Let be an elliptic differential operator, where and are vector bundles over a closed oriented manifold . Suppose and have smooth inner product structures. Let and be the disk and sphere bundle of the cotangent bundle, respectively. Let be the projection. Let be the associated symbol class. Let be its Chern character. Let denote the pullback of the Todd class . Let be the fundamental class.

**Definition. **The topological index of is defined to be .

**Atiyah-Singer Index Theorem. ***, where is the analytical index.*

**Example. ***[Point Case] *Let be a point. Then and are finite dimensional vector spaces. Any non-trivial differential operator is a linear map between them, and hence of order 0. Thus, . Note that is empty set. Recall the definition of Chern character, then we have and . Since is empty set, then and hence . Recall the definition of Todd class, we then have . Therefore, by the definition of topological index, we have . By Atiyah-Singer, .

**Example** *[ Case] *Let , , Then is a first order elliptic operator. We now claim that . We give four different proofs.

*Proof 1.* We compute directly. Note that and that via .

*Proof 2.* Since , we have

*Proof 3.* , since is self-adjoint.

*Proof 4*. We will show that the topological index vanishes whenever is odd. See next example.

**Example** *[Odd Dimensional Case, Theorem 13.12 in Lawson-Michaelson]*

We will show that the topological index of any elliptic differential operator vanishes whenever is odd.

We want to show that , where is an elliptic differential operator of order . Consider the diffeomorphic involution given by . Since

,

it suffices to show . In fact, , since is homotopic to via .

Next, we need to introduce the Thom Isomorphism to talk about the de Rham operator.

Let be an oriented -vector bundle over , with inner product on each fiber. We now give the notion of Thom class and Thom space.

**Definition. ** is a Thom Class of if it restricts to a generator of on each fiber. The quotient is called the Thom space of , and denoted by .

**Thom Isomorphism Theorem. ***The composition is an isomorphism.*

We denote the composition by and denote its inverse by .

**Remark.** has the following two other interpretations.

(1). *Integration over the fiber.* , where is a bundle over .

Let be given and choose and . Associated to these data is a form , defined as follows. Given and a basis , choose lifts such that , for each , and define

Now is defined by

Integrating over the fibers will give the second formulation of the topological index, which is the next theorem. The factor compensates for the difference between the orientation on induced by the one on , and the canonical orientation on inherited from its almost complex structure.

(2) . Also, we can use the second interpretation of to give that formulation.

is the composition (Poincare duality) (Poincare-Lefschetz duality),

Then, we compute

.

**Theorem.** .

We will then give the third formulation of the topological index. To do this we need the notion of Euler class.

**Definition. **The Euler class of an oriented -bundle over , denoted by $latexe(E)$, is the image of the Thom class under the following isomorphism: . We may denote the composition by .

**Theorem.** *[Gysin Sequence] *To any bundle as above there is associated an exact sequence of the form

**Definition. **The Euler characteristic of is defined to be

* *

From now on, we assume .

We want to analyze to give the third formulation of the topological index. For details please see Lawson-Michaelson, P258, Theorem 13.13.

Since – and are inverse to each other, we have . Applying to both sides, we then get Thus we can write , if .

**Theorem. **, if is not zero.

Now, we are trying to apply this formula to the de Rham operator.

**Example.** *[de Rham operator] *Let Then we already know that is an elliptic complex, that is an elliptic operator and that . We want to use the above theorem to show that the Euler charateristic.

For a complex vector bundle , by the splitting principle, we can write as . Then . It follows that Similarly, we have

Back to our example, applying the real splitting principle to , we compute

since for . Note that , and that , then by the theorem above, we obtain

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Let be an -dimensional Riemannian manifold with covariant derivative (on ) , and let be a vector bundle.

** Clifford Module Bundles and a Dirac “Type” Operator**

**Definition** (Clifford module)

A **Clifford module** for a real inner product space is a left module over . Equivalently, there is a -algebra homomorphism given by . Since for any (see the Glossary below), one has that satisfies .

**Definition** (Bundle of Clifford Modules)

A bundle (as above) is a **bundle of Clifford modules** if there is a map of bundles of -algebras such that for any section . In other words, for each , is a Clifford module for .

