8.6 The Hodge dual

中文

Inner product of $p$-vectors

The coupling between linear functionals (1-forms) and vectors, denoted

$$⟨u,\omega ⟩\equiv ⟨\omega ,u⟩=\omega (u)=u^i{w}_i,$$

can be extended to define a product between $p$-vectors $A$ and $p$-forms ${\beta }_{;}$ ,

$$⟨A,\beta ⟩=p!C_1^1{C}_2^2\dots C_p^{p}A\otimes \beta =p!A^{i_1{i}_2\dots i_p}{\beta }_{i_1{i}_2\dots i_p}.\text{\hspace{1em}(8.39)}$$

For each fixed $p$-form $\beta$ the map $A\mapsto ⟨A,\beta ⟩$ clearly defines a linear functional on the vector space ${\Lambda }^p(V)$ .

Exercise

Show that

$$⟨u_1\wedge u_2\wedge \cdots \wedge u_p,\beta ⟩=p!\beta (u_1,u_2,\dots ,u_p).\text{\hspace{1em}(8.40)}$$

Theorem 8.3 · Interior products and multivector contraction

If $A$ is a $p$-vector, $\beta$ a ($p$ 1)-form and $u$ an arbitrary vector, then

$$⟨A,{\mathrm{i}}_u\beta ⟩=⟨u\wedge A,\beta ⟩.\text{\hspace{1em}(8.41)}$$

Proof

For a simple $p$-vector, $A=u_1\wedge u_2\wedge \cdot \cdot \cdot \wedge u_p$ , using Eqs. Eq. (8.40) and Eq. (8.18),

$$\begin{array}{c}⟨u_1\wedge u_2\wedge \cdots \wedge u_p,{\mathrm{i}}_u\beta ⟩=p!({\mathrm{i}}_u\beta )(u_1,u_2,\dots ,u_p)\\ =(p+1)!\beta (u,u_1,\dots ,u_p)\\ =⟨u\wedge u_1\wedge \cdots \wedge u_p,\beta ⟩.\end{array}$$

Since every $p$-vector $A$ is a sum ofsimple $p$-vectors, this generalizes to arbitrary $p$-vectors by linearity. -

If $V$ is an inner product space it is possible to define the inner product of two $p$-vectors $A$ and $B$ to be

$$(A,B)=⟨A,\beta ⟩\text{\hspace{1em}(8.42)}$$

where $\beta$ is the tensor formed from $B$ by lowering indices,

$${\beta }_{i_1{i}_2\dots i_p}=B_{i_1{i}_2\dots i_p}=g_{i_1{j}_1}{g}_{i_1{j}_2}\dots g_{i_p{j}_p}{B}^{j_1{j}_2\dots j_p}.$$

Lemma 8.4 · Inner products of simple multivectors and determinants

Let $X$ and $Y$ be simple $p$-vectors, $X=x_1\wedge x_2\wedge \cdot \cdot \cdot \wedge x_p$ and $Y=y_1\wedge y_2\wedge \cdots \wedge y_p,$ then

$$(X,Y)=\det [x_i\cdot y_j].\text{\hspace{1em}(8.43)}$$

Proof

With respect to a basis $\{e_i\}$

$$\begin{array}{l}(X,Y)=p!x_1^{[i_1}{x}_2^{i_2}\dots x_p^{i_p]}{g}_{i_1{j}_1}{g}_{i_1{j}_2}\dots g_{i_p{j}_p}{y}_1^{[j_1}{y}_2^{j_2}\dots y_p^{j_p]}\\ =p!x_1^{[i_1}{x}_2^{i_2}\dots x_p^{i_p]}{y}_{1i_1}{y}_{2i_2}\dots y_{p{i}_p}\\ ={\sum }_{\sigma }(-1{)}^{\sigma }{x}_1^{i_{\sigma (1)}}{x}_2^{i_{\sigma (2)}}\dots x_p^{i_{\sigma (p)}}{y}_{1i_1}{y}_{2i_2}\dots y_{p{i}_p}\\ =\det [⟨x_i,{\gamma }_j⟩]\text{where}{\gamma }_j=y_{jk}{\epsilon }^k\\ =\det [g_{kl}{x}_i^k{y}_j^l]\\ =\det [x_i\cdot y_j].\end{array}$$

