1
$\begingroup$

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,

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

represents

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

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:

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

represents

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


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 (?))?

$\endgroup$
1
$\begingroup$

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 $1 \wedge 1 \wedge 1 = 1$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.