form of an algebraic group



Given a field KK and an algebraic group GG over KK, and given a field extension kKk \hookrightarrow K, then a kk-form of GG is an algebraic group G kG_k over kk such that its base change to KK yields GG:

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

Typically one requires that GG is also defined over kk, hence one considers kk-forms of the KK-ification of a given kk-form.


Multiplicative group and circle group

Consider the inclusion \mathbb{R} \hookrightarrow \mathbb{C} of the real numbers into the complex numbers and let G=𝔾 m= ×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 𝔾 m\mathbb{G}_m is given by ×\mathbb{R}^\times, and another is given by the circle group S 1=U(1)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:

(x 0 0 y)withxy=1. \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

(x y y x)withx 2+y 2=1. \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

(x y y x)(x+iy 0 0 xiy). \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 01:08:11. See the history of this page for a list of all contributions to it.