form of an algebraic group

- group, ∞-group
- group object, group object in an (∞,1)-category
- abelian group, spectrum
- group action, ∞-action
- representation, ∞-representation
- progroup
- homogeneous space

Given a field $K$ and an algebraic group $G$ over $K$, and given a field extension $k \hookrightarrow K$, then a *$k$-form* of $G$ is an algebraic group $G_k$ over $k$ such that its base change to $K$ yields $G$:

$Spec(K)\times_{Spec (k)} G_k \simeq G
\,.$

Typically one requires that $G$ is also defined over $k$, hence one considers $k$-forms of the $K$-ification of a given $k$-form.

Consider the inclusion $\mathbb{R} \hookrightarrow \mathbb{C}$ of the real numbers into the complex numbers and let $G = \mathbb{G}_m = \mathbb{C}^\times$ be the multiplicative group hence the group of units of $\mathbb{C}$.

Then one $\mathbb{R}$-form, hence a real form of $\mathbb{G}_m$ is given by $\mathbb{R}^\times$, and another is given by the circle group $S^1 = U(1)$.

To see this, realize $\mathbb{C}^\times$ as the group of 2x2 matrices with entries in the complex numbers which are diagonal and of unit determinant:

$\left(
\array{
x & 0
\\
0 & y
}
\right)
\;\;\; with
\;
x y = 1
\,.$

The same prescription over the real numbers yields $\mathbb{R}^\times$ and exhibits it as a real form of $\mathbb{C}^\times$.

On the other hand, realize the circle group as the group of 2x2 real matrices of the form

$\left(
\array{
x & y
\\
-y & x
}
\right)
\;\;\;\;\;
with
\;
x^2 + y^2 = 1
\,.$

One checks that over the complex numbers this is isomorphic to the previous group of diagonal matrices, with the isomorphism being given by

$\left(
\array{
x & y
\\
-y & x
}
\right)
\mapsto
\left(
\array{
x + i y & 0
\\
0 & x- i y
}
\right)
\,.$

(e.g. eom)

Created on July 2, 2014 at 00:30:26. See the history of this page for a list of all contributions to it.