# Lately in Lie Groups

11/12/09

Construction of Lie Algebras from simply laced Dynkin diagrams. (esp. E8). All exceptional ones can be obtained from E8 but taking sub diagrams or taking invariant stuff. Approach to constructing E8 (not in literature, might be errors) (in fact there is)

For a Dynkin diagram of type ADE, there is a vector space . Works for any Dynkin diagram (but today restrict to ADE). Have For each root associated a variable , and an element which is dual to by

Note . acts transitively on the roots, may normalize the inner product so for all roots.

Claim: W acts transitively on the pairs , i.e. . Also, if , then and . (possibly insecurity of this claim, in fact it is false)

At the moment, assume this claim. Know all the inner products between pairs of element in root spaces, up to a scalar. Know , but don’t know the actual multiple in .

Choose , set and scale until get , uniqe choice up to sign. Now have . SImply laced case: iff .

implies $latex (a+b, a+b) = 4$, no roots can have this length.

$latex (a,b) = -2$ ==> $latex a = -b$. The rest of lecture is here, but ultimately there are big errors, so maybe not that useful

Assuming the claim, define g = h \oplus_a \mathbb{C}\cdot X_a

choose fixed a_1, a_2 \in R with (a_1, a_2) = -1. Define bracket: for all . etc.

extend by linearity to antisymmetric form .

Now need to show satisfied Jacobi identity. Again, only have to do that on a basis. If is involved, property of a root system show jacobi holds. Notrivial case for . Only need to check it for , since these span . Let . There exists such that . Permutiting and chaning signs if necessary, can assume (i,e A_3). Last one is because otherwise you would see cycle of edges in a dynkin diagram, or (i.e. A_2 \times A_1)

or (A_1^3), or get latex a = b = c$ so get . In all cases are in a dynkin subalgebra that you know corresponds to a lie algebra, so the Jacobi identity holds.

There also a case (a,b) = -1, (b,c) = -1, (a,c) = -1$, then you’re in A_2, so Jacobi identity still holds.

Note: its important that you can rule out something like (a,b) = -1, (b,c) = -1 but (a,c) = 1$. This is where sign flipping occurs, and assumption that the a,b,c are simple roots (so you can make the no cycle argument)

Example: sl(C,4), roots e_i – ej, i \ne j

a_1 = e_1 – e_2, a_2 = e_2 – e_3

Note [E_{12}, E_{23}] = E_{13}

X_a_1 X_a_2 X_{a_1 + a_2}

checking the unproved claim above

w(e_1 – e_2) = e_i – e_j, w(e_2 – e_3) = e_j – e_h

acts transitively on such pairs

if w(e1-e2) = e2-e3, then w(e2-e3) = e3-ek, e_k \ne e1

Now Proof of the claim:

If (a,b) = -1, then (a,b) maybe be sent to a pair of simple roots by an element of W. Because: just need to show a^\perp and b^\perp are faces of a weyl chamber. Choose \nu \in a^\perp \cap b^\perp. a+b is in chamber that a,b form.

Look at \mu = \nu + \epsilon (a+b), small \epsilon. Then \mu sits in Weyl chamber with a^\perp and b^\perp as faces. (hmm not sure about this)

Now assume a,b are adjacent nodes in a Dynkin diagram correspoinding to simple roots. So have c—–a———-b<

so just need to be able to send one edge to another inside A_3 roots system.

Picture of A_3….

want a,b maps to b,c, ends up being a coxeter element

Back to proof. Now we need to check there is no w \in W, with w(a_1) = a2, w(a_2) = a_1

also a_1 + a_2 \in R and (a_1,a_2) = -1

claim: W_a = stab(a) = < W_b, (a,b) = 0>

stab(a^\perp) = <w_a> \times W_a

assuming have w as above, w(a+b) = a+b, so w \in stab(a+b)

compute stab (a+b)^\perp, break up into Weyl chambers, stabilizer of a face is = <W_a>

hmm something about getting a fundamental domain for (a+b)^\perp

the fundamental domain is made out of n faces of the Weyl chamber, key fact is if a were sent to b, then would fix orthogonal plane (a+b)^\perp, look at projection of a, b to (a+b)^\perp, would have to send projection to projection, but htere lie in same fundemental domain of this frelection group, which is , but just say that there is no Weyl group element sending one chamber to another chamber.

If a, b are permuted , we see a fundamental domain of stab (a+b)^perp) fixed, which give a contradiction that this is a reflection group.

Quick sketch of another method: (Serre’s Method) Described in Fulton & Harris in section 21. Have simple dynkin diagram or rank n

Form free lie algebra H_1, …., H_n, X_1,…., X_n, Y_1,….., Y_n

then divide modulo anti symmetry of the bracket and Jacobi relations. Then divide by relations [H_i, H_j] = 0 for all i,j et.c

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- Published:
- November 19, 2009 / 11:26 am

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- Lie Groups (course)

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