# Abstract Nonsense

## Smooth Manifolds (Pt. III)

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

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Methods to Manufacture Smooth Manifolds

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We now discuss some ways to manufacture new smooth manifolds from new ones.

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The first convenient way to do this is by looking at submanifolds. Now, the interesting thing about the smooth manifold theory is that one doesn’t define the subobjects until much later in one’s learning. This is a huge departure from the rest of mathematics where subobjects are defined almost immediately after the definition of the objects themselves. The reasoning for this is that the notion of what a “submanifold” should be is much more subtle than “a subset which is itself a smooth manifold”. That said, while we can’t define the general notion of submanifolds we can do the next best thing and describe so-called open submanifolds. Indeed, suppose that $M$ is a smooth manifold with smooth structure $\mathscr{A}$ and $N\subseteq M$ be an open subspace of $M$ (in the topological sense). Then, we claim that the set

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$\mathscr{A}_N=\left\{(U\cap N,\varphi\mid_{U\cap N}):(U,\varphi)\in\mathscr{A}\right\}$

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, called the defines a smooth atlas on $N$ of dimension $m$. Indeed, we first must verify that this is, indeed, a continuous atlas of dimension $m$. To see this we begin by noting that each  map $\varphi\mid_{U\cap N}:U\cap N\to \varphi(U\cap N)$ is indeed a homeomorphism and that $U\cap N$, $\varphi(U\cap N)$ are open in their respective spaces. Clearly, by definition of the subspace topology, $U\cap N$ is open in $N$. Now, since $\varphi$ is a homeomorphism we know that $\varphi(U\cap N)$ is an open subset of $\varphi(U)$, and since $\varphi(U)$ is open inside of $\mathbb{R}^m$ we know that $\varphi(U\cap N)$ is indeed open in $\mathbb{R}^m$. Now, evidently $\varphi\mid_{U\cap N}$ is a homeomorphism, being the restriction of a homeomorphism, and mapping onto the image of this restriction. Thus, we may conclude that each $(U\cap N,\varphi\mid_{U\cap N})$ is indeed a chart on $N$. Since $\bigcup (U\cap N)=\bigcup U\cap N=M\cap N=N$ we may conclude that, indeed, $\mathscr{A}_N$ is a continuous atlas for $N$. Thus, to verify that $\mathscr{A}_n$ defines a smooth atlas on $N$ it suffices to check that all of the map $\varphi\mid_{U\cap N}$ are smoothly compatible. But, this follows immediately from the fact that the restriction of a smooth map on Euclidean space is a smooth map. So, with this proven, we can unabashedly call $(N,\mathscr{A}_N)$ an open submanifold of $M$, and say that open submanifolds have the same dimension as their ambient manifolds.

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For example, consider the group $\text{GL}_n(\mathbb{R})$ of all $n\times n$ real invertible matrices. Since the ring $\text{Mat}_n(\mathbb{R})$ of $n\times n$ real matrices can be naturally identified with $\mathbb{R}^{n^2}$ we can think of $\text{GL}_n(\mathbb{R})$ as a subspace of $\mathbb{R}^{n^2}$. What we claim is that, in fact, the general linear group is an open subspace. Indeed, the determinant map $\text{det}:\text{Mat}_n(\mathbb{R})\to\mathbb{R}$ is clearly continuous (being just a polynomial in the entries of the mattrix) and $\text{GL}_n(\mathbb{R})=\det^{-1}(\mathbb{R}-\{0\})$. Thus, we see that $\text{GL}_n(\mathbb{R})$ can be treated as an open submanifold of $\mathbb{R}^{n^2}$.

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The next way one can construct manifolds is by taking products. In particular, suppose that $M_1,\cdots,M_n$ are $n$ smooth manifolds of dimension $m_1,\cdots,m_n$ respectively. We know that if $\mathscr{A}_i$ denotes a topological atlas for $M_i$ then the atlas $\mathscr{A}_1\times\cdots\mathscr{A}_n$ defined by $\left\{(U_1\times\cdots\times U_n:\varphi_1\times\cdots\times\varphi_n:(U_i,\varphi_i)\in\mathscr{A}_i\right\}$ (where $(\varphi_1,\cdots,\varphi_n)(x_1,\cdots,x_n)=(\varphi_1(x_1),\cdots,\varphi_n(x_n))$) defines a topological atlas on $M_1\times\cdots\times M_n$ of dimension $m_1+\cdots+m_n$. Noting that the product of smooth maps between Euclidean spaces is smooth, it’s easy to see that, in fact, $\mathscr{A}_1\times\cdots\times\mathscr{A}_n$ defines a smooth atlas on $M_1\times\cdots\times M_n$. We call the smooth structure it induces the product structure on $M_1\times\cdots\times M_n$, and merely call $M_1\times\cdots\times M_n$ the product manifold.

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Thus, we may give the $n$-torus, $\mathbb{T}^n=(\mathbb{S}^1)^n$ the structure of a compact smooth manifold of dimension $n$. We call this the standard structure on the $n$-torus.

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As a last example, we discuss the notion of the disjoint union of smooth manifolds. In particular, suppose that $M$ and $N$ are smooth manifolds of the same dimension with atlases $\{(U_\alpha,\varphi_\alpha)\}$ and $\{(V_\beta,\psi_\beta)\}$ respectively. We already know that $M\sqcup N$ with $\left\{(U'_\alpha,\varphi'_\alpha)\right\}\cup\{(V'_\beta,\psi_\beta')\}$ is a topological atlas for $M\sqcup N$. That said, all of the charts in this atlas are clearly smoothly compatible since they are either from the same original atlas (in which case they are trivially compatible) otherwise they are from different original atlases in which case they have non-intersecting domains. Thus, we see that this defines a smooth structure on $M\sqcup N$ which we call the disjoint union manifold of $M$ and $N$

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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.