7.9 Differential forms and Hodge theory

7.9.1 Invariant volume elements

We have defined the volume element as a non-vanishing $m$ -form on an $m$ -dimensional orientable manifold $M$ in section 5.5. If $M$ is endowed with a metric $g$ , there exists a natural volume element which is invariant under coordinate transformation. Let us define the invariant volume element by

$${\Omega }_M\equiv \sqrt{|g|}d{x}^1\wedge d{x}^2\wedge \cdots \wedge d{x}^m\text{\hspace{1em}(7.168)}$$

where $g=\det g_{\mu \nu }$ and $x^{\mu }$ are the coordinates of the chart $(U,\phi )$ . The $m$ -form ${\Omega }_M$ is, indeed, invariant under a coordinate change. Let $y^{\lambda }$ be the coordinates of another chart $(V,\psi )$ with $U\cap V\ne \varnothing$ . The invariant volume element is

$$\sqrt{\left|\det \left(\frac{\partial x^{\mu }}{\partial y^{\kappa }}\frac{\partial x^v}{\partial y^{\lambda }}{g}_{\mu v}\right)\right|}d{y}^1\wedge \cdots \wedge d{y}^m$$

in terms of the $y$ -coordinates. Noting that $\mathrm{d}{y}^{\lambda }=(\partial y^{\lambda }/\partial x^{\mu })\mathrm{d}{x}^{\mu }$ , this becomes

$$\begin{array}{rl}& \left|\det \left(\frac{\partial x^{\mu }}{\partial y^{\kappa }}\right)\right|\sqrt{|g|}\det \left(\frac{\partial y^{\lambda }}{\partial x^{\nu }}\right)d{x}^1\wedge d{x}^2\wedge \cdots \wedge d{x}^m\\ & =\pm \sqrt{|g|}d{x}^1\wedge d{x}^2\wedge \cdots \wedge d{x}^m.\end{array}$$

If $x^{\mu }$ and $y^{\kappa }$ define the same orientation, $\det (\partial x^{\mu }/\partial y^{\kappa })$ is strictly positive on $U\cap V$ and ${\Omega }_M$ is invariant under the coordinate change.

Exercise 7.20

Let $\{{\hat{\theta }}^{\alpha }\}=\{e_{\mu }^{\alpha }\mathrm{d}{x}^{\mu }\}$ be the non-coordinate basis. Show that the invariant volume element is written as

$${\Omega }_M=|e|\mathrm{d}{x}^1\wedge \mathrm{d}{x}^2\wedge \cdots \wedge \mathrm{d}{x}^m={\hat{\theta }}^1\wedge {\hat{\theta }}^2\wedge \cdots \wedge {\hat{\theta }}^m\text{\hspace{1em}(7.169)}$$

where $e=\det e_{\mu }^{\alpha }$ .

Now that we have defined the invariant volume element, it is natural to define an integration of $f\in \mathcal{F}(M)$ over $M$ by

$${\int }_{M}f{\Omega }_M\equiv {\int }_{M}f\sqrt{|g|}d{x}^{1}d{x}^2\dots d{x}^m.\text{\hspace{1em}(7.170)}$$

Obviously (7.170) is invariant under a change of coordinates. In physics, there are many objects which are expressed as volume integrals of this type, see section 7.10.

7.9.2 Duality transformations (Hodge star)

As noted in section 5.4, ${\Omega }^r(M)$ is isomorphic to ${\Omega }^{m-r}(M)$ on an $m$ -dimensional manifold $M$ . If $M$ is endowed with a metric $g$ , we can define a natural isomorphism between them called the Hodge * operation. Define the totally anti-symmetric tensor $\epsilon$ by

$${\epsilon }_{{\mu }_1{\mu }_2\dots {\mu }_m}=\{\begin{array}{ll}+1& \text{i f}({\mu }_1{\mu }_2\dots {\mu }_m)\text{i s a n e v e n p e r m u t a t i o n o f}(12\dots m)\\ -1& \text{i f}({\mu }_1{\mu }_2\dots {\mu }_m)\text{i s a n o d d p e r m u t a t i o n o f}(12\dots m)\\ 0& \text{o t h e r w i s e .}\end{array}\text{\hspace{1em}(7.171a)}$$

