Wedge products

The elementary forms may be combined with a wedge product. We note that $\mathrm{d}x\wedge\mathrm{d}y\colon\left(\mathbb{R}^3\right)^2\longrightarrow\mathbb{R}$ and, for example,

$$(\mathrm{d}x\wedge\mathrm{d}y)\left( \begin{pmatrix}u_1\\u_2\\u_3\end{pmatrix}, \begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix} \right) = \det\begin{pmatrix}u_1&v_1\\u_2&v_2\end{pmatrix}$$

and

$$(\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z) \left( \begin{pmatrix}u_1\\u_2\\u_3\end{pmatrix}, \begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}, \begin{pmatrix}w_1\\w_2\\w_3\end{pmatrix} \right) = \det\begin{pmatrix}u_1&v_1&w_1\\u_2&v_2&w_2\\u_3&v_3&w_3\end{pmatrix}$$

Numerically:

as.function(dx ^ dy)(cbind(c(2,3,5),c(4,1,2)))

## [1] -10

Above we see the package correctly giving $\det\begin{pmatrix}2&4\\3&1\end{pmatrix}=2-12=-10$.

The print method

Here I give some illustrations of the package print method.

dx

## An alternating linear map from V^1 to R with V=R^1:
##        val
##  1  =    1

This is somewhat opaque and difficult to understand. It is easier to start with a more complicated example: take $X=\mathrm{d}x\wedge\mathrm{d}y -7\mathrm{d}x\wedge\mathrm{d}z + 3\mathrm{d}y\wedge\mathrm{d}z$:

(X <- dx^dy -7*dx^dz + 3*dy^dz)

## An alternating linear map from V^2 to R with V=R^3:
##          val
##  1 3  =   -7
##  2 3  =    3
##  1 2  =    1

We see that X has three rows for the three elementary components. Taking the row with coefficient $-7$ [which would be $-7\mathrm{d}x\wedge\mathrm{d}z$], this maps $\left(\mathbb{R}^3\right)^2$ to $\mathbb{R}$ and we have

$$(-7\mathrm{d}x\wedge\mathrm{d}z)\left(\begin{pmatrix} u_1\\u_2\\u_3\end{pmatrix}, \begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}\right)= -7\det\begin{pmatrix}u_1&v_1\\u_3&v_3\end{pmatrix}$$

The other two rows would be

$$(3\mathrm{d}y\wedge\mathrm{d}z)\left( \begin{pmatrix}u_1\\u_2\\u_3\end{pmatrix}, \begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix} \right) = 3\det\begin{pmatrix}u_2&v_2\\u_3&v_3\end{pmatrix}$$

and

$$(1\mathrm{d}x\wedge\mathrm{d}y)\left( \begin{pmatrix}u_1\\u_2\\u_3\end{pmatrix}, \begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix} \right) = \det\begin{pmatrix}u_1&v_1\\u_2&v_2\end{pmatrix}$$

Thus form $X$ would be, by linearity

$$X\left( \begin{pmatrix}u_1\\u_2\\u_3\end{pmatrix}, \begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix} \right) = -7\det\begin{pmatrix}u_1&v_1\\u_3&v_3\end{pmatrix} +3\det\begin{pmatrix}u_2&v_2\\u_3&v_3\end{pmatrix} +\det\begin{pmatrix}u_1&v_1\\u_2&v_2\end{pmatrix}.$$

We might want to verify that $\mathrm{d}x\wedge\mathrm{d}y=-\mathrm{d}y\wedge\mathrm{d}x$:

dx ^ dy == -dy ^ dx

## [1] TRUE

Configuring the print method

The print method is configurable and can display kforms in symbolic form. For working with dx dy dz we may set option kform_symbolic_print to dx:

options(kform_symbolic_print = 'dx')

Then the results of calculations are more natural:

dx

## An alternating linear map from V^1 to R with V=R^1:
##  + dx

dx^dy + 56*dy^dz

## An alternating linear map from V^2 to R with V=R^3:
##  + dx^dy +56 dy^dz

However, this setting can be confusing if we work with $\mathrm{d}x^i,i>3$, for the print method runs out of alphabet:

rform()

## An alternating linear map from V^3 to R with V=R^7:
##  +6 dy^dNA^dNA +5 dy^dNA^dNA -9 dNA^dNA^dNA +4 dx^dz^dNA +7 dx^dNA^dNA -3 dy^dz^dNA -8 dx^dNA^dNA +2 dx^dy^dNA + dx^dNA^dNA

Above, we see the use of NA because there is no defined symbol.

hodge(dx^dy + 13*dy^dz)

## An alternating linear map from V^1 to R with V=R^3:
##  +13 dx + dz

hodge(dx)

## [1] 1

hodge(dx,3)

## An alternating linear map from V^2 to R with V=R^3:
##  + dy^dz

options(kform_symbolic_print = NULL)
d(8)

## An alternating linear map from V^1 to R with V=R^8:
##        val
##  8  =    1

save(dx, dy, dz, file="dx.rda")