**Definition** (Dirac type operator)

Let be a Clifford module bundle equipped with a covariant derivative . Let be the map defined by the composition

where is the inverse of the bundle isomorphism , and where is Clifford multiplication. We call such a map (which depends on the Clifford module bundle , and ) **a Dirac type operator**.

If we a fix an orthonormal frame for over some neighborhood and, using the metric , let be the corresponding frame for , we may write this composition locally as

That is, a Dirac type operator is locally of the form

.

**Proposition**

A Dirac type operator is a first order differential operator.

**Proof**

Let and . Using the local description above, we compute:

.

In particular, for , so is -linear and hence in . Thus , and as itself is not -linear, is of order 1.

**Remark**

The above proof extends (by incorporating induction) to show that the composition of a – and an order differential operator is a differential operator of order . Here, and are differential operators of order 1 and 0, respectively.

We next show that is elliptic.

**Lemma** (Symbol of a Dirac type operator)

Let be a Dirac type operator and let . Then the symbol of at is given by

,

where is the dual to determined by the metric, i.e., such that .

**Proof**

Fix , and let be an open neighborhood of in . Using choose an orthonormal frame of with dual frame for . Being a bundle homomorphism (over ), is -linear in the -ordinate. Thus it suffices to verify the proposition for ; that is, we wish to show that for .

Choose a local chart such that ; let . Note that is an orthonormal basis for . Let ; then , so .

Thus (see the Glossary below), as is order , we have

,

as required.

** Corollary **

A Dirac type operator is elliptic.

** Proof **

If then has inverse .

** Remark **

Noting that (the product of linear maps), we observe that . Taking , this — apparently — implies that a Dirac type operator is, at the symbol level, the square root of the Laplacian.

** -grading and “a” Dirac Operator**

A Dirac type operator is formally self-adjoint (so that its index is ) if we impose the following further restrictions on the Clifford-module bundle.

**Definition** (Clifford-Compatible)

Let be a bundle of Clifford modules. We say that is **Clifford compatible** if it is equipped with a metric and a covariant derivative such that

(1) is Riemannian, i.e., for all sections :

, and

(2) for all vector fields and for any section :

.

**Definition** (A Dirac operator)

A differential operator is **a Dirac operator** if

(1) is a Dirac type operator, and

(2) is Clifford compatible.

** Lemma**

If is oriented and is a Dirac operator, then is formally self-adjoint. That is,

,

where (and the integration is with respect to the volume form on ).

** Proof **

Omitted

Consequently we have (see Ning’s blog). To make use of the index, then, we introduce a -grading on Clifford module bundles.

**Definition**

Recall that a Clifford algebra is -filtered — with — and -graded — (even and odd products).

A Clifford module bundle is **-graded** if it decomposes into a direct sum of vector bundles such that for each and , one has .

Such a -graded bundle is **compatible** if this decomposition is both orthogonal with respect to and parallel with respect to the covariant derivative , ie. .

** Example **

If is oriented, the Clifford bundle is a -graded compatible Clifford bundle. Some words which may be connected to verify this: Levi Civita connection, induced connection on , lift to principal spin bundle, induced covariant derivative on associated vector bundle, compatibility with the metric.

** Examples of Dirac Operators**

We now look at four examples of Dirac operators. The first two are familiar; here we reinterpret them in terms of Clifford modules.

**Example 1: The De Rham Operator**

Recall that the filtered algebra has associated graded algebra . (See the Glossary below.)

** Lemma**

Let be the map defined by

.

Then is

(1) an isomorphism of vector spaces

(2) filtration preserving, i.e., , and

(3) -equivariant, i.e., for and .

** Proof**

(1) and (2). Let is an orthonormal basis for . Since (by an equivalent definition of the exterior algebra) , we see that is induced by the map taking to and descending to ; that is, . It follows that is an isomorphism and preserves the grading.

(3) Using that , I feel like we need to be working with here. Please comment!

** Corollary**

The exterior algebra bundle over a(n oriented?) manifold and the clifford bundles are isomorphic as vector bundles.

** Proof **

Let denote the principal -bundle associated to . By parts (1) and (3) of the lemma, the map above induces a vector bundle isomorphism .

