Let $\omega$ be a $p$-form on a Riemannian manifold $M^n$ with metric $g$ and let $\vol_{i_1,\ldots,i_n}=\sqrt{\lvert g\rvert} \epsilon_{i_1,\ldots,i_n}$ be a volume form, where $\epsilon_{i_1,\ldots,i_n}$ is the Levi-Civita symbol with $\epsilon_{1,\ldots,n}=1$. By definition $*\omega$ is the $(n-p)$-form on $M^n$ given by the formula $$(*\omega)_{i_{p+1},\ldots,i_n} = \frac{1}{p!} \vol_{i_1,\ldots,i_n} \omega^{i_1,\ldots,i_p}.$$ I need to show the formula $\alpha \wedge *\beta = g(\alpha,\beta) \vol$ is valid, where $\alpha = \alpha_{i_1,\ldots,i_p}$, $\beta=\beta_{i_1,\ldots,i_p}$ are arbitrary $p$-forms, $g(\alpha,\beta) = \alpha_{i_1,\ldots,i_p}\beta^{i_1,\ldots,i_p}$.

By definition $$ (\alpha \wedge *\beta)_{i_1,\ldots,i_n} = \frac{1}{p!(n-p)!}\delta^{j_1,\ldots,j_n}_{i_1,\ldots,i_n} \alpha_{j_1,\ldots,j_p}(*\beta)_{j_{p+1},\ldots,j_n} = \\ \frac{1}{p!^2(n-p)!}\delta^{j_1,\ldots,j_n}_{i_1,\ldots,i_n} \alpha_{j_1,\ldots,j_p} \vol_{k_1,\ldots,k_p,j_{p+1},\ldots,j_n}g^{k_1,s_1}\ldots g^{k_p,s_p} \beta_{s_1,\ldots,s_p}. $$ And I can't see what can I do with this expression. Please, help me to finish the demonstration.


One method is:

  1. To work with the whole form $\alpha \wedge *\beta$ rather than its components $(\alpha \wedge *\beta)_{i_1\ldots i_n}$.

  2. To use $$\tag{1}\mathrm{d}x^{i_1} \wedge \ldots \wedge \mathrm{d}x^{i_n}~=~\varepsilon^{i_1\ldots i_n} \mathrm{d}x^{1} \wedge \ldots \wedge \mathrm{d}x^{n}~=~\frac{\varepsilon^{i_1\ldots i_n}}{\sqrt{|\det(g_{\cdot\cdot})|}}\Omega,$$ where$^1$ $\varepsilon^{i_1\ldots i_n}$ here denotes the Levi-Civita symbol (as opposed to the Levi-Civita tensor), i.e., $\varepsilon^{1\ldots n}=1$.

  3. To use

$$ \tag{2}\delta_{j_1\ldots j_p}^{i_1\ldots i_p}~=~\frac{1}{(n-p)!}\varepsilon^{i_1\ldots i_pk_{p+1}\ldots k_n} ~\varepsilon_{j_1\ldots j_pk_{p+1}\ldots k_n}. $$

Sketched proof:

$$\alpha \wedge *\beta~=~\frac{1}{p!} \alpha_{i_1\ldots i_p} \mathrm{d}x^{i_1} \wedge \ldots \wedge \mathrm{d}x^{i_p} ~\wedge~\frac{\sqrt{|g|}}{p!(n-p)!} \beta^{j_1\ldots j_p}~\varepsilon_{j_1\ldots j_n}\mathrm{d}x^{j_{p+1}} \wedge \ldots \wedge \mathrm{d}x^{j_n} $$ $$~\stackrel{(1)}{=}~ \frac{1}{(p!)^2 (n-p)!} \alpha_{i_1\ldots i_p}~\beta^{j_1\ldots j_p}~\varepsilon_{j_1\ldots j_n}~ \varepsilon^{i_1\ldots i_pj_{p+1}\ldots j_n}~\Omega $$ $$ \tag{3}~\stackrel{(2)}{=}~\frac{1}{(p!)^2 } \alpha_{i_1\ldots i_p}~\beta^{j_1\ldots j_p}~\delta_{j_1\ldots j_p}^{i_1\ldots i_p}~\Omega ~=~\frac{1}{p! } \alpha_{i_1\ldots i_p}~\beta^{i_1\ldots i_p}~\Omega ~=~g( \alpha,\beta) ~\Omega .$$


$^1$ In the last equality of eq. (1), we allow for simplicity only local coordinate transformations that preserve orientation.

  • $\begingroup$ But then I need to obtain the expression for $*(dx^{i_1}\wedge\ldots\wedge dx^{i_k})$ starting from the given definition via components. I know the answer but it isn't clear how to obtain it (I face with the calculations of the same type as in my post). $\endgroup$ – Appliqué Nov 9 '13 at 22:26
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Nov 10 '13 at 14:33
  • $\begingroup$ Thank you, very interesting why in the Aubin's book "Nonlinear analysis on manifolds" it is written that $g(\alpha,\beta) = \alpha_{i_1,\ldots,i_p} \beta^{i_1,\ldots,i_p}$. It isn't said that the summation is over $i_1 < \ldots < i_p$. $\endgroup$ – Appliqué Nov 13 '13 at 18:44

This calculation in differential forms is quite trivial in geometric algebra. Let $\alpha$ and $\beta$ be $p$-vectors. Let $\epsilon$ be some pesudoscalar.

