Let $q=p^f$ be a prime power such that $q \equiv 1 \pmod 5$. According to the list of irreducible (complex) character degrees of $SL(5, q)$ in Frank Luebeck's homepage (here), $SL(5, q)$ has 20 irreducible characters of degree $1/5(q^2+1)(q^2+q+1)(q+1)^2(q-1)^4$. Using Lusztig's parametrization of irreducible complex characters of groups of Lie type, I want to prove the existence of such an irreducible character for $SL(5, q)$ when $q$ is sufficiently large:
Assume that $\chi \in \mathrm{Irr}(SL(5, q))$ is of degree $1/5(q^2+1)(q^2+q+1)(q+1)^2(q-1)^4$. By Lusztig's result, $\chi$ must belong to a Lusztig's rational serie $\mathcal{E}(SL(5, q), s)$, where $s$ is a semisimple element of $PGL(5, q)$. Moreover, the degree of $\chi$ is given by $\chi(1)=\frac{|SL(5, q)|_{p'}}{|C_{PGL(5, q)}(s)|_{p'}}\psi_s(\chi)(1)$, where $\psi_s(\chi)(1)$ is the degree of a unipotent character of $C_{PGL(5, q)}(s)$ (see Theorem 13.23 and Remark 13.24 of Digne and Michel's book). It is known that if $p$ is sufficiently large (I think $p>5$ would be sufficient), then all the $p'$-degree irreducible characters of $SL(5, q)$ are preciesly the so-called semisimple characters. Therefore, $\chi=\chi_s$ is a semisimple character and $\chi(1)=\frac{|SL(5, q)|_{p'}}{|C_{PGL(5, q)}(s)|_{p'}}$. Thus, the existence of such $\chi$ depends on the existence of a semisimple element $s \in PGL(5, q)$ such that $|C_{PGL(5, q)}(s)|_{p'}=5(1+q+q^2+q^3+q^4)$.
Question: Is there any semisimple element $s$ of $PGL(5, q)$ satisfying $|C_{PGL(5, q)}(s)|_{p'}=5(1+q+q^2+q^3+q^4)$?
Some Thoughts: We deduce from above argument that we must have $|C_{PGL(5, q)}(s)|_{p'}=5(1+q+q^2+q^3+q^4)$. So if $s_1$ is a preimage of $s$ in $GL(5, q)$, the characteristic polynomial of $s_1$ is an irreducible polynomial of degree $5$ in $\mathbb{F}_q[x]$. Hence we have $C_{GL(5, q)}(s_1)\cong GL(1, q^5)$ and $C_{PGL(5, q)}(s)\cong C_{GL(5, q)}(s_1)/Z(GL(5, q))$. But this implies that $|C_{PGL(5, q)}(s)|_{p'}=1+q+q^2+q^3+q^4$ and so $SL(5, q)$ could not contain an irreducible character of degree$1/5(q^2+1)(q^2+q+1)(q+1)^2(q-1)^4$! How can I explain this?
I would be grateful if you could hint me how to find a proper semisimple element corresponds to $\chi$.
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$\begingroup$ Have you asked Frank Lübeck already? $\endgroup$– Stefan Kohl ♦Commented Nov 25, 2016 at 17:48
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$\begingroup$ @StefanKohl Not yet. First I wanted to be sure about the correctness of my arguments. $\endgroup$– user97635Commented Nov 25, 2016 at 17:50
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$\begingroup$ I'm confused about the formulation here. Lusztig's theory based on the Deligne-Lusztig construction allows one to compute (in principle) all the character degrees. However, what Carter calls a "Jordan decomposition" of characters assumes that the ambient algebraic group has a connected center. This is not true for special linear groups but is true for general linear groups. Can you clarify what you are asking for, preferably with reference to Lusztig's papers? $\endgroup$– Jim HumphreysCommented Nov 25, 2016 at 23:12
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1$\begingroup$ @JimHumphreys I think Lusztig's Jordan decomposition of irreducible characters has been generalized also to connected reductive groups with disconnected center in Lusztig's paper "On the representations of reductive groups with disconnected center" Asterisque 168. Unfortunately, I don't have access to the original paper, but this have been covered in standard texts such as Theorem 13.23 and Remark 13.24 of Digne and Michel's book, or more clearly in Theorem 1.73 and Corollary 1.74 of Jay Taylor's thesis tel.archives-ouvertes.fr/tel-00709051/document. $\endgroup$– user97635Commented Nov 26, 2016 at 8:18
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$\begingroup$ @JimHumphreys I added some more description to clarify my question. By the way, you can also take a look at Section 2.1 of Gunter Malle's paper "Height 0 characters of finite groups of Lie type, Represent. Theory (2007)" for a short review of Lusztig's parametrization of characters for an arbitrary connected reductive group. $\endgroup$– user97635Commented Nov 26, 2016 at 9:49
1 Answer
[The author or this question made me aware of this thread, so I send the answer here.]
The description in the question is almost correct, except when it comes to the centralizer of the semisimple element $s$ in $PGL_5(q)$. For $q \equiv 1 \pmod 5$ the characters of $SL_5(q)$ of degree $1/5(q^2+1)(q^2+q+1)(q+1)^2(q−1)^4$ are in Lusztig series corresponding to certain elements $s$ of order $5$ in the dual group $PGL_5(q)$. Preimages $s_1 \in GL_5(q)$ of such $s$ can be found as elements in the cyclic Coxeter torus (a maximal torus of order $q^5-1$) which have an eigenvalue $x$ of order $5(q-1)$. Its other eigenvalues are $x^q, x^{q^2}, x^{q^3}, x^{q^4}$ (the successive quotients of these eigenvalues are a fixed $5$-th root of unity).
Such $s_1 \in GL_5(q)$ are regular and have a torus of order $q^5-1$ as centralizer. Its image $s \in PGL_5(q)$ has a non-connected centralizer generated by a maximal torus and a $5$-cycle in the Weyl group.
To $s_1$ corresponds one irreducible character of $GL_5(q)$ (which is isomorphic to its dual group) and to $s$ corresponds the restriction of this character to $SL_5(q)$ which splits into $5$ irreducible constituents of same degree, these are the characters in question. Since there are $4$ primitive roots of $5$ we get altogether $4 \cdot 5 = 20$ characters of that degree.