Theorem 8.5 · The induced inner product on multivector spaces

The map $A$, $B\mapsto (A,B)$ makes ${\Lambda }^p(V)$ into a real inner product space

Proof

From Eqs. Eq. (8.42) and Eq. (8.39)

$$(A,B)=p!A^{i_1{i}_2\dots i_p}{g}_{i_1{j}_1}{g}_{i_1{j}_2}\dots g_{i_p{j}_p}{B}^{j_1{j}_2\dots j_p}=(B,A)$$

so that $(A,B)$ is a symmetric bilinear function of $A$ and $B$. It remains to show that $\left(.,.\right)$ is non-singular, satisfying (SP3) of Section 5.1.

Let $e_1,e_2,\dots ,e_n$ be an orthonormal basis,

$$e_i\cdot e_j={\eta }_i{\delta }_{ij},{\eta }_i=\pm 1,$$

and for any arbitrary increasing sequences of indices $\mathbf{h}=h_1<h_2<\cdot \cdot \cdot <h_p$ set $e_{\mathbf{h}}$ to be the basis $p$-vector

$$e_{\mathbf{h}}=e_{h_1}\wedge e_{h_2}\wedge \cdots \wedge e_{h_p}.$$

If $\mathbf{h}=h_1<h_2<\cdot \cdot \cdot <h_p$ and $\mathbf{k}=k_1<k_2<\cdot \cdot \cdot <k_p$ are any pair of increasing sequences of indices and $h_i\notin \{k_1,k_2,\dots ,k_p\}$ for some $i$, then, by Lemma 8.4

$$(e_{\mathbf{h}},e_{\mathbf{k}})=\det [e_{h_i}\cdot e_{k_j}]=0$$

since the $i$th row of the determinant vanishes completely. On the other hand, $\mathrm{if}\mathbf{h}=\mathbf{k}$ we have

$$(e_{\mathbf{h}},e_{\mathbf{h}})={\eta }_{h_1}{\eta }_{h_2}\dots {\eta }_{h_p}.$$

In summary,

$$(e_{\mathbf{h}},e_{\mathbf{k}})=\det [e_{h_i}\cdot e_{k_j}]=\{\begin{array}{ll}0& \text{if }\mathbf{h}\ne \mathbf{k},\\ {\eta }_{h_1}{\eta }_{h_2}\dots {\eta }_{h_p}& \text{if }\mathbf{h}=\mathbf{k},\end{array}\text{\hspace{1em}(8.44)}$$

so $E_p=\{e_{\mathbf{h}}|h_1<h_2<\cdot \cdot \cdot <h_p\}$ forms an orthonormal basis with respect to the inner product on ${\Lambda }^p(V)$ . The matrix ofthe inner product is non-singular with respect to this basis since it is diagonal with 1’s along the diagonal.

Exercise

Show tha $(E,E)=(-1{)}^s$ where $E=e_{12,...n}$ and $s$ is the number of signs in $g_{ij}$

An inner product can of course be defined on ${\Lambda }^{\ast p}(V)$ in exactly the same way as for ${\Lambda }^p(V)$ ,

$$(\alpha ,\beta )=⟨A,\beta ⟩=p!A^{i_1{i}_2\dots i_p}{\beta }_{i_1{i}_2\dots i_p}\text{\hspace{1em}(8.45)}$$

where $A$ is the tensor formed from $\alpha$ by raising indices,

$$A^{i_1{i}_2\dots i_p}={\alpha }^{i_1{i}_2\dots i_p}=g^{i_1{j}_1}{g}^{i_1{j}_2}\dots g^{i_p{j}_p}{\alpha }_{j_1{j}_2\dots j_p}.$$

Exercise

Show that Theorem 8.3 can be expressed in the alternative form: for any $p$-form $\alpha$, $(p+1)$ form $\beta$ and vector $u$

$$(\alpha ,{\mathrm{i}}_u\beta )=(\bar{g}u\wedge \alpha ,\beta )\text{\hspace{1em}(8.46)}$$

where $\bar{g}u$ is the 1-form defined in Example 7.8 by lowering the index of $\cdot u^i$ .

