16.2 Properties of exterior derivative

中文

If $\phi :M\to N$ is a smooth map between two differentiable manifolds $M$ and $N$, we define the induced map ${\phi }^{\ast }:{\Lambda }_r(N)\to {\Lambda }_r(M)$ in a similar way to the pullback map, Eq. (15.18):

$$({\phi }^{\ast }\alpha {)}_p((X_1{)}_p,(X_2{)}_p,\dots ,(X_r{)}_p)={\alpha }_{\phi (p)}({\phi }_{\ast }(X_1{)}_p,{\phi }_{\ast }(X_2{)}_p,\dots ,{\phi }_{\ast }(X_r{)}_p).$$

As for covector fields, this map is well-defined on all differential $r$-forms, ${\phi }^{\ast }\alpha$ . The pullback ofa 0-form $f\in \mathcal{F}(N)={\Lambda }_0(N)$ is defined by ${\phi }^{\ast }f=f\circ \phi$ , and it preserves wedge products

$${\phi }^{\ast }(\alpha \wedge \beta )={\phi }^{\ast }\alpha \wedge {\phi }^{\ast }\beta ,$$

which follows immediately from the definition $\alpha {\bigwedge }_{\alpha }\mathbf{\beta }=\mathcal{A}(\alpha \otimes \mathbf{\beta })$

Exercise

Show that the composition of two maps $\phi$ and $\psi$ results in a reverse composition of pullbacks, as in Eq. (15.19), $(\phi \circ \psi {)}^{\ast }={\psi }^{\ast }\circ {\phi }^{\ast }$

Theorem 16.2 · Exterior differentiation commutes with pullback

For any differential form $\alpha \in \Lambda (N)$ , the induced map ${\phi }^{\ast }$ commutes with the exterior derivative,

$$\mathrm{d}{\phi }^{\ast }\alpha ={\phi }^{\ast }\mathrm{d}\alpha .$$

Proof

For a 0-form, $\alpha =f:N\to \mathbb{R}$ , at any point $p\in M$ and any tangent vector $X_p$

$$\begin{array}{rl}& ⟨({\phi }^{\ast }\mathsf{d}f{)}_p,X_p⟩=⟨(\mathsf{d}f{)}_{\phi (p)},{\phi }_{\ast }{X}_p⟩\\ & =({\phi }_{\ast }{X}_p)f\\ & =X_p(f\circ \phi )\\ & =⟨(\mathsf{d}(f\circ \phi ){)}_p,X_p⟩.\end{array}$$

As this equation holds for all tangent vectors $X_p,$ , we have

$${\phi }^{\ast }\mathrm{d}f=\mathrm{d}(f\circ \phi )=\mathrm{d}({\phi }^{\ast }f).$$

For a general $r$-form, it is only necessary to prove the result in any local coordinate chart $(U;x^i).\mathrm{If}\alpha ={\alpha }_{i_1\dots i_r}\mathtt{d}{x}^{i_1}\wedge \cdot \cdot \cdot \wedge \mathtt{d}{x}^{i_r}$ , then

$$\begin{array}{l}{\phi }^{\ast }\mathrm{d}\alpha ={\phi }^{\ast }(\mathrm{d}{\alpha }_{i_1\dots i_r}\wedge \mathrm{d}{x}^{i_1}\wedge \dots \mathrm{d}{x}^{i_r})\\ =\mathrm{d}({\phi }^{\ast }{\alpha }_{i_1\dots i_r})\wedge \mathrm{d}({\phi }^{\ast }{x}^{i_1})\wedge \cdots \wedge \mathrm{d}({\phi }^{\ast }{x}^{i_r})\\ =\mathrm{d}({\phi }^{\ast }\alpha ).\end{array}$$

Applying the definition (15.33) of Lie derivative to the tensor field $\alpha$ and using $\widetilde{{\sigma }_t}={\left({\sigma }_{-t}\right)}^{\ast }$ , where ${\sigma }_t$ 4is a local one-parameter group generating a vector field $X$, it follows from Theorem 16.2 that the exterior derivative and Lie derivative commute,

$${\mathcal{L}}_X\mathrm{d}\alpha =\mathrm{d}{\mathcal{L}}_X\alpha .\text{\hspace{1em}(16.8)}$$