Note that

$${\epsilon }^{{\mu }_1{\mu }_2\dots {\mu }_m}=g^{{\mu }_1{\nu }_1}{g}^{{\mu }_2{\nu }_2}\dots g^{{\mu }_m{\nu }_m}{\epsilon }_{{\nu }_1{\nu }_2\dots {\nu }_m}=g^{-1}{\epsilon }_{{\mu }_1{\mu }_2\dots {\mu }_m}.\text{\hspace{1em}(7.171b)}$$

The Hodge $\ast$ is a linear map $\ast :{\Omega }^r(M)\to {\Omega }^{m-r}(M)$ whose action on a basis vector of ${\Omega }^r(M)$ is defined by

$$\begin{array}{rl}& \ast \left(\mathrm{d}{x}^{{\mu }_1}\wedge \mathrm{d}{x}^{{\mu }_2}\wedge \cdots \wedge \mathrm{d}{x}^{{\mu }_r}\right)\\ & =\frac{\sqrt{|g|}}{(m-r)!}{\epsilon }_{{\nu }_{r+1}\dots {\nu }_m}^{{\mu }_1{\mu }_2\dots {\mu }_r}d{x}^{{\nu }_{r+1}}\wedge \cdots \wedge d{x}^{{\nu }_m}.\end{array}\text{\hspace{1em}(7.172)}$$

It should be noted that $\ast 1$ is the invariant volume element:

$$\ast 1=\frac{\sqrt{|g|}}{m!}{\epsilon }_{{\mu }_1{\mu }_2\dots {\mu }_m}\mathrm{d}{x}^{{\mu }_1}\wedge \cdots \wedge \mathrm{d}{x}^{{\mu }_m}=\sqrt{|g|}\mathrm{d}{x}^1\wedge \cdots \wedge \mathrm{d}{x}^m.$$

For

$$\omega =\frac{1}{r!}{\omega }_{{\mu }_1{\mu }_2\dots {\mu }_r}d{x}^{{\mu }_1}\wedge d{x}^{{\mu }_2}\wedge \cdots \wedge d{x}^{{\mu }_r}\in {\Omega }^r(M)$$

we have

$$\ast \omega =\frac{\sqrt{|g|}}{r!(m-r)!}{\omega }_{{\mu }_1{\mu }_2\dots {\mu }_r}{\epsilon }_{{\nu }_{r+1}\dots {\nu }_m}^{{\mu }_1{\mu }_2\dots {\mu }_r}d{x}^{{\nu }_{r+1}}\wedge \cdots \wedge d{x}^{{\nu }_m}.\text{\hspace{1em}(7.173)}$$

If we take the non-coordinate basis $\{{\theta }^{\alpha }\}=\{e_{\mu }^{\alpha }\mathrm{d}{x}^{\mu }\}$ , the $\ast$ operation becomes

$$\ast \left({\hat{\theta }}^{{\alpha }_1}\wedge \cdots \wedge {\hat{\theta }}^{{\alpha }_r}\right)=\frac{1}{(m-r)!}{\epsilon }_{{\beta }_{r+1}\dots {\beta }_m}^{{\alpha }_1\dots {\alpha }_r}{\hat{\theta }}^{{\beta }_{r+1}}\wedge \cdots \wedge {\hat{\theta }}^{{\beta }_m}\text{\hspace{1em}(7.174)}$$

where

$${\epsilon }_{{\alpha }_1\dots {\alpha }_m}=\{\begin{array}{ll}+1& \text{i f}({\alpha }_1\dots {\alpha }_m)\text{i s a n e v e n p e r m u t a t i o n o f}(12\dots m)\\ -1& \text{i f}({\alpha }_1\dots {\alpha }_m)\text{i s a n o d d p e r m u t a t i o n o f}(12\dots m)\\ 0& \text{o t h e r w i s e}\end{array}\text{\hspace{1em}(7.175)}$$

and the indices are raised by ${\delta }^{\alpha \beta }$ or ${\eta }^{\alpha \beta }$ .

Theorem 7.4

$$\ast \ast \omega =(-1{)}^{r(m-r)}\omega .\text{\hspace{1em}(7.176a)}$$

if $(M,g)$ is Riemannian and

$$\ast \ast \omega =(-1{)}^{1+r(m-r)}\omega \text{\hspace{1em}(7.176b)}$$

if Lorentzian.

