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Articles

Characters of twisted fractional linear groups

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Pages 3940-3959 | Received 03 Nov 2020, Accepted 23 Feb 2022, Published online: 03 May 2022

Abstract

We determine character tables for twisted fractional linear groups that form the “other” family in Zassenhaus’ classification of finite sharply 3-transitive groups.

2020 MATHEMATICS SUBJECT CLASSIFICATION:

1. Introduction and preliminaries

A classification of all finite, sharply 3-transitive permutation groups follows from a classical result by Zassenhaus [Citation17], by which the only two families of such groups are the fractional linear groups PGL(2,q) for any prime power q and their “twisted” companions M(q2) for any odd prime power q. Recall that PGL(2,q) can be introduced as the group of fractional transformations z(az+b)/(cz+d) of the set GF(q){}, where adbc0, with the obvious rules for calculations with . In the case of a finite field F of the form F=GF(q2) for an odd prime power q one may “twist” the fractional transformations by considering the permutations of F{} defined by z(az+b)/(zc+d) if adbcS(F) and z(azσ+b)/(czσ+d) if adbcN(F), where S(F) and N(F) are the sets of non-zero squares and non-squares of F and σ is the unique involutory (Galois) automorphism of F. The collection of all such “untwisted” and “twisted” fractional transformations under composition constitutes the twisted fractional linear group M(q2).

The notation M(q2) comes from the monographs [Citation12, p. 261] and [Citation6, p. 163], and is also used in the textbooks [Citation13, p. 188] and [Citation14, p. 283]; the letter M was introduced in the original article [Citation17, p. 36] as a tribute to Mathieu who discovered the first group in this series (for q = 3, of degree 10). A possible alternative notation could be derived from the existence of just three non-trivial 2-extensions of PSL(2,q2) by outer automorphisms—a diagonal automorphism δ, the Galois automorphism σ, and their product δσ—leading to the groups PGL(2,q2),PΣL(2,q2) and M(q2)PSL(2,q2)δσ, see [Citation16].

While the fractional linear groups PGL(2,q) and their subgroups PSL(2,q) have been studied widely, their twisted versions M(q2) have received comparatively less attention, and this also applies to characters. A number of resources on representations and characters of fractional linear groups are accessible (either in form of explicit tables or in terms of methods of their derivation, cf. [Citation2, Citation9]). Character tables for the “other” family of sharply 3-transitive groups, by contrast, do not appear to be available.

Among numerous applications of characters we mention here the ones related to the theory of regular hypermaps on compact surfaces, or, equivalently, finite groups generated by three involutions. In [Citation1], enumeration of regular hypermaps of a given type on fractional linear groups was obtained with the help of the Frobenius’ character formula [Citation4] for counting tuples of group elements with entries in given conjugacy classes; more applications of this type can be found in [Citation8]. There is, however, no corresponding result for regular hypermaps on twisted fractional linear groups, although enumeration of regular maps (of unspecified type) on these groups can be found in [Citation3].

The purpose of this paper is to calculate character tables for the twisted fractional linear groups M(q2) for any odd prime power q. This will be done in a completely elementary way by means of standard results in representation theory, by inducing characters from subgroups of M(q2) and finding their decomposition into irreducible constituents.

To avoid fractional transformations we will work with a representation of M(q2) used in [Citation3]), which also differs from the one of [Citation16] and which will prove more suitable for our purposes. For F=GF(q2), q an odd prime power, we let J=GL(2,F)⋊ σ, with σ identified with the additive group C2 in the obvious way and with multiplication given by (A,r)(B,s)=(ABσr,r+s), where Bσ is obtained from B by applying σ to every entry. For each AGL(2,F) let ιAC2={0,1} be defined by ιA=0 if det(A)S(F) and ιA=1 if det(A)N(F). The “twisted” subgroup K of J of index 2 is defined by letting K={(A,ιA);AGL(2,F)} with multiplication as before, that is, (A,ιA)(B,ιB)=(ABσιA,ιA+ιB) for every A,BGL(2,F).

Consider now the subgroup K0={(A,0);AGL(2,F),ιA=0} of K of index 2. The center L of K0 consists of pairs (D,0), where DGL(F) is a scalar matrix; note that L is also normal in both K and J. The factor group G=K/L turns out to be isomorphic to M(q2). We will identify G with M(q2) throughout. This way, G can be regarded as a subgroup of index 2 of the group G¯=J/L, and the factor group H=K0/L can be identified with PSL(2,F). Elements (A,ιA)L, that is, cosets {(δA,ιA);δF*} of the factor group G=K/L, will throughout be denoted [A,ιA]; they will be called untwisted if ιA=0 and twisted if ιA=1.

