Assume .

Theorem 1

Every quotient of a torus is a torus.

Analogue for compact Lie groups

A torus is a compact connected abelian Lie group, and .

Subgroups of tori

. If then . We had the notation . Then there exist such that

Every torus is an intersection of codimension subgroups. This gives a discrete family of subgroups (parametrised non-uniquely by ). It is an infinite family: e.g., when , the (proper) subgroups are the th roots of unity (not connected).

Theorem 2

If is a torus and where is a variety, then there exists such that .

Idea

There exists a Zariski open in such that is the same for all . This follows from discreteness of subgroups. Use this with Noetherian induction. All this implies there are finitely many stabilisers (because variety). To find , we need for a finite number . Use linear algebra to complete this.

Theorem 3

In any algebraic group, maximal tori are conjugate.

Remark over

We can talk about a maximal compact subgroup. One has that the flag variety where is a compact torus.

If then .

Compute

is the variety of Borels. acts by .

1

fixes a Borel if and only if . This requires a previous theorem: .

Q: What are the containing ?

Suppose and . Recall the theorem that Borels are conjugate. So there exists such that . So now and . Apply Theorem 3 Theorem 3 to . So there exists such that , i.e., . So . Therefore . Conjugation by doesn’t do anything.

Conclusion: .

Pick such that . One can actually assume that is dominant (as a coweight), i.e., that

We can also further get a for which the above inequality is strict.

Let . What is the tangent space ( is ). We have

We want to decompose this into . Take

()

We had the BB-decomposition locally closed.

Claim

Recall before we had and where consists of all such that .

1

Take

1a: the image the correct place

We have

If then . For general , write and use the same proof.

2: is injective

Otherwise has , in its stabiliser . The claim is that . Compute for :

3: is bijection (Ax-Grothendieck)

We know and (second one due to BB-theory).

Bijective doesn’t imply isomorphism. via . This is a bijection but not an isomorphism. is not normal.

4: Zariski’s main theorem

This implies is an isomorphism. (Bijection to normal target implies isomorphism.)

Implies Bruhat.

Remark: example of disconnected subgroup of unipotent group

Line bundles on ()

Version 1

Take , abstract Cartan subgroup of .

If is a Borel we have

by . This gives a line bundle on whose fibre over is “ by ”.

Version 2

freely (look at first factor) via

Take the quotient space . We get

The fibre over every point is a line (the ). This is a line bundle on .

Exercises

If then . One knows all the line bundles on . Show that every line bundle on arises from this construction.
If then . Show that not every line bundle on arises from this construction.
For , , the above are , so there’s multiplicity, but we do get every line bundle.

2025-08-18