Proof

It is simpler to prove (7.176a) with a non-coordinate basis. Let

$$\omega =\frac{1}{r!}{\omega }_{{\alpha }_1\dots {\alpha }_r}{\hat{\theta }}^{{\alpha }_1}\wedge \cdots \wedge {\hat{\theta }}^{{\alpha }_r}.$$

Repeated applications of $\ast$ on $\omega$ yield

$$\begin{array}{rl}& \ast \ast \omega =\frac{1}{r!}{\omega }_{{\alpha }_1\dots {\alpha }_r}\frac{1}{(m-r)!}{\epsilon }_{{\beta }_{r+1}\dots {\beta }_m}^{{\alpha }_1\dots {\alpha }_r}\\ & \times \frac{1}{r!}{\epsilon }_{{\gamma }_1\dots {\gamma }_r}^{{\beta }_{r+1}\dots {\beta }_m}{\hat{\theta }}^{{\gamma }_1}\wedge \dots \wedge {\hat{\theta }}^{{\gamma }_r}\\ & =\frac{(-1{)}^{r(m-r)}}{r!r!(m-r)!}\sum _{\alpha \beta \gamma }{\omega }_{{\alpha }_1\dots {\alpha }_r}{\epsilon }_{{\alpha }_1\dots {\alpha }_r{\beta }_{r+1}\dots {\beta }_m}{\epsilon }_{{\gamma }_1\dots {\gamma }_r{\beta }_{r+1}\dots {\beta }_m}\\ & \times {\hat{\theta }}^{{\gamma }_1}\wedge \cdots \wedge {\hat{\theta }}^{{\gamma }_r}\\ & =\frac{(-1{)}^{r(m-r)}}{r!}{\omega }_{{\alpha }_1\dots {\alpha }_r}{\hat{\theta }}^{{\alpha }_1}\wedge \cdots \wedge {\hat{\theta }}^{{\alpha }_r}=(-1{)}^{r(m-r)}\omega \end{array}$$

where use has been made of the identity

$$\sum _{\beta \gamma }{\epsilon }_{{\alpha }_1\dots {\alpha }_r{\beta }_{r+1}\dots {\beta }_m}{\epsilon }_{{\gamma }_1\dots {\gamma }_r{\beta }_{r+1}\dots {\beta }_m}{\hat{\theta }}^{{\gamma }_1}\wedge \cdots \wedge {\hat{\theta }}^{{\gamma }_r}=r!(m-r)!{\hat{\theta }}^{{\alpha }_1}\wedge \cdots \wedge {\hat{\theta }}^{{\alpha }_r}.$$

The proof of (7.176b) is left as an exercise to the reader (use $\det \eta =-1$ ).

Thus, we find that $(-1{)}^{r(m-r)}\ast \ast$ (or $(-1{)}^{1+r(m-r)}\ast \ast )$ is an identity map on ${\Omega }^r(M)$ . We define the inverse of ${}^{\ast }$ by

$${\ast }^{-1}=(-1{)}^{r(m-r)}\ast (M,g)\text{i s R i e m a n n i a n}\text{\hspace{1em}(7.177a)}$$

$${\ast }^{-1}=(-1{)}^{1+r(m-r)}\ast (M,g)\text{i s L o r e n t z i a n .}\text{\hspace{1em}(7.177b)}$$

7.9.3 Inner products of $r$ -forms

Take

$$\begin{array}{rl}& \omega =\frac{1}{r!}{\omega }_{{\mu }_1\dots {\mu }_r}d{x}^{{\mu }_1}\wedge \cdots \wedge d{x}^{{\mu }_r}\\ & \eta =\frac{1}{r!}{\eta }_{{\mu }_1\dots {\mu }_r}d{x}^{{\mu }_1}\wedge \cdots \wedge d{x}^{{\mu }_r}.\end{array}$$