2. Conjugacy classes of M(q2)

To be in position to consider characters of the twisted group M(q2)=G=K/L we will need to determine conjugacy classes of G. This was done in [Citation3] for conjugacy of twisted elements with respect to G¯, and it turns out that the detailed analysis therein furnishes all one needs to determine the conjugacy within G, a subgroup of G¯ of index two. We sum up the corresponding results in what follows, using notation and machinery of [Citation3]. For any a,bF we let dia(a,b) and off(a,b) be the 2 × 2 matrix over F with, respectively, the diagonal and off-diagonal entries a, b and with remaining entries equal to zero.

We begin by conjugacy of untwisted elements of G=M(q2), forming the subgroup HPSL(2,q2) of G. Every untwisted element of G can be identified with [A,0] for APSL(2,q2), or, for short, just with A, tacitly assuming that writing APSL(2,q2) means ±A for ASL(2,q2) as usual. Since conjugacy classes in PSL(2,q2)H are well known, the question is which pairs are fused by conjugacy in G. Let ξ be a primitive element of F=GF(q2) and let ζ be a primitive (q2+1)th root of unity in an extension F of F of degree two. Further, let B1 and Bε be the elements of PSL(2,q2) obtained from the identity matrix by replacing the top right 0 with 1 and with some εN(F), respectively. Our calculations will, in spirit, be similar to those in [Citation16, Lemma 4.7].

A nonidentity element APSL(2,q2)H is conjugate in H to either w=off(1,1), or to one of u = B1 and u=Bε, or to a(θ)=dia(0,θ+θ1)+off(1,1) for some θ that is a power of ξ or ζ; observe that one always has θ+θ1F even though ζFF. In the first three cases the elements w, u and u generate a single conjugacy class each. Since the element dia(i,i) for i a primitive 4th root of unity in F conjugates a(θ) to a(θ) and obviously a(θ)=a(θ1), the elements a(θ) generate (q25)/4 distinct conjugacy classes for θ=ξj,1j(q25)/4, and (q21)/4 distinct conjugacy classes for θ=ζj,1j(q21)/4. The corresponding centralizers in G of the above five types of elements have orders q21, q2, q2, (q21)/2, and (q2+1)/2, respectively. (These facts are also included in of Section 3 displaying the irreducible characters of H.)

Table 3. Table of irreducible characters of HPSL(2,q2).

With the help of this we determine conjugacy classes of untwisted elements in the overgroup G of H of index two. Conjugacy in G fuses the classes generated by the untwisted elements [u,0] and [u,0] into a single class (e.g. by conjugating by the twisted element [dia(ε,1),1]), still with centralizer of order q2, while the class generated by [w,0] remains the same under conjugacy in G, with centralizer of twice the original order, that is, 2(q21). To sort out untwisted elements [a(θ),0], observe first that the twisted element [dia(ε,1),1] conjugates [a(θ),0] to [a(θσ),0]. To proceed, we will use the observation from [Citation3] that two untwisted elements [A,0] and [A,0] of H are conjugate by a twisted element if and only if the traces of A and A (defined up to multiplication by −1) satisfy tr(A)=±(tr(A))σ.

This implies that two conjugacy classes in H of untwisted elements [a(θ),0] and [a(θσ),0] are fused by conjugation in G unless [a(θσ),0] itself is contained in the H-conjugacy class of [a(θ),0]. But the latter means that a(θ) and a(θσ) are conjugate elements of PSL(2,q2), which is if and only if the traces of the two elements are the same up to a sign, that is, θσ+(θσ)1=±(θ+θ1). Since taking σ-images means raising in the power of q, this equation reduces to θq+θq(θ+θ1)=0, which is equivalent to (θq+11)(θq11)=0. This happens if and only if θ is one of the (q1)th or (q+1)th root of 1 or −1 in F=GF(q2); in each case there are, respectively, q − 1 and q + 1 such roots in F. These are the situations when the G-conjugacy class of [a(θ),0] coincides with that of [a(θσ),0]; in all the remaining cases, i.e., when θσ=θq{±θ,±θ1}, the two classes are fused by conjugacy in G.

Taking this further, if a representative [a(θ),0] of a non-fused conjugacy class has been chosen for a particular θ we may automatically disregard the values θ and ±θ1. Note that if θ2=1 then the matrices a(θ) and w are conjugate in PSL(2,q2), and if θ2=1 then a(θ) represents the identity element of PSL(2,q2). Excluding the values of θ for which θ4=1, it follows that in the case when q1 mod 4 this leaves, respectively, only (q5)/4,(q1)/4,(q1)/4, and (q1)/4 values of θ such that θq1=1,θq1=1,θq+1=1, and θq+1=1, with the property that the conjugacy classes of elements [a(θ),0] considered above are mutually distinct. Similarly, if q3 mod 4, there are only (q3)/4,(q3)/4,(q3)/4, and (q+1)/4 values of θ with θq1=1,θq1=1,θq+1=1, and θq+1=1, respectively, with distinct conjugacy classes of elements [a(θ),0] as above. In both cases, one has a total of 4q8 such distinct values of θ giving distinct conjugacy classes; recall that the four 4th roots of 1 do not contribute to the classes considered here.

