $\mathbb{RP}^n$ above is called the standard smooth structure on $\mathbb{RP}^n$.

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One may begin to wonder if there are actually different smooth structures on a given manifold–the answer is definitely yes. Indeed, note that if one takes any topological space $M$ and gives a one-chart continuous atlas $\{(M,\varphi)\}$ for $M$, then this atlas is vacuously smooth and thus defines a smooth structure on $M$. For example, $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=x^3$ is a homeomorphism and thus $(\mathbb{R},f)\}$ defines a smooth structure on $\mathbb{R}$. That said, this does not define the same smooth structure as the standard structure since as we have proven earlier this would necessarily make $f$ a diffeomorphism, which it definitely is not (since it’s inverse, $x\mapsto x^{\frac{1}{3}}$ is not differentiable at $0$). That said, in a sense which we shall make clear later, the smooth structures on $\mathbb{R}$ are not actually that different (in the sense of isomorphism in the category of smooth manifolds).

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References:

[1] Lee, John M. Introduction to Smooth Manifolds. New York: Springer, 2003. Print.

[2] Lee, John M. Introduction to Topological Manifolds. New York: Springer, 2000. Print.

[3] Milnor, John Willard, and David W. . Weaver. Topology from the Differentiable Viewpoint. Charlottesville (Va.): University of Virginia, 1969. Print.

August 30, 2012 -