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The following paper gives a classification of the character tables of irreducible representations of $SL(3,GF(q))$ where $q$ is a power of a prime number, and $ GF(q)$ a finite field of $q$ elements.

WILLIAM A. SIMPSON AND J. SUTHERLAND FRAME Can. J. Math., Vol. XXV, No. 3,1973, pp. 486-494 THE CHARACTER TABLES FOR SL(3, q ), SU(3, *•), PSL(3, q), PSU(3, q *)

Here I would to ask "do we have a classification of the character tables of irreducible representations of $SL(3,Z_q)$, where $Z_q=Z/qZ$?"

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    $\begingroup$ Note that the article you mention by Simpson and Frame is accessible online at cms.math.ca/10.4153/CJM-1973-049-7 $\endgroup$ – Jim Humphreys Mar 10 '17 at 20:57
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    $\begingroup$ As @JimHumphreys says below, the answer is certainly "no". So people are led to study numerical invariants. In a remarkabel recent Inventiones paper Aizenbud and Avni study the function $n \mapsto$ (the number of representations in dimension $n$) and show it grows slower than $n^{22}$. See arxiv.org/abs/1307.0371 $\endgroup$ – Uri Bader Mar 11 '17 at 8:31
  • $\begingroup$ This paper relates the representation counting problem to rational singularities of the moduli space of $\text{SL}$-local systems. I really recommend trying to read it. $\endgroup$ – Uri Bader Mar 11 '17 at 8:39
  • $\begingroup$ I don't think the answer is "certainly no", since the OP is about $\mathrm{SL}_3$ and Avni, Klopsch, Onn and Voll have computed the rep zeta function in this case, for residue char bigger than $3$. We may not know the exact character table, but there is no reason to expect this to be impossible to obtain (for large residue char). For all $\mathrm{SL}_n$, $n\geq 3$, the answer is certainly "no". $\endgroup$ – A Stasinski Mar 11 '17 at 21:00
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As was already mentioned, the answer to the question asked is "no, we currently do not have a good classification". Here I wish to describe a successful and interesting recent line of research which does not aim at giving such a classification, rather merely at counting how many representations we do have.

To put things in context, let me first make the following observations:

  • the set of irreducible representations of a product of groups is in bijection with the product of the sets of irreducible representations of the groups.

  • $\text{SL}_n$ over a product of rings (commutative with 1) is isomorphic to the product of $\text{SL}_n$ over the rings.

  • for a natural $m$, $\mathbb{Z}/m\mathbb{Z}$ is the product of the rings $\mathbb{Z}/q\mathbb{Z}$ where $q$ is a prime power.

Thus, the study of the rep theory of $\text{SL}_n(\mathbb{Z}/m\mathbb{Z})$ is naturally reduced to the study of the rep theories of the groups $\text{SL}_n(\mathbb{Z}/p^k\mathbb{Z})$ for prime $p$. Fixing $p$ and varying $k$, these are grouped together as the (continuous) representation theory of $\text{SL}_n(\mathbb{Z}_p)$, where $\mathbb{Z}_p$ is the ring of $p$-adic integers.

For $n=2$ the irreducible representations of $\text{SL}_n(\mathbb{Z}_p)$ are indeed classified, as appeared in the answere by Jim Humphreys. To this answer I wish to add the reference to Uri Onn's paper in which he classifies all irreducible representations over an arbitrary discrete valuation ring. See also a comment below by A Stasinski.

For $n\geq 3$ the representation theory of $\text{SL}_n(\mathbb{Z}_p)$ is much more complicated and a classification is currently out of reach. However we do know that $\text{SL}_n(\mathbb{Z}_p)$ has only finitely many irreducible representations at each dimension $d$. Denoting this number by $r_d$ we are interested in the study of the sequence $(r_d)_{d=1}^\infty$. On this problem there have been in recent years a remarkable progress on which I wish to report, and this is why I am writing this answer.

For $n=3$ it is shown in by Avni-Klopsch-Onn-Voll that the sequence $r_d$ grows similarly to $d^{3/2}$. For a precise statement see their Duke paper. Recently Aizenbud-Avni got a uniform statement for all $n$, that is $r_d$ is $O(d^{22})$. For this see their Inventiones paper. Whether such a uniform bound exists was asked as a question in Larsen-Lubotzky.

I would be happy to tell more about the techniques of these remarkable papers, but this answer is already getting too long. You will have to read through the sources. Let me just mention that in order to study the sequence $(r_d)$ we set the representation zeta function $\zeta(s)=\sum r_d\cdot d^{-s}$. A general result of Jaikin-Zapirain tells us that this function is rational in the variable $p^{-s}$. The representation growth is then related to the study of the poles of this zeta function. Let me also mention that there is a global theory here, which comes from the representation theory of $\text{SL}_n(\mathbb{Z})$.