** Theorem 2.5.12 (Lawson, Michelsohn)**

Under this bundle isomorphism , the de Rham operator corresponds to the Dirac operator .

** Corollary**

Since we have already established (see Hailiang’s(?) blog post) that the Euler characteristic of is equal to , the theorem (along with the grading-preserving property of ) implies we may also compute it as .

** Example 2: The Signature Operator**

We now look to reinterpret the signature operator in terms of Clifford bundles.

Recall (see Hailiang’s blog) in the case that and is even, the Hodge star operator is an involution so we can decompose into the and eigenspaces of . We defined the signature operator . Since , we saw that took to and letting , we found that . The signature of was defined to be the signature of the quadratic form on given by .

In the case that is odd, we had to modify the construction. We complexified, taking , and defined . Then the above paragraph went through with replacing and replacing .

Let be an oriented orthonormal basis for . Let in . Then by the lemma above and the corresponding properties of the volume form in , we obtain that is a basis-independent section of .

** Lemma **

(1) We have

(2) If is even and , then .

** Proof **

(1) We compute . Writing for , one finds that is even if and only if or .

(2) It suffices to verify for . We have and . (Here, a hat indicates that the element be omitted from the product.) Since is even, .

Now, acts on any Clifford module via , and by part (1) of the lemma this defines an involution in the case ; in the case , defines an involution. Compare with the Hodge star operator recalled above. So define

then . Thus if is an oriented manifold and is a Clifford module bundle of , putting , we have

if : , or

if :

** Corollary **

If is even then is -graded, i.e., for and , one has .

** Proof **

By part (2) of the lemma, for we have .

Thus if (i.e. ) then , so .

** Proposition **

If is even and is -graded compatible, then the associated Dirac operator splits as

.

In particular, if , by Theorem 2.5.12 we have .

**Proof**

Since is -graded and is compatible (so in particular, the covariant derivative preserves , the Dirac operator

takes to .

** Example 3: twisted Dirac Operators **

** Preliminary:** If and are vector bundles over with covariant derivatives and , respectively, then the tensor product bundle has covariant derivative

.

**Fact:** If is a compatible -graded Clifford module bundle and is a Riemannian bundle (see Property (1) of a compatible Clifford Bundle above for the definition), then is a compatible -graded Clifford module bundle (with Clifford multiplication for , , ). In the case that , we call the Dirac operator on a **twisted Dirac operator**.

**Fact:** (Apparently from topological K-theory) If the Index theorem holds for any twisted *Signature* operator then it holds for all elliptic differential operators.

** Example 4: Spin Manifolds and The Atiyah-Singer Dirac Operator**

Recall (see Prasit’s blog) that there is an isomorphism . So since

is an module, one has that is an -module via , for and . To avoid confusion, let us call with this module structure .

Now, any -module is isomorphic to , so it follows that any -module is isomorphic to .

Let be a Clifford module bundle. From the above paragraph we see that each fiber (a -module) is isomorphic (via , say) to , where is a copy of . We may then

ask if this splitting extends over the whole bundle; that is, is there a Clifford module bundle and a bundle isomorphism which restricts fiberwise to an isomorphism .

In turns out the answer is a resounding “Yes” if is a spin manifold.

**Definition** (Spin Structure)

Let is an -dimensional vector bundle. A **spin structure** on is a principal -bundle together with a bundle isomorphism . (Then is the associated vector bundle for ). Using classifying space theory, we may reinterpret this

to say that a spin structure on is a lift of the classifying map to .

We may break up the existence of a spin structure into pieces as follows.

After choosing a metric on , we may first reduce the structure group of to . (The only obstruction to doing so is the paracompactness of .) So we’re left to lift a map to .

Since is the universal cover of , we may first try to lift to .

The short exact sequence of groups induces a fibration of classifying spaces

.

It turns out that the map lifts to if and only if the composite is nullhomotopic. Since , there is an element that vanishes if and only if lifts. We call the first Stiefel Whitney class of .

Similarly, the map lifts to if and only if the composition is nullhomotopic; we call the corresponding element in (that vanishes iff lifts) the second Stiefel Whitney class of .

