2
$\begingroup$

Let $M$ be an oriented Riemannian manifold. Let $\nu$ be a differential form of degree $n=\dim M$ defined in the following way: \begin{align} \nu(v_1,\ldots,v_n)(p)&=\pm\sqrt{\det(\langle v_i,v_j \rangle)} \\ &= \text{orient. vol. } \{v_1,\ldots,v_n\},\quad p\in M, \end{align} where $v_1,\ldots,v_n \in T_p(M)$ are linearly independent, and the oriented volume is affected by the sign $+$ or $-$ depending on whether or not the basis $\{v_1,\ldots,v_n\}$ belongs to the orientation of $M$; $\nu$ is called the volume element of $M$. For a vector field $X \in \mathcal X(M)$ define the interior product $i(X)\nu$ of $X$ with $\nu$ as the $(n-1)$-form: $$ i(X)\nu(Y_2,\ldots,Y_n)=\nu(X,Y_2,\ldots,Y_n), \quad Y_2,\ldots,Y_n \in \mathcal X(M). $$ Prove that $d(i(X)\nu)=\text{div} X\nu$.

Hint: Let $p \in M$ and let $E_i$ be a geodesic frame at $p$. Write $X$ as a sum $X=\sum_i f_i E_i$ and let $\omega_i$ be differential forms of degree one defined on a neighborhood of $p$ by $\omega_i(E_j)=\delta_{ij}$. Show that $\omega_i \wedge \cdots \wedge \omega_n$ is a volume form of $\nu$ on $M$. Next put $\theta_i=\omega_1 \wedge \cdots \wedge \hat \omega_i \wedge \cdots \wedge \omega_n$, where $\hat \omega_i$ signifies that the factor $\hat \omega_i$ is not present. $\color{red}{\textrm{Prove that $i(X) \nu = \sum_i (-1)^{i+1} f_i \theta_i$.}}$ It then follows that \begin{align} d(i(X)\nu)&=\sum_i (-1)^{i+1} df_i \wedge \theta_i + \sum_i (-1)^{i+1} f_i \wedge d\theta_i \\ &= (\sum_i E_i(f_i))\nu+\sum_i (-1)^{i+1} f_i \wedge d\theta_i. \end{align} But $d\theta_i=0$ at $p$, since \begin{align} \color{blue}{\textrm d \omega_k(E_i,E_j)} &= \color{blue}{\textrm E_i \omega_k(E_j)-E_j \omega_k(E_i)-\omega_k([E_i,E_j])} \\ &= \omega_k(\nabla_{E_i} E_j - \nabla_{E_j} E_i). \end{align} Therefore $$ d(i(X)\nu)(p)=(\sum_i E_i(f_i)(p))\nu=\text{div} X(p)\nu, $$ and since $p$ is arbitrary, this completes the proof.

I cannot figure out the part---which I highlighted in $\color{red}{\textrm{red}}$ from the hint---where we have to establish $i(X)\nu = \sum_i (-1)^{i+1} f_i \theta_i$. Because I was not sure what I can do with the given $(n-1)$-form $$ i(X)\nu(Y_2,\ldots,Y_n) = \nu(X,Y_2,\ldots,Y_n). $$

Edit: It turns out that I also did not quite understand what I highlighted above in $\color{blue}{\textrm{blue}}$: $$ d \omega_k(E_i,E_j) = E_i \omega_k(E_j)-E_j \omega_k(E_i)-\omega_k([E_i,E_j]). $$ Why is this equality true? I am not sure how was one able to take the exterior derivative of $d\omega_k(E_i,E_j)$. (Though, the equality does keep reminding me of the curvature tensor introduced in Chapter 4 of Do Carmo...)

Other than these, I understand the details of all other steps given in the hint.

$\endgroup$
1
$\begingroup$

Let $\theta_i=\omega_i\wedge\dots\hat\omega_i\wedge\dots\omega_n$, where $\hat\omega_i$ denotes that $\omega_i$ is removed. It suffices to check the desired result on vectors $v_2,\dots,v_n$: \begin{align*} i_X(\nu)(v_2,\dots,v_n) & = \nu(X,v_2,\dots,v_n) \\ & = \sum_if_i\nu(E_i,v_2,\dots,v_n) \\ & = \sum_if_i\omega_1\wedge\dots\wedge\omega_n(E_i,v_2,\dots,v_n) \\ & = \sum_i(-1)^{i+1}f_i\theta_i(v_2,\dots,v_n), \end{align*} where $(-1)^{i+1}$ arises from the number of swaps needed to shift $E_i$ to the $i^{\mathrm{th}}$ coordinate, which is the only permutation with a nonzero evaluation since $\omega_i(E_j)=\delta^i_j$.

The equality in blue follows from the following result: Let $\alpha$ be a smooth 1-form and let $X,Y\in\mathfrak{X}(M)$ be smooth vector field. Then $d\alpha(X,Y)=X\alpha(Y)-Y\alpha(X)-\alpha([X,Y])$.

By definition we have $$d\alpha(X,Y) = (i_X(d\alpha))(Y),$$ and recall Cartan's formula: $$\mathcal{L}_X\alpha = d(i_X\alpha)+i_X(d\alpha),$$ where $\mathcal{L}_X$ is the Lie derivative with respect to $X$. Now we see that \begin{align*} d\alpha(X,Y) & = (i_X(d\alpha))(Y) \\ & = (\mathcal{L}_X\alpha)(Y) - d(i_X\alpha)(Y) \\ & = (\mathcal{L}_X\alpha)(Y) - Y(\alpha(X)) \\ & = \left(\mathcal{L}_X(\alpha(Y)) - \alpha(\mathcal{L}_XY)\right) - Y(\alpha(X)) \\ & = X(\alpha(Y)) - \alpha([X,Y]) - Y(\alpha(X)). \end{align*} Alternatively, if you haven't worked too much with differential forms (in particular with Lie derivatives and Cartan's formula) you should also be able to prove this result by noting that every smooth 1-form $\alpha$ can locally be written as $\alpha = g~df$ where $f$ and $g$ are smooth functions and then showing that both sides of the desired equality are indeed equal.

$\endgroup$
  • $\begingroup$ Why is it not $(-1)^{i-1}$ swaps to shift $E_i$ from the first coordinate to the $i^{\text{th}}$ coordinate? For example, it takes 4 swaps to get from the 1st coordinate to the 5th, am I mistaken? (Of course, I understand that $(-1)^{i-1}=(-1)^{i+1}$.) $\endgroup$ – New day rising May 29 '17 at 20:52
  • 1
    $\begingroup$ You are right that it shifts $i-1$ times! I was just stating that the permutation is where the $(-1)^{i+1}$ comes from... $\endgroup$ – yousuf soliman May 30 '17 at 8:09
  • $\begingroup$ May I ask one more question? There was another step in the hint that I could not justify: how can we show that $$d \omega_k(E_i,E_j) = E_i \omega_k(E_j)-E_j \omega_k(E_i)-\omega_k([E_i,E_j])$$ for $i,j=1,\ldots,n$? (I edited my question post accordingly.) $\endgroup$ – New day rising May 30 '17 at 15:54
  • 1
    $\begingroup$ I just edited my answer to include a possible derivation of the equality in blue. Let me know if there is any confusion or if there is anything else you'd like me to expand upon! $\endgroup$ – yousuf soliman May 30 '17 at 16:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.