The Hodge star operator

Let $V$ be an oriented inner product space and $\{e_1,e_2,\dots ,e_n\}$ be a positively oriented orthonormal basis, $e_i\cdot e_j{\delta }_{ij}{\eta }_i$ , with associated volume element $E=e_{12\dots n}=e_1\wedge e_2\wedge \cdots \wedge e_n$ . For any $A\in {\Lambda }^p(V)$ the map $f_A:{\Lambda }^{n-p}(V)\to$ $R$ defined by $A\wedge B=f_A(B)E$ is linear since

$$f_A(B+aC)E=A\wedge (B+aC)=A\wedge B+aA\wedge C=\left(f_A(B)+a{f}_A(C)\right)E.$$

Thus $f_A$ is a linear functional on ${\Lambda }^{n-p}(V)$ , and as the inner product $\left(.,.\right)$ on ${\Lambda }^{n-p}(V)$ is non-singular there exists a unique $(n-p)$ -vector $A$ such that $f_A(B)=(\ast A,B)$ . The $(n-p)$ -vector $A$ is uniquely determined by this equation, and is frequently referred to as the (Hodge) dual of $A$,

$$A\wedge B=(\ast A,B)E\text{ for all }B\in {\Lambda }^{(n-p)}(V).\text{\hspace{1em}(8.47)}$$

The one-to-one map $\ast :{\Lambda }^p(V)\to {\Lambda }^{n-p}(V)$ is called the Hodge star operator; it assigns an $(n-p)$ -vector to each $p$-vector, and vice versa. This reciprocity is only possible because the dimensions of these two vector spaces are identical,

$$\dim {\Lambda }^p(V)=\dim {\Lambda }^{n-p}(V)=\left(\genfrac{}{}{0ex}{}{n}{p}\right)=\left(\genfrac{}{}{0ex}{}{n}{n-p}\right)=\frac{n!}{p!(n-p)!}.$$

Since $E$ is independent of the choice of positively oriented orthonormal basis, the Hodge dual is a basis-independent concept.

To calculate the components ofthe dual with respect to a positively oriented orthonorma basis $e_1,\dots ,e_n$ , set $\mathbf{i}=i_1<i_2<\cdots <i_p$ and $\mathbf{j}=j_1<j_2<\cdots <j_{n-p}$ . Since

$$e_{\mathbf{i}}\wedge e_{\mathbf{j}}={ϵ}_{i_1{i}_2\dots i_p{j}_i\dots j_{n-p}}E,$$

we have from Eq. (8.47) that

$$(\ast e_{\mathbf{i}},e_{\mathbf{j}})={ϵ}_{i_1{i}_2\dots i_p{j}_i\dots j_{n-p}},$$

and Eq. (8.44) gives

$$\ast e_{\mathbf{i}}={ϵ}_{i_1{i}_2\dots i_p{j}_i\dots j_{n-p}}(e_{\mathbf{j}},e_{\mathbf{j}})e_{\mathbf{j}}\text{\hspace{1em}(8.48)}$$

where $\mathbf{j}=j_1<j_2<\cdots <j_{n-p}$ is the complementary set of indices to $i_1<\cdots <i_p$ . By a similar argument

$$\ast e_{\mathbf{j}}={ϵ}_{j_1{j}_2\dots j_{n-p}{i}_1\dots i_p}(e_{\mathbf{i}},e_{\mathbf{i}})e_{\mathbf{i}},$$

and, temporarily suspending the summation convention,

$$\begin{array}{rl}\ast \ast e_{\mathbf{i}}& ={ϵ}_{i_1{i}_2\dots i_p{j}_i\dots j_{n-p}}{ϵ}_{j_1{j}_2\dots j_{n-p}{i}_1\dots i_{n-p}}(e_{\mathbf{i}},e_{\mathbf{i}})(e_{\mathbf{j}},e_{\mathbf{j}})e_{\mathbf{i}}\\ & =(-1{)}^{p(n-p)}{\eta }_{i_1}{\eta }_{i_2}\dots {\eta }_{i_p}{\eta }_{j_1}{\eta }_{j_2}\dots {\eta }_{j_{n-p}}{e}_{\mathbf{i}}\\ & =(-1{)}^{p(n-p)+s}{e}_{\mathbf{i}}\end{array}$$

