## Big Picture

This section is meant to be read after finishing the exterior derivatives section here.

The de Rham Cohomology and the commutative diagram below are advanced topics in differential geometry, but they offer a nice clean picture of how everything connects together and how to generalize even the Laplace operator to arbitrary dimensions. This is just for intuition, and a few things are specialized for {$\RR^3$} or hidden under the rug to avoid distracting complexity.

**Calculus 3:**scalar {$\ra{\grad}$} (length-)vector {$\ra{\curl}$} (area-)vector {$\ra{\div}$} (volume-)scalar**de Rham Cohomology:**0-form {$\ra{d}$} 1-form {$\ra{d}$} 2-form {$\ra{d}$} 3-form

In {$\RR^3$}, 0-forms are dual to 3-forms (both scalar-like), and 1-forms are dual to 2-forms (both vector-like). So we can have a dual that takes one to the other defined as follows.

**Hodge star/dual {$\hstar$} in {$\RR^3$}:**It is a linear operator that, when applied to a unit-magnitude {$k$}-form {$\phi$}, gives the dual unit-magnitude {$(3-k)$}-form {$\hstar\phi$} such that {$\phi\wedge\hstar\phi = dxdydz$}.- In the right-most column of the differential {$k$}-forms table here, the bases are shown in an order implying the pairing with their dual bases.
- The Hodge star operator essentially picks the
**orthogonal space**(it a generalization of the vector {$\perp$} operator as applied to {$k$}-forms). For example, in {$\RR^2$} (not {$\RR^3$}), the Hodge star maps a unit-magnitude 1-form {$\phi$} to another unit-magnitude 1-form {$\hstar\phi$} rotated by 90-degrees so that {$\phi\wedge \hstar\phi = dxdy$}. This is analogous to the property: {$\vv \cross \vv^{\perp} = 1$} for a unit vector {$\vv$}.

With this dual, we have the following big-picture commutative diagram for {$\RR^3$}

{$
\qquad
\begin{array}{c}
0 & \ra{d} & 0\text{-form} & \ra{d} & 1\text{-form} & \ra{d} & 2\text{-form} & \ra{d} & 3\text{-form} & \ra{d} & 0\\& & \uda{\hstar} & & \uda{\hstar} & & \uda{\hstar} & & \uda{\hstar}& &\\0 & \las{d} & 3\text{-form}& \las{d} & 2\text{-form} & \las{d} & 1\text{-form} & \las{d} & 0\text{-form} & \las{d} & 0
\end{array}
$}

- The horizontal arrows can be reversed by introducing the
**boundary**or**codifferential**operator {$\partial\overset{\text{def}}{=} (-1)^{k}\hstar d\hstar$} (in {$\RR^3$}). This is the same operator that shows up in the generalized Stokes' theorem {$\int_{\Omega} d\omega = \int_{\partial \Omega} \omega$}. (In homological algebra, the {$d$} is the corresponding**co-boundary operator**.) **Verify:**The**Laplace operator**can be produced with {$\grad^2 f = \hstar d \hstar d f$}. (This actually should be obvious given the relationship of {$d$} to grad/curl/div.) More generally, for {$k$}-forms, the Laplace-de Rham operator is defined more generally as {$\Delta\phi = (\partial d + d\partial)\phi$}. What paths do the two parts of this operator correspond to in the commutative diagram above?