NOTES ON STATISTICS, PROBABILITY and MATHEMATICS

The big idea:

There is a (musical) isomorphism between the tangent bundle and the cotangent bundle induced by the metric tensor. See here.

A given vector field can be automatically “converted” into a “work” 1-form field (or differential 1-form field) as in here:

or a “flux” 1-form (orthogonal) to the work one form:

Both forms can be integrated over a path.

Now, differentiating forms will yield the gradient (exterior derivative of 0-forms), curl (exterior derivative of 1-forms) and divergence (exterior derivative of 2-forms).

And the integration of these derivatives will come full-circle in Stokes’ theorem:

\[\int_{\partial M} \omega= \int_{M} d\omega\]

The mechanics of integrating forms:

From here:

Integrating the 1-form

\[\alpha = -y\,dx+ x\,dy\]

along a circle at the origin of radius \(R.\)

Fixing a counterclocwise path

\[\gamma(t)=\begin{bmatrix} R\cos t\\ R\sin t \end{bmatrix}\]

the matrix of derivatives is

\[\gamma'(t)=\begin{bmatrix} -R\sin t\\ R\cos t \end{bmatrix}\] \[\begin{align} \int_\gamma \alpha &=\int_{t=a}^b \alpha_{\gamma(t)}\left(\gamma'(t) \right)\,dt \\[2ex] &=\int_{t=0}^{2\pi}\left(-R\sin t \,dx + R\cos t \,dy \right)\; \begin{bmatrix} -R\sin t\\ R\cos t \end{bmatrix}\, dt \\[2ex] &=\int_{t=0}^{2\pi} R^2 \sin^2t + R^2\cos^2t\,dt \end{align}=\]

From here:

Integrating the 2-form

\[\beta = (x + y)^2\, dx\wedge dy - z\,dy\wedge dz\]

over the upper hemisphere

\[z = \sqrt{c^2 - x^2 - y^2} \geq 0\]

Parameterizing as

\[G\begin{bmatrix} x \\ y \\ \sqrt{c^2 - x^2 - y^2} \end{bmatrix}\]

and the matrix of derivatives is

\[G\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \frac{-x}{\sqrt{c^2 - x^2 - y^2}} & \frac{-y}{\sqrt{c^2 - x^2 - y^2}} \end{bmatrix}\]

Therefore,

\[\begin{align} \int \beta &= \int\int (x+y)^2 \begin{vmatrix} 1&0\\0&1\end{vmatrix}- \sqrt{c^2 - x^2 - y^2}\begin{vmatrix} 0 & 1 \\ \frac{-x}{\sqrt{c^2 - x^2 - y^2}} & \frac{-y}{\sqrt{c^2 - x^2 - y^2}} \end{vmatrix}\;dxdy\\[2ex] &=\int\int_{x^2+y^2+c^2} (x+y)^2 +x \; dxdy \end{align}\]

Exterior Derivative:

An \(m\) differential form in \(\mathbb R^n\) can be expressed in multi-index notation as

\[\begin{align} \omega &= \sum_I f_I \; dx_I \\ &= \sum_{1 \leq i_1 < i_2 \ \dots <i_m} a_{i_1, \dots, i_m} \; dx_1 \wedge dx_2 \wedge \dots \wedge dx_m \\\\ &I=(i_1, i_2, \dots,i_m) \; \mid\; 1 \leq i_1 < i_2 \ \dots <i_m \leq n \end{align}\]

From here and from here:

For an \(\omega\) \(m\)-form on \(\mathbb R^n\):

\[\omega = \sum_I f_I \, dx_I \]

the exterior derivative is

\[d\omega= \sum_I \sum_{j=1}^n \frac{\partial f_I}{\partial x_j}\,dx_j \wedge dx_I \]

Therefore, the output is an \(m+1\) form in \(\mathbb R^n\)

