§ Classification of lie algebras, dynkin diagrams
§ Classification of complex lie algebras
- is a complex vector space with a lie bracket .
- For example, if is a complex Lie group. For a complex manifold, the transition functions are holomorphic.
§ Theorem (Leri)
- Every finite dimensional complex Lie algebra can be decomposed as , where is direct sum, is the semidirect sum.
- is a solvable lie algebra.
- To define solvable, define , , , that is, .
- We have that is a strict subset of .
- If this sequence eventually stabilizes, ie, there is an such that , then is solvable.
- In the decomposition of , the is the solvable part.
- We have , \dots, which are simple. This means that is non-abelian, and contains no non-trivial ideals. An ideal of a lie algebra is a subvevtor space such that . (It's like a ring ideal, except with lie bracket).
- The direct sum of lie algebras is the direct sum of vector spaces with lie bracket in the bigger space given by .
- The semidirect sum as a vector space is . The lie bracket is given by , so is an ideal. (This looks like internal semidirect product).
- It is very hard to classify solvable Lie algebras.
- A lie algebra that has no solvable part, ie can be written as is called as semi-simple .
- It is possible to classify the simple Lie algebras.
- We focus on the simple/semi-simple Lie algebras. Simple Lie algebras are the independent building blocks we classify.
§ Adjoint Map
- Let be a complex lie algebra. Let be an element of the lie algebra.
- Define as . Can be written as . This is the adjoint map wrt .
§ Killing form
- is a bilinear map, defined as .
- See that . the trace will be complex because is complex.
- Since is finite dimensional vector space, is cyclic. So . This means that , or that the killing form is symmetric!
- Cartan criterion: is semi-simple iff the killing form is non-degenerate. That is, implies .
§ Calculation wrt basis: map.
- Consider for actual calculation the components of and with respect to a basis .
- Write down a dual basis .
- We know that by definition.
- We write where the are the structure constants.
- This gives us
- Pull out structure coefficient to get
- Use the fact that and are dual to get
- Contract over repeated index to get :
- This makes sense, since the map is just a fancy way to write the bracket in coordinate free fashion.
§ Calculation wrt basis: Killing form.
- Plug in to become [see that the thing inside the trace is a matrix ]
- Execute trace by setting . This gives us: . This is also easy to calculate from structure coefficients.
- Iff this matrix is non-degenerate, then the lie-algebra is semi-simple.
§ is anti-symmetric with respect to the killing form.
- Recall that is called as an anti-symmetric map wrt a non-degenerate bilinear form iff .
- Fact: is anti-symmetric wrt killing form. For killing form to be non-degenerate we need to be semisimple.
§ Key Definition for classification: Cartan subalgebra
- If is a lie algebra, then the cartan subalgebra denoted by ( is already taken for structure coeff.) is a vector space, and is a maximal subalgebra of such that there exists a basis of that can be extended to a basis of : such that the extension vectors are eigenvectors for any for .
- This means that .
- This can be written as .
- Does this exist?
§ Existence of cartan subalgebra
- Thm Any finite dimensional lie algebra possesses a cartan subalgebra.
- If is simple, then is abelian . That is, .
- Thus, the are simultaneously diagonalized by the since they all commute.
§ Analysis of Cartan subalgebra.
- Since the LHS is linear in , the RHS must also be linear in . But in the RHS, it is only that depends on .
- This means that is a linear map !
- This is to say that is an element of the dual space!
- The elements are called the roots of the Lie algebra.
- This is called as , the root set of the Lie algebra.
§ Root set is closed under negation
- We found that is antisymmetric with respect to killing form.
- Thus, if is a root, is also a root (somehow).
§ Root set is not linearly independent
- We can show that is not LI.
§ Fundamental roots
- Subset of roots such that is linearly independent.
- Let the elements of be called .
- We are saying that such that .
- That is, we can generate the as natural number combinations of , upto an overall global sign factor.
- Fact: such a set of fundamental roots can always be found.
§ complex span of fundamental roots is the dual of the cartan subalgebra
- In symbols, this is .
- They are not a basis of because they are not independent (?)
- is not unique, since it's a basis.
- Real span of fundamental roots: .
- We have that .
- Thus is contained in , which is contained in .
§ Defn: Killing form on
- We restrict to .
- What we want is .
- Define given by .
- is invertible if is non-degenerate.
- The restricted action of on will always spit out real numbers.
- Also, and equal to zero iff .
- See that was non-degenerate, but is a real, bona fide inner product!
- This means we can calculate length and angles of fundamental roots.
§ Recovering from
- How to recover all roots from fundamental roots?
- For any , define the Weyl transformation
- The map is given by .
- This is linear in , but not in .
- Such are called as weyl transformations.
- Define a group generated by the . This is called as the Weyl group.
§ Theorem: Weyl group is generated by fundamental roots
- It's enough to create to generate .
§ Theorem: Roots are prouced by action of Weyl group on fundamental roots
- Any can be produced by the action of some on some .
- So such that .
- This means we can create all roots from fundamental roots: first produce the weyl group, then find the action of the weyl group on the fundamental roots to find all roots.
- The Weyl group is closed on the set of roots, so .
- Consider for .