Abstract Nonsense

Crushing one theorem at a time

Smooth Maps and the Category of Smooth Manifolds (Pt. III)

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

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Now, what we’d like to discuss is the set C^{\infty}(M) of all smooth functions (for us, when we remember, we shall obey history’s tough hand, and call smooth maps into the real line “functions” and reserve this word for this case) M\to\mathbb{R}. It’s easy to see that this is an \mathbb{R}-algebra. Since the set \mathbf{Map}(M,\mathbb{R}) of all functions (not necessarily smooth) M\to\mathbb{R} is an algebra, it really only needs to be checked that constant functions are in C^\infty(M), and that it is closed under addition and multiplication. Clearly constant functions are contained in C^\infty(M), and multiplication and addition are fairly easy. Indeed, it’s trivial that the maps +,\cdot:\mathbb{R}\times\mathbb{R}\to\mathbb{R} given by \cdot(x,y)=xy and +(x,y)=x+y are smooth. From this, we see that if f,g:M\to \mathbb{R} are two smooth maps, then their sum f+g and their product fg (defined pointwise) are smooth since f+g=+\circ(f\times g) and fg=\cdot\circ(f\times g), and the composition of smooth maps is smooth.

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Now, let M be a topological manifold. We can define C(M) to be the set of all continuous maps M\to\mathbb{R}, and similar to the case of smooth maps, it’s easy to show that this is a subalgebra of \text{Map}(M,\mathbb{R}).

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Now, suppose that M and N are smooth manifolds and f:M\to N a smooth map. There is an obvious way to define a map f^\ast:C^\infty(N)\to C^\infty(M) given by g\mapsto g\circ f. This is clearly an algebra map since

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((g+g')\circ f)(x)=(g+g')(f(x))=g(f(x))+g'(f(x))=(g\circ f)(x)+(g'\circ f)(x)

((gg')\circ f)(x)=(gg')(f(x))=g(f(x))g'(f(x))=(g\circ f)(x)(g'\circ f)(x)

(c\circ f)(x)=c(f(x))=c

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where g,g'\in C^\infty(M) and c is a constant map.

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Following the exact same process, we see that if we have a continuous map f:M\to N this induces an algebra map f^\ast:C(N)\to C(M).

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It’s obvious that these association of spaces and maps are  contravariant functors \mathbf{Man}\to\mathbb{R}\text{-}\mathbf{Alg} and \mathbf{TopMan}\to\mathbb{R}\text{-}\mathbf{Alg}. Indeed, it’s easy to see that these are precisely the contravariant Hom functors C^\infty(\bullet,\mathbb{R}) and C(\bullet,\mathbb{R}) respectively.

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Now, we can ask a very interesting question about how C^\infty and C react with one another. In particular, suppose that we have smooth manifolds M and N and a continuous map f:M\to N. We know that this induces an algebra map f^\ast:C(N)\to C(M). Does this map f^\ast restrict down to an algebra map C^\infty(N)\to C^\infty(M)? Of course, this will restrict down to an algebra map C^\infty(N)\to C(M) and so the operative part of this question is whether f^\ast(C^\infty(N))\subseteq C^\infty(M). This is not always true, and in fact, we actually have the following:

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Theorem: Let M and N be smooth manifolds and f:M\to N a continuous map. Then, f^\ast(C^\infty(N))\subseteq C^\infty(M) if and only if f is smooth.

Proof: We have already discussed the fact that if f is smooth, then necessarily the containment condition holds.

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Conversely, suppose that the containment condition holds and let p\in M and let (U,\varphi) and (V,\psi) be charts around p,f(p) respectively such that f(U)\subseteq V. We need to show that \psi\circ f\circ\varphi^{-1} is a smooth map. That said, note that

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\psi\circ f=(\psi_1\circ f)\times\cdots\times(\psi_n\circ f)

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if N is an n-dimensional manifold and \psi_i=\pi_i\circ f. Now, we know that each \psi_i is a smooth map N\to\mathbb{R} and thus we know that \psi_i\circ f is a smooth map M\to\mathbb{R}. Thus, we see that \psi\circ f is the product of smooth maps M\to\mathbb{R} and thus a smooth map M\to\mathbb{R}^n. We see then that \psi\circ f\circ\varphi^{-1}, being equal to (\psi\circ f)\circ\varphi^{-1} is a composition of smooth maps, and thus smooth. \blacksquare

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In fact, it’s easy to see from this that we have the following:

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Theorem: Let M and N be smooth manifolds and f:M\to N a homeomorphism. Then, f^\ast restricts to an isomorphism C^\infty(N)\to C^\infty(M) if and only if f is a diffeomorphism.

Proof: Since the assocation M\mapsto C^\infty(M) and f\mapsto f^\ast is a functor, it follows from first principles that if f is a diffeomorphism then f^\ast is an isomorphism of algebras.

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Conversely, suppose that f^\ast restricts to an isomorphism C^\infty(N)\to C^\infty(M). The fact that f^\ast(C^\infty(N))\subseteq C^\infty(M) tells us that f^\ast is smooth. That said, it’s easy to see (f^{-1})^\ast=(f^\ast)^{-1} and so (f^{-1})^\ast(C^\infty(M))\subseteq C^\infty(N) and so f^{-1} is smooth. \blacksquare

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We thus see that, in a very precise sense, the notion of a differentiable structure on a topological manifold is characterized by its algebra of smooth functions. In fact, some authors define a smooth manifold as being a topological manifold equipped with a particular type of subalgebra of the continuous real-valued functions on that manifold (c.f. [4]).

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[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.

[4] Bredon, Glen E. Topology and Geometry. New York: Springer-Verlag, 1993. Print.



September 3, 2012 - Posted by | Manifold Theory, Topology | , , , , , ,

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