An Introduction to the Generalized Gauss-Bonnet-Chern Theorem

Abstract

This work studies the mathematical structures which are relevant to differentiable manifolds needed to prove the Gauss-Bonnet-Chern theorem. These structures include de Rham cohomology vector spaces of the manifold, characteristic classes such as the Euler class, pfaffians, and some fiber bundles with useful properties. The paper presents a unified approach that makes use of fiber bundles and leads to a non-computational proof of the Gauss-Bonnet-Chern Theorem. It is indicated how it can be generalized to manifolds with boundary.

1. Introduction

One of the great achievements of differential geometry is the Gauss-Bonnet theorem. In its original form, the theorem is a statement about surfaces which connect their geometry in the sense of curvature to the underlying topology of the space, in the sense of the Euler characteristic [1, 2, 3, 4, 5]. The most elementary case of the theorem states that the sum of the angles of a triangle in the plane is $\pi$ radians. If the surface is deformed, the Euler characteristic does not vary as it is a topological invariant, while the curvature at certain points does change [6, 7, 8]. The theorem states that the total integral of the curvature remains the same, no matter how the deformation is performed. If there is a sphere with a ding, its total curvature is $4\pi$ since its Euler characteristic is two. This is the case no matter how big or deep the actual deformation is. A torus has Euler characteristic zero, so its total curvature must also be zero. If the torus carries the usual Riemannian metric from its embedding in ${\mathrm{ℝ}}^3$, then the inside has negative Gaussian curvature, and the outside has positive Gaussian curvature, so the total curvature is zero. It is not possible to specify a Riemannian metric on the torus which has everywhere positive or everywhere negative Gaussian curvature. Manifolds $M$ have dimension $n$ unless stated otherwise [9, 10, 11]. There are many applications of this theorem in both mathematics and mathematical physics such as in gravity [12, 13, 14], string theory [15] and even in the study of Ricci flow [16].

Although the curvature $K$ is defined intrinsically in terms of the metric on the manifold $M$. It can also be defined for $n=2$ extrinsically when the metric on $M$ is induced by an embedding $M\subset {\mathrm{ℝ}}^3$. In fact, it $\nu :M\to S^2$ is the normal map and $\mathit{da}$ is the volume element on $S^2$, then $K\mathit{d\sigma }={\nu }^{\ast }\mathit{da}$ so that

$${\int }_{M}K\mathit{d\sigma }={\int }_M{\nu }^{\ast }\left(\mathit{da}\right)=deg\left(\nu \right)\cdot {\int }_{S^2}\mathit{da}=4\pi \cdot deg\left(\nu \right).$$E1

Without bringing in differential geometric considerations, it is seen to be the case that $deg\left(\nu \right)=\left(1/2\right)\chi \left(M\right)$, where $\chi \left(M\right)$ is the Euler characteristic of $M$. Using this fact in (1), the Gauss-Bonnet theorem for a compact oriented surface $M$, the first version of the theorem is obtained for the case in which the metric on $M$ arises by means of an embedding in ${\mathrm{ℝ}}^3$

$${\int }_{M}K\mathit{d\sigma }=2\mathit{\pi \chi }\left(M\right).$$E2

It is the intention here to state and prove a general version of the theorem which applies to manifolds of even dimension, so a surface with $n=2$ is a special case. An intrinsic proof of the theorem was obtained by Chern $1944$. The kind of argument outlined above was used by Hopf in developing the first generalization of the theorem. To outline the basic idea, consider a compact surface $M^n\subset {\mathrm{ℝ}}^{n+1}$ when $n$ is even. If $d{\mu }_g$ is the volume form on the manifold and ${\mathit{ds}}_n$ denotes the volume element $S^n$, then

$${\int }_{M^n}{K}_{n}d{\mu }_g={\int }_{M^n}{\nu }^{\ast }d{s}_n=\mathrm{vol}\left(S^n\right)\cdot deg\left(\nu \right)=\frac{1}{2}\mathrm{vol}\left(S^n\right)\cdot \chi \left(M^n\right).$$E3

This can be extended to any compact oriented Riemannian $n$-manifold $\left(M^n,g\right)$ which has even dimension, where $K_n$ in a coordinate system is given by

$$K_n=\frac{1}{2^{n/2}n!}\sum _{\begin{array}{c}{i}_1,\dots ,i_n\\ j_1,\dots ,j_n\end{array}}{R}_{i_1{i}_2{j}_1{j}_2}\cdots R_{i_{n-1}{i}_n{j}_{n-1}{j}_n}\frac{1}{\sqrt{g}}{\epsilon }^{i_1\dots i_n}\cdot \frac{1}{\sqrt{g}}{\epsilon }^{j_1\dots j_n}.$$E4

The $\sqrt{g}$(4) is the square root of the determinant of the metric. With $K_n$ given by (4), and ${\mu }_g$ the volume form on the manifold, we are then led to conjecture that

$${\int }_{M^n}{K}_{n}d{\mu }_g=\frac{1}{2}\mathrm{vol}\left(S^n\right)\cdot \chi \left(M^n\right),$$E5

where $M$ is a compact, oriented Riemannian manifold with $n$ even.

It is the objective to look at and study some of the ensuing developments which have led to a much deeper understanding of the foundations which underlie this theorem. It will be seen that this development leads to a completely non-computational proof of this deep theorem.

2. Characteristic classes

When an oriented $n$-dimensional manifold $\left(M,d{\mu }_g\right)$ is compact and closed, with $d{\mu }_g$ is the volume form and $\mu$ the orientation of $M$, so every form has compact support, Stokes theorem leads to the important theorem. Let $\eta$ be any $\left(n-1\right)$-form then

$${\int }_M\mathit{d\eta }={\int }_{\partial M}\eta .$$E6

Therefore an $n$-form $\omega$ on $M$, which is not exact, even though it must be closed as all $n$-forms on $M$ are zero, can be found simply by locating an $\omega$ such that

$${\int }_M\omega \ne 0.$$E7

Such a form always exists, as it is known there is a form $\omega$ such that for $v_1,\dots ,v_n\in M_p$, $\omega \left(v_1,\dots ,v_n\right)>0$ if $\left[v_1,\dots ,v_n\right]={\mu }_p$. If $c:{\left[0,1\right]}^n\to \left(M,\mu \right)$ preserves orientation, $c^{\ast }\omega$ on ${\left[0,1\right]}^n$ is $g{\mathit{dx}}^1\wedge \cdots \wedge {\mathit{dx}}^n$ for some $g>0$ on ${\left[0,1\right]}^n$, hence ${\int }_c\omega >0$. This observation leads to this theorem. A smooth, oriented manifold is not smoothly contractible to a point. In fact, it is the shape of $M$ not the size which determines whether or not every closed form on $M$ is exact. More information about the shape of $M$ can be obtained by analyzing more closely the extent to which closed forms are not exact. So how many non-exact $n$-forms are there on a compact oriented $n$-manifold If $\omega$ is not exact, the same holds for $\omega +\mathit{d\eta }$ for $\eta$ any $\left(n-1\right)$-form $\eta$. Thus it is necessary to regard $\omega$ and $\omega +\mathit{d\eta }$ as equivalent. This suggests an equivalence relation and directs one to think of this in terms of quotient spaces.

For each $k$, $Z^k\left(M\right)$ denotes all closed $k$-forms on $M$ and it is a vector space. The space $B^k\left(M\right)$ of all exact $k$-forms is a subspace since $d^2=0$. The quotient space is called the $k$-dimensional de Rham cohomology vector space of $M$ and is defined to be

$$H^k\left(M\right)=Z^k\left(M\right)/B^k\left(M\right).$$E8

The theorem of de Rham states that the vector space is isomorphic to a vector space defined just in terms of the topology of $M$ called the $k$-dimensional cohomology group of $M$ with real coefficients.

An element of $H^k\left(M\right)$ is an equivalence class $\left[\omega \right]$ of a closed form $\omega$ such that closed forms ${\omega }_1$ and ${\omega }_2$ are equivalent if and only if the difference is exact. In terms of these vector spaces, the Poincaré lemma gives $H^k\left({\mathrm{ℝ}}^n\right)=0$, the vector space consisting of just the zero vector if $k>0$, or $H^k\left(M\right)=0$ if $M$ is contractible and $k>0$. To compute $H^0\left(M\right)$ note $B^0\left(M\right)=0$ as there are no non-zero exact $0$-forms as there are no non-zero minus one forms. Thus $H^0\left(M\right)$ is the same as the vector space of all $C^{\infty }$ functions $f:M\to \mathrm{ℝ}$ with $\mathit{df}=0$. If $M$ is connected, this condition implies $f$ is constant so $H^0\left(M\right)\equiv \mathrm{ℝ}$ and its dimension is the number of components of $M$.

The de Rham cohomolgy vector spaces with compact support $H_c^k\left(M\right)$ are defined similarly to (8), that is, $H_c^k\left(M\right)=Z_c^k\left(M\right)/B_c^k\left(M\right)$, where $Z_c^k\left(M\right)$ is the vector space of all closed $k$-forms with compact support and $B_c^k\left(M\right)$ all $k$-forms $\mathit{d\eta }$ where $\eta$ is a $k$-form with compact support. If $M$ is compact $H_c^k\left(M\right)=H^k\left(M\right)$.