The set S={θF;θq±1±1} is therefore of size (q21)(4q8)4=(q1)(q3); observe that this number is a multiple of 8 as q21 mod 4. The important property of the set S is that it admits a partition into subsets of size 8 of the form {±θ,±θ1,±θq,±θq}; the fact that all these 8 elements are distinct follows from the way S has been defined. Let S be an arbitrary but fixed set of distinct representatives of the partition just described, so that |S|=|S|/8=(q1)(q3)/8, and for a fixed primitive element ξF let U={j;1jq22;ξjS}. Similarly, consider the set T={θF{±1};θq2+1=1} with |T|=q21, which is again a multiple of 8 (exclusion of those θ for which θ2=±1 now reduces to leaving out ±1 only); all θT are powers of the primitive (q2+1)th root of unity ζ introduced earlier. The set T also admits a partition into 8-element subset of exactly the same shape as before, and we let T be any fixed set of distinct representatives of this partition, with |T|=(q21)/8. Finally, let V={j;1jq22;ζjT}.

Using the above facts and working out the values of θ in the case θq±1=±1 as powers of ξ we arrive at for the 1+(q+1)2/4 conjugacy classes of untwisted elements of G.

Table 1. Table of conjugacy classes of untwisted elements of G=M(q2).

We proceed by determining conjugacy classes of twisted elements in G=M(q2). By [Citation3, Propositions 5 and 6], every twisted element [A,1]G is conjugate in G to an element of the form [B,1] such that B=dia(θ,1) or B=off(θ,1) for some θN(F). Conjugacy of twisted elements of G in the overgroup G¯ was further investigated in detail in Propositions 7 and 8 and Theorem 1 of [Citation3]. An inspection of the proofs of the three results with emphasis on comparison of conjugacy in G¯ and G leads to the following.

Proposition 1.

Let ξ be a primitive element of F and let [A,1] be a twisted element of G. Then, exactly one of the following two cases occur:

  1. There exists exactly one odd j{1,2,,q2} such that [A,1] is conjugate in G to [B,1] with B=dia(ξj,1), of order 2(q1)/gcd{q1,j}. The stabilizer of [B,1] in G is isomorphic to the cyclic group C2(q1) generated by (conjugation by) a twisted element [P,1]G with P=dia(ξj,1) for j=j12(j1)(q+1).

  2. There exists exactly one odd j{1,2,,q} such that [A,1] is conjugate in G to [C,1] with C=off(ξj,1), of order 2(q+1)/gcd{q+1,j}. The stabilizer of [C,1] in G is isomorphic to the cyclic group C2(q+1) generated by (conjugation by) a twisted element [Q,1]G with Q=off(ξj,1) for j=j+12(j1)(q1).

Proof.

As indicated, a proof can be obtained in an almost verbatim way from the statements and the proofs of Propositions 7 and 8, and Theorem 1, of [Citation3]. For readers interested in checking the details we mark here the differences in these proofs that are significant for distinction between conjugacy in G and G¯.

Let ξ be a primitive element of F. In the cases 1 and 2 of Proposition 7, twisted elements [B,1] and [B,1] with B=dia(θ,1) and B=dia(θ,1) for θ,θN(F) are conjugate in G¯ but not in G if, and only if, the ratio θ/θ in the case 1, and the product θθ in the case 2, have the form ξt(q1) for t odd. Similarly, in cases 1 and 2 of Proposition 8, elements [B,1] and [B,1] for B=off(θ,1) and B=off(θ,1) are conjugate in G¯ but not in G if, and only if, θ/θ in the case 1, and θθ in the case 2, have the form ξt(q+1) for t odd (i.e., they are non-squares in the unique subfield of F of order q). All the remaining facts in the proofs of Propositions 7 and 8 apply to conjugacy in G.

Oddness of the values of t above has the following two consequences in the proof of Theorem 1 of [Citation3] for restriction to conjugacy in G. First, the congruences appearing in parts 1 and 2 of the proof have to be taken mod 2(q1) and 2(q+1) and not just mod (q1) and (q+1), respectively, leading to upper bounds for j in parts 1 and 2 of our proposition that are two times larger than the bounds in [Citation3, Theorem 1]. Second, parts 3 and 4 of the statement of Theorem 1 in [Citation3] refer to special cases arising due to the presence of exceptional conjugating elements in the corresponding parts of the proof. These need not be considered in our proposition, because the exceptional conjugating elements turn out to lie outside G. Again, the remaining arguments apply to conjugacy in G. □

By Proposition 1, there are a total of q conjugacy classes of twisted elements in G, consisting of (q1)/2 classes of “diagonal type” described in part 1, and (q+1)/2 classes of “off-diagonal type” from part 2, with centralizers of order 2(q1) and 2(q+1), respectively. The explicit form of the twisted representatives from Proposition 1 then immediately gives our displaying conjugacy classes of twisted elements in G.