So for just \(f \, dx_I,\)

\[d(f\; dx_I)= \frac{\partial f}{\partial x_1}\,dx_1 \wedge dx_I + \frac{\partial f}{\partial x_2}\,dx_2 \wedge dx_I+\cdots + \frac{\partial f}{\partial x_n}\,dx_n \wedge dx_I\]

with \(dx_I = dx_{i_1}\wedge \dots \wedge dx_{i_m}\) and the multi-index \(I =(i_{1},\dots, i_m)\)

The exterior derivative \(\rm d\) of a \(0\)-form or function \(\phi\) is (in \(3\)-dimensions for ease of notation):

\[d\phi=\frac{\partial \phi}{\partial x}dx + \frac{\partial \phi}{\partial y}dy+ \frac{\partial \phi}{\partial z}dz\]

This takes a vector field \(d\phi(X)=D_X\phi\) and gives the directional derivative along that vector field.

How does it relate to the gradient? \((d\phi)^\sharp = \nabla \phi\) (sharp), and \((\nabla \phi)^\flat = d\phi\) (flat).

The exterior derivative of a \(1\)-differential form (a \(1\)-form is just a fixed value in front of the elementary forms \(dx_i\), whereas a “differential form” implies a function in front of the elementary forms - i.e. a field of forms) \(\omega = F_x \,dx + F_y \,dy + Fz \,dz,\) where \(F_x, F_y, F_z\) are the corresponding functions (the sub-index here is not meant to indicate a partial derivative) in \(3\) dimensions is

\[\begin{align} \frac{\partial F_x}{\partial x} \; dx \wedge dx + \frac{\partial F_y}{\partial x} \; dx \wedge dy + \frac{\partial F_z}{\partial x} \; dx \wedge dz \\[2ex] + \frac{\partial F_x}{\partial y} \; dy \wedge dx + \frac{\partial F_y}{\partial y} \; dy \wedge dy + \frac{\partial F_z}{\partial y} \; dy \wedge dz \\[2ex] + \frac{\partial F_x}{\partial z} \; dz \wedge dx + \frac{\partial F_y}{\partial z} \; dz \wedge dy + \frac{\partial F_z}{\partial z} \; dz \wedge dz \end{align} \\[2ex] = \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\; dy \wedge dz \quad+\quad \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\; dz \wedge dx \quad + \quad\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\; dx \wedge dy \]

which looks a lot like the curl (from here):

\[\nabla \times {\bf F}=\begin{bmatrix} \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\\ \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\\ \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \end{bmatrix}\]

We can see that they are related because \(\star d\omega = \nabla \times {\bf F}\) and \(\nabla \times {\bf F} = (\star dF^\flat)^\sharp.\) We can see the first relationship to the Hodge star in noticing that the complementary of \(dx\wedge dy\) is \(dz\), and the rows of the expression for the curl right above correspond to the vector components \(\partial / \partial x, \partial/\partial y,\partial/\partial z\) respectively.

What is the relation to Green’s theorem? The circulation form of Green’s theorem states

\[\int_{\partial D} \underset{1-\text{form}}{\underbrace{F_x \,dx + Fy\,dy}}=\int\int_D\underset{z \text{ component curl}}{\underbrace{\frac{\partial F_y}{\partial x}- \frac{\partial F_x}{\partial y}}}\, dA\] The integrand in the RHS is also the \(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\; dx \wedge dy\) component of the exterior derivative. Green’s theorem is just in the plane but the analogy with Stokes’ theorem is clear:

\[\int_{\partial D} \omega = \int_D d\omega\]

Similarly, the flux form of Green’s theorem can be derived by remembering that the flux is orthogonal to the original field, and that, again, we are on a plane with Green’s theorem. The form to integrate is not \(F_x \, dx + F_y \, dy,\) but rather \(F_x \, dy - F_y \,dx.\) And going through the exterior derivative, and obtaining the \(dx \wedge dy\) component results in:

\[\int_{\partial D} F_x \, dy - F_y \,dx= \int\int_{D} \underset{\text{divergence}}{\underbrace{\frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y}}} \;dA\]

