nLab form of an algebraic group

Idea

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.

Examples

Multiplicative group and circle group

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)

Related concepts

• relation between compact Lie groups and reductive algebraic groups

References

• David Herzog, Forms of algebraic groups, Proceedings of the American Mathematical Society Vol. 12, No. 4 (Aug., 1961), pp. 657-660 (JSTOR, pdf)

• eom, Form of an algebraic group