# M721: Index Theory

## Two beautiful theorems about C(X)

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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 ${Swan}$. $\Box$

Written by prasit0605

September 5, 2012 at 1:33 am

Posted in Uncategorized