Edit: As remarked by Alexansder Stasinski (thanks!), for $n=3$ (but not for higher $n$'s) the representation zeta function is explicitly computed in the Avni-Klopsch-Onn-Voll paper alluded to above (see Theorem E) and one should be able get out of it the exact numbers of irreducible representation of $\text{SL}_3(\mathbb{Z}/q\mathbb{Z})$ at each dimension, which is closer to answering the OP question. One can also hope to get a full answer using Kirillov orbit method. However, things get messy for higher $n$'s.

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  • $\begingroup$ This is interesting (and cautionary for the OP, who started out with a more naive-looking question). By the way, to save space with citations, try typesettting this way: <a href="http:// ....."> label </a>. And in your second sentence, the word is "successful", while near the end you want "poles". $\endgroup$ – Jim Humphreys Mar 11 '17 at 20:02
  • $\begingroup$ Regarding $\mathrm{SL}_2(\mathbb{Z}/p^r)$, the classification is due to Kutzko and Shalika for $p\neq 2$. The case $p=2$ is much more difficult and due to Nobs and Wolfart. The reps of $\mathrm{SL}_2(\mathbb{F}_q[t]/t^r)$, where $q$ is even is still not completely known. $\endgroup$ – A Stasinski Mar 11 '17 at 21:22
  • $\begingroup$ @Uri Bader "$SL_n$ over a product of rings (commutative with 1) is isomorphic to the product of $SL_n$ over the rings." Do you have any references for the above statement? $\endgroup$ – Xiao-Gang Wen Mar 11 '17 at 22:37
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    $\begingroup$ @Xiao-GangWen I don't have a specific reference, but I guess any reasonable book having the phrase "Algebraic K-Theory" in the title will have it. What you need to observe is that the determinant map $M_n(R\times S)\to R\times S$ satisfies $\det(x+y)=\det(x)+\det(y)$ for $x\in M_n(R)$, $y\in M_n(S)$. This becomes clear when you observe that $\det$ commutes with ring homomorphism and consider both projections. $\endgroup$ – Uri Bader Mar 12 '17 at 7:46
  • $\begingroup$ @JimHumphreys, thanks! I always wanted to do this, but never bothered to learn how to. $\endgroup$ – Uri Bader Mar 12 '17 at 7:59
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This computation could in principle be done using Clifford theory. Clifford theory tells you how to describe the representation theory of a group $G$ given that it can be described as an extension

$$1 \to N \to G \to H \to 1$$

and $SL_d(\mathbb{Z}/p^n\mathbb{Z})$ can be described as an extension

$$1 \to N \to SL_d(\mathbb{Z}/p^n \mathbb{Z}) \to SL_d(\mathbb{Z}/p^{n-1}\mathbb{Z}) \to 1$$

where $N$ is the kernel of the reduction $\bmod p^{n-1}$ map. It consists precisely of elements of $SL_d(\mathbb{Z}/p^n\mathbb{Z})$ congruent to $I \bmod p^{n-1}$, or equivalently of the form $I + p^{n-1} M$. Every such matrix is invertible over $\mathbb{Z}/p^n\mathbb{Z}$ and has determinant $1$ iff $\text{tr}(M) \equiv 0 \bmod p$, and multiplying two such matrices even shows that $N$ is isomorphic to the additive group of such matrices, hence

$$N \cong (\mathbb{Z}/p\mathbb{Z})^{d^2-1}.$$

If you like, you can think of $N$ as $\mathfrak{sl}_d(\mathbb{Z}/p\mathbb{Z})$.

Clifford theory simplifies substantially when $N$ is abelian, so this is helpful, although I think the detailed analysis will still be difficult to carry out, and doing it this way you'll have to repeat the analysis $n-1$ times to reduce to the case of $SL_d(\mathbb{Z}/p\mathbb{Z})$.