Note (from here): As for the LHS of the equation above, with \(\partial D\) traversed counterclockwise by \(\vec r=\langle f(t), g(t) \rangle\) the tangent is

\[T'=\frac{\langle f'(t), g'(t) \rangle}{\Vert \vec r'(t)\Vert}\]

and the normal

\[n'=\frac{\langle g'(t), - f'(t) \rangle}{\Vert \vec r'(t)\Vert}\]

and

\[\begin{align} \int_{\partial D} \vec F \cdot \vec n\, ds &= \int_{\partial D} \vec F \cdot \frac{\langle g'(t), - f'(t) \rangle}{\Vert \vec r'(t)\Vert}\Vert \vec r'(t)\Vert \, dt \\[2ex] &=\int_{\partial D}\langle F_x,F_y \rangle \cdot \langle g'(t), - f'(t) \rangle dt \\[2ex] &= \int_{\partial D} F_x \, g'(t) dt- \int_{\partial D}F_y \, f'(t) dt \\[2ex] &=\int_{\partial D} F_x \, dy - \int_{\partial D} F_y \, dx\\[2ex] &=\int_{\partial D} F_x \, dy - F_y \, dx \end{align}\]

Finally, the exterior derivative of a \(2\)-differential form \(\beta=F\, dx\wedge dy + G\, dx\wedge dz + H\, dy\wedge dz\) is

\[d\beta = \frac{\partial F}{\partial z}\; dz\wedge dx \wedge dy + \frac{\partial F}{\partial y}\; dy\wedge dx \wedge dz + \frac{\partial F}{\partial x}\; dx\wedge dy \wedge dz=\left( \frac{\partial F}{\partial x} - \frac{\partial F}{\partial y}+\frac{\partial F}{\partial z} \right) dx\wedge dy \wedge dz\]

That minus sign is annoying, but it is still true that the exterior derivative of the Hodge star (see here) of a \(1\)-form \(\omega = F_x \,dx + F_y \,dy + Fz \,dz\) is

\[\begin{align} d(\star w)&=d(\star( F_x \,dx + F_y \,dy + F_z \,dz))\\[2ex] &=d(F_x\, dy \wedge dz + F_y\, dz\wedge dx + F_z \, dx\wedge dy)\\[2ex] &=dF_x\wedge dy \wedge dz + dF_y\wedge dz\wedge dx + dF_z\wedge dx \wedge dy\\[2ex] &=\frac{\partial F_x}{\partial x}dx\wedge dy \wedge dz + \frac{\partial F_y}{\partial y}dy\wedge dz \wedge dx + \frac{\partial F_z}{\partial z}dz\wedge dx \wedge dy \\[2ex] &=\left(\frac{\partial F_x}{\partial x}+ \frac{\partial F_y}{\partial y}+ \frac{\partial F_z}{\partial z}\right) dx\wedge dy\wedge dz \end{align}\]

NOTE on the Hodge star operator: \(dy\wedge dz\) is the Hodge star of \(dx\) because \(dx \wedge dy \wedge dz\), i.e. \(dx \wedge \star dx = dx\wedge dy\wedge dz\) returns the elementary \(n\) form. Likewise for \(dy\) the Hodge star is \(dz\wedge dx\) because \(dy \wedge dz \wedge dx= dx \wedge dy \wedge dz\) since commuting two elements doesn’t change the sign.

We have then that \(\star d \star \omega= \nabla \cdot X,\) where \(\nabla \cdot X\) is the divergence of a vector field. Notice that \(\omega = X^\flat.\) Therefore the divergence is \(\star d \star X^\flat\).

