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In trying to understand the exterior or wedge product following Prof. Shifrin's series (Chapter 8 - here, here and here), and connecting it with the more general tensor algebra operations, I would like to make sure that the following is correct:

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

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

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 $\mathrm dx_{12}$ 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 $\mathrm dx_{13}$

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

with $\mathrm dx_{23}$

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

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 $\small\vec q =\begin{bmatrix}10^1&10^2&10^3\end{bmatrix}^\top:$

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


Attempt at mending after Professor Shifrin's correction:

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

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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    $\begingroup$ Not worth an entire answer, but when you write "dual vector space [ . . .] is a linear map", don't you mean "a vector space of linear maps"? $\endgroup$ – PersonX Apr 3 '18 at 20:06
  • $\begingroup$ @PersonX Yes, thank you for your help reading this long question. I figured that if nobody points out any major flaws it might even help someone searching the web for this topic. $\endgroup$ – Antoni Parellada Apr 3 '18 at 20:08
  • $\begingroup$ You have a bunch of typos. You mean to have $dx_{ij}$ everywhere, which of course is (my) shorthand for $dx_i\wedge dx_j$. I also don't particularly like using square brackets around the $p$ and $q$ (as that suggests all sorts of other things in mathematics. An element of $\Lambda^k(\Bbb R^{n*})$ should be applied to an ordered $k$-tuple of vectors. $\endgroup$ – Ted Shifrin Apr 3 '18 at 21:21
  • $\begingroup$ Thank you for looking at it. The typos you are referring to (please feel free to edit) are in the English explanations (common language) or do you mean that $dx_{ij}$ should have been used instead of $\mathrm dx_{ij}$? $\endgroup$ – Antoni Parellada Apr 3 '18 at 21:30
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    $\begingroup$ Oh, I don't care about $d$ versus d. No, you dropped the $x$'s and typed $d_{13}$ and $\mathbb d_{23}$. $\endgroup$ – Ted Shifrin Apr 3 '18 at 21:44
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Everything looks fine up until the end, where you should have, for example, $dx_{12}(p,q) = \left|\begin{matrix}-1&10 \\-\pi & 100\end{matrix}\right| = -100+10\pi$. Fitting things into the tensor algebra picture is quite simple: $$dx_{12} = dx_1\wedge dx_2 = dx_1\otimes dx_2 - dx_2\otimes dx_1.$$

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  • $\begingroup$ Now, the vectors p and q are being fed into the $\Lambda^2$ form $\psi \wedge \varphi.$ The determinants of the submatrices were already calculated. At that point, I wanted to figure out the expected effect of this multilinear form on the two vectors p and q. Can you please look at it again, and perhaps include the whole calculation so that there is no miscommunication? $\endgroup$ – Antoni Parellada Apr 3 '18 at 21:34
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    $\begingroup$ I'm talking about the last line of your computation. Minors show up in two places, both in wedging $\psi$ and $\phi$ together and in evaluating on the vectors. As I typed, by definition, $dx_{12}(p,q)$ computes the $12$-minor of the $3\times 2$ matrix with columns $p$ and $q$. What I'm talking about occurs when you compute $-14 dx_{12}(p,q)$. It should be $-14(-100+10\pi)$. $\endgroup$ – Ted Shifrin Apr 3 '18 at 21:46

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