where $s$ is the number of 1’s in $g_{ij}$ . The coefficient $s$ can also be written as $\begin{array}{r}s=\frac{1}{2}(n-t)\end{array}$ where $t$ is the index of the metric. As any $p$-vector $A$ is a linear combination of the $e_{\mathrm{i}}$ , we have the identity

$$\ast \ast A=(-1{)}^{p(n-p)+s}A.\text{\hspace{1em}(8.49)}$$

Theorem 8.6 · Identities for the Hodge star operator

For any $p$-vectors $A$ and $B$ we also have thefollowing identities:

$$A\wedge \ast B=B\wedge \ast A=(-1{)}^s(A,B)E,\text{\hspace{1em}(8.50)}$$

and

$$(\ast A,\ast B)=(-1{)}^s(A,B).\text{\hspace{1em}(8.51)}$$

Proof

The first part of Eq. (8.50) follows from

$$A\wedge \ast B=(\ast A,\ast B)E=(\ast B,\ast A)E=B\wedge \ast A.$$

The second part of Eq. (8.50) follows on using Eqs. Eq. (8.16) and Eq. (8.49),

$$\begin{array}{l}A\wedge \ast B=(-1{)}^{p(n-p)}\ast B\wedge A\\ =(-1{)}^{p(n-p)}\ast B\wedge A\\ =(-1{)}^{p(n-p)}(\ast \ast B,A)E\\ =(-1{)}^s(A,B)E.\end{array}$$

Using Eqs. Eq. (8.47) and Eq. (8.50) we have

$$(\ast A,\ast B)E=A\wedge \ast B=(-1{)}^s(A,B)E$$

and Eq. (8.51) follows at once since $E\ne 0$

The component prescription for the Hodge dual in an o.n. basis is straightforward, and is left as an exercise

$$\ast A^{j_1\dots j_{n-p}}=\frac{(-1{)}^s}{(n-p)!}{ϵ}^{i_1\dots i_p{j}_1\dots j_{n-p}}{A}_{i_1\dots i_p}.\text{\hspace{1em}(8.52)}$$

Writing this equation as the component form of the tensor equation,

$$\ast A^{j_1\dots j_{n-p}}=\frac{(-1{)}^{s}n!}{(n-p)!}{E}^{i_1\dots i_p{j}_1\dots j_{n-p}}{A}_{i_1\dots i_p},$$

and using Eq. (8.30), we can express the Hodge dual in an arbitrary basis:

$$\ast A^{j_1\dots j_{n-p}}=\frac{(-1{)}^s}{(n-p)!\sqrt{|g|}}{ϵ}^{i_1\dots i_p{j}_1\dots j_{n-p}}{A}_{i_1\dots i_p}.\text{\hspace{1em}(8.53)}$$

Exercise

Show that on lowering indices, Eq. (8.53) can be written

$$\ast A_{j_1\dots j_{n-p}}=\frac{\sqrt{|g|}}{(n-p)!}{ϵ}_{i_1\dots i_p{j}_1\dots j_{n-p}}{A}^{i_1\dots i_p}.\text{\hspace{1em}(8.54)}$$

Example 8.6

Treating 1 as the basis 0-vector, we obtain from Eq. (8.48) that

$$\ast 1={ϵ}_{12\dots n}(E,E)E=(-1{)}^{s}E.$$

Conversely the dual of the volume element $E$ is 1,

$$\ast E={ϵ}_{12\dots n}1=1.$$

These two formulae agree with the double star formula Eq. (8.49) on setting $p=n\mathrm{or}p=0$

Example 8.7

In three-dimensional cartesian tensors, $s=0$ and all indices are in the sub script position:

$$\begin{array}{c}(\ast 1{)}_{ijk}={ϵ}_{ijk},\\ A=A_i{e}_i⟹(\ast A{)}_{ij}=\frac{1}{2!}{ϵ}_{kij}{A}_k=\frac{1}{2}{ϵ}_{ijk}{A}_k,\\ A=A_{ij}{e}_i\otimes e_j⟹(\ast A{)}_i={ϵ}_{jki}{A}_{jk}={ϵ}_{ijk}{A}_{jk},\\ \ast E=\frac{1}{3!}{ϵ}_{ijk}{ϵ}_{ijk}=1.\end{array}$$

