Representation Theory and Lie Algebras Final Exam 2013 The first four problems are immediate if you have attended exercise sessions. The next two are also quite straightforward, but require some calculations (made easier in case you remember identities for binomial coefficients and induced characters). The last one is the easiest one of all, once you know the idea! Problem 1. Let G be a finite group, with complex irreducible representations of degrees 1,1,1,3,3,3,3. • Find the order of G and of its commutator subgroup [G, G]. • Find the number and orders of the conjugacy classes of G. At this point it would be straightforward to identify the group and to construct its character table, but this is not assumed. Problem 2. Below we reproduce a partial character table of a group, with two columns and four rows missing. 1

1

4

6

4

1A 2A 3A 4A 6A χ1

1

1

1

1

1

χ2

1

1

ω

1

ω

χ4

2

0

1

Here, as usual, ω = − 12 + i

√ 3 . 2

−2 −1

Complete the table. Explain each step.

Hint 1. It is much easier, than it seems, you’ve repeatedly seen all necessary ideas in the class. Hint 2. Actually, you know the group, but it is much easier to proceed formally using the properties of character tables, rather than to start with the group itself. Problem 3. Recall that A5 has 5 irreducible representations of degrees 1,3,3,4 and 5. Let π be the 5-dimensional irreducible representation. Decompose π ⊗ π into irreducibles. Hint. If you think, you might need it, you can use the character table of A5 , but this means doing it the hard way. All you have to know is that A5 has two classes of 5-elements. If you still do not see it, first try to decompose ρ ⊗ ρ, where ρ is the 4-dimensional irreducible representation. 1

2

Problem 4. Let µ ¶ 0 1 e= , 0 0

µ ¶ 1 0 h= , 0 −1

µ f=

0 0 1 0

¶

be the standard base of sl(2, K). Verify that the normalised Casimir element ef + f e + h2 /2 is a central element of the universal enveloping algebra U (sl(2, K)) Hint. You can express everything in the PBW-base of the universal enveloping algebra. Easy, but exceedingly boring, use the fearful symmetry instead (“Tiger, Tiger, burning bright...”) Problem 5*. Define the Zassenhaus algebra W (1, m) over a field K of characteristic p > 0 as the pm -dimensional algebra with the base Ei , −1 ≤ i ≤ pm − 2, and multiplication given by the formula µµ ¶ µ ¶¶ i+j+1 i+j+1 [Ei , Ej ] = − Ei+j . j i • Prove that this multiplication indeed makes W (1, m) into a Lie algebra. • Prove that for p > 2 this algebra is simple. • What happens in characteristic 2? Hint 1. The Witt algebra, that you’ve seen in the class, is W (1, 1), but there the base was choosen differently. Namely, as ea , a ∈ Fpm , with multiplication table [ea , eb ] = (a − b)ea+b . One can prove that these algebras are isomorphic, and then repeat the same calculations we’ve seen in the class. But for a beginner that wouldn’t be a smart idea at all , since it is substantially harder to guess the transformation formulae, then to honestly perform all calculations from scratch in the new base. Hint 2. I am not assuming you remember combinatorial identities µ for ¶ n binomial coefficients. But I assume that you do remember that m equals the number of m-element subsets of an n-element set. For instance, let k ≤ m ≤ n and you have to choose an m-element subset Y inside an n-element set X, and then a k-element subset Z inside Y . There are two different procedures to do that, but they should give the same number of pairs (Y, Z). The resulting equality, known as the electorial identity, is all you need.

3

Problem 6*. Show that the group B(2, q) of upper triangular matrices of degree 2 over the field Fq of q elements has • (q − 1)2 irreducible complex representations of degree 1, and • (q − 1) irreducible complex representations of degree q − 1. Hint 1. This group is a direct product of a group of order q − 1 and a group of order q(q − 1). Hint 2. At this point it is much easier, if not strictly necessary, to use induction. Problem 7*. Let π be a faithful irreducible complex representation of G with character χ. Suppose that χ only takes m distinct values. Show that every complex irreducible character of G occurs in at least one of the 1, π, π ⊗2 , . . . , π ⊗(m−1) . Hint. Vandermonde matrix with distinct columns is usually invertible. DON’T PANIC!