The geometric product $\alpha \beta \epsilon$ has as its grade-$n$ component

$$\langle (\alpha \beta )\epsilon\rangle_n = \langle \alpha \beta \rangle_0 \epsilon = (\alpha \cdot \beta) \epsilon$$

But associativity of the geometric product tells us we can group $\beta \epsilon$ instead to get

$$\langle \alpha (\beta \epsilon) \rangle_n = \alpha \wedge (\beta \epsilon)$$

Alas, while I'm sure that this solution can be translated into differential forms parlance, the key ingredient--the geometric product and the use of grade projection--is something that would be pretty onerous in itself to translate.


The question is already solved but I just wanted to add a proof alogn the same lines of the original post, using (abstract) index notation.

In my notation, the volume form is $\mu$.

\begin{align*} (\alpha\wedge{\star}\beta)_{a_1\dots a_n} &= \frac{n!}{k!(n-k)!}(\alpha\otimes{\star}\beta)_{[a_1\dots a_n]}\tag{1}\\ &= \frac{n!}{k!(n-k)!}\alpha_{[a_1\dots a_k}({\star}\beta)_{a_{k+1}\dots a_n]}\tag{2}\\ &= \frac{n!}{k!(n-k)!}\frac{1}{k!}\alpha_{[a_1\dots a_k} \beta^{b_1\dots b_k} \mu_{|b_1\dots b_k | a_{k+1}\dots a_n]}\tag{3}\\ &= \frac{1}{k!(n-k)!}\frac{1}{k!}\alpha_{c_1\dots c_k} \beta^{b_1\dots b_k} \mu_{b_1\dots b_k c_{k+1}\dots c_n}\delta^{c_1\dots c_n}_{a_1\dots a_n}\tag{4}\\ &= \frac{\operatorname{sgn}(g)}{k!(n-k)!}\frac{1}{k!}\alpha_{c_1\dots c_k} \beta^{b_1\dots b_k} \mu_{b_1\dots b_k c_{k+1}\dots c_n}\mu^{c_1\dots c_n}\mu_{a_1\dots a_n}\tag{5}\\ &= \frac{1}{k!(n-k)!}\frac{1}{k!}\alpha_{c_1\dots c_k} \beta^{b_1\dots b_k} \delta^{c_1\dots c_n}_{b_1\dots b_k c_{k+1}\dots c_n}\mu_{a_1\dots a_n}\tag{6}\\ &= \frac{1}{k!}\frac{1}{k!}\alpha_{c_1\dots c_k} \beta^{b_1\dots b_k} \delta^{c_1\dots c_k}_{b_1\dots b_k}\mu_{a_1\dots a_n}\tag{7}\\ &= \frac{1}{k!}\alpha_{b_1\dots b_k} \beta^{b_1\dots b_k}\mu_{a_1\dots a_n}\tag{8}\\ &= {g}(\alpha,\beta)\mu_{a_1\dots a_n}\tag{9} \end{align*}

Where I've used the following identities

  1. $\displaystyle(\omega\wedge\eta)_{a_1\dots a_{p+q}} =\frac{(p+q)!}{p!q!}(\omega\otimes\eta)_{[a_1\dots a_{p+q}]}$
  2. $\displaystyle(\omega\otimes\eta)_{a_1\dots a_{p}b_1\dots b_q} = \omega_{a_1\dots a_p}\eta_{b_1\dots b_q}$
  3. $\displaystyle(\star\omega)_{a_{k+1}\dots a_n} = \frac{1}{k!} \omega^{a_1\dots a_k} \mu_{a_1\dots a_k a_{k+1}\dots a_n}$
  4. $\displaystyle\omega_{[a_1\dots a_k]} = \frac{1}{k!}\delta^{b_1\dots b_k}_{a_1\dots a_k}\omega_{b_1\dots b_k}$
  5. $\displaystyle\delta^{a_1\dots a_n}_{b_1\dots b_n} = \operatorname{sgn}(g)\mu^{a_1\dots 1_n}\mu_{b_1\dots b_n}$
  6. Same as 5, but in the other direction.
  7. $\displaystyle\delta^{a_1 \dots a_p c_{1} \dots c_q}_{b_1 \dots b_p c_{1} \dots c_q} = \frac{(n-p)!}{(n-(p+q))!} \, \delta^{a_1 \dots a_p}_{b_1 \dots b_p}$
  8. Same as 4, but in the other direction.
  9. $\displaystyle{g}(\omega,\eta) = \frac{1}{k!}\omega^{a_1\dots a_k}\eta_{a_1\dots a_k}$

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