Dimension of an algebraic group: Difference between revisions

Latest revision as of 18:29, 1 January 2012

The dimension of an algebraic group over a field is defined in the following equivalent ways:

No.	Shorthand	Definition
1	algebraic variety	its dimension as an algebraic variety over the field over which it is defined
2	formal group law	the dimension of the formal group law associated with the algebraic group
3	Lie algebra dimension	the dimension (as a vector space over the field) of the Lie algebra of the algebraic group.

The dimension is an invariant under any isomorphism of algebraic groups. It is possible, however, to have the same abstract group arising as algebraic groups of different dimensions over a given field.
Given an algebraic group $G$ over a field extension $L$ of $K$ , $G$ naturally acquires the structure of an algebraic group over $K$ . The dimension of $G$ over $K$ is the product of the dimension of $G$ over $L$ and the degree of the extension $L / K$ . Further information: formula for dimension for change of base field of algebraic group
The dimension of an algebraic group equals the dimension of its connected component of identity.

The trivial group, viewed as an algebraic group over any field, is zero-dimensional.
The additive group of a field, as well as the multiplicative group of a field, are both one-dimensional as algebraic groups under the usual structure.
The additive group of a $n$ -dimensional vector space has dimension $n$ as an algebraic group.
The general linear group over a field of degree $n$ has dimension $n^{2}$ as an algebraic group.

Operation	Input groups and their orders	Output group and its order	Proof and comment
external direct product of two algebraic groups	$G$ has dimension $m$ , $H$ has dimension $n$	$G \times H$ has dimension $m + n$	dimension of direct product is sum of dimensions; the same formula works for internal direct product, which is equivalent to external direct product.
external direct product of finitely many algebraic groups	$G_{1}, G_{2}, \dots, G_{n}$ with orders $m_{1}, m_{2}, \dots, m_{n}$ respectively	$G_{1} \times G_{2} \times \dots \times G_{n}$ has order $\sum_{i = 1}^{n} m_{i} = m_{1} + m + 2 + \dots + m_{n}$	dimension of direct product is sum of dimensions; same formula works for internal direct product
external semidirect product of two groups	$G$ , dimension $m$ , $H$ , dimension $n$ , acting on it via algebraic automorphisms	$G ⋊ H$ has dimension $m + n$	dimension of semidirect product is sum of dimensions; same formula works for internal semidirect product
group extension	closed normal subgroup $N$ , dimension $m$ , quotient group $G / N$ , dimension $n$	$m + n$	dimension of extension is sum of dimensions of normal subgroup and quotient

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 * The dimension is an invariant under any [[isomorphism of algebraic groups]]. It is possible, however, to have the same abstract group arising as algebraic groups of different dimensions over a given field.
 * Given an algebraic group <math>G</math> over a field extension <math>L</math> of <math>K</math>, <math>G</math> naturally acquires the structure of an algebraic group over <math>K</math>. The dimension of <math>G</math> over <math>K</math> is the product of the dimension of <math>G</math> over <math>L</math> and the [[Galois:degree of a field extension|degree]] of the extension <math>L/K</math>. {{further|[[formula for dimension for change of base field of algebraic group]]}}
+* The dimension of an algebraic group equals the dimension of its [[connected component of identity]].
 ==Particular cases==