The exterior product $\omega \wedge \ast \eta$ is an $m$ -form:

$$\begin{array}{rl}& \omega \wedge \ast \eta =\frac{1}{(r!{)}^2}{\omega }_{{\mu }_1\dots {\mu }_r}{\eta }_{{\nu }_1\dots {\nu }_r}\frac{\sqrt{|g|}}{(m-r)!}{\epsilon }_{{\mu }_{r+1}\dots {\mu }_m}^{{\nu }_1\dots {\nu }_r}\\ & \times \mathrm{d}{x}^{{\mu }_1}\wedge \cdots \wedge \mathrm{d}{x}^{{\mu }_r}\wedge \mathrm{d}{x}^{{\mu }_{r+1}}\wedge \cdots \wedge \mathrm{d}{x}^{{\mu }_m}\\ & =\frac{1}{r!}\sum _{\mu v}{\omega }_{{\mu }_1\dots {\mu }_r}{\eta }^{v_1\dots v_r}\frac{1}{r!(m-r)!}{\epsilon }_{v_1\dots v_r{\mu }_{r+1}\dots {\mu }_m}\\ & \times {\epsilon }_{{\mu }_1\dots {\mu }_r{\mu }_{r+1}\dots {\mu }_m}\sqrt{|g|}d{x}^1\wedge \cdots \wedge d{x}^m\\ & =\frac{1}{r!}{\omega }_{{\mu }_1\dots {\mu }_r}{\eta }^{{\mu }_1\dots {\mu }_r}\sqrt{|g|}d{x}^1\wedge \cdots \wedge d{x}^m.\end{array}\text{\hspace{1em}(7.178)}$$

This expression shows that the product is symmetric:

$$\omega \wedge \ast \eta =\eta \wedge \ast \omega .\text{\hspace{1em}(7.179)}$$

Let $\{{\hat{\theta }}^{\alpha }\}$ be the non-coordinate basis and

$$\omega =\frac{1}{r!}{\omega }_{{\alpha }_1\dots {\alpha }_r}{\hat{\theta }}^{{\mu }_1}\wedge \cdots \wedge {\hat{\theta }}^{{\alpha }_r}$$

$$\eta =\frac{1}{r!}{\eta }_{{\alpha }_1\dots {\alpha }_r}{\hat{\theta }}^{{\alpha }_1}\wedge \cdots \wedge {\hat{\theta }}^{{\alpha }_r}.$$

Equation (7.178) is rewritten as

$$\omega \wedge \ast \eta =\frac{1}{r!}{\omega }_{{\alpha }_1\dots {\alpha }_r}{\eta }^{{\alpha }_1\dots {\alpha }_r}{\hat{\theta }}^1\wedge \cdots \wedge {\hat{\theta }}^m.\text{\hspace{1em}(7.180)}$$

Since $\alpha \wedge \ast \beta$ is an $m$ -form, its integral over $M$ is well defined. Define the inner product $(\omega ,\eta )$ of two $r$ -forms by

$$\begin{array}{rl}& (\omega ,\eta )\equiv \int \omega \wedge \ast \eta \\ & =\frac{1}{r!}{\int }_M{\omega }_{{\mu }_1\dots {\mu }_r}{\eta }^{{\mu }_1\dots {\mu }_r}\sqrt{|g|}d{x}^1\dots d{x}^m.\end{array}\text{\hspace{1em}(7.181)}$$

Since $\omega \wedge \ast \eta =\eta \wedge \ast \omega$ , the inner product is symmetric,

$$(\omega ,\eta )=(\eta ,\omega ).\text{\hspace{1em}(7.182)}$$

If $(M,g)$ is Riemannian, the inner product is positive definite,

$$(\alpha ,\alpha )\ge 0.\text{\hspace{1em}(7.183)}$$

where the equality holds only when $\alpha =0$ . This is not true if $(M,g)$ is Lorentzian.

7.9.4 Adjoints of exterior derivatives

Definition 7.6

Let $\mathsf{d}:{\Omega }^{r-1}(M)\to {\Omega }^r(M)$ be the exterior derivative operator. The adjoint exterior derivative operator ${\mathrm{d}}^{†}:{\Omega }^r(M)\to {\Omega }^{r-1}(M)$ is defined by

$${\mathrm{d}}^{†}=(-1{)}^{mr+m+1}\ast \mathrm{d}\ast \text{\hspace{1em}(7.184a)}$$

if $(M,g)$ is Riemannian and

$${\mathrm{d}}^{†}=(-1{)}^{mr+m}\ast \mathrm{d}\ast \text{\hspace{1em}(7.184b)}$$

if Lorentzian, where $m=\dim M$ .