Theorem 2.1. (The Poincaré-Duality Theorem) If $M$ is a connected, oriented $n$-manfold of finite type, then the map

$$\mathrm{\Pi }:H^k\left(M\right)\to H_c^{n-k}\left(M\right)$$E9

is an isomorphism for all $k$. $\square$

This theorem eventually motivates the introduction of the Euler characteristic for any smooth connected oriented manifold $M$. Consider then a smooth $k$-dimensional vector bundle $\xi =\pi :E\to M$ over $M$. Orientations $\mu$ for $M$, and $\nu$ for $\xi$ give an orientation $\mu \oplus \nu$ for the $\left(n+k\right)$-manifold $E$, since $E$ is locally a product. Let $\left\{U_1,\dots ,U_r\right\}$ be a cover of $M$ by geodesically convex sets so small that each bundle $\xi$ restricted to $U_i$ is trivial. Then $\left\{{\pi }^{-1}\left(U_1\right),\dots ,{\pi }^{-1}\left(U_r\right)\right\}$ turns out to be a nice cover for $E$, so it is a manifold of finite type. For the section and projection maps $s,\pi$, $\pi \circ s=I$ on $M$ and $s\circ \pi$ is smoothly homotopic to the identity of $E$, so the map ${\pi }^{\ast }:H^l\left(M\right)\to H^l\left(E\right)$ is an isomorphism for all $l$. The reason for mentioning (6) and Theorem 2.1 is that it shows there is a unique class $U\in H_c^k\left(E\right)$ such that

$${\pi }^{\ast }:\mu \cup U=\mu \oplus \nu \subset H_c^{n+k}\left(E\right).$$E10

This class is called the Thom class of $\xi$.

A theorem states that if $\left(M,\mu \right)$ is a compact oriented, connected manifold $\xi =\pi :E\to M$ an oriented $k$-plane bundle over $M$ orientation $\nu$, the Thom class $U$ is the unique element of $H_c^k\left(E\right)$ such that for all $p\in M$, and $j_p:F_p\to E$ the inclusion map, we have $j_p^{\ast }U={\nu }_p$. This condition has the implication that ${\int }_{\left(F_p,{\nu }_p\right)}{j}_p\omega =1$, where $U$ is the class of closed form $\omega$.

The Thom class $U$ of $\xi =\pi :E\to M$ can now be used to determine an element of $H^k\left(M\right)$. Let $s:M\to E$ be any section. There is always one, any two are clearly homotopic. Define the Euler class $\chi \left(E\right)\subset H^k\left(M\right)$ of $\xi$ by

$$\chi \left(\xi \right)=s^{\ast }U.$$E11

If $\xi$ has a non-zero section $s:M\to E$ and $\omega \in C_c^k\left(M\right)$ represents $U$, a suitable multiple $c\cdot s$ of $s$ takes $M$ to the complement of support $\omega$, so in this case, $\chi \left(\xi \right)={\left(c\cdot s\right)}^{\ast }U=0$.

The term Euler class is connected with the special case of the bundle $\mathit{TM}$ which has sections which are vector fields on $M$. If $X$ is a vector field on $M$ having an isolated zero at some point $p$, $X\left(p\right)=0$, but $X\left(q\right)\ne 0$ for $q\ne p$ in a neighborhood of $p$. An index of $X$ at $p$ can be defined. Suppose $X$ is a vector field on an open set $U\subset {\mathrm{ℝ}}^n$ with an isolated zero at $0\in U$. Define $f_X:U\to \left\{0\right\}\to S^{n-1}$ by $f_X\left(p\right)=X\left(p\right)/\mid X\left(p\right)\mid$. If $i:S^{n-1}\to U$ is $i\left(p\right)=\mathit{\epsilon p}$ mapping $S^{n-1}$ into $U$, then the map $f_X\circ i:S^{n-1}\to S^{n-1}$ has a certain degree independent of $\epsilon$ for small $\epsilon$, since maps $i_1,i_2:S^{n-1}\to U$ correspinding to ${\epsilon }_1,{\epsilon }_2$ will be smoothly homotopic. This degree is called the index of $X$ at $0$. Consider a diffeomorphism $h:U\to V\subset {\mathrm{ℝ}}^n$ with $h\left(0\right)=0$, so $h_{\ast }X$ is the vector field on $V$ such that $\left(h_{\ast }X\right)\left(y\right)=h_{\ast }\left(X_{h^{-1}\left(y\right)}\right)$. So $0$ is an isolated zero of $h_{\ast }X$. It can be shown, if $h:U\to V\subset {\mathrm{ℝ}}^n$ is a diffeomorphism with $h\left(0\right)=0$ and $X$ has an isolated zero at $0$, the index of $h_{\ast }X$ at $0$ equals the index of $X$ at $0$.

As a consequence of this, an index of a vector field on a mainifold can be defined. If $X$ is a vector field on $M$, with isolated zero at $p\in M$, choose a coordinate system $\left(x,U\right)$ such that $x\left(p\right)=p$ and define the index of $X$ at $p$ to be the index of $x_{\ast }X$ at $0$.

Theorem 2.2. Let $M$ be a compact, connected manifold with orientation $\mu$, also an orientation for the tangent bundle $\xi =\pi :\mathit{TM}\to M$. Let $X:M\to \mathit{TM}$ be a vector field with only a finite number of zeros and let $\sigma$ be the sum of indices of $X$ at these zeros. Then

$$\chi \left(\xi \right)=\sigma \cdot \mu \in H^0\left(M\right).$$E12

$$\square$$

3. Pfaffians

An intrinisic expression along with one in a coordinate system for the function $K_n$ on a compact, oriented Riemannian manifold of even dimension has been given already. Another more important way of expressing $K_n$ involves the curvature form ${\mathrm{\Omega }}_j^i$ for a positively oriented orthonormal moving frame $X_1,\dots ,X_n$ on $M$. In terms of these forms, the $n$-form $K_{n}d{\mu }_g$, the one to be integrated, can be written down. A sum over permutations such as

$$\sum _{\pi \in S_n}B\left(X_{\pi \left(1\right)},\dots ,X_{\pi \left(n\right)}\right),$$E13

can be written just as well as

$$\sum _{j_1,\cdots ,j_n}{\epsilon }^{j_1,\cdots ,j_n}B\left(X_{j_1},\dots ,X_{j_n}\right).$$E14

Suppose this is the $n$-fold wedge product

$${\mathrm{\Omega }}_{i_2}^{i_1}\wedge \cdots \wedge {\mathrm{\Omega }}_{i_n}^{i_{n-1}}.$$E15

Since the ${\mathrm{\Omega }}_j^i$ are $2$-forms, using the definition of wedge product,

$$\begin{array}{c}{\mathrm{\Omega }}_{i_2}^{i_1}\wedge \cdots \wedge {\mathrm{\Omega }}_{i_n}^{i_{n-1}}\left(X_1,\dots ,X_n\right)=\frac{\left(2+\cdots +2\right)!}{2!\cdots 2!}\cdot \frac{1}{n!}\sum _{j_1,\cdots j_n}{\epsilon }^{i_1\cdots i_p}{\mathrm{\Omega }}_{i_2}^{i_1}\left(X_{j_1},X_{j_2}\right)\cdots {\mathrm{\Omega }}_{i_n}^{i_{n-1}}\left(X_{j_{n-1}},X_{j_n}\right)\\ =\frac{1}{2^{n/2}}\sum _{j_1,\cdots j_n}{\epsilon }^{j_1,\cdots ,j_n}⟨R\left(X_{j_1},X_{j_2}\right)X_{j_2},X_{j_1}⟩\cdots ⟨R\left(X_{j_{n-1}},X_{j_n}\right)X_{j_n},X_{j_{n-1}}⟩\\ =\frac{1}{2^{n/2}}\sum _{j_1,\cdots ,j_n}{R}_{i_1{i}_2{j}_1{j}_2}\cdots R_{i_{n-1}{i}_n{j}_{n-1}{j}_n}.\end{array}$$E16

Comparing this to (4), it may be concluded that

$$K_n=\frac{1}{2^{n/2}n!}\sum _{i_1,\cdots ,i_n}{\epsilon }^{i_1,\cdots ,i_n}{2}^{n/2}{\mathrm{\Omega }}_{i_2}^{i_1}\wedge \cdots \wedge {\mathrm{\Omega }}_{i_n}^{i_{n-1}}\left(X_1,\cdots ,X_n\right).$$E17

When (17) is multiplied by the volume form $d{\mu }_g$, it becomes

$$K_{n}d{\mu }_g=\frac{1}{n!}\sum _{i_1,\cdots ,i_n}{\mathrm{\Omega }}_{i_2}^{i_1}\wedge \cdots \wedge {\mathrm{\Omega }}_{i_n}^{i_{n-1}}.$$E18

By (18) the form on the right does depend on the choice of the positively oriented orthonormal frame, $X_1,\dots ,X_n$. There is a direct way to get this algebraically.

