This site is being phased out.

Codifferential

From Mathematics Is A Science
Jump to navigationJump to search

Consider this Hodge duality diagram, where $\star$ stands for Hodge duality of differential forms: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccccccc} &C ^{0}& \ra{d} & C ^1 &\ra{d} &... & \ra{d}& C ^k & \ra{d} & C ^{k+1} & \ra{d} &...& \ra{d} & C ^{n}\\ &\da{\star} & \ne & \da{\star} & & & & \da{\star} & \ne & \da{\star} & & & & \da{\star} \\ &C ^{n}& \la{d} & C ^{n-1}& \la{d}&... & \la{d}& C ^{n-k}& \la{d}& C ^{n-(k+1)}& \la{d} &...& \la{d} & C ^{0} \\ \end{array} $$

The codifferential $$\delta =d^*:C^k(M)\rightarrow C^{k-1}(M)$$ is the operator adjoint of the exterior derivative (aka the differential) $$d:C^k(M)\rightarrow C^{k+1}(M).$$ The degree goes down not up.

We understand that the codifferential is the adjoint of $d$ with respect to an inner product: $$ < \varphi,\delta \psi > = < d\varphi,\psi >,$$ where $\psi$ is a $(k+1)$-form and $\varphi$ a $k$-form.

This identity comparable to the Stokes' theorem, where the boundary operator $\partial$ is adjoint of $d$ with respect to the "evaluation": $$ < \varphi,\partial \sigma> = < d\varphi,\sigma>,$$ where $\sigma$ is a $(k+1)$-chain and $\psi$ a $k$-form.

The codifferential can also be defined as $$\delta = (-1)^{nk + n + 1}s\, {\star d\star} = (-1)^k\,{\star d\star},$$ where $\star$ is the Hodge star operator. Here $s=1$ for Riemannian manifolds.

Since the exterior derivative satisfies $d^2=0$, the codifferential has the sameproperty. Indeed: $$\delta^2 = s^2{\star d{\star {\star d{\star}}}} = (-1)^{k(n-k)} s^3{\star d^2\star} = 0. $$

So, the codifferential turns the cochain complex into a chain complex seen in the first row of this commutative diagram: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccccccc} &C ^{0} & \la{\delta} & C ^1 & \la{\delta} &...& \la{\delta} & C ^k & \la{\delta} & C ^{k+1} & \la{\delta} &...& \la{\delta} & C ^{n}\\ &\ua{\star} & &\da{\star}\ua{} & & & &\da{\star}\ua{} & &\da{\star}\ua{} & & & & \da{\star} \\ &C ^{n}& \la{d} & C ^{n-1}& \la{d}&...& \la{d}& C ^{n-k}& \la{d}& C ^{n-(k+1)}& \la{d} &...& \la{d} & C ^{0} \\ \end{array} $$ We can compare these: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccccccc} & C _{0}(K)& \la{\partial}& ... & \la{\partial}& C _{k}(K)& \la{\partial} & C _{k+1}(K)& \la{\partial} &...& \la{\partial} & C _{n}(K) \\ & C ^{0}(K)& \ra{d} & ... & \ra{d}& C ^k(K) & \ra{d} & C ^{k+1}(K) & \ra{d} &...& \ra{d} & C ^{n}(K)\\ & C ^{0}(K)& \la{\delta}& ... & \la{\delta}& C ^{k}(K)& \la{\delta} & C ^{k+1}(K)& \la{\delta} &...& \la{\delta} & C ^{n}(K) \\ \end{array} $$

Properties. For these operators as operators on $\Omega ^k(M)$ we have

  • $\star d=(-1)^{k+1}\delta\star $;
  • $\star \delta=(-1)^kd\star $;
  • $\star \star =(-1)^{k(n-k)}Id$.

The Laplace-de Rham operator (aka the Laplacian) is given by $$\Delta=\delta d + d\delta.$$

All definitions apply equally to continuous and discrete differential forms.