Table 2. Table of conjugacy classes of twisted elements of G=M(q2).

3. Preliminary results on characters of M(q2)

Summing up the results of and , the group G=M(q2) splits into a total of (q+1)(q+5)/4 conjugacy classes. Note that in G the number of conjugacy classes of twisted elements, q, is by an order of magnitude smaller than the number of conjugacy classes of untwisted elements, q2/4, so that one may attempt to determine the character table of G by inducing characters from the index-two subgroup HPSL(2,q2) of G. The character table of PSL(2,q2) is known and we will reproduce here a modification of the one from [Citation9, pp. 147–148], see .

The symbols B1, Bε, w, a(ξj), and a(ζk) used in for representatives of conjugacy classes are the same as explained at the beginning of Section 2, and |CH| is the order of the centralizer of the corresponding element in H. The irreducible characters are ι (the trivial one), St (the Steinberg permutation character), ρ, ρ,ρ for 1(q25)/4, and πm for 1m(q21)/4. As before, ξ and ζ are a primitive element of F and a primitive (q2+1)st root of 1 in F, and the powers j and k in ξj and ζk are bounded by 1j(q25)/4 and 1k(q21)/4. Finally, in a somewhat nonstandard notation but with i denoting the complex imaginary unit as usual, α=exp(4πi/(q21)) and β=exp(4πi/(q2+1)) are complex primitive roots of unity of order (q21)/2 and (q2+1)/2, respectively.

Observe that PSL(2,q2) has a total of (q2+5)/2 conjugacy classes. This is roughly twice the number of conjugacy classes of M(q2) for large q. An explanation offered by the previous two sections is that G-conjugation fuses “most” pairs of H-conjugacy classes of untwisted elements in H but there are only q conjugacy classes of twisted elements of G.

The degrees of irreducible characters of PSL(2,q2) follow from , and those of GM(q2) have been determined in [Citation16]. For convenience we display both in tabular form, with the proviso that degree 20 cannot occur in the exceptional case when q = 3.

For reference to standard concepts and results in the theory of group characters we will use the monograph [Citation7]. We will focus on results on characters of a group G with a normal subgroup H of index 2; up to the end of this section the pair G, H may be arbitrary but later we will return to our situation of G and H standing for M(q2) and PSL(2,q2).

The restriction χH of a character χ of G to H is the character of H defined by χH(g)=χ(g) if gH and χH(g)=0 for gGH. For the reverse direction of inducing characters of G from those of H, let CH(h) and CG(h) denote the centralizers of an element hH in H and G. If φ is a character of an index-two subgroup H of G, the corresponding induced character φG of G is given as follows (for some fixed gGH): φG(h)={φ(h)+φ(ghg1) if hH and CH(h)=CG(h);2φ(h) if hH and CH(h)CG(h);0 if hGH. We note that if CH(h)=CG(h), then conjugacy in G fuses a pair of distinct H-conjugacy classes C ∋ h and C ∋ ghg1 in H to a single G-conjugacy class (still in H); each such unordered pair {C,C} will be called a fusion pair.

For brevity we will refer to the value of the standard inner product χ,χG for a character χ of G as the norm of χ, denoted by ||χ||G. A similar notation will be used for the norm of characters of H, and we will drop the subscript if the group is clear from the context. It is well known that χ is irreducible if and only if it has norm 1.

The following is a short summary of some results and observations of [Citation7, Chapters 5 and 6] we will need later. As usual, we let λ denote the alternating character of G, with values 1 on elements of H and −1 on elements in GH.

Lemma 1.

Let H be a normal subgroup of G of index 2 and let φ be an irreducible character of H. Then, ||φG||{1,2}, with ||φG||=1 if and only if φG is an irreducible character of G, and ||φG||=2 if and only if for every fusion pair (C,C) of H-conjugacy classes in H the values of φ on CC are constant (i.e., if φ is G-invariant); in such a case φG=χ+χλ for some irreducible character χ of G. □

Under the hypotheses of Lemma 1 this means that determination of the character table of G out of that of H reduces to understanding decomposition of characters induced by the G-invariant irreducible characters of H. This is what will be done in the next sections for HPSL(2,q2) and GM(q2).