In summary, we have the following diagram (for a Riemannian manifold),

$$\begin{array}{ccc}{\Omega }^{m-r}(M)& \stackrel{(-1{)}^{mr+m+1}d}{\to }& {\Omega }^{m-r+1}(M)\\ \ast & & \downarrow \ast \\ {\Omega }^r(M)& \stackrel{d^{†}}{\to }& {\Omega }^{r-1}(M).\end{array}\text{\hspace{1em}(7.185)}$$

The operator ${\mathrm{d}}^{†}$ is nilpotent since $\mathrm{d}$ is: ${\mathrm{d}}^{†2}=\ast \mathrm{d}\ast \ast \mathrm{d}\ast \propto \ast {\mathrm{d}}^2\ast =0$ .

Theorem 7.5

Let $(M,g)$ be a compact orientable manifold without a boundary and $\alpha \in {\Omega }^r(M),\beta \in {\Omega }^{r-1}(M)$ . Then

$$(\mathrm{d}\beta ,\alpha )=(\beta ,{\mathrm{d}}^{†}\alpha ).\text{\hspace{1em}(7.186)}$$

Proof

Since both $\mathrm{d}\beta \wedge \ast \alpha$ and $\beta \wedge \ast {\mathrm{d}}^{†}\alpha$ are $m$ -forms, their integrals over $M$ are well defined. Let $\mathrm{d}$ act on $\beta \wedge \ast \alpha$ ,

$$\mathrm{d}(\beta \wedge \ast \alpha )=\mathrm{d}\beta \wedge \ast \alpha -(-1{)}^r\beta \wedge \mathrm{d}\ast \alpha .$$

Suppose $(M,g)$ is Riemannian. Noting that $\mathrm{d}\ast \alpha$ is an $(m-r+1)$ -form and inserting the identity map $(-1{)}^{(m-r+1)[m-(m-r+1)]}\ast \ast =(-1{)}^{mr+m+r+1}\ast \ast$ , in front of $\mathrm{d}\ast \alpha$ in the second term, we have

$$\mathrm{d}(\beta \wedge \ast \alpha )=\mathrm{d}\beta \wedge \ast \alpha -(-1{)}^{mr+m+1}\beta \wedge \ast (\ast \mathrm{d}\ast \alpha ).$$

Integrating this equation over $M$ , we have

$$\begin{array}{rl}{\int }_M\mathrm{d}\beta \wedge \ast \alpha -{\int }_M\beta \wedge \ast [(-1{)}^{mr+m+1}\ast \mathrm{d}\ast \alpha ]& ={\int }_M\mathrm{d}(\beta \wedge \ast \alpha )\\ & ={\int }_{\partial M}\beta \wedge \ast \alpha =0\end{array}$$

where the last equality follows by assumption. This shows that $(\mathrm{d}\beta ,\alpha )=(\beta ,{\mathrm{d}}^{†}\alpha )$ . The reader should check how the proof is modified when $(M,g)$ is Lorentzian.

7.9.5 The Laplacian, harmonic forms and the Hodge decomposition theorem

Definition 7.7

The Laplacian $\Delta :{\Omega }^r(M)\to {\Omega }^r(M)$ is defined by

$$\Delta ={\left(\mathrm{d}+{\mathrm{d}}^{†}\right)}^2={\mathrm{dd}}^{†}+{\mathrm{d}}^{†}\mathrm{d}.\text{\hspace{1em}(7.187)}$$

As an example, we obtain the explicit form of $\Delta :{\Omega }^0(M)\to {\Omega }^0(M)$ . Let $f\in \mathcal{F}(M)$ . Since ${\mathrm{d}}^{†}f=0$ , we have

$$\begin{array}{rl}\Delta f& ={\mathrm{d}}^{†}\mathrm{d}f=-\ast \mathrm{d}\ast \left({\partial }_{\mu }f\mathrm{d}{x}^{\mu }\right)\\ & =-\ast \mathrm{d}\left(\frac{\sqrt{|g|}}{(m-1)!}{\partial }_{\mu }f{g}^{\mu \lambda }{\epsilon }_{\lambda v_2\dots v_m}\mathrm{d}{x}^{v_2}\wedge \cdots \wedge \mathrm{d}{x}^{v_m}\right)\\ & =-\ast \frac{1}{(m-1)!}{\partial }_{\nu }[\sqrt{|g|}{g}^{\lambda \mu }{\partial }_{\mu }f]{\epsilon }_{\lambda {\nu }_2\dots {\nu }_m}d{x}^{\nu }\wedge d{x}^{{\nu }_2}\wedge \cdots \wedge d{x}^{{\nu }_m}\\ & =-\ast {\partial }_v[\sqrt{|g|}{g}^{v\mu }{\partial }_{\mu }f]g^{-1}d{x}^1\wedge \cdots \wedge d{x}^m\\ & =-\frac{1}{\sqrt{|g|}}{\partial }_v[\sqrt{|g|}{g}^{v\mu }{\partial }_{\mu }f].\end{array}\text{\hspace{1em}(7.188)}$$