For any vector field $X$ define the interior product ${\mathfrak{i}}_X:{\Lambda }_r(M)\to {\Lambda }_{(r-1)}(M)$ ) as in Section 8.4,

$${\mathrm{i}}_X\alpha =r{C}_1^1(X\otimes \alpha ),\text{\hspace{1em}(16.9)}$$

or equivalently, for arbitrary vector fields $X_1,X_2,\dots ,X_r$

$$({\mathrm{i}}_{X_1}\alpha )(X_2,\dots ,X_r)=r\alpha (X_1,X_2,\dots ,X_r).\text{\hspace{1em}(16.10)}$$

By Eq. (8.20) ${\mathrm{i}}_X$ is an antiderivation – for any differential $r$-form $\alpha$ and arbitrary differentia form $\beta$

$${\mathrm{i}}_X(\alpha \wedge \beta )=({\mathrm{i}}_X\alpha )\wedge \beta +(-1{)}^r\alpha \wedge ({\mathrm{i}}_X\beta ).\text{\hspace{1em}(16.11)}$$

Exercise

Show that for any pair of vector fields $X$ and $Y,{\mathbf{i}}_X\circ {\mathbf{i}}_Y=-{\mathbf{i}}_Y\circ {\mathbf{i}}_X$

Theorem 16.3 · Cartan’s formulas

(Cartan) If $X$ and $Y$ are smooth vectorfields on a differentiable manifold $M$ and $\omega$ is a differential 1-form then

$${\mathrm{i}}_{[X,Y]}={\mathcal{L}}_X\circ {\mathrm{i}}_Y-{\mathrm{i}}_Y\circ {\mathcal{L}}_X,\text{\hspace{1em}(16.12)}$$

$${\mathcal{L}}_X={\mathrm{i}}_X\circ \mathrm{d}+\mathrm{d}\circ {\mathrm{i}}_X,\text{\hspace{1em}(16.13)}$$

$$\mathrm{d}\omega (X,Y)=\frac{1}{2}(X(⟨Y,\omega ⟩)-Y(⟨X,\omega ⟩)-⟨[X,Y],\omega ⟩).\text{\hspace{1em}(16.14)}$$

Proof

The first identity follows essentially from the fact that the Lie derivative ${\mathcal{L}}_X$ commutes with contraction operators, ${\mathcal{L}}_X{C}_i^i=C_i^i{\mathcal{L}}_X$ (see Problem 15.24). Thus for an arbitrary $r$-form $\alpha$, using the Leibnitz rule (15.35) gives

$$\begin{array}{rl}& \\ & {\mathcal{L}}_X({\mathrm{i}}_Y\alpha )=r{C}_1^1{\mathcal{L}}_X(Y\otimes \alpha )\\ & =r{C}_1^1[({\mathcal{L}}_{X}Y)\otimes \alpha +Y\otimes {\mathcal{L}}_X\alpha ]\\ & ={\mathrm{i}}_{[X,Y]}\alpha +{\mathrm{i}}_Y({\mathcal{L}}_X\alpha )\end{array}$$

as required.

To show Eq. (16.13) set $K_X$ to be the operator $K_X=i_{X}o\mathbf{d}+\mathbf{d}\circ i_X:{\Lambda }_r(M)\to {\Lambda }_r(M)$ Using the fact that both $i$ and $d$ are antiderivations, Eqs. Eq. (16.11) and Eq. (16.3), it is straightforward to show that $K_X$ is a derivation,

$$K_X(\alpha \wedge \beta )=K_X\alpha \wedge \beta +\alpha \wedge K_X\beta$$

for all differential forms $\alpha$ and $\beta .$ . From ${\mathrm{d}}^2=0$ the operator $K_X$ commutes with $d$,

$$K_X\circ \mathrm{d}={\mathrm{i}}_X\circ {\mathrm{d}}^2+\mathrm{d}\circ {\mathrm{i}}_X\circ \mathrm{d}=\mathrm{d}\circ {\mathrm{i}}_X\circ \mathrm{d}=\mathrm{d}\circ K_X.$$

If $\alpha$ is a 0-form $\alpha =f$ then ${\mathrm{i}}_{X}f=0$ by definition, and

$$K_{X}f={\mathrm{i}}_X(\mathrm{d}f)+{\mathrm{di}}_{X}f=⟨\mathrm{d}f,X⟩=Xf={\mathcal{L}}_{X}f.$$

Hence, since $K_X$ commutes both with $d$ and ${\mathcal{L}}_X$ ,

$$K_X\mathrm{d}f=\mathrm{d}{K}_{X}f=\mathrm{d}({\mathcal{L}}_{X}f)={\mathcal{L}}_X\mathrm{d}f.$$