4. Inducing characters from PSL(2,q2) to M(q2)

We begin this section by identifying the irreducible characters of G of degree 1 and q2, the unique corresponding characters of H being the trivial character ι and the Steinberg character St. Recall that the (irreducible) Steinberg character of a finite 2-transitive permutation group is evaluated at any permutation in the group by subtracting 1 from the number of fixed points of the permutation. The standard actions of the groups HPSL(2,q2) and G=M(q2), on the q2+1 projective points are both 2-transitive, and even sharply 3-transitive in the case of G. It is easy to verify that the values of the Steinberg character of G, which we will denote Σ, are the same as those of St on conjugacy classes of untwisted elements, and + 1 and −1 on conjugacy classes of twisted diagonal and twisted off-diagonal elements, respectively.

Lemma 2.

Let φ be an irreducible character of H. If deg(φ)=1, then φG=ι+λ, and ι with λ are the only irreducible characters of G of degree 1. If deg(φ)=q2, then one has φG= Σ + Σλ, where Σ is the Steinberg permutation character, and Σ with Σλ are the only irreducible characters of G of degree q2.

Proof.

By in both cases we have φG=χ+χ for irreducible characters χ and χ of G, so that Lemma 1 applies and φG=χ+χλ. The rest is a consequence of uniqueness of irreducible characters of H of degree 1 and q2. □

Table 4. Degrees of irreducible characters of HPSL(2,q2) and GM(q2).

With the help of Lemma 2 and known facts from character theory we are in position to determine the number of irreducible characters of G of degrees appearing .

Lemma 3.

The number of irreducible characters of G=M(q2) of a given degree are as follows:

Proof.

By Lemma 2 we know that the number of irreducible characters of G of degree 1 and q2 is 2 in both cases. Further, by there are only three remaining degrees of irreducible characters of G, namely, 2(q21),q2+1 and 2(q2+1); let c1,c2,c3, respectively, be the numbers of such characters. Now, 2+2+c1+c2+c3=(q+1)(q+5)/4 is the number of conjugacy classes in G, and as the sum of squares of all character degrees is equal to |G|2, we also have 2×12+2×q4+c1×4(q21)2+c2×(q2+1)2+c3×4(q2+1)2=q2(q41) which can be shown to be equivalent to (1) ((c2+4c3)(q2+1)+2)(q2+1)=(q21)2(q24c1)(1) Since (q21)/2 and (q2+1)/2 are relatively prime and the second one is odd, by the factorization appearing in (Equation1) the number (q2+1)/2 must divide the (odd) number q24c1, and so q2+1 is a divisor of 2(q24c1). This, however, is possible only if the two numbers are equal, that is, q2+1=2(q24c1), which is if and only if c1=(q21)/8. Having determined the value of c1 we are left with a system of two equations in two unknowns and it is easy to check that its unique solution if c2=2q3 and c3=(q1)(q3)/8.

We are in position to give substantial information about inducing irreducible characters from H to G. Its statement includes remarks on integrality of values of the characters, and refers to the notation used in the table of conjugacy classes of untwisted elements of M(q2) (, in particular, the sets U and V) and in of characters of PSL(2,q2); the first two assertions from Lemma 2 are included for completeness.

Proposition 2.

Let φ be an irreducible character of HPSL(2,q2) with the induced character φG of GM(q2). Then, exactly one of the following cases occurs:

  1. deg(φG)=2 and φG=ι+λ, with ι and λ being the only irreducible characters of G of degree 1; they have integral values and are non-zero on GH;

  2. deg(φG)=2q2 and φG= Σ + Σλ, where Σ is the Steinberg permutation character, and Σ with Σλ are the only irreducible characters of G of degree q2; they are again integral and have non-zero values on GH;

  3. deg(φG)=2(q21) and φG is one of the (q21)/8 irreducible characters of G of degree 2(q21) such that (φG)H=πm+πmq for mV; these characters φG are all real and with all-zero values on GH;

  4. deg(φG)=q2+1 and φG is a unique irreducible character of G of degree q2+1 with (φG)H=ρ+ρ; it is integral and identically zero on GH;

  5. deg(φG)=2(q2+1) and φG is one of the (q1)(q3)/8 irreducible characters of G of degree 2(q2+1) such that (φG)H=ρ+ρq for U; these φG are all real and all-zero on GH;

  6. deg(φG)=2(q2+1) and φG=χ+χλ for q – 2 pairs χχλ of irreducible characters of G of degree q2+1, with χH=ρ for the (q3)/2 values =r(q+1)/2 such that 1r(q3)/2, and the (q1)/2 values =s(q1)/2,1s(q1)/2; there is a total of 2q4 such distinct irreducible characters of G and each of these have at least one non-zero value on GH.

Proof.

As noted, we may skip the first two items, and among the remaining ones we begin with (Equation4). If deg(φG)=q2+1, then φG is necessarily an irreducible character of G such that (φG)H is the sum of two distinct irreducible characters of H of degree (q2+1)/2. But by there are only two such characters of H, namely, ρ and ρ, and one may check that they both induce the same character of G, giving the conclusion of (Equation4).