Exercise 7.21

Take a one-form $\omega ={\omega }_{\mu }\mathrm{d}{x}^{\mu }$ in the Euclidean space $({\mathbb{R}}^m,\delta )$ . Show that

$$\Delta \omega =-\sum _{\mu =1}^m\frac{{\partial }^2{\omega }_v}{\partial x^{\mu }\partial x^{\mu }}\mathrm{d}{x}^v.$$

Example 7.16

In example 5.11, it was shown that half of the Maxwell equations are reduced to the identity, $\mathrm{d}F={\mathrm{d}}^{2}A=0$ , where $A=A_{\mu }\mathrm{d}{x}^{\mu }$ is the vector potential one-form and $F=\mathrm{d}A$ is the electromagnetic two-form. Let $\rho$ be the electric charge density and $j$ the electric current density and form the current one-form $j={\eta }_{\mu \nu }{j}^{\nu }\mathrm{d}{x}^{\mu }=-\rho \mathrm{d}t+j\cdot \mathrm{d}x$ . Then the remaining Maxwell equations become

$${\mathrm{d}}^{†}F={\mathrm{d}}^{†}\mathrm{d}A=j.\text{\hspace{1em}(7.189a)}$$

The component expression is

$$\nabla \cdot \mathbf{E}=\rho \nabla \times \mathbf{B}-\frac{\partial \mathbf{E}}{\partial t}=\mathbf{j}.\text{\hspace{1em}(7.189b)}$$

The vector potential $A$ has a large number of degrees of freedom and we can always choose an $A$ which satisfies the Lorentz condition ${\mathrm{d}}^{†}A=0$ . Then (7.189a) becomes $({\mathrm{dd}}^{†}+{\mathrm{d}}^{†}\mathrm{d})A=\Delta A=j$ .

Let $(M,g)$ be a compact Riemannian manifold. The Laplacian $\Delta$ is a positive operator on $M$ in the sense that

$$(\omega ,\Delta \omega )=(\omega ,\left(d^{†}d+d{d}^{†}\right)\omega )=(d\omega ,d\omega )+\left(d^{†}\omega ,d^{†}\omega \right)\ge 0\text{\hspace{1em}(7.190)}$$

where (7.183) has been used. An $r$ -form $\omega$ is called harmonic if $\Delta \omega =0$ and closed (coclosed) if $\mathrm{d}\omega =0$ ( ${\mathrm{d}}^{†}\omega =0$ ). The following theorem is a direct consequence of (7.190).

Theorem 7.6

An $r$ -form $\omega$ is harmonic if and only if $\omega$ is closed and coclosed.

An $r$ -form $\omega$ is called coexact if it is written globally as

$${\omega }_r={\mathrm{d}}^{†}{\beta }_{r+1}\text{\hspace{1em}(7.191)}$$

where ${\beta }_{r+1}\in {\Omega }^{r+1}(M)$ [cf a form ${\omega }_r\in {\Omega }^r(M)$ is exact if ${\omega }_r=\mathrm{d}{\alpha }_{r-1},$ ${\alpha }_{r-1}\in {\Omega }^{r-1}(M)]$ . We denote the set of harmonic $r$ -forms on $M$ by ${\mathrm{Harm}}^r(M)$ and the set of exact $r$ -forms (coexact $r$ -forms) by $\mathrm{d}{\Omega }^{r-1}(M)({\mathrm{d}}^{†}{\Omega }^{r+1}(M))$ [Note: The set of exact $r$ -forms has been denoted by $B^r(M)$ so far.]

