Borel-Morozov theorem

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Suppose G is a linear algebraic group over an algebraically closed field K. Then, the following are true:

  1. Any two Borel subgroups of G are conjugate subgroups.
  2. Any closed connected solvable algebraic subgroup of G is contained in a Borel subgroup.

Related facts

Breakdown over a non-algebraically closed field

See Borel-Morozov theorem fails for non-algebraically closed field.

Similar facts

Facts with similar proofs

Facts used

  1. Equivalence of definitions of Borel subgroup
  2. Borel fixed-point theorem


We prove version (1). Version (2) is equivalent to it.

Given: A linear algebraic group G over an algebraically closed field K. Borel subgroups B_1, B_2 of G.

To prove: B_1 is conjugate to B_2.


Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 The quotient variety G/B_1 is a complete algebraic variety. Fact (1) B_1 is Borel in G.
2 B_2 is a connected solvable linear algebraic group over the algebraically closed field K. B_2 is a Borel subgroup of the linear algebraic group G
K is algebraically closed
follows from definition of Borel subgroup
3 Consider the action of B_2 on G/B_1 by left multiplication, obtained by restricting the action of G. This is an algebraic group action (i.e., it is regular in the algebraic group sense). The action is obtained by restricting a group multiplication, which is algebraic.
4 The action in Step (3) has a fixed point, say a coset gB_1. In other words, B_2(gB_1) = gB_1. Fact (2) Steps (1)-(3) Step-fact combination direct
5 B_2 \le gB_1g^{-1} Step (4) multiply both sides of the equality in step (4) on the right by g^{-1}, and get B_2(gB_1g^{-1}) = gB_1g^{-1}. Since both B_2 and gB_1g^{-1} are subgroups, this gives the subgroup containment.
6 B_2 = gB_1g^{-1}, so B_1,B_2 are conjugate subgroups. B_1,B_2 are Borel subgroups Step (5) Since B_1 is Borel, so is gB_1g^{-1}. B_2 is also Borel. Thus, both are maximal among connected closed solvable subgroups of G, and hence, by the containment of Step (5), they must be equal.