Next, we turn our attention to (Equation6), where φG is assumed to be reducible and of degree 2(q2+1). By Lemma 1, we have φG=χ+χλ for some irreducible character χ of G of degree q2+1, with χH also irreducible. This means that χH must be one of the characters ρ for a suitable , and the same applies to φ, of course. But ||φG||=2, so that by Lemma 1 the character φ=ρ must be constant on any fusion pair of H-conjugacy classes. By the findings of Section 2 applied to this case, the H-conjugacy classes generated by the elements [a(ξj),0] and [a(ξjq),0] for j{1,,(q25)/4} form a fusion pair if they are distinct. The previous condition therefore means that the values of ρ and ρq on every such pair of classes must be the same (so that we may ignore distinctness here). By this translates, for a fixed , to the equality of the complex numbers αj+αj and αjq+αjq for all j as above.

A simple calculation (as in Section 2) reveals that the two complex numbers coincide if and only if (αj(q+1)1)(αj(q1)1)=0 (or, equivalently, αjq{αj,αj}). To fulfill the condition on the constant value on fusion classes we are looking for the values of {1,,(q25)/4} for which the last equation holds for every j{1,,(q25)/4}, which happens if and only if is a multiple of (q+1)/2 or (q1)/2. In our range 1(q25)/4 given by this yields the q − 2 values of appearing in the statement of (Equation6), and hence also the q − 2 possibilities for (φG)H=ρ+ρq=2ρ. The latter give 2(q2) irreducible characters χ and χλ of degree q2+1, all with some non-zero value on GH, and such that χH=(χλ)H=ρ. Since the total number of irreducible characters of G of degree q2+1 is 2q3 by Lemma 3 and one of such characters was identified in part (Equation4), the total number of irreducible characters of G referred to in (Equation6) is 2q4, as claimed.

We continue with (Equation5), assuming that φG is irreducible and has degree 2(q2+1). The restriction of φ to H cannot be irreducible, so that (φG)H is a sum of two distinct irreducible characters of H of degree q2+1. Such characters are all of the form ρ for suitable values of , and φ in this case must also be equal to one of these. Further, our assumption on the degree of φG together with Lemma 1 imply that there must be at least one fusion pair of H-conjugacy classes such that φ is not constant on the pair. The fusion pairs here are generated by the same classes as in (Equation6), and our calculations in the proof of (Equation6) imply that a fusion pair on which φ is not constant exists if and only if is not one of the values listed in the conclusion of (Equation6). One may check that this reduces to the condition U for the set U introduced in Section 2, with (φG)H=ρ+ρq for U. (We note again that all the accompanying calculations are analogous to those in Section 2 where the condition θq{±θ,±θ1} was considered. Here the condition on α translates to αjq{αj,αj} and leads to the same conclusion that U because the expression (φG)H=ρ+ρq is invariant under the substitutions and q; the ±1 term is absorbed by the factor 4 in α=exp(4πi/(q21)).)

By it may be verified that if φ=ρ for U, then (φG)H=ρ+ρq, so that ρ and ρq induce the same character of G. Referring again to and one obtains this way a total of |U|=(q1)(q3)/8 irreducible characters φG of G whose restriction to H has the form ρ+ρq for U. All such induced characters are distinct, and by Lemma 3 there cannot be any other irreducible character of G of degree 2(q2+1).

Finally, let us consider (Equation3), assuming that deg(φG)=2(q21). By , φG must be an irreducible character of G, and as its restriction to H cannot be irreducible, (φG)H is a sum of two distinct irreducible characters of H of degree q21. The latter are all of the form πm for suitable values of m, and φ must also be of this form. In Section 2, we showed that the H-conjugacy classes of the elements [a(ζj),0] and [a(ζjq),0] form a fusion pair for every jV. One may check that the values of any given πm are non-constant on at least one of these fusion pairs, which conforms to the last part of Lemma 1. Further, a direct verification against shows that if φ=πm, then (φG)H=πm+πmq. Thus, both πm and πmq determine the same induced character of G, and by and they give rise to (q21)/8 irreducible characters of G of the form πm+πmq corresponding to the values mV. It may be verified that these induced characters are distinct, and by Lemma 3 there is no other irreducible character of G of degree 2(q21).

As regards remarks on integral and real values, most of them are obvious, and note that φG is always identically zero on GH. This completes the proof. □

An inspection of the numbers in Proposition 2 reveals that in the cases (Equation1)–(Equation5) there are, respectively, 2+2+(q21)/8+1+(q1)(q3)/8 real characters, and all of them have been completely determined. This makes a total of (q22q+21)/4 irreducible characters of G that are all real. Subtracting this from the number of all irreducible characters of G leaves us only with the 2q4 ones described in part (Equation6) of Proposition 2, that may assume non-real values, but only on GH as all the characters of H are real.

