# Pushforward of vector field and its divergence

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 $$\operatorname{div}$$?

• 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. Apr 3 at 9:44

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 $$\operatorname{div} Y = (\operatorname{div} X) \circ \phi^{-1} .$$

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.