**Definition** (Spin manifold)

We will call an oriented manifold (so ) a **spin manifold** if its tangent bundle admits a spin structure (i.e., ). It can be shown that this is equivalent to the existence of a trivialization of over the -skeleton of . (Compare with the fact that is orientable if and only if is trivializable over the -skeleton.)

**Definition** (The Atiyah-Singer Dirac Operator)

Suppose has a spin structure with principal -bundle (associated to ). Since acts on on the left and (where has general element with , ) is a subgroup of the group of units , we may define

the -bundle associated to by . Since is a module, is a Clifford module bundle.

Some words: By lifting the Levi-Civita connection on one obtains a connection on , and hence (see who’s blog?) a covariant derivative on which makes it Clifford compatible

as a graded Clifford module bundle.

We may then define the ** Atiyah-Singer Dirac Operator** by

.

Some more words:

If then is called a harmonic spinor.

If has positive scalar curvature, then is injective. So if we have ways to compute the index (using the ASHI theorem, for example), we may be able to deduce that does not admit a metric of positive scalar curvature.

**Glossary**

(to include links to other blog posts)

** Differential Operator (global definition) **

If and are vector bundles over (of the same dimension), we define the family of differential operators of order from to by

with . In particular, .

** Symbol of a differential operator**

(cf. Juanita’s blog) We recall the definition of the symbol of an order- differential operator . Denote , , so . Let . Let . The symbol of at is the homomorphism defined by

where , and is such that .

** Covariant Derivative**

A covariant derivative on a vector bundle is a map , where is an -linear map satisfying the Leibnitz rule for and .

** Graded Algebras **

(cf Prasit’s blog) A -algebra is -graded if such that . A -algebra is filtered over

if such that and . A graded algebra defines a filtered algebra by taking , and conversely a filtered algebra defines a graded algebra by taking (with ).

** Tensor Algebra and the Clifford Algebra **

If is a vector space, the tensor algebra has multiplication defined by concatenation, i.e., . Thus is a graded algebra with , and filtered with .

Recall that with . (Here, is a quadratic form; sometimes it is convenient to refer instead to the associated symmetric bilinear form .) For any , so maps to under the quotient . This sets up a natural identification between and . Whence the associated graded algebra for is isomorphic to . In particular, they are isomorphic as vector spaces, with dimension .

** Clifford bundle**

The Clifford bundle has fibers (), where is a Riemannian metric on . (The latter isomorphism is given by identifying an orthonormal (with respect to ) basis for with the standard generators .) Just as is the -bundle associated to the orthogonal frame bundle (of the tangent bundle) over , is the associated -bundle to .

]]>As a provisional definition, *clifford algebra*over a field can be defined as

where,

Easy to see that has dimension . can also be thought of as

where as vector space, is the tensor algebra, is same as above, except that ‘s are standard basis for . This observation leads to a more general definition of *clifford algebra*, where is a vector space equipped with a symmetric bilinear form

Definition 1Let be a vector space with symmetric biliear form and quadratic form . Then theclifford algebraover can be defined as

Remark 1These are some of the properties that enjoys

- There is a natural inclusion of .
- If then the exterior algebra
- Let denote the
clifford multiplication( induced by the tensor product of ), thenUniversal Property :Let , where is a -algebra, such that , then there exists an unique map such that , that is the following diagram commutes- A map , such that , extends to a -algebra homomorphismThus the orthogonal group has an action on

Proposition 2is a filtered algebra whose associate graded is

Before proving the theorem, recall the following definition

Definition 3If is a -algebra then a filtration of is sequence of subspaces

Theassociate gradedof is defined as

*Proof:* Let be the quotient map

Define, ( fold tensor product). Define filtration on by setting

Define filtration be the filtration on the clifford algebra. Note . Hence, in the associated graded . On the other hand the relation prevails in the associated graded. Hence the associate graded is isomorphic to .

Remark 2is a -graded algebra.

Definition 4Recall,. Define,

and

On we have an involution map, which is induced by the involution on given by,

If then

Let and , then observe

Lemma 5There exist short exact sequences

and

where is the map which sends

Let be a field. Recall, tensor product of -algebras and is a -algebra, denoted by and multiplication is given by

moreover if denotes the set of all matrices. Then we have the following isomorphism

Define

Remark 3If , then

This follows from the fact that the quadratic forms and induces isomorphic innerproduct structure on where the isomorphism sends

Theorem 6If , then we have the following isomorphisms

Corollary 7(Bott Periodicity) As a consequence of (iv) we have

if even, and

if is odd.