We now address the question of possible character values that are not real. A well-known result in the theory of group characters is that, for an element g of a group, the values of all irreducible characters of the group evaluated at g are real if and only if g is conjugate to g1 in the group, or, equivalently, the conjugacy class of g is closed under inversion in the group; such conjugacy classes are called real. By every conjugacy class of untwisted elements in G=M(q2) is real. For conjugacy of twisted elements we have, in the notation of Proposition 1:

Lemma 4.

The conjugacy class of the twisted element [dia(ξj,1),1] in G=M(q2) for odd j, 1jq2, is real if and only if q3 mod 4 and j=(q1)/2. The conjugacy class of [off(ξj,1),1] for odd j, 1jq, is real if and only if q1 mod 4 and j=(q+1)/2.

Proof.

Let d(j)=[dia(ξj,1),1] and d(j)=[off(ξj,1),1]. Observe first that d(j)1=d(jq) and d(j)1=d(jq). By Proposition 1 the elements d(j) and d(jq) are in the same conjugacy class in G for odd j, 1jq2, if and only if jjq mod 2(q1). This can only happen if q3 mod 4, and then the congruence is equivalent to j(q+1)/40 mod (q1)/2. As the modulus of the last congruence is relatively prime to (q+1)/4 it follows that this only leaves us with the value j=(q1)/2 in our range for j. The analysis for the elements d(j) and d(jq) is analogous. □

This gives, in both cases mod 4, exactly q − 1 conjugacy classes of twisted elements in G that are not real. By the earlier remarks we also know that there are at most 2q4 irreducible characters of G (those from part (Equation6) of Proposition 2) that are not real.

5. Representations for the remaining characters

In the previous section, we have almost completely determined the character table of G=M(q2) and we have been left with 2q4 “missing” irreducible characters, which are the only ones that may assume non-real values on q − 1 conjugacy classes of G formed by elements of GH. These characters have been referred to in part (Equation6) of Proposition 2 and from now on we will denote them by χ and χλ, with (χ)H=(χλ)H=ρ, for the total of q − 2 values of in the set L=L+L, where L+={r(q+1)/2;1r(q3)/2} and L={s(q1)/2;1s(q1)/2}. We will determine these characters in the next section; here we first derive some related representations by extending one-dimensional representations of a suitable subgroup of H. The method is an adaptation of derivation of principal series representations for two-dimensional special linear groups (cf. [Citation2, p. 232]).

Let Hupp be the subgroup of H < G, HPSL(2,q2), induced by upper-triangular matrices with determinant 1. Explicitly, if ξF*=F{0} is a fixed primitive element as before and if h(u, d) is the 2 × 2 matrix with first and second row of the form (ξu,d) and (0,ξu) for an arbitrary non-negative integer u<(q21)/2 and any dF, then (2) Hupp={[h(u,d),0]G;0u<(q21)/2,dF};(2) observe that |H|=q2(q21)/2. It is well known (see e.g. [Citation2, p. 232] adapted to the projective case) that for every L (and, in fact, for every integer but this will not be needed here) the assignment (3) Φ:[h(u,d),0]exp(4πiq21u)(3) defines a one-dimensional non-real representation of Hupp. To extend such a representation to one of the entire group G=M(q2) we proceed as ibid by first constructing a suitable set of coset representatives of Hupp in G. To do this, for 0t<q21 we introduce 2 × 2 lower-triangular matrices m(t) and m(t), as well as matrices m() and m(), as follows: (4) m(t)=(10ξt1),m(t)=(ξ0ξt1),m()=(0110),m()=(0ξ10).(4) Using the 2q2 matrices from (Equation4) we introduce 2(q2+1) elements of G by letting xt=[m(t),0],andyt=[m(t),1]for0t<q21,together withx=[m(),0],y=[m(),1],x=[I,0],andy=[dia(ξ,1),1]; note that x is the unit element of G. It may be checked that this set of nq=2(q2+1) elements of G is a left transversal for the subgroup Hupp.

With the help of this transversal we will now extend any one-dimensional representations Φ{Φ;L} of Hupp described in (Equation3) to an nq-dimensional representation ΦG of G; the method originates from [Citation5]. Before doing so we will make an agreement about indexation. Let Ind={0,1,,q22,,}, where the entries 0,1,,q22 are considered mod q21, and let Ind={z;zInd}. To describe nq×nq matrices we will use the nq indices from the set IndInd equipped with the linear ordering (5) 0<1<<(q22)<0<1<<(q22)<<<<.(5)

Invoking now [Citation2, Lemma 9.1] adapted to our situation and using the introduced notation, an nq-dimensional representation ΦG is obtained by assigning, to every element gG the nq×nq matrix ΦG(g) whose (a, b)th entry for a,bIndInd is determined by the following rules (using the convention that (z)=z for our indices): ΦG(g)a,b={Φ(xagxb1) if gH,a,bInd and xagxb1Hupp;Φ(yagyb1) if gH,a,bInd and yagyb1Hupp;Φ(xagyb1) if gGH,aInd,bInd and xagyb1Hupp;Φ(yagxb1) if gGH,aInd,bInd and yagxb1Hupp;0 in all other cases .