Suppose $A$ is an $n\times n$ matrix $A=\left(a_{\mathit{ij}}\right)$ with $n=2m$ even. Define the Pfaffian, $\mathrm{Pf}\left(A\right)$ of $A$ to be

$$\mathrm{Pf}\left(A\right)=\frac{1}{2^{m}m!}\sum _{i_1,\cdots ,i_n}{a}_{i_1{i}_2}\cdots a_{i_{n-1}{i}_n}.$$E19

Note that ${\epsilon }^{i_1\cdots i_n}$ does not change when any permutation of the pairs $\left(i_{2l-1},i_{2l}\right)$. For any set $S=\left\{\left(h_1,k_1\right),\cdots ,\left(h_m,k_m\right)\right\}$ of pairs of integers between $1$ and $n$, let us define

$$\epsilon \left(S\right)={\epsilon }^{h_1{k}_1\cdots h_m{k}_m}.$$E20

It is not necessary to specify any ordering of pairs $\left(h_i,k_i\right)$ in $S$. Also a permutation of the pairs $\left(i_{2l-1},i_{2l}\right)$ does not change the factor $a_{i_1{i}_2}\cdots a_{i_{n-1}{i}_n}$. So for each $P$ above define $a_S=a_{h_1{k}_1}\cdots a_{h_m{k}_m}$. If $P$ is the collection of all such $S$, we clearly have

$$\mathrm{Pf}\left(A\right)=\frac{1}{2^m}\sum _{S\in P}\epsilon \left(S\right)a_S.$$E21

Theorem 3.1. Let $n=2m$ then for all $n\times n$ matrices $A$ and $B$,

$$\mathrm{Pf}\left(B^t\mathit{AB}\right)=\left(detB\right)\cdot \mathrm{Pf}\left(A\right).$$E22

and $B^t$ denotes the transpose. If $B\in \mathit{SO}\left(n\right)$ then

$$\mathrm{Pf}\left(B^{-1}\mathit{AB}\right)=\mathrm{Pf}\left(A\right).$$E23

Proof:

$$\begin{array}{c}{2}^m\cdot m!\mathrm{Pf}\left(B^t\mathit{AB}\right)=\sum _{i_1\cdots i_n}{\epsilon }^{i_1\cdots i_n}\sum _{j_1\cdots j_n}\left(b_{j_1{i}_1}{a}_{j_1{j}_2}{b}_{j_2{i}_2}\right)\cdots \left(b_{j_{m-1}{i}_{m-1}}{a}_{j_{m-1}{j}_n}{b}_{j_n{i}_n}\right)\\ =\sum _{j_1\cdots j_n}\left[\sum _{i_1\cdots i_n}{\epsilon }^{j_1\cdots j_n}{b}_{j_1{i}_1}\cdots b_{j_n{i}_n}\right]a_{j_1{j}_2}\cdots a_{j_{n-1}{j}_n}=\sum _{j_1,\cdots ,j_n}{\epsilon }^{j_1,\cdots ,j_n}det\left(B\right)a_{j_1{j}_2}\cdots a_{j_{n-1}{j}_n}\\ =2^{m}m!\left(detB\right)\mathrm{Pf}\left(A\right).\end{array}$$E24

$$\square$$

This theorem was stated for matrices of real numbers, but $\mathrm{Pf}\left(A\right)$ can be defined provided the entries of $A$ are in some commutative algebra over $\mathrm{ℝ}$.

Consider again a positively oriented orthogonal moving frame $X_1,\dots ,X_n$ on $M$, with curvature forms ${\mathrm{\Omega }}_j^i$. For each $p\in M$, the direct sum $A=\mathrm{ℝ}\oplus {\mathrm{\Omega }}^2\left(M_p\right)\oplus {\mathrm{\Omega }}^2\left(M_p\right)\oplus \cdots$ is a commutative algebra over $\mathrm{ℝ}$ under the operation $\wedge$. So one can consider $\mathrm{Pf}\left(\mathrm{\Omega }\left(p\right)\right)$, where $\mathrm{\Omega }\left(p\right)$ is an $n\times n$ matrix of connection $2$-forms at $p$

$$\mathrm{Pf}\left({\mathrm{\Omega }}_p\right)=\frac{1}{2^{m}m!}\sum _{i_1,\cdots ,i_n}{\epsilon }^{i_1,\cdots ,i_n}{\mathrm{\Omega }}_{i_1}^{i_1}\wedge \cdots \wedge {\mathrm{\Omega }}_{i_n}^{i_{n-1}}\left(p\right).$$E25

If $X^{\prime }=X\cdot a$ is another positively oriented orthonormal moving frame then $a\left(p\right)\in O\left(n\right)$ and the corresponding curvature forms satisfy ${\mathrm{\Omega }}^{\prime }=a^{-1}\mathrm{\Omega }a$. Then Theorem 3.1 implies that

$$\mathrm{Pf}\left({\mathrm{\Omega }}^{\prime }\left(p\right)\right)=\mathrm{Pf}\left(a^{-1}\left(p\right)\mathrm{\Omega }a\left(p\right)\right)=\mathrm{Pf}\left(\mathrm{\Omega }\left(p\right)\right)$$E26

so the form ${\sum }_{i_1,\cdots ,i_n}{\epsilon }^{i_1,\cdots ,i_n}{\mathrm{\Omega }}_{i_2}^{i_1}\wedge \cdots \wedge {\mathrm{\Omega }}_{i_n}^{i_{n-1}}$ is well defined.

4. Bundles of paticular importance

Projective $n$-space ${\mathrm{\mathbb{P}}}^n$ can be defined as the set of all pairs $\left(-p,p\right)$ for $p\in S^n\subset {\mathrm{ℝ}}^{n+1}$ or the set of line through $0$ in ${\mathrm{ℝ}}^{n+1}$, since each lines intersects $S^n$ through two anti-podal points. A Grassmannian manifold $G_n\left({\mathrm{ℝ}}^n\right)$ is the set of all $n$-dimensional subspaces of ${\mathrm{ℝ}}^N$ with $N>0$. Over the Grassmannian manifold $G_n\left({\mathrm{ℝ}}^N\right)$, there is a natural $n$-dimensional bundle ${\zeta }^n\left({\mathrm{ℝ}}^n\right)$ constructed as follows. The total space of the bundle $E\left({\zeta }^n\left({\mathrm{ℝ}}^N\right)\right)$ is the subset of $G_n\left({\mathrm{ℝ}}^N\right)\times {\mathrm{ℝ}}^N$ consisting of all pairs

$$\left(W,w\right)\in G_n\left({\mathrm{ℝ}}^N\right)\times {\mathrm{ℝ}}^M,w\in W.$$E27

The projection map which takes $E\left({\zeta }^n\left({\mathrm{ℝ}}^N\right)\right)\to G_n\left({\mathrm{ℝ}}^N\right)$ is $\pi \left(\left(W,w\right)\right)=W$. The fibre ${\pi }^{-1}\left(W\right)$ over $W$ of $G_n\left({\mathrm{ℝ}}^N\right)$ will be $W$ itself or, more explicitly, $\left\{\left(W,w\right):w\in W\right\}$. A vector space structure is defined on ${\pi }^{-1}\left(W\right)$ by using the vector space structure on $W\subset {\mathrm{ℝ}}^N$; if $a$ is a scalar, then $\left(W,w_1\right)+\left(W,w_2\right)=\left(W,w_1+w_2\right)$ and $a\left(W,w\right)=\left(W,\mathit{aw}\right)$. Also ${\zeta }^n\left({\mathrm{ℝ}}^N\right)$ satisfies the local triviality condition.

For $M>N$ there is a natural map $\alpha :G_n\left({\mathrm{ℝ}}^N\right)\to G_n\left({\mathrm{ℝ}}^M\right)$, as an $n$-dimensional subspace of ${\mathrm{ℝ}}^N$ can be considered an $n$-dimensional subspace of ${\mathrm{ℝ}}^M$. There is clearly a map $\overline{\alpha }:E\left({\zeta }^n\left({\mathrm{ℝ}}^N\right)\right)\to E\left({\zeta }^n\left({\mathrm{ℝ}}^M\right)\right)$ such that $\left(\overline{\alpha },\alpha \right)$ is a bundle map from ${\zeta }^n\left({\mathrm{ℝ}}^N\right)$ to ${\zeta }^n\left({\mathrm{ℝ}}^M\right)$ and thus ${\zeta }^n\left({\mathrm{ℝ}}^N\right)\eqsim {\alpha }^{\ast }\left({\zeta }^n\left({\mathrm{ℝ}}^M\right)\right)$.

In algebraic topology, one often considers the union $G_0\left({\mathrm{ℝ}}^{\infty }\right)$ of the increasing sequence $G_n\left({\mathrm{ℝ}}^{n+1}\right)\subset G_n\left({\mathrm{ℝ}}^{n+1}\right)$$\subset \cdots$ with weak topology; that is, a set $U\in G_n\left({\mathrm{ℝ}}^{\infty }\right)={\cup }_l{G}_n\left({\mathrm{ℝ}}^{n+l}\right)$ is open if and only if $U\cap G_n\left({\mathrm{ℝ}}^{n+l}\right)$ is open in $G_n\left({\mathrm{ℝ}}^{n+l}\right)$ for all $l$. There is a natural $n$-dimensional bundle ${\zeta }^n$ over $G_n\left({\mathrm{ℝ}}^{\infty }\right)$ defined in a way similar to ${\zeta }^n\left({\mathrm{ℝ}}^N\right)$ such that the following properties are maintained: $\left(i\right)$ for every bundle $\xi$ over a paracompact space $X$, there is a map $f:X\to G_n\left({\mathrm{ℝ}}^{\infty }\right)$ such that $\xi \simeq f^{\ast }\left({\zeta }^n\right)$. $\left(\mathit{ii}\right)$ if $f_0,f_1:X\to G_n\left({\mathrm{ℝ}}^{\infty }\right)$ are maps of a paracompact space $X$ into $G_n\left({\mathrm{ℝ}}^{\infty }\right)$ with $f_0^{\ast }:{\zeta }^n\simeq f_1^{\ast }{\zeta }^n$ then $f_1\simeq f_0$.

