Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
It is $\Bbb R\wedge\Bbb R$, no? –  Berci Oct 20 '12 at 1:35
    
It is R in degree 0 + R in degree 1 + nothing in all higher degrees. –  zyx Oct 20 '12 at 1:53
1  
@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). –  zyx Oct 20 '12 at 1:57
    
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. –  Qiaochu Yuan Oct 20 '12 at 4:24
1  
@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. –  zyx Oct 20 '12 at 18:53

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.