Let , e.g., .

Every smooth connected solvable subgroup of stabilises a complete flag.

Equivalently, it is conjugate to a subgroup of the form

A complete flag is

where each is a subspace of and .

The “set” of complete flags of a vector space is in fact a projective algebraic variety:

The above Theorem 1Theorem 1 is implied by: A smooth connected linear algebraic group acting on a proper variety has a fixed point.

Induct on dimension. We have for .

WLOG, is the closure of an orbit. If is affine, and points in are fixed points.

A Borel subgroup of is a maximal smooth connected solvable subgroup of .

In ,

You can’t add any more elements to while leaving it connected solvable by Theorem 1Theorem 1: you must stabilise a complete flag.

In , i.e., such that , ⋰⋰, we claim that

is Borel.

Sketch of proof/exercise Why is the above Borel? Let :

Exercise: The only complete flag stabilised by

is the standard one. This implies that the claimed Borel is Borel.

Pick a Borel subgroup . This has a unipotent radical , the maximal unipotent normal subgroup of (unique). E.g., in the examples above

Define . This is called the “abstract” Cartan, or the maximal torus (Ch??? Ginzberg p.137). This group is a “canonical” torus ( over an algebraically closed field) associated to . This is “canonical” because any two Borels are conjugate: given and , there exists such that . Since the inside is canonical, we have via conjugation by . Now we want to know if there are other isomorphisms possible: take , which induces . We want the map on the quotient to be the identity. By 2.2. in Two theorems about BorelsTwo theorems about Borels, , so is conjugation by an element of . But is abelian so the map is the identity.

Pick a maximal torus. Given a , there exist many choices of . This choice gives an isomorphism which depends on .

Let and . This lies in two Borels and they give different isomorphisms .

For ,

For ,

Here, .

A weight of a torus is a homomorphism (of algebraic groups) . (For example, , is not algebraic.) Fixing an isomorphism gives an isomorphism (from FactFact)

A coweight is an algebraic group homomorphism and

Further, there is the following duality:

where , which gives a perfect pairing.

A (root) is a non-zero weight appearing in the adjoint representation. The adjoint representation is . We use the notation .

One asks the question (where , so we are in )? This computation is easiest done when and .

For , we need to be invertible in , which is always true. So is all matrices.

For , we need . The linear term in the expansion of the LHS yields . This is the condition for . If we write everything in block matrix form:

this gives , , .

For , let

Then

This is a common eigenvector for all . So is a root (). Using additive notation and the standard basis of , the roots are .

For , take for

Then . If instead we take

This s a type root system.