For this reason ${\zeta }^n$ is called the universal$n$-dimensional bundle and $G_n\left({\mathrm{ℝ}}^{\infty }\right)$, is called the classifying space for $n$-dimensional bundles since equivalence classes of $n$-dimensional bundles over $X$ are classified by homotopy classes of maps of $X$ into $G_n\left({\mathrm{ℝ}}^{\infty }\right)$. Now $G_n\left({\mathrm{ℝ}}^{\infty }\right)$ is not a manifold so we continue to use the bundles ${\zeta }^n\left({\mathrm{ℝ}}^N\right)$, which are usually called universal bundles.

An orientation for a vector space $V$ is an equivalence class of ordered bases for $V$ where $\left(v_1,\dots ,v_n\right)\sim \left(w_1,\dots ,w_n\right)$ if and only if $\left(a_{\mathit{ij}}\right)$ defined by $w_i={\sum }_j{a}_{\mathit{ji}}{v}_j$ has $det\left(a_{\mathit{ij}}\right)>0$. There are only two such equivalence classes $\eta$ and $-\eta$. An oriented vector space is a pair $\left(V,\eta \right)$, where $\eta$ is an orientation for $V$.

An orientation for a bundle $\xi =\pi :E\to X$ is a collection $\eta =\left\{{\eta }_x\right\}$ of orientations for the fibres ${\pi }^{-1}\left(x\right)$ which satisfy an obvious compatibility requirement, while an oriented bundle is a pair $\left(\xi ,\eta \right)$, where $\eta$ is an orientation for $\xi$. Orientation $\eta$ of $\xi$ gives another $-\eta =\left\{-\eta \right\}$ if $X$ is connected. This is the only other one for $\xi$. Define $\left({\xi }_1\oplus {\xi }_2,{\mu }_1\oplus {\mu }_2\right)$ to be the sum ${\xi }_1\oplus {\xi }_2$ with the indicated orientation.

Suppose $\xi =\pi :E\to M$ is a smooth oriented $n$-dimensional vector bundle over a smooth manifold $M$ of any dimension. The Euler class $\chi \left(\xi \right)\in H^n\left(M\right)$ was defined by first defining the Thom class $U\left(\xi \right)\in H_c^n\left(E\right)$. It can be proved $U\left(\xi \right)$ is the unique class whose restriction to each ${\pi }^{-1}\left(p\right)$ is the generator ${\nu }_p=H_c^n\left({\pi }^{-1}\left(p\right)\right)$ determined by the orientation. This result leads directly into the next theorem.

Theorem 4.1. Let $\xi =\pi :E\to M$ be a smooth manifold where $M^{\prime }$ is also a compact manifold. If $E$ is the total space of $f^{\ast }\xi$ and $\stackrel{\sim }{f}:E^{\prime }\to E$ is a bundle map covering $f$,

$${\stackrel{\sim }{f}}^{\ast }\left(U\left(\xi \right)\right)=U\left(f^{\ast }\xi \right)\in H_c^n\left(E^{\prime }\right).$$E28

Proof: Note $\stackrel{\sim }{f}$ has the property inverse of a compact set is compact, so ${\stackrel{\sim }{f}}^{\ast }$ takes $H_c^n\left(E\right)$ to $H_c^n\left(E^{\prime }\right)$. Let $f^{\ast }\xi$ be ${\pi }^{\prime }:E^{\prime }\to M^{\prime }$. If $p\in M^{\prime }$ is any point, and $i_{p^{\prime }}:{\pi }^{\text{'}-1}\left(p\right)\to E^{\prime }$ is the inclusion map, then

$$i_{p^{\prime }}^{\ast }\stackrel{\sim }{f}U\left(\xi \right)={\left(\stackrel{\sim }{f}\circ i_{p^{\prime }}\right)}^{\ast }U\left(\xi \right).$$E29

Recall how $f^{\ast }\xi$ is defined then ${\left(\stackrel{\sim }{f}\circ i_{p^{\prime }}\right)}^{\ast }U\left(\xi \right)$ must be the generator of $H_c^n\left({\pi }^{\text{'}-1}\left(p^{\prime }\right)\right)$, since $i_{f\left(p^{\prime }\right)}^{\ast }U\left(\xi \right)$ is the generator of $H_c^n({\pi }^{-1}\left(f\left(p^{\prime }\right)\right)$. This shows ${\stackrel{\sim }{f}}^{\ast }U\left(\xi \right)$ must be $U\left(f^{\ast }\xi \right)$. $\square$

The Euler class $\chi \left(\xi \right)$ was defined as $s^{\ast }U\left(\xi \right)$ for any section $s$ of $\xi$. Suppose $s=0$ is the zero section, which is chosen. It can be shown that

$$f^{\ast }\chi \left(\xi \right)=\chi \left(f^{\ast }\xi \right)\in H^n\left(M^{\prime }\right).$$E30

A consequence of (30) is important as it gives the following.

Theorem 4.2. If $n$ is even, then

$$\chi \left({\stackrel{\sim }{Q}}^n\left({\mathrm{ℝ}}^N\right)\right)\ne 0,N>n.$$E31

Proof: Since $S^n\subset {\mathrm{ℝ}}^N$ for $N>n$, we have a bundle map $\left(\stackrel{\sim }{f},f\right):{\mathit{TS}}^n\to E\left({\stackrel{\sim }{\zeta }}^n\left({\mathrm{ℝ}}^N\right)\right)$ so

$$\chi \left({\mathit{TS}}^n\right)=f^{\ast }\chi \left({\stackrel{\sim }{\zeta }}^n\left({\mathrm{ℝ}}^N\right)\right).$$E32

However, it is known to be the case that $\chi \left({\mathit{TS}}^n\right)$ is $\chi \left(S^n\right)$ times the fundamental class of $S^n$ and $\chi \left(S^n\right)=2\ne 0$. $\square$

5. A unique one-form constructed from the curvature

This is an important characteristic class which is important and plays a significant role. Consider principal bundles associated with a smooth oriented $n$-dimensional vector bundle $\xi$ over a smooth manifold $M$. There is the principal bundle of frames $F\left(\xi \right)$ of $E$. If $\xi$ has a Riemannian metric $⟨,⟩$, the bundle $O\left(E\right)$ of orthonormal frames can be considered, which is a principal bundle with group $O\left(n\right)$. Since only paracompact $M$ are considered here, there is an Ehresmann connection $\omega$ on $O\left(E\right)$. Thus $\omega$ is a matrix of one-forms $\left({\omega }_j^i\right)$ on $O\left(E\right)$ with values in $o\left(n\right)$, the curvature form $\mathrm{\Omega }=\mathit{D\omega }$ is a matrix of two-forms $\left({\mathrm{\Omega }}_j^i\right)$ with values in $o\left(n\right)$. A connection $\omega$ on $O\left(E\right)$ is equivalent to a covariant derivative operator on $E$ compatible with the metric, and for a general $\xi$ over $M$, there will be many connections compatible with the metric. One can not be singled out by requiring a symmetric connection which only makes sense for the tangent bundle. As $\xi$ is oriented, we can also consider the bundle $\mathit{SO}\left(E\right)$ of positively oriented frames. If $X$ is connected, it is simply one of the two components of $O\left(E\right)$, with group $\mathit{SO}\left(n\right)$ and Lie algebra $o\left(n\right)$. A connection $\omega$ on $\mathit{SO}\left(n\right)$ again has values in $o\left(n\right)$ as does the matrix of two-forms $\mathrm{\Omega }$.

If we specialize to the case of a smooth oriented $n$-dimensional vector bundle $\xi =\pi :E\to M$ over $M$, with $n=2m$ even. If $⟨,⟩$ is a Riemannian metric for $\xi$ and $\omega$ is a connection on the corresponding principal bundle $\overline{\omega }:\mathit{SO}\left(E\right)\to M$, consider the $n$-form which is defined on $\mathit{SO}\left(E\right)$

$$2m\cdot m!\mathrm{Pf}\left(\mathrm{\Omega }\right)=\sum _{i_1,\cdots ,i_n}{\epsilon }^{i_1,\cdots ,i_n}{\mathrm{\Omega }}_{i_2}^{i_1}\wedge \cdots \wedge {\mathrm{\Omega }}_{i_n}^{i_{n-1}}.$$E33

The following is an invariant formulation of a previous theorem.