The concept of vector product of any two vectors $\mathbf{u}\times \mathbf{v}$ is the dual of the wedge product, for

$$\mathbf{u}\equiv u=u_i{e}_i,\mathbf{v}\equiv v=v_j{e}_j⟹(u\wedge v{)}_{ij}=\frac{1}{2}(u_i{v}_j-u_j{v}_i),$$

whence

$$\begin{array}{rl}\ast (u\wedge v{)}_i& ={ϵ}_{ijk}(u\wedge v{)}_{jk}\\ & =\frac{1}{2}{ϵ}_{ijk}(u_j{v}_k-u_k{v}_j)\\ & ={ϵ}_{ijk}{u}_j{v}_k\\ & =(\mathbf{u}\times \mathbf{v}{)}_i.\end{array}$$

Example 8.8

In four-dimensional Minkowski space, with $s=1$ , the formulae for duals of various $p$-vectors are

$$\begin{array}{l}(\ast 1{)}^{ijkl}=-\frac{1}{4!}{ϵ}^{ijkl}=-\frac{1}{24}{ϵ}^{ijkl},\\ A=A^i{e}_i⟹(\ast A{)}^{ijk}=\frac{-1}{3!}{ϵ}^{lijk}{A}_l=\frac{1}{6}{ϵ}^{ijkl}{A}_l,\\ B=B^{ijk}{e}_i\otimes e_j\otimes e_k⟹(\ast B{)}^i=-{ϵ}^{jkli}{B}_{jkl}={ϵ}^{ijkl}{B}_{jkl},\\ F=F^{ij}{e}_i\otimes e_j⟹(\ast F{)}^{ij}=\frac{-1}{2!}{ϵ}^{klij}{F}_{kl}=-\frac{1}{2}{ϵ}^{ijkl}{F}_{kl},\\ \ast E=-{ϵ}^{ijkl}\frac{-1}{4!}{ϵ}_{ijkl}=\frac{4!}{4!}=1.\end{array}$$

Note that if the components $F^{ij}$ are written out in the following ‘electromagnetic form’, the significance of which will become clear in Chapter 9,

$$[F^{ij}]=\left(\begin{array}{cccc}0& B_3& -B_2& -E_1\\ -B_3& 0& B_1& -E_2\\ B_2& -B_1& 0& -E_3\\ E_1& E_2& E_3& 0\end{array}\right)⟹[F_{ij}]=\left(\begin{array}{cccc}0& B_3& -B_2& E_1\\ -B_3& 0& B_1& E_2\\ B_2& -B_1& 0& E_3\\ -E_1& -E_2& -E_3& 0\end{array}\right)$$

then the dual tensor essentially permutes electric and magnetic fields, up to a sign change,

$$\ast F^{ij}=\left(\begin{array}{cccc}0& -E_3& E_2& -B_1\\ E_3& 0& -E_1& -B_2\\ -E_2& E_1& 0& -B_3\\ B_1& B_2& B_3& 0\end{array}\right).$$

The index lowering map $\bar{g}:V\to V^{\ast }$ can be uniquely extended to a linear map from $p$-vectors to $p$-forms,

$$\bar{g}:{\Lambda }^p(V)\to {\Lambda }^{\ast p}(V)$$

by requiring that

$$\bar{g}(u\wedge v\wedge \cdots \wedge w)=\bar{g}(u)\wedge \bar{g}(v)\wedge \cdots \wedge \bar{g}(w).$$

In components it has the expected effect

$$(\bar{g}A{)}_{i_1{i}_2\dots i_p}=g_{i_1{j}_1}{g}_{i_2{j}_2}\dots g_{i_p{j}_p}{A}^{j_1{j}_2\dots j_p}$$

and from Eqs. Eq. (8.42) and Eq. (8.45) it follows that

$$(\bar{g}(A),\bar{g}(B))=(A,B).\text{\hspace{1em}(8.55)}$$

If the Hodge star operator is defined on forms by requiring that it commutes with the lowering operator $\bar{g}.$

$$\ast \bar{g}(A)=\bar{g}(\ast A)\text{\hspace{1em}(8.56)}$$

then we find the defining relation for the Hodge star of a $p$-form $\alpha$ in a form analogous to Eq. (8.47),

$$\alpha \wedge \beta =(-1{)}^s(\ast \alpha ,\beta )\Omega .\text{\hspace{1em}(8.57)}$$

The factor $(-1{)}^s$ that enters this equation is essentially due to the fact that the ‘lowered’ basis vectors $\bar{g}(e_i)$ have a different orientation to the dual basis ${\epsilon }^i$ if $s$ is an odd number, while they have the same orientation when $s$ is even.

