Irreducible representations of the unitriangular group - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T18:56:08Z http://mathoverflow.net/feeds/question/68207 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68207/irreducible-representations-of-the-unitriangular-group Irreducible representations of the unitriangular group trew 2011-06-19T12:43:07Z 2011-08-14T12:21:29Z <p>Hi, I wonder how much is known about the irreducible representations of the nxn unitriangular group over a finite field with q elements.<br> 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</p> http://mathoverflow.net/questions/68207/irreducible-representations-of-the-unitriangular-group/68211#68211 Answer by Jack Schmidt for Irreducible representations of the unitriangular group Jack Schmidt 2011-06-19T13:59:04Z 2011-06-19T13:59:04Z <p>Let <em>q</em> = <em>p</em><sup><em>k</em></sup> be a prime power, <em>n</em> be a positive integer, and <em>U</em> be a Sylow <em>p</em>-subgroup of GL(<em>n</em>, <em>q</em>).</p> <p>For <em>n</em> = 1, <em>U</em> = 1 is the trivial group and its character table is known (just the identity/principal character).</p> <p>For <em>n</em> = 2, <em>U</em> = <em>q</em> is elementary abelian of order <em>q</em>, and its character table is known (just <em>q</em> distinct linear characters).</p> <p>For <em>n</em> = 3, <em>U</em> is similar to an extra-special <em>q</em>-group, and the same calculation works to find its ordinary character table. A sketch: <em>U</em>/[<em>U</em>, <em>U</em>] is elementary abelian of order <em>q</em><sup>2</sup>, giving <em>q</em><sup>2</sup> (known) linear characters. Since |<em>U</em>|−[<em>U</em> : [<em>U</em>, <em>U</em>]] = <em>q</em><sup>3</sup>−<em>q</em><sup>2</sup> = <em>q</em><sup>2</sup>(<em>q</em>−1), and the degrees of the irreducible characters are powers of <em>q</em> and their squares sum to |<em>U</em>|, one must have the remaining characters have degree <em>q</em> and there are q−1 of them. Since <em>U</em> is monomial, each of these characters is induced from a (non-principal) linear character on a subgroup of index <em>q</em> (which must be abelian). Hence each such character vanishes off of Z(<em>U</em>), and acts in the expected way on its center. That is, take any non-identity irreducible character θ of Z(<em>U</em>), there are <em>q</em>−1 of these, and then define χ(<em>g</em>) = q⋅θ(<em>g</em>) if <em>g</em> in Z(<em>U</em>) and χ(<em>g</em>) = 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.</p> <p>I think prime powers <em>q</em> for higher <em>n</em> work similarly, but I'm not very familiar with even the prime case for <em>n</em> ≥ 4.</p> http://mathoverflow.net/questions/68207/irreducible-representations-of-the-unitriangular-group/68219#68219 Answer by L Spice for Irreducible representations of the unitriangular group L Spice 2011-06-19T16:34:21Z 2011-06-19T16:41:40Z <p>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 <em>supercharacters</em>. As far as I know, the original definitions are due to <a href="http://www.sciencedirect.com/science/article/pii/S0021869385711878" rel="nofollow">André</a> (who calls them ‘basic characters’) and Yan, and there is an explicit <a href="http://www.sciencedirect.com/science/article/pii/S0021869301987344" rel="nofollow">supercharacter table</a>.</p> http://mathoverflow.net/questions/68207/irreducible-representations-of-the-unitriangular-group/68227#68227 Answer by Jeff Adler for Irreducible representations of the unitriangular group Jeff Adler 2011-06-19T18:37:25Z 2011-06-19T18:37:25Z <p>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.</p> <p>More generally, Boyarchenko has recently <a href="http://arxiv.org/abs/1003.2742" rel="nofollow">shown</a> 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.</p> http://mathoverflow.net/questions/68207/irreducible-representations-of-the-unitriangular-group/68242#68242 Answer by Igor Pak for Irreducible representations of the unitriangular group Igor Pak 2011-06-19T22:51:49Z 2011-06-19T22:51:49Z <p>I think you might enjoy Kirillov's <a href="http://bit.ly/ki7ACg" rel="nofollow">survey article</a> which describes <a href="http://en.wikipedia.org/wiki/Orbit_method" rel="nofollow">Orbit method</a> approach in this particular case. Also, if I recall correctly, <a href="http://math.ucsd.edu/~eariasca/papers/UniRW.pdf" rel="nofollow">this article</a> 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. </p> http://mathoverflow.net/questions/68207/irreducible-representations-of-the-unitriangular-group/72866#72866 Answer by trew for Irreducible representations of the unitriangular group trew 2011-08-14T12:21:29Z 2011-08-14T12:21:29Z <p>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 <a href="http://fourier.math.uoc.gr/~marial/uni1.published.pdf" rel="nofollow">http://fourier.math.uoc.gr/~marial/uni1.published.pdf</a> 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 $\psi^{G/Z(G)}(1)=q^{3} &lt; q^{4} =q^{2} \phi(1)$. 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.</p>