To work out the case when the underlying field is . For any field we have the following isomorphisms.

Lemma 8For any field

*Proof:* Let denote the standard basis of and cannonical generatoring set of the *Clifford algebra* .

- To get the first isomorphism we simply produce a map given by sendingandIt is easy to check that the above map is an isomorphism.
- is similar to .

One can explicitly check some of the lower dimension cases( ). Then one can repeatedly use the isomorphisms in previous lemma. One has to work upto dimension when , before one sees the patern, which is called the *Bott periodicity*. TSome of the calculations are as follows calculations are as follows

- In general one gets,

Putting all these observations together we get

Theorem 9TheBott periodicityin case of real number looks like

**Lemma 10** * As -algebras Proof:The isomorphism is given explicitly by the map induced by sending*

*Easy to check that this is an isomorphism of algebras. *

**Hodge Star Operator**

**Lemma 1** There is a unique map s.t. for any

**PROOF**. (Uniqueness) suppose we have another map , then

for every

so for every ,i.e.

(Existence) Fix an oriented orthonormal basis , for

we define

We have and

Suppose is an oriented closed Riemannian Manifold, are forms,define the inner product by

Using integration by parts and stokes theorem,we have the following equalities:

hence we yield

**Lemma 2** The formal adjoint of is ,i.e..

**Exercise** Define ,show that

**Corollary**

**Harmonic Form and Signature**

If is an and char,then

where denote the eigenspace of

If ,,then

**Theorem 1 **

**PROOF** using corollary,we could define the following bilinear form:

Let be the eigenspace of . For ,,we have:

Hence there is a decomposation

and is positive definite on and negative definite on

Using Hodge-de Rham isomorphism

the above non degenerate bilinear form is equivalent to the intersection form:

So we have

**Signature Operator**

If ,then

Let be the complex-valued forms,define

we have and

so if we write ,where denotes the -eigenspace of ,then interchanges , due to its anti-commutitivity with , i.e.:

**Definition** is called the *signature operator*

**Theorem 2 **

**PROOF ** we have following facts:

- is elliptic,hence is elliptic,so are and . and are finite
- is self-adjoint,so
- =.so consists of harmonic forms for the eigenvectors of
- for

Using these facts,we yield:

]]>**Definition. **Let and be inner product spaces. Let be a linear map. Then a linear map is called the formal adjoint of if for any and any .

**Lemma. ** (1) If a formal adjoint exists, it is unique. (2) If , then exists.

**Example.** The map defined by summing the coodinates has no formal adjoint, where is the colimit of .

If and are Hilbert spaces, then we have

**Theorem.** Any continuous linear map of Hilbert spaces has a formal adjoint.

**Example. ** Let be a differential operator. Suppose and have smooth inner product structures, then we have the “-inner products” on and , given by . Then has a formal adjoint . If we write locally as , then .

An elliptic operator is *self-adjoint* if .

We can now state the Fundamental Theorem of Elliptic Operators. Later in this entry we will give some corollaries and much later in the course we will outline a proof using the method of elliptic regularity.

**Fundamental Theorem of Elliptic Operators.** For a self-adjoint elliptic operator , there is an orthogonal decomposition with finite-dimensional.

It is important here that the manifold is closed.

** The algebraic Hodge theorem **

Suppose now we have (co)chain complex over or :

Give each an inner product. Assume each has a formal adjoint . Define the Laplacian . Then we have

**Lemma.** iff and .

*Proof. *Suppose . Then we have

and hence .

**Theorem. **Let be a (co)chain complex over a field . Then there exist decompositions such that the (co)chain complex can be written as

When or and is finite dimensional for each , setting and , the theorem above becomes a corollary of the following Algebraic Hodge Theorem:

**Algebraic Hodge Theorem.** Let be a (co)chain complex over or . Suppose that has inner product for each and that formal adjoint exists for each . Let , then

(1) TFAE: (a) , (b) and , (c) .

(2) .

(3) If is finite dimensional, then .

