Divergence-Free Vector Fields
The quality of being divergence-free greatly restricts the number of possible vector fields and also the dynamically accessible states of the flow map. It is however a weaker property then being Hamiltonian. We first define what it means to be divergence-free:
A vector field $X:\mathcal{M}\to{}T\mathcal{M}$ defined on a $d$-dimensional Riemannian manifold $(\mathcal{M}, g)$ is called divergence-free if $\forall{}z\in\mathcal{M}$ we have:
\[\mathrm{div} (X)= \sum_{i = 1}^d {\frac {1}{\rho }}{\frac {\partial \left({\frac {\rho }{\sqrt {g_{ii}}}}{\hat {X}}^{i}\right)}{\partial z^{i}}}= \sum_{i = 1}^d {\frac {1}{\sqrt {\det g}}}{\frac {\partial \left({\sqrt {\frac {\det g}{g_{ii}}}}\,{\hat {X}}^{i}\right)}{\partial z^{i}}} = 0,\]
for some parametrization of a neighborhood around $z$.
If we do not deal with a general Riemannian manifold but simply with a vector space, the divergence for $X:\mathbb{R}^d\to\mathbb{R}^d$ is usually written as
\[ \mathrm{div}(X) = \nabla\cdot{}X = \sum_{i=1}^d\frac{\partial{}X_i}{\partial{}z_i},\]
where $z$ are global coordinates.
We further define what it means to be volume-preserving:
We call a map $\phi:\mathcal{M}\to\mathcal{M}$ volume-preserving if for all volume elements $V\subset\mathcal{M}$ we have that $\mathrm{vol}(V) = \mathrm{vol}(\phi(V)),$ where vol is a measure of a volume in a Riemannian manifold.
If we deal with vector spaces instead of more general manifolds the property of being volume-preserving in some domain $\mathcal{D}\subset\mathbb{R}^d$ can be expressed as
\[ \det\nabla_z\varphi^t = 1 \quad \forall{}z\in\mathcal{D}, t\in[0, T].\]
So the columns of the Jacobian of $\varphi^t$ span a volume element of size 1 for each $z$ and each $t$.
We can proof the theorem:
The flow of a divergence-free vector field is volume-preserving.
Here we only proof this statement for the case of a vector space. A proof of the more general statement can be found in standard textbooks on differential geometry[1], e.g. [15, 16].
Proof
We refer to the flow of $X$ by $\varphi^t:\mathbb{R}^d\to\mathbb{R}^d$ and have the following property:
\[ \frac{d}{dt}\nabla\varphi^t(z) = \nabla{}X(\varphi^t(z))\nabla\varphi^t(z).\]
Note that we used the convention $[\nabla{}X(z)]_{ij} = \partial/\partial{}z_jX_i$ here. This expression for $d/dt\nabla{}\varphi^t(x)$ further implies:
\[ \mathrm{Tr}\left( (\nabla\varphi^t(z))^{-1}\frac{d}{dt}\nabla\varphi^t(z) \right) = \mathrm{Tr}\left( (\nabla\varphi^t(z))^{-1}\nabla{}X(\varphi^t(z))\nabla\varphi^t(z) \right) = \mathrm{Tr}(\nabla{}X(\varphi^t(z))) = 0,\]
where we have used $\mathrm{Tr}(ABC) = \mathrm{Tr}(BCA)$ in the second equality, and $\mathrm{Tr}(\nabla{}X) = \sum_{i=1}^d\partial{}X_i/\partial{}z_i = \mathrm{div}(X)$ and the divergence-freeness of $X$ in the third equality. We further have
\[ \mathrm{Tr}(A^{-1}\dot{A}) = \frac{\frac{d}{dt}\mathrm{det}(A)}{\mathrm{det}(A)},\]
which can be derived from the classical result $\mathrm{det}(\mathrm{exp}(A)) = \mathrm{exp}(\mathrm{Tr}(A)).$ Hence we have
\[ \frac{d}{dt}\mathrm{det}(\nabla\varphi^t(z)) = 0,\]
and the result is proved.
It is a classical result that all Hamiltonian vector fields are divergence-free, so volume-preservation is weaker than preservation of symplecticity [27].
References
- [16]
- S. I. Richard L. Bishop. Tensor Analysis on Manifolds (Dover Publications, Mineola, New York, 1980).
- [15]
- S. Lang. Fundamentals of differential geometry. Vol. 191 (Springer Science & Business Media, 2012).
- [27]
- V. I. Arnold. Mathematical methods of classical mechanics. Vol. 60 of Graduate Texts in Mathematics (Springer Verlag, Berlin, 1978).
- 1Together with a precise definition of Riemannian integration and the volume form introduced above.