The exterior algebra formed from a vector space $V$ with vector basis $\{e_1,e_2,e_3\}$ will have basis

$$\begin{align} &\Delta^0 V=\langle 1 \rangle\\ &\Delta^1 V=\langle e_1,e_2,e_3 \rangle\\ &\Delta^2 V=\langle e_1\wedge e_2, e_1 \wedge e_3, e_2 \wedge e_3 \rangle\\ &\Delta^3 V=\langle e_1\wedge e_2 \wedge e_3 \rangle\\ \end{align}$$

I presume...

Mathworld states

The alternating algebra, also called the exterior algebra, $\Delta^* V$ is a $2^n$ dimensional algebra. In Wolfram Language, an element of the alternating algebra can be represented by an $n$-nested binary list. For example,

$$\{\{\{\color{blue}1,\color{red}2\},\{0,0\} \},\{\{\color{orange}3,0\},\{\color{magenta}4,\color{brown}5\}\}\}$$


$$\begin{align} \color{blue} 1 \;e_1 \wedge e_2 \wedge e_3\quad+\\ \color{red}2\; e_1 \wedge e_3\quad+\\ \color{orange}3\;e_2\wedge e_3\quad +\\ \color{magenta}4\; e_3\quad +\\ \color{brown}5\; 1\quad \quad \end{align}$$

The color-coding and $\LaTeX$ transcription is mine in the quote, trying to reflect my problem understanding this.

The actual formulation in the quote is:

$$\{\{\{1,2\},\{0,0\} \},\{\{3,0\},\{4,5\}\}\}$$


$$e_1 \wedge e_2 \wedge e_3\,+\, 2\; e_1 \wedge e_3\, +\, 3\;e_2\wedge e_3\, +\, 4\; e_3\, +\, 5 $$

For instance, $\{\color{blue}1,\color{red}2\}$ (iff the color-coding makes sense) combines the coefficient $1$ of $e_1\wedge e_2\wedge e_3,$ composed of the maximum combination of basis vectors from $V$ with the coefficient for a wedge of just $2$ basis vectors, i.e. $e_1 \wedge e_3.$ Likewise, I don't see, either, what the next binary element $\{0,0\}$ corresponds to.

What is the equivalence between the WL notation and the different possible wedge products, and how are the coefficients usually "stored" (presumably a regular matrix, provided they are not functions (?))?


I think they've made a mistake, writing $e_1 \wedge e_3$ when they meant $e_1 \wedge e_2$. At least, that's the only way I can make sense of the pattern.

Then the coefficients are leaves in the binary tree obtained by successively making the choices (“$e_1$ or $1$”) $\wedge$ (“$e_2$ or $1$”) $\wedge$ (“$e_3$ or $1$”). So the first number in the (flattened) list is the coefficent of $e_1 \wedge e_2 \wedge e_3$, next comes the coefficent of $e_1 \wedge e_2 \wedge 1 = e_1 \wedge e_2$, and so on, until you get to the last number which is the coefficient of $1 \wedge 1 \wedge 1 = 1$.

  • $\begingroup$ That would make sense. Thank you! I will leave this open for a while, and come back to it to close if there are no other takes. $\endgroup$ – Antoni Parellada May 23 '18 at 17:41
  • $\begingroup$ Too little to make it a separate question... If you don't mind it... in an expression such as $A_{\alpha\beta\gamma\delta}\;\mathrm dx^\alpha \wedge \mathrm dx^\beta \wedge \mathrm dx^\gamma \wedge \mathrm dx^\delta$ the term $A_{\alpha\beta\gamma\delta}$ would be just a number - not a matrix - correct? This is the second part of my OP question. I presume that multilinearity would bring all the coefficients for each wedged component to the front, where they get multiplied. $\endgroup$ – Antoni Parellada May 23 '18 at 17:54
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    $\begingroup$ @MathAsFun: Yes, that's right. (Usually, at least. Who knows what strange things some people might do, but normally the coeffeicients are just numbers: real, complex, whatever.) $\endgroup$ – Hans Lundmark May 23 '18 at 19:16

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