Wedge or Exterior Product:

An \(m\) differential form in \(\mathbb R^n\) can be expressed in multi-index notation as

\[\begin{align} \omega &= \sum_I f_I \; dx_I \\ &= \sum_{1 \leq i_1 < i_2 \ \dots <i_m} a_{i_1, \dots, i_m} \; dx_1 \wedge dx_2 \wedge \dots \wedge dx_m \\\\ &I=(i_1, i_2, \dots,i_m) \; \mid\; 1 \leq i_1 < i_2 \ \dots <i_m \leq n \end{align}\]

In \(\mathbb R^5\), we can construct \(1\)-forms using the \(dx\) elementary forms of the dual basis \((1,3),(1,2,3,4,5)\) or \((2,4,5):\)

\[\begin{align} \color{blue}{dx_1} && dx_2 &&\color{blue}{dx_3} && dx_4 && dx_5 \\ \color{orange}{dx_1} && \color{orange}{dx_2} && \color{orange}{dx_3} && \color{orange}{dx_4} && \color{orange}{dx_5} \\ dx_1 && \color{red}{dx_2} && {dx_3} && \color{red}{dx_4} && \color{red}{dx_5} && \\ \end{align}\]

These are not the multi-indices above, which correspond to wedged dual basis elements.

Here are three \(1\)-forms:

\[\begin{align}\omega_1 &= -7\,\color{blue}{dx_1} + 3 \,\color{blue}{dx_3}\\ \omega_2 &=-1 \,\color{orange}{dx_1} + \color{orange}{dx_2} + \color{orange}{dx_3}-4\,\color{orange}{d_4} + 5\,\color{orange}{dx_5}\\ \omega_3 &=-2 \,\color{red}{dx_2} + \color{red}{dx_4} - \color{red}{dx_5} \end{align} \]

A \(3\)-form results from wedging \(\omega_1 \wedge \omega_2\wedge \omega_3.\)

It acts on three vectors in \(\mathbb R^5,\) such as

\[\begin{align}\small v_1 &=\begin{bmatrix}1&&2 &&-1&& 0 &&5\end{bmatrix}\\ v_2 &=\begin{bmatrix}0&&0 &&1&& 10 &&1\end{bmatrix}\\ v_3 &=\begin{bmatrix}7&&1 &&0&& 3 &&2\end{bmatrix} \end{align}\]

as

\[\small\omega_1 \wedge \omega_2\wedge \omega_3\;(v_1,v_2,v_3)=\begin{vmatrix} (-7)\cdot1 +3\cdot(-1)=\color{blue}{-10}&&(-1)\cdot1+1\cdot2+1\cdot(-1)+(-4)\cdot0+5\cdot5=\color{orange}{25} && (-2)\cdot2+1\cdot0+(-1)\cdot5=\color{red}{-9}\\(-7)\cdot0+3\cdot1=\color{blue}{3}&&(-1)\cdot0+1\cdot0+1\cdot1+(-4)\cdot10+5\cdot1=\color{orange}{-34} &&(-2)\cdot0+1\cdot10+(-1)\cdot1=\ \color{red}{9}\\(-7)\cdot7+3\cdot0=\color{blue}{-49}&&(-1)\cdot7+1\cdot1+1\cdot0+(-4)\cdot3+5\cdot2=\color{orange}{-8} && (-2)\cdot1+1\cdot3+(-1)\cdot2=\color{red}{-1}\end{vmatrix}=3200\]

Trying to express it as in the multi-index notation above invoves distributing the wedged \(1\)-forms:

