M721: Index Theory

Ruminations of a Graduate Class

Two beautiful theorems about C(X)

with one comment

Let {X} be a topological space. Let {C(X)} be the set of continuous functions from {X} to {\mathbb{R}}. {C(X)} can also be thought of as set of smooth sections of the trivial bundle { X \times \mathbb{R} \longrightarrow X}. Anyway, we get a contravariant functor

\displaystyle C: Top \longrightarrow Ring

where {Top} is the category of {topological} {spaces}. {Ring} is the category of {rings} with {1} and every {ring} {homomorphism} sends {1} to {1}. The two beautiful {theorems} to be discussed here are the following. (Hewitt)For {X}, {Y} compact Hausdorff there is a bijection

\displaystyle Top(X, Y) \longrightarrow Ring(C(Y), C(X))

(Swan)If {X} is compact Hausdorff then taking sections gives a bijection from

\displaystyle \lbrace \text{ isomorphism class of vector bundles over } X \rbrace \longrightarrow \lbrace \text{finitely generated projective C(X)-modules} \rbrace

These two beautiful theoremshave some remarkable consequences If {X}, {Y} are compact and Hausdorff then,

\displaystyle X \cong Y \Longleftrightarrow C(X) \cong C(Y)

{Theorem} \textup{2} leads to the following result in {K}{Theory}

\displaystyle K^{0}(X) \cong K_{0}(C(X))

{Theorem} {1} is a consequence of the following {Lemma} Let {X} be compact, Hausdorff topological space

  1. For {x_{0} \in X},\displaystyle M(x_{0}) = \lbrace f \in C(X): f(x_{0}) = 0 \rbrace

    is a maximal ideal,

  2. If {M \lhd C(X)} is a maximal ideal, then {\exists !} {x_{0} \in X} such that {M = M(x_{0})},
  3. {MaxSpec(C(X)) \cong X} where {MaxSpec(R)} is the set of all maximal ideals of a ring {R} equipped with {Zariski} topology. The isomorphism takes {M(x_{0})} to {x_{0}}

Proof:

  1. Clearly {M(x_{0})} is maximal as {C(X) / M(x_{0}) \simeq \mathbb{R}} which is a field.
  2. Notice, if {f \notin M(x_{0})}, then {f(x_{0}) \neq 0} If {I} is an {ideal} such that {I \nsubseteq M(x_{0})} for all {x_{0}} in {X} then for every {x \in X}, {\exists} {f_{x} \in I} such that {f_{x}(x) \neq 0}. Each {f_{x}} there exists {U_{x} \ni x} such that {f_{x}(t) \neq 0} {\forall t \in U_{x}}. Since {X} is compact {\lbrace U_{x_{0}}, \ldots U_{x_{n}} \rbrace} cover {X}. Using {bump} {functions} {b_{i}} which do not vanish on {U_{x_{i}}} respectively, define\displaystyle f = |f_{x_{0}}|b_{0} + \ldots +|f_{x_{n}}|b_{n}

    . Observe, {f(x) \neq 0} {\forall x \in X}. Define {g(x) = \frac{1}{f(x)}}. Clearly {g(x) \in C(X)} and {f.g = 1}. Thus {I = C(X)}. Thus the only maximal ideals of {C(X)} is of the form {M(x_{0})} for some {x_{0} \in X}.

  3. For any ideal {I} of a ring {R} define,\displaystyle V(I) = \lbrace M \text{ maximal in } R : I < M \rbrace

    {V(I)} is the basis for all {closed} sets in the space {MaxSpec(R)} under {Zariski} topology. The map\displaystyle F : X \longrightarrow MaxSpec(X)

    which sends\displaystyle x \longmapsto M(x_{0})

    is already a bijection. All we need to show is\displaystyle C \text{ closed} \Leftrightarrow F(C) \text{ closed}

    {(\Rightarrow):}
    IF {C} closed then define {I_{C} = \lbrace f \in C(X): f(C) = 0 \rbrace} and {F(C) = V(I_{C}) = \bigcup_{x \in C} M(x)} \vspace{5pt}
    {(\Leftarrow):}
    IF {C} be a basic closed set in {MaxSpec(C(X))}, ie, {C= V(I)} for some {I \in C(X)} then, define\displaystyle D = \lbrace x \in X: f(x) = 0 \forall f \in I \rbrace

    . Then {D} is clearly a closed set and clearly {D = F^{-1}(C)}.

\Box Proof: (of Theorem 1) In fact the {C} gives the map between the {Hom} sets of the respective category.
One-one
Let {f: X \longrightarrow Y}. Then

\displaystyle C(f)(g) =f^{*}g = g \circ f

, where {g \in C(Y)}. If {C(f) = C(f')} then

{\Rightarrow g \circ f = g \circ f' \forall g \in C(Y)}
{\Rightarrow f(x) = g(x)} by using bump functions near each point
Onto
Given a map {F:C(Y) \longrightarrow C(X)}, we induce a map

\displaystyle \overline{F} : Spec(C(X))\longrightarrow Spec(C(Y))

By {Lemma} \textup{5} we get a map

\displaystyle \overline{F}: X \longrightarrow Y

It is clear that {\overline{F}^{*} = F}. \Box Proof: ( sketch of proof of theorem 2)
Notice that Let {G} be the map
G: {\lbrace} isomorphism class of vector bundles over {X} {\rbrace \longrightarrow \lbrace} finitely generated C(X)-modules {\rbrace}
where given a vector bundle {\xi}
{G(\xi)} = {\lbrace} smooth sections of {\xi \rbrace}.
Since {X} is compact, any vector bundle {\xi} is a {subbundle} of a trivial bundle of finite dimension, ie {X \times \mathbb{R}^{n}}. Hence {G(\xi)} is a sub-module of {\bigoplus_{1}^{n} C(X)} due to the following isomorphism.
{ \tau: \bigoplus_{1}^{n} C(X) \cong \lbrace } smooth sections on the trivial bundle {X \times \mathbb{R}^{n} \rbrace }
Thus {G(\xi)} is a finitely generated module. Moreover every bundle {\xi} of finite dimension over a compact space has a complement, say {\xi^{-1}}, hence {G(\xi) \oplus G(\xi^{-1})=\bigoplus_{1}^{n} C(X)}. Hence its projective. Given a finitely generated projective module over {C(X)}, say {M}, find {n} and a {C(X)} module {N}, such that

\displaystyle M \oplus N = \bigoplus_{1}^{n} C(X)

Then define {G^{-1}(M) = \tau^{-1}(M)}. This is a {vector} {bundle} over {X}. The proof is non-trivial and is a {theorem} of {Swan}. \Box

Advertisements

Written by prasit0605

September 5, 2012 at 1:33 am

Posted in Uncategorized

One Response

Subscribe to comments with RSS.

  1. The smooth sections of the trivial line bundle over X correspond to C^\infty(X), not C(X). However, the same statement where ‘smooth’ is replaced by ‘continuous’ works.

    rallymoore

    October 24, 2012 at 8:11 pm


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: