## Two beautiful theorems about C(X)

Let be a topological space. Let be the set of continuous functions from to . can also be thought of as set of smooth sections of the trivial bundle . Anyway, we get a contravariant functor

where is the category of . is the category of with and every sends to . The two beautiful to be discussed here are the following. *(Hewitt)*For , compact Hausdorff there is a bijection

*(Swan)*If is compact Hausdorff then taking sections gives a bijection from

These two beautiful *theorems*have some remarkable consequences If , are compact and Hausdorff then,

\textup{2} leads to the following result in –

is a consequence of the following Let be compact, Hausdorff topological space

- For ,
is a maximal ideal,

- If is a maximal ideal, then such that ,
- where is the set of all maximal ideals of a ring equipped with topology. The isomorphism takes to

*Proof:*

- Clearly is maximal as which is a field.
- Notice, if , then If is an such that for all in then for every , such that . Each there exists such that . Since is compact cover . Using which do not vanish on respectively, define
. Observe, . Define . Clearly and . Thus . Thus the only maximal ideals of is of the form for some .

- For any ideal of a ring define,
is the basis for all sets in the space under topology. The map

which sends

is already a bijection. All we need to show is

IF closed then define and \vspace{5pt}

IF be a basic closed set in , ie, for some then, define. Then is clearly a closed set and clearly .

*Proof:* *(of Theorem 1)* In fact the gives the map between the sets of the respective category.

**One-one**

Let . Then

, where . If then

by using bump functions near each point

**Onto**

Given a map , we induce a map

By \textup{5} we get a map

It is clear that . *Proof:* *( sketch of proof of theorem 2)*

Notice that Let be the map

G: isomorphism class of vector bundles over finitely generated C(X)-modules

where given a vector bundle

= smooth sections of .

Since is compact, any vector bundle is a of a trivial bundle of finite dimension, ie . Hence is a sub-module of due to the following isomorphism.

smooth sections on the trivial bundle

Thus is a finitely generated module. Moreover every bundle of finite dimension over a compact space has a complement, say , hence . Hence its projective. Given a finitely generated projective module over , say , find and a module , such that

Then define . This is a over . The proof is non-trivial and is a of .

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.

rallymooreOctober 24, 2012 at 8:11 pm