Question

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?

Answer 1

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,

Answer 2

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.