On applying the derivation property we obtain $K_X(g\mathrm{d}f)={\mathcal{L}}_X(g\mathrm{d}f)$ and the required identity holds for any 1-form $\omega ,$ as it can be expressed locally in a coordinate chart at any point as $\omega =w_i\mathrm{d}{x}^i$ . The argument may be generalized to higher order $r$-forms to show that the operators ${\mathcal{L}}_X$ and $K_X$ are identical on all of ${\Lambda }_r(M)$

The final identity Eq. (16.14) is proved on applying Eq. (16.13) to a 1-form $\omega ,$

$$⟨Y,{\mathcal{L}}_X\omega ⟩=⟨Y,{\mathrm{i}}_X(\mathrm{d}\omega )+\mathrm{d}({\mathrm{i}}_X\omega )⟩$$

and using the Leibnitz rule for the Lie derivative,

$${\mathcal{L}}_X(⟨Y,\omega ⟩)-⟨{\mathcal{L}}_{X}Y,\omega ⟩={\mathrm{i}}_X\mathrm{d}\omega (Y)+Y({\mathrm{i}}_X\omega ).$$

Setting $r=1$ and $\alpha =\omega$ in Eq. (16.10),

$$X(⟨Y,\omega ⟩)-⟨{\mathcal{L}}_{X}Y,\omega ⟩=2\mathrm{d}\omega (X,Y)+Y(⟨X,\omega ⟩)$$

from which Eq. (16.14) is immediate.

If $\alpha$ is an $r$-form on $M$, a formula for $\alpha (X_1,X_2,\dots ,X_{r+1})$ ) that generalizes Eq. (16.14) is left to the reader (see Problem 16.5).

Problems

Problem 16.4

Let $\phi :{\mathbb{R}}^2\to {\mathbb{R}}^3$ be the map

$$(x,y)\to (u,v,w)\text{ where }u=\sin (xy),v=x+y,w=2.$$

For the 1-form $\omega =w_1\mathrm{d}u+w_2\mathrm{d}v+w_3$ $\mathrm{d}w$ on ${\mathbb{R}}^3$ evaluate ${\phi }^{\ast }\omega$ . For any function $f:{\mathbb{R}}^3\mathbb{R}$ verify Theorem 16.2, that $\mathbf{d}({\phi }^{\ast }f)={\phi }^{\ast }\mathbf{d}f$

Problem 16.5

If $\alpha$ is an $r$-form on a differentiable manifold $M,$ , show that for any vector field $X_1,X_2,\dots X_{r+1}$

$$\begin{array}{l}\mathrm{d}\alpha (X_1,X_2,\dots ,X_{r+1})=\frac{1}{r+1}[{\sum }_{i=1}^{r+1}(-1{)}^{i+1}{X}_i\alpha (X_1,X_2,\dots ,{\hat{X}}_i,\dots ,X_{r+1})\\ +{\sum }_{i=1}^r{\sum }_{j=i+1}^{r+1}(-1{)}^{i+j}\alpha ([X_i,X_j],\dots ,{\hat{X}}_i,\dots ,{\hat{X}}_j,\dots ,X_{r+1})]\end{array}$$

where $\widehat{X_i}$ signifies that the argument $X_i$ is to be omitted. The case $r=0$ simply asserts that $df(X)=Xf,$ , while Eq. (16.14) is the case $r=1$ . Proceed by induction, assuming the identity is true for all $(r-1)$ -forms, and use the fact that any $r$-form can be written locally as a sum of tensors of the type $\omega \wedge \beta$ where $\omega$ is a 1-form and $\beta$ an $r$-form.

Problem 16.6

Show that the Laplacian operator on ${\mathbb{R}}^3$ may be defined by

$$\mathrm{d}\ast \mathrm{d}\varphi ={\nabla }^2\varphi \mathrm{d}x\wedge \mathrm{d}y\wedge \mathrm{d}z=\left(\frac{{\partial }^2\varphi }{\partial x^2}+\frac{{\partial }^2\varphi }{\partial y^2}+\frac{{\partial }^2\varphi }{\partial z^2}\right)\mathrm{d}x\wedge \mathrm{d}y\wedge \mathrm{d}z$$

where is the Hodge star operator of Section 8.6.

Use this to express the Laplacian operator in spherical polar coordinate $(r,\theta ,\varphi )$