(4) If for any , then there are orthogonal decompositions

*Proof.* (1) (a) (b) (c) (a).

(2) Let , then .

(3) Show the inclusion in (2) is an equality by counting dimensions.

(4) It suffices to show the following orthogonal decomposition:

However easily we have

By checking the decomposition diagram above, we can obtain:

**Corollary.** is an isomorphism, where

**Corollary.** iff is an isomorphism for any .

** Wrapping up **

**Corollary. ** *Algebraic Wrapping up. For as above: * is an isomorphism. Hence is an isomorphism for all p.

This corollary “wraps up” a (co)chain complex into a single map.

Next, we consider wrapping up an ellliptic complex.

**Definition.** An e*lliptic complex of differential operators* is a cochain complex of differential operators

so that for all the associated symbol complex is exact.

If we define the symbol of differential operator of order by , then we have .

**Proposition.** Let be an elliptic complex of differential operators. Give and metrics for each . Then is an elliptic operator.

*Proof.* For any , since the complex is elliptic, we have the exact sequence

Thus, is an isomorphism.

Finally note that .

** Consequences of the fundamental theorem.**

We deduce the following corollaries of the Fundamental Theorem, the Algebraic Hodge Theorem, and Wrapping Up.

**Corollary.** Let be an elliptic complex of differential operators.

(1) For any , is finite dimensional.

(2) For any , .

(3) .

**Corollary.** If is an elliptic differential operator, then we have isomorphisms . Hence the kernel and the cokernel of an elliptic differential operator are finite dimensional.

**Corollary.** If is self-adjoint, then .

**Review of Local Definitions**

Let’s start by recalling that a differential operator of order on the manifold is is defined by:

where is smooth and if , then

Let and . The Symbol of , denoted by is then

A differential operator is said to be elliptic if for all and every we have that is invertible.

**Global definition of the Symbol**

Consider a globally defined differential operator

for and we want to define a linear map

in a coordinate free way.

With this in mind let and choose:

1. such that

2. such that

Then we define

Notice that even though this is a coordinate free definition of the symbol, it is still unclear how it changes in and . We will later see that is actually smooth on . Before this, we should prove that this definition is in fact independent on the choice of and .

* does not depend on *

**Claim 1** If is a smooth function such that and , then

*Proof:* For any differential operator , any section and any function ,

Setting we have

Induction on the order of and (3) will give us the result:

Let , then by definition

and so

Now assume the claim is true for every differential operator of order less than and suppose . By definition,

Thus, by induction

and notice that (3) gives us

so that

* does not depend on *

**Claim 2** Let be such that , then

*Proof:* It is easier if we use the easy direction of Peetre’s Theorem so that we can use the fact that is local, that is

equivalently

equivalently

So, since , we have as sought.

Let us finish the section with a short remark:

is homogeneous of degree in . That is, for every ,

Proof: Simply take instead of in the definition for .

**Local=Global**

**Lemma 1** For a differential operator of order , the two definitions of symbol coincide under the identification given by

*Proof:* Let . The function satisfies the conditions stated in the coordinate free definition of .
Let be the constant section , that is, for every .

Then (2) reads

where by (1)

Notice that here

since is a constant section.

Also notice that

1. for every :
This is because there is always a factor of in the expression for whenever .

2. :
This is a simple calculation.

Consolidating all the information we conclude

**Symbol as a section**

By consolidating definitions (*) and (1) of we get . Here is the bundle map and we are just looking at the diagram

To be explicit, if , then with . So

that is, and we are using the identification .

Smoothness follows from the smoothness of the local definition and the fact that both definitions coincide locally.

Finally, let

then we have

**Proposition 2** There is an exact sequence

Notice that this proposition (re)captures the fact that the symbol of an operator only `sees’ the `top’ degree of the operator.

**Fundamental Theorem of Elliptic Operators**

Now that we have a global definition of the symbol of a differential operator, we can state what it means for a differential operator to be elliptic. Namely, is elliptic if for every (i.e is in the complement of the zero section of the cotangent bundle), the map is invertible. The most important result involving elliptic operators is the following theorem:

**Theorem 3** Fundamental Theorem of Elliptic Operators

If is an elliptic differential operator over a compact manifold , then both and are finite dimensional vector spaces.

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 .

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