Theorem 7.7 · Hodge decomposition theorem

Let $(M,g)$ be a compact orientable Riemannian manifold without a boundary. Then ${\Omega }^r(M)$ is uniquely decomposed as

$${\Omega }^r(M)=\mathrm{d}{\Omega }^{r-1}(M)\oplus {\mathrm{d}}^{†}{\Omega }^{r+1}(M)\oplus {\mathrm{Harm}}^r(M).\text{\hspace{1em}(7.192a)}$$

[That is, any $r$ -form ${\omega }_r$ is written globally as

$${\omega }_r=\mathrm{d}{\alpha }_{r-1}+{\mathrm{d}}^{†}{\beta }_{r+1}+{\gamma }_r\text{\hspace{1em}(7.192b)}$$

where ${\alpha }_{r-1}\in {\Omega }^{r-1}(M),{\beta }_{r+1}\in {\Omega }^{r+1}(M)$ and ${\gamma }_r\in {\mathrm{Harm}}^r(M).]$

If $r=0$ , we define ${\Omega }^{-1}(M)=\{0\}$ . The proof of this theorem requires the results of the following two easy exercises.

Exercise 7.22

Let $(M,g)$ be as given in theorem 7.7. Show that

$$\left(\mathrm{d}{\alpha }_{r-1},{\mathrm{d}}^{†}{\beta }_{r+1}\right)=\left(\mathrm{d}{\alpha }_{r-1},{\gamma }_r\right)=\left({\mathrm{d}}^{†}{\beta }_{r+1},{\gamma }_r\right)=0.\text{\hspace{1em}(7.193)}$$

Show also that if ${\omega }_r\in {\Omega }^r(M)$ satisfies

$$\left(\mathrm{d}{\alpha }_{r-1},{\omega }_r\right)=\left({\mathrm{d}}^{†}{\beta }_{r+1},{\omega }_r\right)=\left({\gamma }_r,{\omega }_r\right)=0\text{\hspace{1em}(7.194)}$$

for any $\mathrm{d}{\alpha }_{r-1}\in \mathrm{d}{\Omega }^{r-1}(M)$ , ${\mathrm{d}}^{†}{\beta }_{r+1}\in {\mathrm{d}}^{†}{\Omega }^{r+1}(M)$ and ${\gamma }_r\in {\mathrm{Harm}}^r(M)$ , then ${\omega }_r=0$

Exercise 7.23

Suppose ${\omega }_r\in {\Omega }^r(M)$ is written as ${\omega }_r=\Delta {\psi }_r$ for some ${\psi }_r\in {\Omega }^r(M)$ . Show that $({\omega }_r,{\gamma }_r)=0$ for any ${\gamma }_r\in {\mathrm{Harm}}^r(M)$ . The proof of the converse ‘if ${\omega }_r$ is orthogonal to any harmonic $r$ -form, then ${\omega }_r$ is written as $\Delta {\psi }_r$ for some ${\psi }_r\in {\Omega }^r(M)$ ’ is highly technical and we just state that the operator ${\Delta }^{-1}$ (the Green function) is well defined in the present problem and ${\psi }_r$ is given by ${\Delta }^{-1}{\omega }_r$ .

Let $P:{\Omega }^r(M)\to {\mathrm{Harm}}^r(M)$ be a projection operator to the space of harmonic $r$ -forms. Take an element ${\omega }_r\in {\Omega }^r(M)$ . Since ${\omega }_r-P{\omega }_r$ is orthogonal to ${\mathrm{Harm}}^r(M)$ , it can be written as $\Delta {\psi }_r$ for some ${\psi }_r\in {\Omega }^r(M)$ . Then we have

$${\omega }_r=\mathrm{d}\left({\mathrm{d}}^{†}{\psi }_r\right)+{\mathrm{d}}^{†}\left(\mathrm{d}{\psi }_r\right)+P{\omega }_r.\text{\hspace{1em}(7.195)}$$

This realizes the decomposition of theorem 7.7.

7.9.6 Harmonic forms and de Rham cohomology groups

We show that any element of the de Rham cohomology group has a unique harmonic representative. Let $[{\omega }_r]\in H^r(M)$ . We first show that ${\omega }_r\in {\mathrm{Harm}}^r(M)\oplus \mathrm{d}{\Omega }^{r-1}(M)$ . According to (7.192b), ${\omega }_r$ is decomposed as ${\omega }_r={\gamma }_r+\mathrm{d}{\alpha }_{r-1}+{\mathrm{d}}^{†}{\beta }_{r+1}$ . Since $\mathrm{d}{\omega }_r=0$ , we have

$$0=\left(\mathrm{d}{\omega }_r,{\beta }_{r+1}\right)=\left(\mathrm{d}{\mathrm{d}}^{†}{\beta }_{r+1},{\beta }_{r+1}\right)=\left({\mathrm{d}}^{†}{\beta }_{r+1},{\mathrm{d}}^{†}{\beta }_{r+1}\right).$$

