Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi, I wonder how much is known about the irreducible representations of the nxn unitriangular group over a finite field with q elements.
I know that all characterdegrees are a power of q and all degrees which occur are known.But what is known about the irreducible representations or the complete charactertable at least for small values of n?For example is the charactertable for n=3 known? Thanks for helping

share|improve this question
add comment

5 Answers 5

Not much—the theory of individual irreducible representations is a 'wild' problem, in some technical sense (that I don't know). My understanding, which comes entirely from informal conversations with Nat Thiem, is that the state of the art is to lump together representations until you get more nicely behaved objects called supercharacters. As far as I know, the original definitions are due to André (who calls them ‘basic characters’) and Yan, and there is an explicit supercharacter table.

share|improve this answer
add comment

Let q = pk be a prime power, n be a positive integer, and U be a Sylow p-subgroup of GL(n, q).

For n = 1, U = 1 is the trivial group and its character table is known (just the identity/principal character).

For n = 2, U = q is elementary abelian of order q, and its character table is known (just q distinct linear characters).

For n = 3, U is similar to an extra-special q-group, and the same calculation works to find its ordinary character table. A sketch: U/[U, U] is elementary abelian of order q2, giving q2 (known) linear characters. Since |U|−[U : [U, U]] = q3q2 = q2(q−1), and the degrees of the irreducible characters are powers of q and their squares sum to |U|, one must have the remaining characters have degree q and there are q−1 of them. Since U is monomial, each of these characters is induced from a (non-principal) linear character on a subgroup of index q (which must be abelian). Hence each such character vanishes off of Z(U), and acts in the expected way on its center. That is, take any non-identity irreducible character θ of Z(U), there are q−1 of these, and then define χ(g) = q⋅θ(g) if g in Z(U) and χ(g) = 0 otherwise. Personally I just execute the definition of induced character and then check the norm of χ is (χ,χ)=1, but I think there are cleverer ways.

I think prime powers q for higher n work similarly, but I'm not very familiar with even the prime case for n ≥ 4.

share|improve this answer
add comment

An earlier answer tells you what happens when $n=3$. This is a special case of the Heisenberg group (at least in odd characteristic; not sure otherwise), any exposition of which might be illuminating if you're looking for more context.

More generally, Boyarchenko has recently shown that all representations of "nilpotent algebra" groups, of which unitriangular groups are a special case, are induced from one-dimensional representations of "nilpotent algebra" subgroups. While this fact is not a magical tool for computing the full character table (a wild problem, as LS says), it's pretty interesting, and might allow you to work out the $n=4$ case if you were interested in such an exercise.

share|improve this answer
    
thanks,i know that result.it would be interesting to know what has been done for n=4,before i try ;) I know for example that all characterdegrees with their multiplicity is known.can someone by the way say what exactly "wild" means? –  trew Jun 19 '11 at 19:24
1  
Since this question is about finite algebra groups, it should be pointed out that the result that says that any irrep of an algebra group over a finite field is induced from a 1-dim'l rep of an algebra subgroup is due to Halasi. Boyarchenko has given a new proof valid also for algebra groups over local fields. –  A Stasinski Jun 19 '11 at 22:14
    
A somewhat relevant question I'm curious about: if we work over $\mathbb{R}$ instead of a finite field, the problem of determining the irreps of the group remains just as "wild"? –  Mark Jun 19 '11 at 22:26
    
@AStansinski: You're right, I should have mentioned Halasi, but have local fields on the brain. –  Jeff Adler Jun 20 '11 at 3:01
    
@trew, I believe that the sense in which the classification is wild is that discussed in mathoverflow.net/questions/10481/…. –  L Spice Jun 20 '11 at 3:58
add comment

I think you might enjoy Kirillov's survey article which describes Orbit method approach in this particular case. Also, if I recall correctly, this article by Ery Arias-Castro, Persi Diaconis and Richard Stanley gives a very readable introduction to the state of art on the conjugacy classes and characters.

share|improve this answer
add comment

Here is what I found out about the characters when n=4.I dont know if thats interesting and how to get the actuall irreducible characters then.Maybe someone has an idea: There are $ q^{3} $ linear and from http://fourier.math.uoc.gr/~marial/uni1.published.pdf there are $q^{3}-q $ characters of degree q and $q(q-1)$ characters of degree $q^{2}$. Now lets look at the characters of $G/Z(G)=1+J/J^{3}$ ,where Z(G) is the center of the group and J are the lower triangular matrices with zeros on the diagonal.This is again an algebragroup,so we have: $q^{5}=q^{3}+aq^{2}+bq^{4}$,where a is the number of the degree q characters of G/Z(G) and b the number of degree $q^{2}$ characters. Assume $ \phi$ is a degree $q^{2}$ character in G/Z(G),then $ [ \phi_{Z(G/Z(G))} , \psi] \neq 0$ ,for a linear character $\psi $ of Z(G/Z(G)).But then since $\psi$ is G/Z(G) invariant as a character of the center and using Clifford: $\phi_{Z(G/Z(G))}=q^{2} \psi$ and then $ \psi^{G/Z(G)} = q^{2} \phi +...$ which is not possible because of $ \psi^{G/Z(G)}(1)=q^{3} < q^{4} =q^{2} \phi(1)$. So we have b=0 and a=q^{3}-q from $q^{5}=q^{3}+aq^{2}$.So all the degree q characters of G are also the degree q characters of G/Z(G). Let now $\chi$ be a character of degree q of G/Z(G) then we can choose a linear character $\psi$ of Z(G/Z(G)) with $[\chi_{Z(G/Z(G))} ,\psi] \neq 0 $ and again as above: $ \psi^{G/Z(G)} = q^{2} \chi +...$ Since $(1_{Z(G/Z(G))})^{G/Z(G)}$ has all linear characters as constituents $\psi^{G/Z(G)}$ can only have degree q irreducible constituents.So for all(there are $q^{2}-1$ such) the nontrivial linear characters $\psi_k $ of Z(G/Z(G)),we have : $ (\psi_k)^{G/Z(G)} = q \sum\limits_{i=1}^{q} {\chi_i}$. Similary one can show that for all nontrivial linear characters $\vartheta_k $ of Z(G) one has: $(\vartheta_k)^{G}=q \sum\limits_{i=1}^{q} {\phi_i} $,where $\phi_i$ are degree-$q^{2}$ characters of G.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.