I am working on a project about Spectral Geometry. One of the main goals of the project is to be able to define the Laplacian on a Riemannian Manifold.

As such, Let $(M,g)$ be a Riemannian Manifold, then the Laplacian is given by $\triangle_g= -\operatorname{div}_g\circ \operatorname{grad}_g$. Given a vector field X on M, we define the divergence as being $$d(\iota_X \omega)=(\operatorname{div}_g X) \omega,$$ where $\omega$ is the volume form and $\iota_X \omega(Y_1, ..., Y_{n-1}) = \omega (X, Y_1, ..., Y_{n-1})$. I have already shown that given coordinates $x_1,...,x_n$ then we can write the divergence of $X=\sum_{i=1}^n X_i \frac{d}{dx_i}$ as:

$$\frac{1}{\sqrt{\det g}}\sum_{i=1}^n\frac{\partial}{\partial x_i}(X_i\sqrt{\det g })$$

I can clearly see that this definition $d(\iota_X \omega)=(\operatorname{div}_g X) \omega$ is independent of the choice of coordinates system. However, if instead of been given that first definition I was only given the last one, how can I prove the independence?

I have tried this:

Let $x_i$ and $y_i$ be two charts of $M$. Let $g$ be the metric of $M$ with respect to the coordinates $x_i$ and $\widetilde{g}$ the one with respect to $y_i$. Then thinking in terms of matrices we have that $\widetilde{g}=J^t g J$, where $J$ is the jacobian associated with the change of variables from $x_i$ to $y_i$ so we get that:

$$\sqrt{\det(\widetilde{g})}=\sqrt{\det (g)} \times \det (J)$$

Let $X=\sum_{i=1}^n X_i \frac{d}{dx_i}$ be a vector field in $x_i$ coordinates then it can be written as $\sum_{i,a=1}^n X_i \frac{dy_a}{dx_i}\frac{d}{dy_a}$ so let $ X_i \frac{dy_a}{dx_i}= Y_i$.Then , \begin{align*} \frac{1}{\sqrt{\det \widetilde{g}}}\sum_{i=1}^n \frac{\partial}{\partial y_i}(Y_i\sqrt{\det \widetilde{g}})&=\frac{1}{\sqrt{\det g} \det J}\sum_{i=1}^n \frac{\partial}{\partial y_i}(Y_i\sqrt{\det g} \det(J))\\ &=\frac{1}{\sqrt{\det g} \det J}\sum_{i,a=1}^n \frac{\partial x_a}{\partial y_i}\frac{\partial}{\partial x_a}(X_i\frac{\partial y_a}{\partial x_i}\sqrt{\det g}\det(J))\\ &=\frac{1}{\sqrt{\det g} \det J}\sum_{i,a=1}^n \det(J)\frac{\partial y_a}{\partial x_i} \frac{\partial x_a}{\partial y_i}\frac{\partial}{\partial x_a}(X_i\sqrt{\det g})+ \frac{1}{\sqrt{\det g} \det J}\sum_{i,a=1}^n X_i\sqrt{\det(g)} \frac{\partial x_a}{\partial y_i}\frac{\partial}{\partial x_a}(\frac{\partial y_a}{\partial x_i}\det J)\\ \end{align*}

Now I am stuck and not sure how to proceed. On the left side, I have what I want but on the right side, I don't know how to proceed. Any ideas?

  • 1
    $\begingroup$ $\newcommand{\div}{\operatorname{div}}$ The divergence of a vector field arises naturally from integration by parts. If you write the integral $$\int \langle df, X\rangle\,dV_g$$ in local coordinates and integrate by parts, you'll find that $$ \int f\div X\,dV_g ,$$ where $\div X$ is given by your second definition. You can also do the integration by parts abstractly without using coordinates, and that gives the first definition. $\endgroup$
    – Deane
    Jul 23 at 18:10

Before we start, let's use $a, b, \cdots$ for indices of $x's$ and $i, j, \cdots$ for indices of $y$'s. Also we use the Einstein summation convention, that repeated indices are summed over from $ 1$ to $n$.

\begin{align*} \frac{1}{\sqrt{\det \widetilde{g}}}& \frac{\partial}{\partial y^i}(Y^i\sqrt{\det \widetilde{g}})\\ & =\frac{1}{\sqrt{\det {g}} \det J} \frac{\partial x^b}{\partial y^i}\frac{\partial}{\partial x^b}(X^a \frac{\partial y^i}{\partial x^a} \sqrt{\det {g}} \det J) \\ &= \frac{1}{\sqrt{\det {g}} \det J} \frac{\partial x^b}{\partial y^i}\frac{\partial y^i}{\partial x^a} \frac{\partial}{\partial x^b}(X^a \sqrt{\det {g}} \det J) \\ &\ \ \ \ + \frac{1}{\sqrt{\det {g}} \det J} \frac{\partial x^b}{\partial y^i}\frac{\partial^2 y^i}{\partial x^a \partial x^b} (X^a \sqrt{\det {g}} \det J) \\ &= \frac{1}{\sqrt{\det {g}} \det J} \frac{\partial}{\partial x^a}(X^a \sqrt{\det {g}} \det J) + \frac{\partial x^b}{\partial y^i}\frac{\partial^2 y^i}{\partial x^a \partial x^b} X^a \\ &= \frac{1}{\sqrt{\det {g}}} \frac{\partial}{\partial x^a}(X^a \sqrt{\det {g}}) \\ &\ \ \ + \frac{1}{ \det J} X^a \frac{\partial}{\partial x^a}( \det J) + \frac{\partial x^b}{\partial y^i}\frac{\partial^2 y^i}{\partial x^a \partial x^b} X^a \\ &= \frac{1}{\sqrt{\det {g}}} \frac{\partial}{\partial x^a}(X^a \sqrt{\det {g}}) \\ &\ \ \ + X^a \left( \frac{\partial}{\partial x^a}(\log \det J) + \frac{\partial x^b}{\partial y^i}\frac{\partial^2 y^i}{\partial x^a \partial x^b} \right) \\ \end{align*}

Thus it suffices to check $$ \frac{\partial}{\partial x^a}(\log \det J) + \frac{\partial x^b}{\partial y^i}\frac{\partial^2 y^i}{\partial x^a \partial x^b} = 0.$$

We rewrite it as

$$ \frac{\partial}{\partial x^a}(\log \det J)= \frac{\partial}{\partial x^a}\left(\frac{\partial x^b}{\partial y^i}\right)\frac{\partial y^i}{ \partial x^b} $$

using $$\frac{\partial x^b}{\partial y^i}\frac{\partial^2 y^i}{\partial x^a \partial x^b} = - \frac{\partial}{\partial x^a}\left(\frac{\partial x^b}{\partial y^i}\right)\frac{\partial y^i}{ \partial x^b}$$

but this is just the famous Jacobi formula, since $J = (\frac{\partial x}{\partial y})$.

  • 1
    $\begingroup$ It's probably worth mentioning that strictly speaking, one has $\sqrt{\det \tilde{g}}=\sqrt{\det g}\cdot |\det J|$. From here, one can of course restrict the domains of coordinate charts so that the determinant maintains a constant sign, so that we can write this as $\sigma\cdot \det J$ for some $\sigma\in \{-1,1\}$ throughout. Then, it's a nice feature in this expression that the signs cancel because it appears once in the numerator and once in the denominator. This can be seen as the coordinate-reason for why the divergence is well-defined even on a non-orientable manifold. $\endgroup$
    – peek-a-boo
    Jul 23 at 19:30

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