# Understanding Exterior algebra

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 $$e_i \wedge e_{jk} = e_{ijk}, \ i \neq j \neq k$$

• $e_{\emptyset}$ is $1$. – 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.
• I see, thank you for the answer. Just a quick question, do you define the binomial coefficient as 0 when $k > n?$ – Oskar Feb 1 '20 at 10:59
• 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. – Eric Canton Feb 1 '20 at 13:40