\[\begin{align} \omega_1 \wedge \omega_2\wedge \omega_3 &=\Tiny\left( -7\, dx_1+ 3\,dx_3\right)\wedge \left(-1\,dx_1 + dx_2 + dx_3-4\,dx_4 + 5\,dx_5 \right) \wedge \left(-2\,dx_2 + dx_4 - dx_5 \right)\\ &\Tiny=( -7 \, dx_1\wedge dx_2 - 7\, dx_1\wedge dx_3 +28\, dx_1 \wedge dx_4 -35\,dx_1\wedge dx_5 +3 dx_1\wedge dx_3\\ &\Tiny -3\,dx_2\wedge dx_3 - 12\,dx_3\wedge dx_4) + 15 \, dx_3 \wedge dx_5\\ &\Tiny\wedge \Tiny \left(-2\,dx_2 + dx_4 - dx_5 \right)=\\ & \small -8\,dx_1\wedge dx_2 \wedge dx_3\\ & \small + 49\,dx_1\wedge dx_2 \wedge dx_4 \\ & \small - 63\, dx_1\wedge dx_2 \wedge dx_5 \\ &\small -4\, dx_1\wedge dx_3 \wedge dx_4 \\ & \small + 4 \, dx_1\wedge dx_3 \wedge dx_5\\ & \small +7 \, dx_1\wedge dx_4 \wedge dx_5\\ &\small + 21\, dx_2\wedge dx_3 \wedge dx_4\\ &\small - 27\, dx_2\wedge dx_3 \wedge dx_5\\ &\small - 3\, dx_3\wedge dx_4 \wedge dx_5 \end{align}\]

which applied to the vectors above:

\[\Tiny -8 \begin{vmatrix} 1& 2 & -1 \\ 0&0&1\\7&1&0 \end{vmatrix} +49\begin{vmatrix} 1&2&0 \\ 0&0&10\\7&1&3 \end{vmatrix} -63\begin{vmatrix} 1&2&5 \\ 0&0&1\\7&1&2 \end{vmatrix} -4 \begin{vmatrix} 1&-1&0 \\ 0&1&10\\7&0&3 \end{vmatrix} + 4 \begin{vmatrix} 1&-1&5 \\ 0&1&1\\7&0&2 \end{vmatrix} +7 \begin{vmatrix} 1&0&5 \\ 0&10&1\\7&3&2 \end{vmatrix} +21 \begin{vmatrix} 2&-1&0 \\ 0&1&10\\1&0&3 \end{vmatrix} -27 \begin{vmatrix} 2&-1&5 \\ 0&1&1\\1&0&2 \end{vmatrix} -3 \begin{vmatrix} -1&0&5 \\ 1&10&1\\0&3&2 \end{vmatrix}=3200\]

m = matrix(c(-10,25,-9,3,-34,9,-49,-8,-1),3,byrow=T)
m
##      [,1] [,2] [,3]
## [1,]  -10   25   -9
## [2,]    3  -34    9
## [3,]  -49   -8   -1
det(m)
## [1] 3200
one = matrix(c(1,2,-1,0,0,1,7,1,0),3,byrow=T)
two = matrix(c(1,2,0,0,0,10,7,1,3),3,byrow=T)
three = matrix(c(1,2,5,0,0,1,7,1,2),3,byrow=T)
four = matrix(c(1,-1,0,0,1,10,7,0,3),3,byrow=T)
five = matrix(c(1,-1,5,0,1,1,7,0,2),3,byrow=T)
six = matrix(c(1,0,5,0,10,1,7,3,2),3,byrow=T)
seven = matrix(c(2,-1,0,0,1,10,1,0,3),3,byrow=T)
eight = matrix(c(2,-1,5,0,1,1,1,0,2),3,byrow=T)
nine = matrix(c(-1,0,5,1,10,1,0,3,2),3,byrow=T)

-8*det(one)  +49*det(two)  -63*det(three) -4*det(four) +4*det(five) +7*det(six) +21*det(seven) -27*det(eight) -3*det(nine)
## [1] 3200

Here is another example:

\[\begin{align}\omega_1 &= dx + dy + dz\\ \omega_2 &=2\, dx - 3 \, dy\\ \omega_3 &=dx + 2\, dz \end{align} \]

A \(3\)-form can be the result of wedging \(\omega_1 \wedge \omega_2\wedge \omega_3.\)