This is satisfied if and only if ${\mathrm{d}}^{†}{\beta }_{r+1}=0$ . Hence, ${\omega }_r={\gamma }_r+\mathrm{d}{\alpha }_{r-1}$ . From (7.195) we have

$${\omega }_r=P{\omega }_r+\mathrm{d}\left({\mathrm{d}}^{†}\psi \right)=P{\omega }_r+{\mathrm{dd}}^{†}{\Delta }^{-1}{\omega }_r.\text{\hspace{1em}(7.196a)}$$

${\gamma }_r\equiv P{\omega }_r$ is the harmonic representative of $[{\omega }_r]$ . Let ${\tilde{\omega }}_r$ be another representative of $[{\omega }_r]$ : ${\tilde{\omega }}_r-{\omega }_r=\mathrm{d}{\eta }_{r-1}$ , ${\eta }_{r-1}\in {\Omega }^{r-1}(M)$ . Corresponding to (7.196a), we have

$${\tilde{\omega }}_r=P{\tilde{\omega }}_r+\mathrm{d}\left({\mathrm{d}}^{†}{\Delta }^{-1}{\tilde{\omega }}_r\right)=P{\omega }_r+\mathrm{d}(\dots )\text{\hspace{1em}(7.196b)}$$

where the last equality follows since $\mathrm{d}{\eta }_{r-1}$ is orthogonal to ${\mathrm{Harm}}^r(M)$ and hence its projection to ${\mathrm{Harm}}^r(M)$ vanishes. (7.196a) and (7.196b) show that $[{\omega }_r]$ has a unique harmonic representative $P{\omega }_r$ .

This proof shows that $H^r(M)\subset {\mathrm{Harm}}^r(M)$ . Now we prove that $H^r(M)\supset {\mathrm{Harm}}^r(M)$ . Since $\mathrm{d}{\gamma }_r=0$ for any ${\gamma }_r\in {\mathrm{Harm}}^r(M)$ , we find that $Z^r(M)\supset {\mathrm{Harm}}^r(M)$ . We also have $B^r(M)\cap {\mathrm{Harm}}^r(M)=\varnothing$ since $B^r(M)=\mathrm{d}{\Omega }^{r-1}(M)$ , see (7.192a). Thus, every element of ${\mathrm{Harm}}^r(M)$ is a non-trivial member of $H^r(M)$ and we find that ${\mathrm{Harm}}^r(M)$ is a vector subspace of $H^r(M)$ and hence ${\mathrm{Harm}}^r(M)\subset H^r(M)$ . We have proved:

Theorem 7.8 · Hodge's theorem

On a compact orientable Riemannian manifold $(M,g)$ , $H^r(M)$ is isomorphic to ${\mathrm{Harm}}^r(M)$ :

$$H^r(M)\cong {\mathrm{Harm}}^r(M).\text{\hspace{1em}(7.197)}$$

The isomorphism is provided by identifying $[\omega ]\in H^r(M)$ with $P\omega \in {\mathrm{Harm}}^r(M)$ .

In particular, we have

$$\dim {\mathrm{Harm}}^r(M)=\dim H^r(M)=b^r\text{\hspace{1em}(7.198)}$$

$b^r$ being the Betti number. The Euler characteristic is given by

$$\chi (M)=\sum (-1{)}^r{b}^r=\sum (-1{)}^r\dim {\mathrm{Harm}}^r(M)\text{\hspace{1em}(7.199)}$$

see theorem 3.7. We note that the LHS is a topological quantity while the RHS is an analytical quantity given by the eigenvalue problem of the Laplacian $\Delta$ .

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来源 PDF SHA-256：fcd24c3371d092a8c699ccf5cf0765462924b6c48412804231076e8084b1940d

OCR 来源 SHA-256：0b6324f01172a3548cd34382e82fd8bfae574602b6b2cfaa357ba96bc2c35478

OCR 产物 SHA-256：35c8bc2e7ea21f27627c6b2c0a0eb0edc8fcbb833d75abeade639afab92146f2