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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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  • $\begingroup$ It is $\Bbb R\wedge\Bbb R$, no? $\endgroup$ Commented Oct 20, 2012 at 1:35
  • $\begingroup$ It is R in degree 0 + R in degree 1 + nothing in all higher degrees. $\endgroup$ Commented Oct 20, 2012 at 1:53
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    $\begingroup$ @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). $\endgroup$ Commented Oct 20, 2012 at 1:57
  • $\begingroup$ 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. $\endgroup$ Commented Oct 20, 2012 at 4:24
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    $\begingroup$ @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. $\endgroup$ Commented Oct 20, 2012 at 18:53

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As you mention, we can easily see that $\Bbb R(\epsilon)\cong\bigwedge\Bbb R$, for example by considering the construction of $\bigwedge\Bbb R$ as the quotient of the tensor algebra $\operatorname{T}(\Bbb R)$ of the vector space $\Bbb R^1$ over $\Bbb R$ by elements of the form $v\otimes v.$

On the other hand, considering the construction of $\operatorname{Sym}(\Bbb R)$ as a quotient of $\operatorname{T}(\Bbb R)$ by elements of the form $v\otimes w-w\otimes v,$ we see that $\operatorname{Sym}(\Bbb R)\cong\Bbb R[x]$ is the polynomial ring over the reals in one variable. Moreover, $\Bbb R(\eta)\cong\Bbb R[x]/(x^2-1),$ which is certainly a different ring, although we can view this as a quotient ring of the symmetric algebra.

Does this help at all? I feel like I may just be stating the obvious. I kind of like to think of $\Bbb R(\epsilon)$ as a degeneration of $\Bbb R(\eta)$ under the family $\Bbb R(\eta_t)$ where $\eta_t^2=t\in \Bbb R.$ $\Bbb R(i)$ is then another member of this family, at $t=-1$. In particular, we find that one special member of these quotients of the symmetric algebra is the exterior algebra, though I'm not sure how meaningful this is.

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