It acts on three vectors in \(\mathbb R^3,\) such as

\[\begin{align}\small v_1 &=\begin{bmatrix}2&1 &0\end{bmatrix}\\ v_2 &=\begin{bmatrix}-1&3&-2\end{bmatrix}\\ v_3 &=\begin{bmatrix}1&0&1\end{bmatrix} \end{align}\]

\[\omega_1 \wedge \omega_2\wedge \omega_3\;(v_1,v_2,v_3)=\begin{vmatrix}3&&1 && 2\\0&&-11&& -5\\2&&2 && 3\end{vmatrix}=-35\]

det(M <- matrix(c(3,1,2,0,-11,-5,2,2,3),3,byrow=T))
## [1] -35
M
##      [,1] [,2] [,3]
## [1,]    3    1    2
## [2,]    0  -11   -5
## [3,]    2    2    3

If instead we distribute as

\[\begin{align} (dx + dy + dz) \wedge (2dx - 3dy) \wedge (dx+2dz) &= (-3\,dx\wedge dy - 2\,dx\wedge dy -2\,dx\wedge dz+3\,dy\wedge dz )\wedge (dx+2dz)\\ &= (-5\,dx\wedge dy -2\,dx\wedge dz+3\,dy\wedge dz)\wedge (dx+2dz)\\ &= -10\, dx\wedge dy \wedge dz + 3 dx \wedge dy \wedge dz \\ &= - 7\, dx\wedge dy \wedge dz \end{align}\]

-7 * det(Q <- matrix(c(2,1,0,-1,3,-2,1,0,1),3,byrow=T))
## [1] -35
Q
##      [,1] [,2] [,3]
## [1,]    2    1    0
## [2,]   -1    3   -2
## [3,]    1    0    1

In reference to the explanation of the exterior or wedge product by Prof. Shifrin’s series (Chapter 8 - here, here and here), and connecting it with the more general tensor algebra operations, here is a summary:

The dual vector space \(V^*=(\mathbb R^n)^*\) is a vector space of linear maps \(\mathbb R^n \to \mathbb R\) that can be represented by a \(1\times n\) matrix (i.e. row vector). It is isomorphic to \(\mathbb R^n,\) and forms a vector space, which basis can be expressed as \(\mathrm e^1,\mathrm e^2,\dots,\mathrm e^n,\) or as \(\mathrm dx_1, \mathrm dx_2, \dots ,\mathrm dx_n,\) such that \(\mathrm dx_i(\vec v)=\vec {\mathrm e}_i\cdot \vec v.\)

This dual vector space constitutes the specific case of the vector space of alternating multilinear maps, denoted \(\Lambda^k(\mathbb R^n)^*\) when \(k=1.\) For example, in 3-dimensional Euclidean space, i.e. \(n=3,\) a possible form would be \(7\mathrm dx_2=7\mathrm dy=7\mathrm e^2,\) and if this is fed a vector \(\vec v=\begin{bmatrix}-7&8&2\end{bmatrix}^\top,\) the result will be simply the dot product \(\small\begin{bmatrix}0&7&0\end{bmatrix}\begin{bmatrix}-7\\8\\2\end{bmatrix}=56.\)

As a vector space, these forms can be added (and scalar multiplied), so that \(4\mathrm dx +7 \mathrm dy=\begin{bmatrix}4&7&0\end{bmatrix},\) an expression naturally belonging to the same vector space \(\Lambda^1(\mathbb R^3)^*,\) and also accepting vectors in \(\mathbb R^3\) to produce a real number through the dot product operation.

