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In Dual Number it said that "It may also be defined as the exterior algebra of a one-dimensional vector space with $\varepsilon$ as its basis element." But I can't find the detailed rigorous expresion anywhere. I learned in Grassmann Number that ：

"Formally, let $V$ be an $n$-dimensional complex vector space with basis $\theta_i,i=1,\ldots,n$ The Grassmann algebra whose Grassmann variables are $\theta_i,i=1,\ldots,n$ is defined to be the exterior algebra of $V$, namely $\Lambda(V)=\mathbb{C}\oplus V\oplus(V\wedge V)\oplus(V\wedge V\wedge V)\oplus\cdots\oplus(V\wedge V\wedge\cdots\wedge V) \equiv \mathbb{C} \oplus \Lambda^1V \oplus \Lambda^2V \oplus \cdots \oplus \Lambda^nV$

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If we apply the definition of Grassmann number to a one-dimension vector space $V$ with a basis of $\{\varepsilon\}$, we can immediately obtain $\mathbb{C}\oplus V$, my question is how can we obtain $\varepsilon^2=0$ from this definition ? Certainly it can be obtained from anticommunity that $\varepsilon\wedge\varepsilon=-\varepsilon\wedge\varepsilon$, but $\mathbb{C}\oplus V$ does not consist of $V\wedge V$ since the number of directsum terms depends on the dimension of the vector space $V$.

Or maybe I misunderstood the sentence "It may also be defined as the exterior algebra of a one-dimensional vector space with $\varepsilon$ as its basis element." in wikipedia of dual number. So what's rigorous definition of dual number described in exterior algebra ?

Richard Mahler
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1 Answers1

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Short answer.

Prosaically, the algebra of dual numbers $(\Bbb{D},+,\cdot)$ may be reinterpreted as the exterior algebra $(\Lambda(V),+,\wedge)$ of a one-dimensional real vector space $V$, in the sense that the "usual" multiplication is replaced by the exterior product, which is anticommutative for 1-vectors, as you recalled, in such a way that (pure) dual numbers vanish when squared.

Long answer.

As you stated, the exterior algebra of a $n$-dimensional vector space $V$ is given by $\Lambda(V) = \bigoplus_{k=0}^n \Lambda^nV$, whose direct sum stops after $n$. Actually, the spaces $\Lambda^kV$ with $k > n$ exist formally, but they are trivial, i.e. $\Lambda^kV = 0 \;\forall k > n$, that is why they are usually omitted in the definition of the exterior algebra $\Lambda(V)$. Why so?

Let's recall that the (canonical) basis of the space $\Lambda^kV$ is given by $\{e_{i_1} \wedge \ldots \wedge e_{i_k}\}_{1 \le i_1 < \ldots < i_k \le n}$, where $\{e_1,\ldots,e_n\}$ represents the basis of $V$. Thus, one basis vector $e_i$ can appear no more than once, otherwise the resulting wedge product will vanish, since the exterior product is alternating and a fortiori anticommutative for 1-vectors. In consequence, one understands that it is impossible to construct a non-zero $k$-vector when $k > n$.

In conclusion, when coming back to the case of dual numbers $\Lambda(V) = \Bbb{R} \oplus V$, with $\varepsilon \in V$ being the basis of the one-dimensional vector space $V$, one has $\varepsilon^2 := \varepsilon \wedge \varepsilon \in \Lambda^2V = \{0\}$, hence $\varepsilon^2 = 0$.

Abezhiko
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