Theorem 5.1. There is a unique $n$-form $\mathrm{\Lambda }$ on $M$ such that

$${\overline{\omega }}^{\ast }\left(\mathrm{\Lambda }\right)=\sum _{i_1,\cdots ,i_n}{\epsilon }^{i_1,\cdots ,i_n}{\mathrm{\Omega }}_{i_2}^{i_1}\wedge \cdots \wedge {\mathrm{\Omega }}_{i_n}^{i_{n-1}}=2^{m}m!\mathrm{Pf}\left(\mathrm{\Omega }\right).$$E34

Proof: Let $X_1,\cdots ,X_n\in M_p$$Y_1,\cdots ,Y_n\in \mathit{SO}{\left(E\right)}_u$ be tangent vectors such that $\pi Y_i\to X_i$, and choose some $u\in {\omega }^{-1}\left(p\right)$. Then form $\mathrm{\Lambda }$ must satisfy

$$\mathrm{\Lambda }\left(X_1,\cdots ,X_n\right)=2^m\cdot m!\mathrm{Pf}\left(\mathrm{\Omega }\right)\left(Y_1,\cdots ,Y_n\right)$$E35

This suffices to give uniqueness. If it can be shown this $\mathrm{\Lambda }$ in (35) is well-defined, then existence can be established.

Suppose different tangent vectors $Z_1,\cdots ,Z_n$ are taken such that ${\overline{\omega }}_x{Z}_i=X_i$. Since ${\overline{\omega }}_x\left(Y_i-Z_i\right)=0$, all $Y_i-Z_i$ are vertical. However, $\mathrm{\Omega }\left(Y,Z\right)=0$ if either $Y$ or $Z$ is vertical. Consequently,

$$\begin{array}{c}\mathrm{Pf}\left(Y_1,\cdots ,Y_n\right)=\mathrm{Pf}\left(\mathrm{\Omega }\right)\left(Z_1,Y_2,\cdots ,Y_n\right)=\mathrm{Pf}\left(\mathrm{\Omega }\right)\left(Z_1,Z_2,Y_3,\cdots ,Y_n\right)\\ =\mathrm{Pf}\left(\mathrm{\Omega }\right)\left(Z_1,\cdots ,Z_n\right).\end{array}$$E36

This means the definition of $\mathrm{\Lambda }$ does not depend on the $Y_i$ selected. Suppose a different $\overline{u}\in {\overline{\omega }}^{-1}\left(p\right)$ is chosen. Then $\overline{u}=R_A\left(u\right)=u\cdot A$ for some $A\in \mathit{SO}\left(n\right)$, and so let ${\overline{Y}}_i\in \mathit{SO}{\left(E\right)}_{\pi }$ be given by ${\overline{Y}}_i=R_{A\ast }{Y}_i$ and

$$\begin{array}{c}\mathrm{Pf}\left(\mathrm{\Omega }\right)\left({\overline{Y}}_1,\cdots ,{\overline{Y}}_n\right)=\mathrm{Pf}\left(\mathrm{\Omega }\right)\left(R_{A\ast }{Y}_1,\cdots ,R_{A\ast }{Y}_n\right)=\mathrm{Pf}\left(R_A^{\ast }\mathrm{\Omega }\right)\left(Y_1,\cdots ,Y_n\right)\\ =\mathrm{Pf}\left(A^{-1}\mathrm{\Omega }A\right)\left(Y_1,\cdots ,Y_n\right)=\mathrm{Pf}\left(\mathrm{\Omega }\right)\left(Y_1,\cdots ,Y_n\right).\end{array}$$E37

$$\square$$

Theorem 5.2. The unique $n$-form $\mathrm{\Lambda }$ in (35) is closed, $d\mathrm{\Lambda }=0$.

Proof: Suppose $X_1,\cdots ,X_{n+1}\in M_p$ be given and choose $u\in {\overline{\omega }}^{-1}p)$ and $Y_1,\cdots ,Y_{n+1}\in \mathit{SO}{\left(E\right)}_u$ with ${\overline{\omega }}_x{Y}_i=X_i$ and ${\mathit{hY}}_i$ the horizontal component of $Y_i$. Then working out $d\mathrm{\Lambda }$

$$\begin{array}{c}d\mathrm{\Lambda }\left(X_1,\cdots ,X_{n+1}\right)=d\mathrm{\Lambda }\left({\overline{\omega }}_x{Y}_1,\cdots ,{\overline{\omega }}_x{Y}_{n+1}\right)=d\mathrm{\Lambda }\left({\overline{\omega }}_x{\mathit{hY}}_1,\cdots ,{\overline{\omega }}_x{\mathit{hY}}_{n+1}\right)\\ =\left({\overline{\omega }}^{\ast }d\mathrm{\Lambda }\right)\left({\mathit{hY}}_1,\cdots ,{\mathit{hY}}_n\right)=d\left({\overline{\omega }}^{\ast }\mathrm{\Lambda }\right)\left({\mathit{hY}}_1,\cdots ,{\mathit{hY}}_{n+1}\right)\\ =2^m\cdot m!d\left(\mathrm{Pf}\left(\mathrm{\Omega }\right)\right)\left({\mathit{hY}}_1,\cdots ,{\mathit{hY}}_{n+1}\right)=2^m\cdot m!D\left(\mathrm{Pf}\left(\mathrm{\Omega }\right)\right)\left(Y_1,\cdots ,Y_{n+1}\right).\end{array}$$E38

However, $D\mathrm{\Omega }=0$ by Bianchi’s identity and a consequence of this is that (38) vanishes. $\square$

This result applies automatically when $\xi$ is the tangent bundle. The implication of this is that the $n$-form $\mathrm{\Lambda }$ determines a de Rham cohomolgy class $\left[\mathrm{\Lambda }\right]\in H^n\left(M\right)$ of $M$. The form $\mathrm{\Lambda }$ itself depends on the oriented $n$-dimensional bundle $\xi =\pi :E\to M$ over $M$ as well as the choice of metric for $\xi$ and connection $\omega$ on the corresponding bundle $\mathit{SO}\left(E\right)$.

Theorem 5.3. The cohomology class $\left[\mathrm{\Lambda }\right]$ is independent of both the metric and the connection $\omega$.

Proof: Suppose two metrics $⟨,⟩$, ${⟨,⟩}^{\prime }$ are given for $\xi$. Then the corresponding principal bundles $\mathit{SO}\left(E\right)$ and $S{O}^{\prime }\left(E\right)$ are equivalent. If $\stackrel{\sim }{f}:S{O}^{\prime }\left(E\right)\to \mathit{SO}\left(E\right)$ is a fiber preserving diffeomorphism which commutes with the action $\mathit{SO}\left(n\right)$ and $\omega$ a connection on $\mathit{SO}\left(E\right)$. Then ${\omega }^{\prime }={\stackrel{\sim }{f}}^{\ast }\omega$ is a connection on $\mathit{SO}\left(E\right)$. Corresponding curvature forms satisfy ${\mathrm{\Omega }}^{\prime }={\stackrel{\sim }{f}}^{\ast }\mathrm{Pf}\left(\mathrm{\Omega }\right)$ so $\mathrm{Pf}\left({\mathrm{\Omega }}^{\prime }\right)={\stackrel{\sim }{f}}^{\ast }\mathrm{Pf}\left(\mathrm{\Omega }\right)$. The corresponding forms $\mathrm{\Lambda }$ and ${\mathrm{\Lambda }}^{\prime }$ are in fact equal. It suffices to show any two connection differential forms ${\omega }_0$, ${\omega }_1$ on the same $\mathit{SO}\left(E\right)$ generate forms ${\mathrm{\Lambda }}_0$, ${\mathrm{\Lambda }}_1$ whose difference is exact. If $\pi :M\times \left[0,1\right]\to M$ is the projection $\pi \left(p,t\right)=p$, consider the bundle ${\pi }^{\ast }\mathit{SO}\left(\xi \right)$ over $M\times \left[0,1\right]$. Induced connections are ${\pi }^{\ast }{\omega }_0$ and ${\pi }^{\ast }{\omega }_1$ on ${\pi }^{\ast }\mathit{SO}\left(\xi \right)$. Let $\tau :M\times \left[0,1\right]\to \left[0,1\right]$ defined here as $\tau \left(p,t\right)=t$ and define a connection

$$\omega =\left(1-\tau \right){\pi }^{\ast }{\omega }_0+{\mathit{\tau \pi }}^{\ast }{\omega }_1$$E39

on ${\pi }^{\ast }\mathit{SO}\left(\xi \right)$ with $\mathrm{\Omega }$ the connection form. If $i_t$ maps $M$ to $M\times \left[0,1\right]$ and is defined as $i_t\left(p\right)=\left(p,t\right)$, then $i_0^{\ast }\left(\omega \right)$ can be identified with ${\omega }_0$ and $i_1^{\ast }\left(\omega \right)$ with ${\omega }_1$. By Theorems 5.1 and 5.2, which hold for manifolds with and without boundary, there is a closed $n$-form $\mathrm{\Lambda }$ on $M\times \left[0,1\right]$ which pulls back to $2^{m}m!\mathrm{Pf}\left(\mathrm{\Omega }\right)$ on the total space of ${\pi }^{\ast }\mathit{SO}\left(\xi \right)$. A theorem states for any $k$-form $\omega$ on $M\times \left[0,1\right]$, $i_1^{\ast }\omega -i_0^{\ast }\omega =d\left(\mathit{I\omega }\right)-I\left(\mathit{d\omega }\right)$. So if $\mathit{d\omega }=0$, this implies $i_1^{\ast }\omega -i_0^{\ast }\omega =d\left(\mathit{I\omega }\right)$. Substituting the form $\mathrm{\Lambda }$ in place of $\omega$ into this, it follows that ${\mathrm{\Lambda }}_1-{\mathrm{\Lambda }}_0$ is exact. $\square$

Thus every smooth oriented smaooth bundle $\xi$ over $M$ of even fibre dimension $n$ determines a de Rham cohomology class $C\left(\xi \right)=\left[\mathrm{\Lambda }\right]\in H^n\left(M\right)$ and $C\left(\xi \right)=C\left(\eta \right)$ if $\xi \simeq \eta$. It may be asked how does the object $C\left(\xi \right)$ behave with respect to $f^{\ast }$.