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    $\begingroup$ The matrix $M$ is not arbitrary, consider the simplest case of $2{\times}2$ matrices, then $\det(1+p^{n-1}M)\equiv1\pmod{p^n}$ means that $\operatorname{tr}(M)\equiv0\pmod{p}$. $\endgroup$ – Andrei Smolensky Mar 9 '17 at 9:05
  • $\begingroup$ @Qiaochu Yuan Thank you very much for the answer. I wonder if the extension $1 \to N \to SL_d(\mathbb{Z}/p^n \mathbb{Z}) \to SL_d(\mathbb{Z}/p^{n-1}\mathbb{Z}) \to 1 $ split or not. If it splits, then $SL_d(\mathbb{Z}/p^n \mathbb{Z}) =N\times SL_d(\mathbb{Z}/p^{n-1}\mathbb{Z}) $, and we can use the character tables of $N$ and $SL_d(\mathbb{Z}/p^{n-1}\mathbb{Z}) $ to construct that of $SL_d(\mathbb{Z}/p^n\mathbb{Z}) $. But if it does not spit, can we still use the character table of $N$ and $SL_d(\mathbb{Z}/p^{n-1}\mathbb{Z}) $ to construct that of $SL_d(\mathbb{Z}/p^n\mathbb{Z}) $. $\endgroup$ – Xiao-Gang Wen Mar 9 '17 at 12:50
  • $\begingroup$ @Andrei: yes, you're right. In fact the condition is trace zero for all values of $d$. $\endgroup$ – Qiaochu Yuan Mar 9 '17 at 23:09
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    $\begingroup$ @Xiao-Gang: it doesn't split in that sense, but you can still use Clifford theory to answer the question (in principle), although the answer is complicated. The easiest special case of Clifford theory to understand is that of a semidirect product $G \cong A \rtimes H$ where $A$ is abelian; in this case $H$ acts on the group of characters $\hat{A}$ of $A$ and the representation theory of $G$ breaks up into the representation theories of the stabilizers of this action. This should be familiar to physicists from the case that $G$ is the Poincare group; the stabilizers are the "little groups." $\endgroup$ – Qiaochu Yuan Mar 9 '17 at 23:15
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    $\begingroup$ Qiaochu Yuan, you are definitley right saying that applying Clifford Theory is the first step in attacking this problem. However, a glance look at the Avni-Klopsch-Onn-Voll paper mentioned in my answer shows that this is only the tip of the iceberg, and there is a way to go from there. $\endgroup$ – Uri Bader Mar 12 '17 at 9:19
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The answer to your stated question is certainly "no", which I can support indirectly by reference to the simpler case of rank 1 groups.

The basic question here is natural but is already quite difficult even in rank 1: study the representation theory of a given group scheme over a ring of $p$-adic integers (or more generally, the ring of integers of a local field) by working out the representations of the finite groups over finite residue class rings .

This type of question has a fairly long history, so it may be worthwhile to follow the paper trail. There are for example a number of relevant papers in the rank 1 case, including one by Kutzko here and a little later by A. Nobs and J. Wolfart here and here.

It's important here to keep an open mind about techniques not encountered in the study of matrix groups over finite fields. That study was done in a special case in the Simpson-Frame work (which has some minor errors), building on the recursive combinatorial determination of characters of finite general linear groups by Green in 1955. There was also the work of his student Srinivasan on $Sp_4$ and Chang-Ree on groups of type $G_2$ over finite fields, and ultimately the far more sophistiated work that flowed out of the 1976 Deligne-Lusztig construction of virtual characters. By now the theory over finite fields has acquired considerable detail through the efforts of Lusztig and others. As far as I can tell, there is still no overall program for constructing the representations (or even the characters) of the finite groups of Lie type over rings such as $\mathbb{Z}/q \mathbb{Z}$ which are not fields.

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    $\begingroup$ Kloosterman studied the case of $SL_2(\mathbf{Z}/p^n\mathbf{Z})$ in "The behaviour of general theta functions under the modular group and the characters of binary modular congruence groups I, II," Ann. of Math. 47 (1946), 317–447; see also the account of Springer in his survey of Kloosterman's work ams.org/notices/200008/fea-springer.pdf $\endgroup$ – Denis Chaperon de Lauzières Mar 10 '17 at 6:08
  • $\begingroup$ @Denis: Thanks for the reminder of earlier work. It's not easy to give a complete survey, but I wanted to emphasize that even in rank 1 there is a continuing issue about how much can be written down explicitly. The underlying question is certainly natural in the context of $p$-adic groups. $\endgroup$ – Jim Humphreys Mar 10 '17 at 14:13
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    $\begingroup$ There is also this paper by Uri Onn, which classifies all irreps of $\text{GL}_2$ over a discrete valuation ring, and has a nice introduction which regards the history of the problem, as well as a discussion of the higher rank case: arxiv.org/pdf/math/0611383.pdf. $\endgroup$ – Uri Bader Mar 11 '17 at 8:04
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    $\begingroup$ I think it is misleading to say that the answer is certainly "no". See my comment to the OP. I don't think we currently know the character table of any $\mathrm{SL}_3(\mathbb{Z}/p^r)$, $r\geq 2$, but for $p$ large enough all the dimensions of the irreps and their numbers are known. A construction of the irreps can be approached via a Kirillov orbit method. $\endgroup$ – A Stasinski Mar 11 '17 at 21:14

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