The Unapologetic Mathematician

Line Integrals

We now define some particular kinds of integrals as special cases of our theory of integrals over manifolds. And the first such special case is that of a line integral.

Consider an oriented curve $c$ in the manifold $M$. We know that this is a singular $1$-cube, and so we can pair it off with a $1$-form $\alpha$. Specifically, we pull back $\alpha$ to $c^*\alpha$ on the interval $[0,1]$ and integrate.

More explicitly, the pullback $c^*\alpha$ is evaluated as

$\displaystyle\left[c^*\alpha\left(\frac{d}{dt}\right)\right](t_0)=\left[\alpha_{c(t_0)}\right]\left(c_{*t_0}\frac{d}{dt}\bigg\vert_{t_0}\right)$

That is, for a $t_0\in[0,1]$, we take the value $\alpha_{c(t_0)}\in\mathcal{T}^*_{c(t_0)}M$ of the $1$-form $\alpha$ at the point $c(t_0)\in M$ and the tangent vector $c'(t_0)\in\mathcal{T}_{c(t_0)}M$ and pair them off. This gives us a real-valued function which we can integrate over the interval.

So, why do we care about this particularly? In the presence of a metric, we have an equivalence between $1$-forms $\alpha$ and vector fields $F$. And specifically we know that the pairing $\alpha_{c(t)}\left(c'(t)\right)$ is equal to the inner product $\langle F(c(t)),c'(t)\rangle$ — this is how the equivalence is defined, after all. And thus the line integral looks like

$\displaystyle\int\limits_c\alpha=\int\limits_{[0,1]}\langle F(c(t)),c'(t)\rangle\,dt$

Often the inner product is written with a dot — usually called the “dot product” of vectors — in which case this takes the form

$\displaystyle\int\limits_{[0,1]}F(c(t))\cdot c'(t)\,dt$

We also often write $ds=c'(t)\,dt$ as a “vector differential-valued function”, in which case we can write

$\displaystyle\int\limits_{[0,1]}F\cdot ds$

Of course, we often parameterize a curve by a more general interval $I$ than $[0,1]$, in which case we write

$\displaystyle\int\limits_IF\cdot ds$

This expression may look familiar from multivariable calculus, where we first defined line integrals. We can now see how this definition is a special case of a much more general construction.

October 21, 2011

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.

October 21, 2011