Theorem 5.4. Let $\xi =\pi :\xi \to M$ be a smooth oriented bundle over $M$ with fibre dimension $n$ even, let $f:M^{\prime }\to M$ be a smooth map. Then

$$C\left(f^{\ast }\xi \right)=f^{\ast }C\left(\xi \right)\in H^n\left(M^{\prime }\right).$$E40

Proof: The total space of $f^{\ast }\xi$ is called $E^{\prime }$. Let $\stackrel{\sim }{f}:E^{\text{'}}\to E$ be the bundle map covering $f$. If $⟨,⟩$ is a metric on $E$, then ${\stackrel{\sim }{f}}^{\ast }⟨,⟩$ is a metric on $E$. There is an equivalence $\overline{f}:\mathit{SO}\left(E^{\prime }\right)\to \mathit{SO}\left(E\right)$ covering $f$ with ${\overline{\omega }}^{\text{'}}$ taking $\mathit{SO}\left(E^{\prime }\right)$ to $M^{\prime }$ and $\overline{\omega }$ mapping $\mathit{SO}\left(E\right)$ to $M$.

If $\omega$ is a connection on $\mathit{SO}\left(E\right)$, then ${\overline{f}}^{\ast }\left(\omega \right)$ is a connection on $\mathit{SO}\left(E^{\prime }\right)$. It is seen that the corresponding connection forms satisfy ${\mathrm{\Omega }}^{\prime }={\stackrel{\sim }{f}}^{\ast }\mathrm{\Omega }$. Aa a result, we have

$$\mathrm{Pf}\left({\mathrm{\Omega }}^{\prime }\right)=\mathrm{Pf}\left({\overline{f}}^{\ast }\mathrm{\Omega }\right)={\overline{f}}^{\ast }\mathrm{Pf}\left(\mathrm{\Omega }\right).$$E41

For $n$-forms $\mathrm{\Lambda }$ on $M$ given by Theorem 5.1, we then have

$${\overline{\omega }}^{\text{'}\ast }\left(f^{\ast }\mathrm{\Lambda }\right)={\overline{f}}^{\ast }{\overline{\omega }}^{\ast }\mathrm{\Lambda }=2^m\cdot m!{\overline{f}}^{\ast }\mathrm{Pf}\left(\mathrm{\Omega }\right)=2^m\cdot m!\mathrm{Pf}\left({\mathrm{\Omega }}^{\prime }\right).$$E42

This means $f^{\ast }\mathrm{\Lambda }$ must be the $n$-form ${\mathrm{\Lambda }}^{\prime }$ on $M^{\prime }$ given in (31). $\square$

When $\xi$ is a smooth oriented bundle of odd fibre dimension, the definition of $C$ may be extended. It would be remarkable if it were the case that $C\left(\xi \right)$ were always a constant multiple of $\chi \left(\xi \right)$. To this end, the following theorem is needed.

Theorem 5.5. Let ${\xi }_i={\pi }_i:E_i\to M$ for $i=1,2$ be smooth oriented vector bundles over $M$ of fibre dimension $n_1$ and $n_2$. If $n_i=2m_i$, then

$$C\left({\xi }_1\oplus {\xi }_2\right)=\frac{\left(m_1+m_2\right)!}{m_1!m_2!}C\left({\xi }_1\right)\cup C\left({\xi }_2\right).$$E43

If $n_1$ or $n_2$ is odd, this reduces to $C\left({\xi }_1\oplus {\xi }_2\right)=0$.

Proof: Pick two metrics which are Riemannian for each ${\xi }_i$ and set $⟨\cdot ,\cdot ⟩={⟨\cdot ,\cdot ⟩}_1\oplus {⟨\cdot ,\cdot ⟩}_2$ on ${\xi }_1\oplus {\xi }_2=\pi :E\to M$. Let ${\overline{\omega }}_i:\mathit{SO}\left(E_i\right)\to M$ and $\overline{\omega }:\mathit{SO}\left(E\right)\to M$ be the corresponding principal bundles. Over $M$ consider the product principal bundle $Q=\mathit{SO}\left(E_1\right)\ast \mathit{SO}\left(E_2\right)$ with corresponding group $\mathit{SO}\left(n_1\right)\times \mathit{SO}\left(n_2\right)\subset \mathit{SO}\left(n_1+n_2\right)$ whose fiber over $p\in M$ is the direct product ${\overline{\omega }}_1^{-1}\times {\overline{\omega }}_2^{-1}\left(p\right)$_, so this bundle is a subset of $\mathit{SO}\left(E\right)$.

Let ${\rho }_i$ be the projection maps for $Q$ which project this down onto either of its factors. If ${\omega }_i$ are connections on $\mathit{SO}\left(E_i\right)$, with curvature forms ${\mathrm{\Omega }}_i$, then

$${\rho }_1^{\ast }{\omega }_1\oplus {\rho }_2^{\ast }{\omega }_2=\left(\begin{array}{cc}{\rho }_1^{\ast }{\omega }_1& 0\\ 0& {\rho }_2^{\ast }{\omega }_2\end{array}\right)$$E44

is a connection on $Q$ and the curvature form is

$$\mathrm{\Omega }={\rho }_1^{\ast }{\mathrm{\Omega }}_1\oplus {\rho }_2^{\ast }{\mathrm{\Omega }}_2=\left(\begin{array}{cc}{\rho }_1^{\ast }{\mathrm{\Omega }}_1& 0\\ 0& {\rho }_2^{\ast }{\mathrm{\Omega }}_2\end{array}\right)$$E45

The connection $\overline{\omega }$ can be extended uniquely to a connection $\stackrel{\sim }{\omega }$ on $\mathit{SO}\left(E\right)$. The requirement $\stackrel{\sim }{\omega }\left(\sigma \left(M\right)\right)=M$ determines $\stackrel{\sim }{\omega }$ at the new vertical vectors, hence $\stackrel{\sim }{\omega }$ is determined at all points of $Q$, and then at all points of $\mathit{SO}\left(E\right)$ by the requirement $\stackrel{\sim }{\omega }\left(R_A^{\ast }Y\right)=\mathit{Ad}\left(A^{-1}\right)\stackrel{\sim }{\omega }\left(Y\right)$.

At any point $e\in Q$_, the horizontal vectors for $\stackrel{\sim }{\omega }$ are the same as that for $\overline{\omega }$. At $E$, it holds that $\overline{\mathrm{\Omega }}=\stackrel{\sim }{\mathrm{\Omega }}$ for tangent vectors to $Q$ which implies, using $\mathrm{Pf}\left(A\oplus B\right)=\mathrm{Pf}\left(A\right)\cdot \mathrm{Pf}\left(B\right)$, that

$$\mathrm{Pf}\left(\stackrel{\sim }{\mathrm{\Omega }}\right)=\mathrm{Pf}\left(\overline{\mathrm{\Omega }}\right)=\mathrm{Pf}\left({\rho }_1^{\ast }{\mathrm{\Omega }}_1\right)\wedge \mathrm{Pf}\left({\rho }_2^{\ast }{\mathrm{\Omega }}_2\right)={\rho }_1^{\ast }\mathrm{Pf}\left({\mathrm{\Omega }}_1\right)\wedge {\rho }_2^{\ast }\left({\mathrm{\Omega }}_2\right).$$E46

Consequently, if ${\mathrm{\Lambda }}_i$ are the forms given by (34), then at $e$ it must hold that on tangent vectors to $Q$

$$\begin{array}{c}{\overline{\omega }}^{\ast }\mathrm{\Lambda }=2^{m_1+m_2}\left(m_1+m_2\right)!\mathrm{Pf}\left(\stackrel{\sim }{\mathrm{\Omega }}\right)=\frac{\left(m_1+m_2\right)!}{m_1!m_2!}{2}^{m_1}{m}_1!{\rho }_1^{\ast }\mathrm{Pf}\left({\mathrm{\Omega }}_1\right)\wedge 2^{m_2}{m}_2!{\rho }_2^{\ast }\mathrm{Pf}\left({\mathrm{\Omega }}_2\right)\\ =\frac{\left(m_1+m_2\right)!}{m_1!m_2!}{\rho }_1^{\ast }{\overline{\omega }}_1^{\ast }{\mathrm{\Lambda }}_1\wedge {\rho }_2^{\ast }{\overline{\omega }}_2^{\ast }{\mathrm{\Lambda }}_2=\frac{\left(m_1+m_2\right)!}{m_1!m_2!}{\overline{\omega }}_1^{\ast }{\mathrm{\Lambda }}_1\wedge {\overline{\omega }}_2^{\ast }{\mathrm{\Lambda }}_2.\end{array}$$E47

