Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose $\mathcal{A}$ is a $\mathbf{C}$-algebra then an integral form would be a subring $\mathcal{B} \subset \mathcal{A}$ such that the canonical map $\mathcal{B} \otimes_{\mathbf{Z}} \mathbf{C} \rightarrow \mathcal{A}$ is a bijection.

For some algebras there is an obvious integral form in the following sense: there is a preferred $\mathbf{C}$-basis for $\mathcal{A}$ and the $\mathbf{Z}$-span of that basis is $\mathcal{B}$. Now my question is do we have examples where $\mathcal{B}$ is non-obvious? In other words the basis coming from $\mathcal{B}$ would look very strange for those who only work with $\mathcal{A}$. Is there such an example where $\mathcal{A}$ is commutative?

share|improve this question
    
Minor quibble: I think you want $\mathcal B$ to be something like free as $\mathbb Z$-module. Otherwise $\mathcal B$ could be for instance a $\mathbb Q$-vector space (note that for a $\mathbb Q$-algebra $\mathcal B$ we have $\mathcal B\bigotimes_{\mathbb Z}\mathbb C=\mathcal B\bigotimes_{\mathbb Q}\mathbb C$. –  Torsten Ekedahl Jul 24 '11 at 4:51
add comment

3 Answers

There are many important integral forms with a "strange" appearance. Namely, we have 1) Kostant's form for a finite-dimensional complex simple Lie algebra, see the classical Humphrey's book; 2) Garland's form for the loop algebra of a finite-dimensional complex simple Lie algebra $g$, see the proper Garland's paper. In this case appear some elements given by the coefficient of some series given by the exponential of a suitable generating series with coeficients in the loop Cartan subalgebra of $g$.

For commutative cases we probably have those expected elements from a $\mathbb{C}$-basis, since the "strange" elements always come from brackets in the non-commutative case!

Hope help you,

share|improve this answer
add comment

Finite dimensional algebras (over alg. closed fields) which are of finite representation type have multiplicative bases, that is, bases such that the product of two of its elements is either an element of the basis or zero. This is a difficult and beautiful theorem of R. Bautista, P. Gabriel, A. Roiter and L. Salmerón.

This gives rather special integral forms for such algebras and, in general, it is quite non-obvious how to get them.

share|improve this answer
add comment

I got some "strange" integral forms when I generalized Garland's integral forms for the loop algebras to the map algebras of complex simple Lie algebras. The map algebras are a generalization of the loop algebras where we replace $\mathbb{C}[t^{\pm1}]$ by any algebra A. My integral forms are valid when A has a basis which is closed under multiplication. Thanks Mariano for letting me know that they are valid in more cases than I previously thought.

share|improve this answer
add comment

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.