Problems

Problem 8.6

Show that the quantity $⟨A,\beta ⟩$ defined in Eq. (8.39) vanishes for all $p$-vectors $A$ if and only if $\beta =0$ . Hence show that the correspondence between linear functionals on $\Lambda (V)$ and $p$-forms is one-to-one,

$$({\Lambda }^p(V){)}^{\ast }\cong {\Lambda }^{\ast p}(V).$$

Problem 8.7

Show that the interior product between basis vectors $e_i$ and ${\epsilon }^{i_1{i}_2\dots i_r}$ is given by

$${\mathrm{i}}_{e_i}{\epsilon }^{i_1\dots i_r}=\{\begin{array}{ll}0& \text{ if }i\notin \{i_1,\dots i_r\},\\ (-1{)}^{a-1}{\epsilon }^{i_1\dots i_{a-1}{i}_{a+1}\dots i_r}& \text{ if }i=i_a.\end{array}$$

Problem 8.8

Prove Eq. (8.52).

Problem 8.9

Every $p$-form $\alpha$ can be regarded as a linear functional on ${\Lambda }^p(V)$ through the action $\alpha (A)=⟨A,\alpha ⟩$ . Show that the basis ${\epsilon }^{\mathbf{i}}$ is dual to the basis $e_{\mathrm{j}}$ where $\mathbf{i}=i_1<i_2<\cdot \cdot \cdot <i_p,\mathbf{j}=j_1<j_2<\cdots <j_p,$

$$⟨e_{\mathbf{j}},{\epsilon }^{\mathbf{i}}⟩={\delta }_{\mathbf{j}}^{\mathbf{i}}\equiv {\delta }_{j_1}^{i_1}{\delta }_{j_2}^{i_2}\dots {\delta }_{j_p}^{i_p}.$$

Verify that

$$\sum _{i_1<i_2<\cdots <i_p}\cdots \sum e_{i_1{i}_2\dots i_p}{\epsilon }^{i_1{i}_2\dots i_p}=\dim {\Lambda }^p(V).$$

Problem 8.10

Show that if $u$, $v$ and $w$ are vectors in an $n$-dimensional real inner product space then

(a) $(u\wedge v,u\wedge v)=(u\cdot u)(v\cdot v)-(u\cdot v{)}^2.$

(b) $u\wedge \ast (v\wedge w)=(u\cdot w)\ast v-(u\cdot v)\ast w.$

(c) Which identities do these equations reduce to in three-dimensional cartesian vectors?

Problem 8.11

Let $g_{ij}$ be the Minkowski metric on a four-dimensional space, having index 2 (so tha there are three signs and $\mathrm{one}-\mathrm{sign})$ .

(a) By calculating the inner products $(e_{i_1{i}_2},e_{j_1,j_2})$ , using Eq. (8.44) show that there are three 1’s 1’s in these inner products, and the index of the inner product defined on the six-dimensional space of bivectors ${\Lambda }^2(V)$ is therefore 0.

(b) What is the index ofthe inner product on ${\Lambda }^2(V)$ if $V$ is $n$-dimensional and $g_{ij}$ has index $t$? [Ans.: $\frac{1}{2}(t^2-n).]$

Problem 8.12

Show that in an arbitrary basis the component representation of the dual of a $p$-form $\alpha$ is

$$(\ast \alpha {)}_{j_1{j}_2\dots j_{n-p}}=\frac{\sqrt{|g|}}{(n-p)!}{ϵ}_{i_1\dots i_p{j}_1{j}_2\dots j_{n-p}}{\alpha }^{i_1\dots i_p}.\text{\hspace{1em}(8.58)}$$

Problem 8.13

If $u$ is any vector, and $\alpha$ any $p$-form show that

$${\mathrm{i}}_u\ast \alpha =\ast (\alpha \wedge \bar{g}(u)).$$