# The Unapologetic Mathematician

## The Codifferential

From our calculation of the square of the Hodge star we can tell that the star operation is invertible. Indeed, since $*^2=(-1)^{k(n-k)}\lvert g_{ij}\rvert$ — applying the star twice to a $k$-form in an $n$-manifold with metric $g$ is the same as multiplying it by $(-1)^{k(n-k)}$ and the determinant of the matrix of $g$ — we conclude that $*^{-1}=(-1)^{k(n-k)}\lvert g^{ij}\rvert*$.

With this inverse in hand, we will define the “codifferential” $\displaystyle\delta=(-1)^k*^{-1}d*$

The first star sends a $k$-form to an $n-k$-form; the exterior derivative sends it to an $n-k+1$-form; and the inverse star sends it to a $k-1$-form. Thus the codifferential goes in the opposite direction from the differential — the exterior derivative.

Unfortunately, it’s not quite as algebraically nice. In particular, it’s not a derivation of the algebra. Indeed, we can consider $fdx$ and $gdy$ in $\mathbb{R}^3$ and calculate \displaystyle\begin{aligned}\delta(fdx)&=-*d*(fdx)=-*d(fdy\wedge dz)=-*\frac{\partial f}{\partial x}dx\wedge dy\wedge dz=-\frac{\partial f}{\partial x}\\\delta(gdy)&=-*d*(gdy)=-*d(gdz\wedge dx)=-*\frac{\partial g}{\partial y}dy\wedge dz\wedge dx=-\frac{\partial g}{\partial y}\end{aligned}

while \displaystyle\begin{aligned}\delta(fgdx\wedge dy)&=-*d*(fgdx\wedge dy)\\&=-*d(fgdz)\\&=-*\left(\left(\frac{\partial f}{\partial x}g+f\frac{\partial g}{\partial x}\right)dx\wedge dz+\left(\frac{\partial f}{\partial y}g+f\frac{\partial g}{\partial y}\right)dy\wedge dz\right)\\&=\left(\frac{\partial f}{\partial x}g+f\frac{\partial g}{\partial x}\right)dy-\left(\frac{\partial f}{\partial y}g+f\frac{\partial g}{\partial y}\right)dx\end{aligned}

but there is no version of the Leibniz rule that can account for the second and third terms in this latter expansion. Oh well.

On the other hand, the codifferential $\delta$ is (sort of) the adjoint to the differential. Adjointness would mean that if $\eta$ is a $k$-form and $\zeta$ is a $k+1$-form, then $\displaystyle\langle d\eta,\zeta\rangle=\langle\eta,\delta\zeta\rangle$

where these inner products are those induced on differential forms from the metric. This doesn’t quite hold, but we can show that it does hold “up to homology”. We can calculate their difference times the canonical volume form \displaystyle\begin{aligned}\left(\langle d\eta,\zeta\rangle-\langle\eta,\delta\zeta\rangle\right)\omega&=d\eta\wedge*\zeta-\eta\wedge*\delta\zeta\\&=d\eta\wedge*\zeta-(-1)^k\eta\wedge**^{-1}d*\zeta\\&=d\eta\wedge*\zeta-(-1)^k\eta\wedge d*\zeta\\&=d\left(\eta\wedge*\zeta\right)\end{aligned}

which is an exact $n$-form. It’s not quite as nice as equality, but if we pass to De Rham cohomology it’s just as good.