I was introduced to Exterior Algebra in a very abstract way, and I am wondering if I got it right, concrete.

Consider $ R^3 $ and let $ V = \{e_1,e_2,e_3 \} $ be basis for it.

The Exterior space, $ \bigwedge^3V $, is of dimension $ 2^n = 8 $ and a basis consists of the vectors

  • $ e_1,e_2,e_3, $
  • $ e_{12} = e_1 \wedge e_2, \ e_{13} = e_1 \wedge e_3, \ e_{23} = e_2 \wedge e_3 $
  • $ e_{123} = e_1 \wedge e_2 \wedge e_3 $

Q: I should get one more basis vector in $ \bigwedge^3 V $, and the only one I can think of is $ e_\varnothing $? If it is $ e_\varnothing $, is $ e_\varnothing = 0?$ Why do we have to include the zero vector in the basis?

Q: Furthermore, since $ A \wedge A = 0 $, the only operations with the given basis we can do thats not zero is $$ e_i \wedge e_j = e_{ij}, \ i \neq j $$ and $$ e_i \wedge e_{jk} = e_{ijk}, \ i \neq j \neq k$$

  • $\begingroup$ $e_{\emptyset}$ is $1$. $\endgroup$ – Charlie Frohman Jan 31 '20 at 19:57

One starting comment: instead of $\wedge^3 V$ you want $\wedge^3 \mathbb{R}^3$, since exterior products apply to vector spaces, not sets of basis vectors for them.

The basis for $\wedge^3 \mathbb{R}^3$ is indexed by subsets of $\{1,2,3\}$ of cardinality 3, so there's one basis vector: what you have called $e_{123}$. Thus, it's a one-dimensional vector space. More generally, the dimension of $\wedge^d \mathbb{R}^n$ is the binomial coefficient $\binom{n}{d}$.

The exterior algebra does indeed have the 8 basis vectors you listed above, although $e_\varnothing = 1 \in \mathbb{R}$; this is the "empty" exterior power, so you can think of this as analogous to $c^0 = 1$ for $c \in \mathbb{R}$ (take $c \ne 0$ if you want $0^0$ to be something besides 1 of course!). The exterior algebra contains all of the exterior powers $\wedge^d \mathbb{R}^3$ for $d \ge 0$. For $d > 3$ this is zero, by the binomial dimension calculation earlier.

  • $\begingroup$ I see, thank you for the answer. Just a quick question, do you define the binomial coefficient as 0 when $ k > n?$ $\endgroup$ – Oskar Feb 1 '20 at 10:59
  • $\begingroup$ Yes. The binomial coefficient $\binom{n}{k}$ is the number of subsets of $\{1, 2, ... n\}$ of size $k$. The formula for binomial coefficients in terms of factorials is useful, but not definitional, from this perspective. $\endgroup$ – Eric Canton Feb 1 '20 at 13:40
  • $\begingroup$ I see. Thank you! $\endgroup$ – Oskar Feb 1 '20 at 13:54

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