1
$\begingroup$

Motivated by classical formulas $L_{X}=d\circ i_{X}+i_{X}\circ d$ and $L_{X} \Omega=Div(X) \Omega$ and the essential role of the diff operator $d$ in definition of divergence, we define some quantities, say $\overline{Div}$, based on the adjoint operator $d^{*}=\pm *d*$, where $*$ is the Hodge star operator.

In this way our main question is that:

What are some geometric or physical interpretations for $\overline{Div}$? What are some calculus identities for this quantity?In particular is it true that for a closed manifold $M$, with volum form $\Omega$, we have $\int_{M} \overline{Div}(X)\Omega=0$?

Moreover what is the dynamical interpretation of $\overline{Div}(X)=0$. This is motivated by classical case: If $Div(X)=0$ then $X$ has no an attractor, since the flow of $X$ generates a one parameter family of volume preserving diffeomorphisms. So we ask: Is there a vector field $X$ which has a compact attractor invariant set but $\overline{Div}(X)$ is identically zero?

  1. For a vector field $X$ on a $2$ dimensional surface with volum form $\Omega$ define:

$$\overline{Div}(X)=(i_{X}\circ d^{*}+d^{*}\circ i_{X})(\Omega)$$

  1. A vector field $X$ on a Riemannian manifold $(M,g)$ defines a one form $\alpha$. Now $\overline{Div}(X)$ is defined as a unique function with $$\alpha \wedge d^{*}(\Omega)=\overline{Div}(X). \Omega $$

  2. For a symplectic manifold $(M,\omega)$, $\overline{Div}$ is the unique function with $$ (i_{X}\circ d^{*}+d^{*}\circ i_{X})(\Omega)\wedge \omega=\overline{Div}(X).\Omega$$ where $\Omega$ is the corresponding volume form.

$\endgroup$
2
  • 3
    $\begingroup$ $d^*$ involves a (pseudo-)Riemannian metric. Whereas $d$ is invariant under pullback by all mappings, in particular under diffeomorphisms, and commutes with with infinitesimal diffeomorphisms $L_X$, $d^*$ is only invariant under isometries of the Riemannian metric used. So the integral in your main question only vanishes for Killing vector fields $X$, in general. $\endgroup$ Dec 29, 2014 at 17:31
  • $\begingroup$ @PeterMichor thank you very much for your interesting comment. $\endgroup$ Dec 31, 2014 at 5:14

1 Answer 1

2
$\begingroup$

The divergence operator is simply the exterior derivative operator acting on $(n-1)$-forms, and the divergence theorem is Stokes theorem. This can be seen by just writing everything in local coordinates. Any other version is just using an additional geometric structure to identify an $(n-1)$-form with a vector field or 1-form.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.