This implies that

$$\mathrm{\Lambda }=\frac{\left(m_1+m_2\right)!}{m_1!m_2!}{\mathrm{\Lambda }}_1\wedge {\mathrm{\Lambda }}_2.$$E48

$$\square$$

Corollary 5.1. If the oriented bundle $\xi =\pi :E\to M$ has a nowhere zero section $s$, then

$$C\left(\xi \right)=0.$$E49

Proof: Let $E_1\subset E$ be written

$${\cup }_{p\in M}\mathrm{ℝ}\cdot s\left(p\right),$$E50

and let $E_2\subset E$ be the orthogonal complement

$${\cup }_{p\in M}{\left(\mathrm{ℝ}\cdot s\left(p\right)\right)}^{\perp }$$E51

with respect to some Riemannian metric on $E$. Then ${\xi }_1={\pi }_1\mid E_1:E_1\to M$ is an oriented one-dimensional bundle. Consequently, ${\xi }_2={\pi }_2\mid E_2:E_2\to M$ is also an oriented bundle since $\xi$ is oriented. Clearly $\xi \simeq {\xi }_1\oplus {\xi }_2$. An application of the previous result shows that $C\left(\xi \right)=0$. $\square$

This theorem is almost enough to characterize $\chi$ as we can now show the statement which relates $C\left(\xi \right)$ and the Euler class.

Corollary 5.2. If $\xi =\pi :E\to M$ is a smooth vector bundle of fibre dimension $n$ over a compact oriented manifold $M$, then the class $C\left(\xi \right)\in H^n\left(M\right)$ is a multiple of the Euler class $\chi \left(\xi \right)$.

Proof: Suppose $S$ is the sphere bundle $S=\left\{e\in E:⟨e,e⟩=1\right\}$_, which is constructed with respect to some Riemannian metric on $E$. Let ${\pi }_0:S\to X$ be the restriction $\pi \mid S$. The bundle ${\pi }_0^{\ast }\xi$ has a nowhere zero section. Corrollary 5.1 and Theorem 5.4 then yield

$${\pi }_0^{\ast }C\left(\xi \right)=C\left({\pi }_0^{\ast }\xi \right)=0.$$E52

However, there is a theorem which states a class $\alpha \in H^n\left(M\right)$ satisfies ${\pi }_0^{\ast }\alpha =0$ if and only if $\alpha$ is a multiple of $\chi \left(\xi \right)$. It can now be inferred that $C\left(\xi \right)$ is a multiple of the Euler class $\chi \left(\xi \right)$. $\square$

6. The Gauss-Bonnet-Chern theorem

If Corollary 5.2 is applied to the tangent bundle of a compact oriented manifold $M$ of dimension $n$ which is even, the class $C\left(\mathit{TM}\right)\in H^n\left(M\right)$ is some multiple of the Euler class $\chi \left(\mathit{TM}\right)$. This fact is not so interesting because $H^n\left(M\right)$ is one-dimensional since it means $C\left(\mathit{TM}\right)=0$ if $\chi \left(\mathit{TM}\right)=0$. The corollary does lead to something interesting when applied to the universal bundle.

Theorem 6.1 For every even $n$, there is a constant ${\beta }_n$ such that

$$C\left(\xi \right)={\beta }_n\chi \left(\xi \right).$$E53

for all smooth oriented $n$-dimensional bundles $\xi$ over compact oriented manifolds. In this sense, it is universal.

Proof: Begin with the bundles $\stackrel{\sim }{z}{\mathit{eta}}_n\left({\mathrm{ℝ}}^N\right)$ for $N>n$. Corollary 5.2 implies there are constants ${\beta }_{n,N}$ such that

$$C\left({\stackrel{\sim }{\zeta }}^n\left({\mathrm{ℝ}}^N\right)\right)={\beta }_{n,N}\chi \left({\stackrel{\sim }{\zeta }}^n\left({\mathrm{ℝ}}^N\right)\right)\in H^n\left({\stackrel{\sim }{\zeta }}_n\left({\mathrm{ℝ}}^N\right)\right).$$E54

If $j:{\stackrel{\sim }{G}}_n\left({\mathrm{ℝ}}^N\right)\to {\stackrel{\sim }{G}}_n\left({\mathrm{ℝ}}^M\right)$ is the natural inclusion, then $j^{\ast }\left({\stackrel{\sim }{\zeta }}^N\left({\mathrm{ℝ}}^N\right)\right)\simeq {\stackrel{\sim }{\zeta }}^n\left({\mathrm{ℝ}}^N\right)$. Equation (30) and Theorem 5.4 yield

$$C\left({\stackrel{\sim }{\zeta }}^n\left({\mathrm{ℝ}}^N\right)\right)=j^{\ast }C\left({\stackrel{\sim }{\zeta }}^n\left({\mathrm{ℝ}}^M\right)\right)=j^{\ast }\chi \left({\stackrel{\sim }{\zeta }}^n\left({\mathrm{ℝ}}^M\right)\right).$$E55

Thus, (54), (55) give

$${\beta }_{n,N}\chi \left({\stackrel{\sim }{\zeta }}^n\left({\mathrm{ℝ}}^N\right)\right)={\beta }_{n,M}\chi \left({\stackrel{\sim }{\zeta }}^n\left({\mathrm{ℝ}}^N\right)\right).$$E56

Since $\chi \left({\stackrel{\sim }{\zeta }}^n\left({\mathrm{ℝ}}^N\right)\right)\ne 0$ by Theorem 4.2, this implies that ${\beta }_{n,N}={\beta }_{n,M}$ for all $M,N>1$. This common number is called ${\beta }_n$, and we have

$$C\left({\stackrel{\sim }{\zeta }}^n\left({\mathrm{ℝ}}^N\right)\right)={\beta }_n\chi \left({\stackrel{\sim }{\zeta }}^n\left({\mathrm{ℝ}}^N\right)\right).$$E57

However it is known that any smooth oriented $n$-dimensional bundle $\xi$ over a compact manifold $M$ is equivalent to $f^{\ast }\left({\stackrel{\sim }{\zeta }}^n\left({\mathrm{ℝ}}^N\right)\right)$ for some smooth map $f:M\to \stackrel{\sim }{G}\left({\mathrm{ℝ}}^N\right)$, then

$$C\left(\xi \right)=C\left(f^{\ast }{\stackrel{\sim }{\zeta }}^n\left({\mathrm{ℝ}}^N\right)\right)=f^{\ast }C\left({\stackrel{\sim }{\zeta }}^n\left({\mathrm{ℝ}}^N\right)\right)={\beta }_n{f}^{\ast }\chi \left({\stackrel{\sim }{\zeta }}^n\left({\mathrm{ℝ}}^N\right)\right)={\beta }_n\chi \left(\xi \right).$$E58

$$\square$$

The constant ${\beta }_N$ is universal in nature and it may be asked whether it can be computed. Since it has this universality property, it suffices to compute this constant for a special case where the calculation is easier and in turn implies another application of the next theorem.

Theorem 6.2 For integer $n=2m$, the constant ${\beta }_n$ in Theorem 6.1 is

$${\beta }_n=\frac{n!}{2}V\left(S^n\right)={\pi }^m{2}^{n}m!.$$E59

If $\left(M,⟨,⟩\right)$ is a compact manifold of even dimension $n=2m$ then

$${\int }_M{K}_{n}d{\mu }_g=\frac{{\pi }^m{2}^{n}m!}{n!}\chi \left(M\right).$$E60

Proof: Let $\xi$ be the tangent bundle $\mathit{TM}$ of a compact oriented manifold of dimension $n$. Now (17) gives a formula for $K_{n}d{\mu }_g$, where ${\mathrm{\Omega }}_j^i$ are curvature forms for some positively oriented orthonormal moving frame. This implies that the form $\mathrm{\Lambda }$ in Theorem 5.1 for the bundle $\mathit{SO}\left(\xi \right)=\mathit{SO}\left(\mathit{TM}\right)$ is

$$\mathrm{\Lambda }=n!K_{n}d{\mu }_g.$$E61

If $\kappa$ is the fundamental class of $M$ then

$$\left({\int }_M{K}_{n}d{\mu }_g\right)\kappa =\frac{1}{n!}\left({\int }_M\mathrm{\Lambda }\right)\kappa =\frac{1}{n!}C\left(\xi \right)=\frac{{\beta }_n}{n!}\chi \left(\xi \right)=\frac{{\beta }_n}{n!}\chi \left(M\right)\cdot \kappa .$$E62

Hence, equating the coefficients of $\kappa$ on both sides,

$${\int }_M{K}_{n}d{\mu }_g=\frac{{\beta }_n}{n!}\chi \left(M\right).$$E63

Consider the case of a specific manifold $M=S^n$ in (63), where $K_n=1$ so the left side of (63) reduces to $V\left(S^n\right)$

$$V\left(S^n\right)=\frac{{\beta }_n}{n!}\chi \left(S^n\right)=\frac{2{\beta }_n}{n!}.$$E64

