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Let $X:U_1 \rightarrow \mathbb{R}^2$ be a smooth vector field defined in an open subset of $\Bbb{R}^2$ and $\phi:U_1\rightarrow U_2$ a diffeomorphism between open subsets of $\Bbb{R}^2$. Let $Y = \phi_*X$, ie, $$Y(p) = d\phi_{\phi^{-1}(p)}(X(\phi^{-1}(p))),\qquad p \in U_2.$$

Here, we are considering that if $X(x,y) = (P(x,y), Q(x,y))$, where $P:U_1 \to \Bbb{R}$ and $Q:U_1\to \Bbb{R}$ are smooth functions, then $$ \operatorname{div}(X) := \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}. $$

Is there any relation between $\operatorname{div}(X)$ and $\operatorname{div}(Y)$? Can I get $\operatorname{div}(Y)$ in terms of $\phi$ and $\operatorname{div}$?

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  • $\begingroup$ To take some of the flamboyance out of the expression $$d\phi_{\phi^{-1}(p)}(X(\phi^{-1}(p)))$$ the most obvious choice of a concrete $\phi$ you could study is the coordinate transformation from polar to Cartesian. It is easy to derive the divergence of $Y$ in polar coordinates and you can check your results in a lot of MSE posts. $\endgroup$
    – Kurt G.
    Apr 3 at 9:44

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Recall the given a volume $\mu$, the divergence of a vector field $X$ with respect to $\mu$ is characterized by $$\mathcal L_X \mu = (\operatorname{div} X) \mu .$$ Now, suppose $\phi : M \to N$ is (local) map between manifolds equipped respectively with volume forms $\mu, \nu$, so that we can identify $\det \phi$ with a scalar function on $M$ characterized by $\phi^* \nu = (\det \phi) \mu$. Then \begin{align*} \phi^* ((\operatorname{div} Y) \nu) &= \phi^* (\mathcal L_Y \nu) \\ &= \phi^* (\mathcal L_{\phi_* X} \nu) \\ &= \mathcal L_X (\phi^* \nu) \\ &= \mathcal L_X ((\det \phi) \mu) \\ &= (\det \phi) \mathcal L_X \mu + (X \cdot \det \phi) \mu \\ &= (\det \phi) (\operatorname{div} X) \mu + (X \cdot \det \phi) \mu \\ &= ((\det \phi) (\operatorname{div} X) + X \cdot \det \phi) \mu \\ \end{align*} If $\phi$ is an orientation-preserving diffeomorphism, then pulling back both sides by $\phi^{-1}$ (noting that $(\phi^{-1})^* \mu = ((\det \phi)^{-1} \circ \phi^{-1}) \nu$) and clearing the common factors $\nu$leaves the identity $$\boxed{\operatorname{div} Y = (\operatorname{div} X + X \cdot (\log \det \phi)) \circ \phi^{-1}}$$ of functions on $N$. In particular, if $\phi$ is an isomorphism of volume forms, i.e., if $\phi^* \nu = \mu$, then $$\operatorname{div} Y = (\operatorname{div} X) \circ \phi^{-1} .$$

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Let $M,N$ be smooth manifolds and let $\phi:M\rightarrow N$ such that $\phi$ is a diffeomorphism. Consider the vector field $X\in\Gamma(TM)$. Because $\phi$ is a diffeomorphism we can safely define the pushforward of the vector field $X$ as;

$$Y(f)=d\phi(X)(f)=X(f\circ\phi)=X^i\frac{\partial}{\partial x^i}(f\circ\phi)$$

where $f\in C^{\infty}(N)$.

Therefore, we have;

$$Y(f)=X^i(\frac{\partial f}{\partial\phi^m}\cdot\frac{\partial \phi^m}{\partial x^i})=\bigg[X^i\frac{\partial \phi^m}{\partial x^i}\bigg]\frac{\partial}{\partial\phi^m}(f)$$

We may now consider the divergence of our vector fields. The Voss-Weyl formula for divergence is;

$$\text{div}(X)=\frac{1}{\rho}\frac{\partial(\rho X^i)}{\partial x^i}$$

where $\rho$ is equal to the volume element on the manifold. So, we will have;

$$\text{div}(X)=\text{div}\bigg(X^i\frac{\partial}{\partial x^i}\bigg)=\frac{1}{\rho_{\\_M}}\frac{\partial(\rho_{\\_M} X^i)}{\partial x^i}$$

$$\text{div}(Y)=\text{div}\bigg(\bigg[X^i\frac{\partial \phi^m}{\partial x^i}\bigg]\frac{\partial}{\partial\phi^m}\bigg)=\frac{1}{\rho_{\\_N}}\frac{\partial(\rho_{\\_N} X^i\frac{\partial \phi^m}{\partial x^i})}{\partial \phi^m}$$

In general, There is very little relation between these two expressions. The divergence of a vector field depends on the volume form, which is a nowhere zero $n$-form on an oriented $n$- manifold. Of course, most diffeomorphisms will change the volume form.

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