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Consider the commutative, unital algebras $\mathbb{R}(i), \mathbb{R}(\epsilon)$ and $\mathbb{R}(\eta)$, where the adjunctions satisfy $i^{2} = -1, \epsilon^{2} = 0$ and $\eta^{2} = 1$ (but $i, \epsilon$ and $\eta$ are not elements of $\mathbb{R}$). Since the operations of addition and multiplication are continuous in the corresponding product topologies, these algebras are examples of topological rings of hypercomplex numbers.

It is clear that $\mathbb{R}(i) \cong \mathbb{C}$ and, with a little work, one can prove $\mathbb{R}(\epsilon) \cong \bigwedge \mathbb{R}$, the exterior algebra of the vector space $\mathbb{R}$ (over the field $\mathbb{R}$) and also $\mathbb{R}(\eta) \cong \mathbb{R} \oplus \mathbb{R}$, where the explicit bijection is a lift of the map $a + b \eta \mapsto (a+b, a- b)$. The latter two are not fields because they contain non-trivial nilpotent elements, e.g., $b\epsilon $ and $\frac{1}{2}(1-\eta)$.

Since there is clearly some relationship between the algebras, does $\mathbb{R}(\eta)$ admit an interpretation in terms of $S(\mathbb{R})$, the symmetric algebra of the vector space $\mathbb{R}$ over the field $\mathbb{R}$ or some similar structure?

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    It is $\Bbb R\wedge\Bbb R$, no?2012-10-20
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    It is R in degree 0 + R in degree 1 + nothing in all higher degrees.2012-10-20
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    @02138, R looks like a red herring here, Koszul duality is over any field and your question makes sense over C for which the first and third cases are the same (or part of the same family, at least).2012-10-20
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    I don't really see where the duality is. I also don't think this question is well-specified; for example, I don't understand what it would take to answer "no" to this question.2012-10-20
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    @QiaochuYuan: surely it is a reference to Koszul duality of symmetric and exterior algebras. You could say that the question is whether a deformation of the duality is known in this case.2012-10-20
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    @zyx: okay, but I don't see where the symmetric algebra appears here.2012-10-20

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