Since the volume $V\left(S^n\right)$ is known to be ${\pi }^m{2}^{n+1}m!/n!$, solve (64) for ${\beta }_n$ in terms of $V\left(S^n\right)$ ,

$${\beta }_n=\frac{n!}{2}V\left(S^n\right)={\pi }^m{2}^{n}m!.$$E65

This value of ${\beta }_n$ can be put back into (60) and for the manifold $M$, it follows that

$${\int }_M{K}_{n}d{\mu }_g=\frac{{\pi }^m{2}^{n}m!}{n!}\chi \left(M\right).$$E66

$$\square$$

7. The theorem for manifolds with boundary

Theorem 5.5 played a large part in the proof of (64). It allowed us to state that if $\xi =\pi :E\to M$ is an oriented $n$-dimensional vector bundle with sphere bundle ${\pi }_0:S_0\to M$_, then ${\pi }_0^{\ast }C\left(\xi \right)=0$ was a large part of the proof of (). If $\mathrm{\Lambda }$ is the $n$-form on $M$ representing $C\left(\xi \right)$, then the $n$-form ${\pi }_0^{\ast }\mathrm{\Lambda }$ on $S$ is exact ${\pi }_0^{\ast }\mathrm{\Lambda }=d\mathrm{\Phi }$ for some $\left(n-1\right)$-form on $S$. Suppose $\xi ={\mathit{TM}}^{\ast }$ and let $X$ be a unit vector bundle on $M$ which has an isolated singularity at $p\in M$ Let $B\left(\epsilon \right)$ be a closed ball of radius $\epsilon$ around $p$ and set $M_{\epsilon }=M\backslash int\left(B\left(\epsilon \right)\right)$ where $int$ denotes the interior. Then $X\left(M_{\epsilon }\right)$ is a manifold with boundary, the image of $M_{\epsilon }$ under the section $X:M\backslash \left\{p\right\}\to S$. Consequently

$${\int }_M\mathrm{\Lambda }={\int }_{M-\left\{p\right\}}\mathrm{\Lambda }=\underset{\epsilon \to 0}{lim}{\int }_{M_{\epsilon }}\mathrm{\Lambda }=\underset{\epsilon \to 0}{lim}{\int }_{M_{\epsilon }}{X}^{\ast }\left({\pi }_0^{\ast }\mathrm{\Lambda }\right)=\underset{\epsilon \to 0}{lim}{\int }_{X(M_{\epsilon }}{\pi }_0^{\ast }\mathrm{\Lambda }$$

$$=\underset{\epsilon \to 0}{lim}{\int }_{X\left(M_{\epsilon }\right)}d\mathrm{\Phi }=\underset{\epsilon \to 0}{lim}{\int }_{\partial X\left(M_{\epsilon }\right)}\mathrm{\Phi }.$$E67

If $\mathrm{ind}\left(X,p\right)$ is the index of $X$ at $p$

$${\int }_M\mathrm{\Lambda }=\mathrm{ind}\left(X,p\right){\int }_{{\pi }_0^{-1}\left(p\right)}\mathrm{\Phi }=\chi \left(M\right){\int }_{{\pi }_0^{-1}\left(p\right)}\mathrm{\Phi }.$$E68

Since $n=2m$ we also have the Gauss-Bonnet-Chern Theorem 6.2

$${\int }_M\mathrm{\Lambda }={\int }_{M}n!K_{n}d{\mu }_g={\pi }^{m}m!2^n\chi \left(M\right),$$E69

we finally obtain

$${\int }_{{\pi }_0^{-1}\left(p\right)}\mathrm{\Phi }={\pi }^{m}m!2^n.$$E70

Let $\left(M,\partial M\right)$ be a compact orientable manifold with boundary with Euler characteristic $\chi \left(M\right)=dim{H}^0\left(M\right)-dim{H}^1\left(M\right)+\cdots$. A compact oriented manifold $M^2$ can be constructed, the double of $M$, by taking two disjoint copies of $M$ and identifying corresponding points of $\partial M$.

Theorem 7.1 The Euler characteristic of the manifold $M^2$ is given by

$$\chi \left(M^2\right)=2\chi \left(M\right)-\chi \left(\partial M\right).$$E71

Proof: Let $U$ and $V$ be open neighborhoods of the two copies of $M$ in $M^2$ such that $H^k\left(U\right)\equiv H\left(V\right)\equiv H^k\left(M\right)$ for all $k$ and $H^k\left(U\cap V\right)\equiv H^k\left(\partial M\right)$ for all $k$. So there is the sequence $0\to H^0\left(M^2\right)\to \cdots \to H^k\left(M^2\right)\to H^k\left(U\right)\oplus H^k\left(V\right)\to \cdots \to H^k\left(U\cap V\right)\to H^{k+1}\left(M^2\right)$. When the sequence is exact, a theorem can be applied to obtain the result. $\square$

This is very interesting since it claims different things depending on whether the dimension $n$ of $M$ is even or odd. When $n$ is odd $\chi \left(M^2\right)=0$ hence $\chi \left(M\right)=1/2\chi \left(\partial M\right)$ which implies $\chi \left(\partial M\right)$ must be even. But when $n$ is even, $\chi \left(\partial M\right)=0$, so the previous theorem implies

$$2\chi \left(M\right)=\chi \left(M^2\right).$$E72

Corollary 7.1 Let $M$ be a compact orientable manifold with boundary of even dimension $n$. Let $X$ be a vector field on $M$ with only finitely many zeros all in $M\backslash \partial M$ such that $X$ is outward pointing on $\partial M$. The sum of indices of $X$ is $\chi \left(M\right)$.

Proof: Modify $X$ near $\partial M$ so it is an outward pointing unit normal $\nu$ on the boundary and so there are no new zeros. Then there is a vector field on $M^2$ which looks like $X$ on one copy of $M$ and $-X$ on the other. Since $n$ is even, the index $-X$ of an isolated zero is the same as the index of $X$ at that zero. The Theorem of Poincaré-Hopf on the sum of indices of $X$ gives twice the sum of the indices of $X$ equals $\chi \left(M^2\right)=2\chi \left(M\right)$ by (72). $\square$

Theorem 7.2 Let $M$ be a compact oriented Riemannian manifold with boundary of even dimension $n=2m$, tangent bundle $\pi :\mathit{TM}\to M$ and associated sphere bundle ${\pi }_0=\pi \mid S:S\to M$. Let $\omega$ be a connection on the principal bundle $\overline{\omega }:\mathit{SO}\left(\mathit{TM}\right)\to M$, with curvature form $\mathrm{\Omega }$. Let $\mathrm{\Lambda }$ be the unique $n$-form on $M$ with

$${\overline{\omega }}^{\ast }\mathrm{\Lambda }=\sum {\epsilon }^{i_1,\cdots ,i_n}{\mathrm{\Omega }}_{i_2}^{i_1}\wedge \cdots \wedge {\mathrm{\Omega }}_{i_n}^{i_{n-1}}=2^{m}m!\mathrm{Pf}\left(\mathrm{\Omega }\right),$$E73

and $\mathrm{\Phi }$ an $\left(n-1\right)$-form on $S$ with ${\pi }_0^{\ast }\mathrm{\Lambda }=d\mathrm{\Phi }$. Let $\nu :\partial M\to S$ be the outward pointing unit normal on $\partial M$. Then

$${\int }_M{K}_{n}d{\mu }_g=\frac{1}{n!}{\int }_M\mathrm{\Lambda }=\frac{{\pi }^{m}m!2^n}{n!}\chi \left(M\right)+\frac{1}{n!}{\int }_{\partial M}{\nu }^{\ast }\mathrm{\Phi }.$$E74

Proof: Extend $\nu$ to a vector field $X$ on $M$ with only finitely many zeros $p_1,\dots ,p_k\in M\backslash \partial M$. Let $B_i\left(\epsilon \right)$ be the closed balls of radius $\epsilon$ centered at $p_i$ which are disjoint from each other and from $\partial M$. Put $M_{\epsilon }=M\backslash {\cup }_{i=1}^{k}int{B}_i\left(\epsilon \right)$. Now integrate the form $\mathrm{\Lambda }$ over $M$ and use (70)

$$\begin{array}{c}{\int }_M\mathrm{\Lambda }=\underset{\epsilon \to 0}{lim}{\int }_{\partial X\left(M_{\epsilon }\right)}\mathrm{\Phi }={\int }_{\nu \left(\partial M\right)}\mathrm{\Phi }+\sum _{i=1}^k\underset{\epsilon \to 0}{lim}{\int }_{\partial B_i\left(\epsilon \right)}\mathrm{\Phi }\\ ={\int }_{\partial M}{\nu }^{\ast }\mathrm{\Phi }+{\pi }^{m}m!2^n\sum _{i=1}^M{\left(\mathrm{ind}X\right)}_{p_i}={\int }_{\partial M}{\nu }^{\ast }\mathrm{\Phi }+{\pi }^{m}m!2^n\chi \left(M\right).\end{array}$$E75

The last line makes use of Corollary 7.1. $\square$

Theorem 7.2 presents one way in which Theorem 6.2 can be generalized to the case of manifolds with boundary. At this point an interpretation for the first term in (75) is not easy to provide. It is required to obtain an explicit $\mathrm{\Phi }$ such that ${\pi }_0^{\ast }\mathrm{\Lambda }=d\mathrm{\Phi }$. In fact such a $\mathrm{\Phi }$ can be constructed.