用户: Eee/Gauss-Bonnet-Chern theorem

This note is copied from my note of independent study. This note is so complicated that I have no energy to change the calculation, especially the name of some named formulas.

1Differential forms on Riemannian manifold

First we review some basic definition and properties about Riemann geometry, we use to denote the set of differential forms and to denote the set of smooth vector fields.

Defition 1.1. Given a Riemannian manifold with Levi-Civita connection , choose a point and a neighborhood , assume is a local frame field on with as its dual frame field, then we have

We define where . Then are called the connection forms. We set and also call it the connection form.

Define curvature , where are vector fields on and set then is a 2-form on are are called the curvature forms. We set and also call it the curvature form.

Then we shall prove the structure equations.

Proposition 1.2. (1)

Proof. For any , we have

Notice that the Levi-Civita connection is torsion-free, then we have then we get the formula .

Proposition 1.3. (2)Componentwise, this is

Proof. For any , we have

Note that which gives the formula .

Proposition 1.4. The first Bianchi identity: (3)

Proof. From , we have which proves the first Bianchi identity.

Proposition 1.5. The second Bianchi identity: (4)Componentwise, that is

Proof. From , we have which gives the second Bianchi identity in the matrix form.

Proposition 1.6. Both and are anti-symmetry matrix with respect to the Levi-Civita connection, namely and .

Proof. Since the Levi-Civita connection is compatible with the metric, then we have where is an arbitrary smooth vector field on . This shows the anti-symmetry of .

For , we have which shows the anti-symmetry of .

Then we show how behave when the frame undergoes a orthonormal transformation.

Proposition 1.7. In a neighborhood of in which the same system of coordinates is valid, which is still denoted by , let be changed to according to the orthonormal transformation: , where is an orthonormal matrix, whose element are functions of the coordinates.

Suppose be formed from the frames in the same way as are formed from . Then we have (5)where denotes the transposition of .

Proof. For any , then we have namely Then we have

For curvature form, we have

Notice that and , then we have and , substitute into the last long formula, we get

2Gauss-Bonnet-Chern theorem.

Defition 2.1. The dimension of is denoted by and called the dimensional Betti number of . is usually called the Euler number or Euler characteristic or Euler-Poincaré characteristic. For an dimensional manifold , we have

Theorem 2.2. The Euler characteristic of an odd-dimensional closed manifold is .

Theorem 2.3. Gauss-Bonnet-Chern theorem.

Let be an oriented ()-dimensional closed manifold. Then for the curvature form of the Levi-Civita connection, we have where

Lemma 2.4. By unit sphere bundle , we mean the sub-bundle of the tangent bundle of formed by its unit vectors. We denote the projection from to by . We say that is a closed differentiable manifold.

Proof. Clearly, it has no boundary. It suffices to show that is compact.

We introduce a lemma. For any , assume that is a compact subset of . If is a neighborhood of , then there exist neighborhoods of and respectively, s.t. (Choose small enough such that ).

For any , is an inner point of , then we could find the neighborhoods of and , s.t. (Let small enough such that ). Since is an open covering of , by the compactness of , there exists a finite sub-covering . Set , , then and are the neighborhoods of and respectively which satisfy . Then we compete the proof of the lemma.

Back to the proof of compactness of . Let be an open covering of . For any point , is compact. Since that is an open covering of , it has a finite sub-covering whose union is a neighborhood of . By lemma, there exists neighborhood of such that . Then we get is covered by finite open sets in . Notice that is an open covering of compact space , then it has a finite sub-covering , i.e. . Note that every is covered by finite open sets in , then we get is covered by finite open sets in , which means that has a finite sub-covering of . Then we’ve proved that is compact.

Proof. We give the proof of S.S. Chern and fulfill the details.

First from Definition 2.8, we say that is intrinsic.

We define in a continuous field of unit vectors with a point of as the only singular point.1 For a non-singular point , there exists a neighborhood s.t. . We set , then we have

Note that we have locally. Then we consider as a manifold, setting . The equality is from the truth that is decided by if the function is fixed.

Then we set , where . We have then (6)

From , we see (7)

Differential , we get namely, (8)

As to the effect of a change of frame on the components , it is evidently given by the equation

We now construct the following two sets of differential forms: (9)

(10)where .

The forms are of degree and of degree , and we remark that differs from only by a numerical factor.

From the behave of , and under the change of frame, we see that and are intrinsic and are therefore defined over the entire Riemannian manifold but . For details, we fix , then we choose the , , row of to construct a new matrix, denoted by . State that if there exist s.t. . Otherwise, we use to denote the permutation from to , then .

Notice that then we just need to consider the case that are pairwise different. Then Thus, we conclude that is intrinsic. Similarly, is intrinsic.

Using the property of skew-symmetry of the symbol of , we can write substitute from formula (1.1.4), (3.2.1) and (3.2.3), the resulting expression for will then consist of terms of two kinds, those involving and those not. We collect the terms not involving , which are

(11)

Note that

Then equation (3.2.6) changes into (12)Since and are intrinsic, this expression is intrinsic.

Then we consider a special connection. In a neighborhood of we can choose a family of frames such that at , (This process is "equivalent" to the use of geodesic coordinates in tensor notation.)

Notice that the difference of and formula (3.2.7) contains a factor in each of its terms, then the difference becomes at , namely for this particular family of frames. It follows that they are identical, since both expression are intrinsic and the point is arbitrary.

We fix and consider the case , where , where we exchange and at the third row. Then we take , we get

We fix and consider the case , where , where we exchange and at the third row. Then we take , we get

Thus we have

To transform the expression (3.2.7) we shall introduce the abbreviations

Note that where . Then we rewrite (13)where .

Then we need to calculate . From , we see

Notice that

For , we have

For , where .

For , where .

Thus, we have which gives (14)

From equation , we see

For the first term ,

For the second term ,

Notice that , then for ,

For , where , we have

For , where , we have

From the calculation above, we get and hence (15)

From the formula (3.2.8), (3.2.9) and (3.2.10), we see rewrite in the form (16)where .

From the formula (3.2.11) we can solve in terms of . The result is easily found to be where . Then we get (17)

In particular, let , we have It follows that is the exterior derivative of a form : (18)where (19)

Basing on the formula (3.2.13) we shall give a proof of Gauss-Bonnet theorem, under the assumption that is a closed orientable Riemannian manifold.

By Poincaré-Hopf theorem, the index of the field at is equal to , the Euler-Poincaré characteristic of .

This vector field could be treated as a smooth section of the sphere bundle except the point . This vector field defines in a submanifold , which has as boundary , where is the -dimensional cycle formed by all the unit vectors through . Remark that could be any numbers satisfying .

The integral of over is equal to the same over .

Applying Stokes’s theorem, we get therefore

Note that Since the variables of the integral domain are merely, we could ignore the terms who has in the integration. Thus, (20)

From the definition of we have

Substitute , we get we open the brackets and rearrange it into two kinds, those involving and those not. We only care the part which doesn’t contain , namely due to the variables of the integral domain are merely . By basic knowledge of multivariable calculus, we see the last sum is the volume element of the -dimensional unit sphere.

The volume of a -dimensional unit sphere is the surface area of a -dimensional unit ball, which is , therefore Substitute this into (3.2.15), we get .