The wedge product of two such forms in \(\Lambda^1(\mathbb R^3)^*\) will result in an element in \(\Lambda^2(\mathbb R^3)^*.\) For example, taking the last element of \(\Lambda^1(\mathbb R^3)^*\) in the preceding example, i.e. \(\psi =4\mathrm dx_1 +7 \mathrm dx_2,\) and wedging it with \(\varphi =2\mathrm dx_1 +5 \mathrm dx_3,\) also in the same vector space \(\Lambda^1(\mathbb R^3)^*,\) we obtain

\[\psi\wedge\varphi=4\mathrm dx_1 +7 \mathrm dx_2 \wedge 2\mathrm dx_1 +5 \mathrm dx_3=-14\;\mathrm dx_{12}+20\;\mathrm d_{13}+35\; \mathbb d_{23}\tag 1\]

Critically, this is the same as picturing the two forms in \(\Lambda^1(\mathbb R^3)^*,\) i.e. \(4\mathrm dx_1 +7 \mathrm dx_2=\color{red}{\begin{bmatrix}4&7&0\end{bmatrix}}^\top\) and \(2\mathrm dx_1 +5 \mathrm dx_3=\color{blue}{\begin{bmatrix}2&0&5\end{bmatrix}}^\top\) as

\[\begin{bmatrix} \color{red}4&\color{red}7&\color{red}0\\ \color{blue}2& \color{blue} 0 & \color{blue}5 \end{bmatrix}\]

and realizing that \(\mathrm d_{12}\) is simply the determinant of the first two columns (minor of the submatrix):

\[\det{\begin{bmatrix} \color{red}4&\color{red}7\\ \color{blue}2& \color{blue} 0 \end{bmatrix}}=-14\]

and that \(\mathrm d_{13}\)

\[\det{\begin{bmatrix} \color{red}4&\color{red}0\\ \color{blue}2& \color{blue} 5 \end{bmatrix}}=20\]

with \(\mathrm d_{23}\)

\[\det{\begin{bmatrix} \color{red}7&\color{red}0\\ \color{blue}0& \color{blue} 5 \end{bmatrix}}=35\]

We could express this as

\[\psi\wedge\varphi=\sum \text {minors}_{ij}\; \mathrm d_{ij}\]

The wedge product can be carried out between vector spaces \(\Lambda^l(\mathbb R^n)^*\wedge \Lambda^k(\mathbb R^n)^*\in \Lambda^{l+k}(\mathbb R^n)^*.\)

This all comes together when two vectors in \(\mathbb R^3\) are fed into the expression in Eq (1), for example \[\small\vec p =\begin{bmatrix}-1&-\pi&-\sqrt 2\end{bmatrix}^\top\] and \[\small\vec q =\begin{bmatrix}10^1&10^2&10^3\end{bmatrix}^\top:\]

\[\begin{align}\left(\psi \wedge \varphi\right)(p,q) &= -14\,\mathrm dx_{12}(p,q) + 20 \,\mathrm dx_{13}(p,q) + 35\, \mathrm dx_{23}(p,q)\\[2ex] &=-14 \,\begin{vmatrix}-1&10 \\-\pi & 10^2\end{vmatrix}+ 20\, \begin{vmatrix}-1 & 10 \\-\sqrt 2 & 10^3\end{vmatrix} + 35\, \begin{vmatrix}-\pi & 10^2 \\-\sqrt 2 & 10^3\end{vmatrix} \end{align}\]

As for the equivalence to tensor algebra pointed out in the answer

\[\begin{align} \mathrm dx_{12}(p,q)&=\mathrm dx_1 \wedge \mathrm dx_2 (p,q)\\[2ex] &= \left(\mathrm dx_1 \otimes \mathrm dx_2 - \mathrm dx_2\otimes \mathrm dx_1\right)(p,q)\\[2ex] &= (\mathrm dx_1 \otimes \mathrm dx_2 )(p,q)-(\mathrm dx_2\otimes \mathrm dx_1)(p,q)\\[2ex] &=\mathrm dx_1(p)\,\mathrm dx_2(q) - \mathrm dx_2(p) \,\mathrm dx_1(q)\\[2ex] &= (-1)(10^2) - (-\pi)(10)\\[2ex] &=\begin{vmatrix}-1&10 \\-\pi & 10^2\end{vmatrix}. \end{align}\]

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NOTE: These are tentative notes on different topics for personal use - expect mistakes and misunderstandings.