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Let $(M, g)$ be a Riemannian manifold and $X, Y$ vector fields on a compact subset $D$ of $M$ such that $Y$ is divergence free. I would like to show that $$\int_D g(\nabla_Y X, X) dV = 0 \tag{1}$$ where $\nabla$ is the Levi-Civita connection on $M$ and $dV$ the Riemannian volume form.

In order to show this I am thinking of somehow integrating by parts. I know on a Riemannian manifold there exists the formula $$\int_Mg(\text{grad}(f), X) dV = -\int_Mf \text{div}(X) dV + \int_{\partial M} fg(X,N) d\tilde{V}$$ for any $f \in C^\infty(M)$ and where $N$ is the outward-pointing unit normal vector field along $\partial M$ and $\tilde{g}$ is the induced Riemannian metric on $\partial M$. Is there a similar formula where instead of the gradient we have the covariant derivative as in (1)?

Alternatively, I know that for any $f \in C^\infty(D)$ we have $$\int_D L_Yf dV = 0$$ where $L_Y$ is the Lie derivative taken along $Y$. Is the Lie derivative here in anyway related to the integrand in (1)?

CBBAM
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1 Answers1

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Your proposed equality (1) is false, but what we can say is the following. By metric-compatibility of the Levi-Civita connection, undoing the product rule gives \begin{align} g(\nabla_YX,X)&=\nabla_Yf\tag{@}\\ &=\mathcal{L}_Yf, \end{align} where $f:=\frac{1}{2}g(X,X)$. Next, note that \begin{align} (\mathcal{L}_Yf)dV&=\mathcal{L}_Y(f\,dV)-f\mathcal{L}_Y(dV)\\ &=\mathcal{L}_Y(f\,dV)-0\tag{$Y$ divergence-free}\\ &=d(f\,Y\lrcorner\,dV)+0\tag{Cartan’s formula} \end{align} Hence, by Stokes’ theorem, \begin{align} \int_Dg(\nabla_YX,X)\,dV&=\int_Dd\left(f\,Y\lrcorner\,dV\right)=\int_{\partial D}fY\lrcorner\,dV=\int_{\partial D}f\langle Y,N\rangle\,dA, \end{align} where the last equal sign uses equation (*) from here… really it’s just playing around with the definition of the surface area form. Note that this integral doesn’t necessarily vanish.

Another way of saying this, without the language of differential forms, is that from (@), we have by definition of gradient of a function that \begin{align} \nabla_Yf&=\langle\text{grad}(f),Y\rangle\tag{definition}\\ &=\text{div}(fY)-f\,\text{div}(Y)\tag{product rule}\\ &=\text{div}(fY).\tag{$Y$ divergence-free} \end{align} So, integrating over $D$ and using the divergence theorem, we get \begin{align} \int_Dg(\nabla_YX,X)\,dV=\int_{\partial D}f\langle Y,N\rangle\,dA. \end{align} But again, the surface integral on the right is not necessarily $0$, unless you make more assumptions about $Y$ and/or $X$. For example if $X$ has constant norm, so $f$ is constant, then from the beginning, $\nabla_Yf=0$ so its integral is $0$.

peek-a-boo
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  • Thank you for your answer. I had asked a question about the last equality in another question in which one of the answers gave a proof (https://math.stackexchange.com/questions/4790341/the-integral-of-the-lie-derivative-over-a-divergence-free-vector-field-is-zero), is this incorrect? If this is correct, then I had thought we may use (@) to prove the first equality (1). – CBBAM Oct 21 '23 at 02:16
  • I am also fairly new to Riemannian geometry so I apologize if these questions are basic, but the compatibility equation I am familiar with is that $Yg(X,X) = 2g(\nabla_Y X)$ which is different from your (@). Does this imply $Yg(X,X) = \nabla_Y g(X,X)$? Also, how did you establish that the Lie derivative is equal to the covariant derivative, i.e. $\nabla_Y f = L_y f$? Is this a well known identity? – CBBAM Oct 21 '23 at 02:18
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    for a smooth real valued function $f$ and any smooth vector field $Y$, we have $Yf=\nabla_Yf=\mathcal{L}_Yf= (df)(Y)$, all by definition. – peek-a-boo Oct 21 '23 at 02:33
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    @CBBAM what you wrote here is different from what is written there. Here, you said $\int_{D}L_Y(f,dV)=0$, but it is not. The correct term is what I have written, it is a surface integral. In the link, you’re presumably talking about a compact Riemannian manifold $M$ without boundary, so $\partial M=\emptyset$ i.e the result is $0$; but if $M$ has non-empty boundary, then once again, you need to take that into account. – peek-a-boo Oct 21 '23 at 02:36
  • Thank you again, this is extremely helpful. If I had instead wrote $\int_D (L_Yf) dV$ would the result then hold? The textbook I am following does not include any parentheses but I assume this is what they mean. – CBBAM Oct 21 '23 at 02:38
  • well, $Y$ is divergence-free so $(L_Yf),dV=L_Y(f,dV)$ as we have seen here and in your link, so no it doesn’t make a difference. The boundary term is always there. – peek-a-boo Oct 21 '23 at 02:39
  • The only information I have about the boundary is that $Y$ is tangent to $\partial M$. – CBBAM Oct 21 '23 at 02:40
  • ok, so perfect, look closely at the formula I gave and explain why the integral is $0$. – peek-a-boo Oct 21 '23 at 02:41
  • I see now, this is because $N$ is defined to be normal to the boundary, so $g(Y,N)$ is of course zero. Thank you for all your help! Also, do you have any resources you can recommend for this kind of material (e.g. weak forms of PDEs on manifolds and general PDE theory on manifolds)? – CBBAM Oct 21 '23 at 02:43
  • btw if you really think about it, nothing here is too abstract. If you’re confused, then always work out the algebra first in $M=\Bbb{R}^n$ and $g$ the usual metric tensor. Then covariant derivatives become usual directional derivatives, divergence is defined as usual, the usual product rules hold etc. So, do the algebra there first (where you can be sure what you’re doing is right), then see if things generalize. – peek-a-boo Oct 21 '23 at 02:45
  • Let us continue this discussion in chat. – peek-a-boo Oct 21 '23 at 02:45
  • Sorry if this is a silly question, but I have been working out some of the details in your answer and how did you get the second equality? I have used Stokes' theorem but how do we know that the pull back of the second integrand is equal to the third integrand under the inclusion map $\partial M \rightarrow M$? – CBBAM Oct 23 '23 at 23:04
  • To specify, wouldn't we need to have $i^*(f)$ in the third integrand instead of just $f$, where $i$ is the inclusion map? – CBBAM Oct 23 '23 at 23:34
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    @CBBAM yes, but after a while people just stop writing the inclusion maps everywhere. – peek-a-boo Oct 23 '23 at 23:54
  • Thank you again! – CBBAM Oct 24 '23 at 00:54