By part (Equation6) of Proposition 2, each induced representation ΦG for Φ{Φ;L+L} is reducible and splits into two irreducible representations of dimension q2+1 each. The missing irreducible characters χ and χλ on GH for L+L are given by traces of these representations, or, equivalently, by traces corresponding to the two G-invariant (q2+1)-dimensional subspaces of the representation ΦG.

For evaluation of these traces (which we will do in the next section) we will need explicit knowledge of ΦG-images of a generating set of G=M(q2) and its subgroup HPSL(2,q2). It is well known that HPSL(2,q2) is generated by the two elements g1=[dia(ξ,ξ1),0] and g2=[dia(1,0)+off(1,1),0], of order (q21)/2 and 3, respectively. In accordance with , as representatives of conjugacy classes of M(q2)H we will take the elements g3=g3(j)=[dia(ξj,1),1] and g4=g4(j)=[off(ξj,1),1] for odd positive integers jq2 and jq, respectively. For the ΦG-images of these elements we obtain:

Proposition 3.

For every L the values of the induced representation ΦG at the elements g1, g2, g3, and g4 are nq×nq unitary matrices with entries as follows, where γ=exp(2πi/(q21))=exp(πir/(q1)), and indices in IndInd distinct from ,,, are understood mod q21:

  1. If g=g1=[dia(ξ,ξ1),0], then ΦG(g)a,b={γ2 if (a,b){(,),(t,t+2),0tq22},γ2q if (a,b){(,),(t,(t+2q)),0tq22},γ2 if (a,b)=(,),γ2q if (a,b)=(,),0 in all other cases .

  2. If g=g2=[dia(1,0)+off(1,1),0], then, letting q¯=(q21)/2, ΦG(g)a,b={γ2a if (a,b)=(t,f(t))or(t,f(t)),0tq22andtq¯,1 if (a,b)or(a,b)isintheset{(q¯,),(,),(,q¯)},0 in all the remaining cases,  where the function f on residue classes t mod q21 and tq¯ is given by ξt+ξf(t)=1.

  3. If g=g3=g3(j)=[dia(ξj,1),1] for an odd positive integer jq2, then ΦG(g)a,b={γj1 if a=t and b=(t+j),0tq22,or(a,b)=(,),γjq+1 if a=t and b=t+jq,0tq22,or(a,b)=(,),γj1 if (a,b)=(,),γjq+1 if (a,b)=(,),0 in all other cases.

  4. If g=g4=g4(j)=[off(ξj,1),1] for an odd positive integer jq, then ΦG(g)a,b={(1)γj2t1 if a=t and b=(tj),0tq22,(1)γjq2t+1 if a=t and b=tjq,0tq22,(1)γj1 if (a,b)=(,),(1)γj1 if (a,b)=(,),(1)γjq+1 if (a,b)=(,),(1)γjq+1 if (a,b)=(,),0 in all the remaining cases.

Proof.

For illustration we will only deal with the entire part (b) and the second items of both (c) and (d), as the verification in all the remaining cases is analogous; the matrices are unitary by inspection and hence so is the entire induced representation ΦG. For all calculations we note that ΦG was introduced before the statement of Proposition 3, preceded by an exposition of the associated matrices in (Equation4) together with the elements xa and yb.

Part (b). Letting g=g2=[dia(1,0)+off(1,1),0], for a,b{0,1,,t22} one has ΦG(g)a,b=Φ(xagyb1) if xagyb1Hupp; otherwise ΦG(g)a,b=0. By the rules (explained in Section 1) for calculations in our group G, the inverse of yt=[m(t),1] is yt1=[n(t)σ,1] where n(t) is the 2 × 2 matrix with rows (ξ1,0) and (ξt1,1). Following these rules (and slightly abusing the notation and using a,b{,} also as exponents at ξ), the product X=xagxb1 evaluates to X=(10ξa1)(1110)(10ξb1)=(ξb+11ξa+ξa+b+1ξa). The condition XHupp is equivalent to ξa+ξb=1, which, for aq¯, determines b as a function f(a) from the equation ξa+ξf(a)=1. In such a case the main diagonal of X consists of the elements 1+ξf(a)=ξa and ξa, so that by (Equation3) the value of ΦG(g)a,f(a) is equal to γ2a. One may check that for the pair (a,b),0a,bq22, evaluation of the product yagyb1 reduces to determining membership of m(a)gn(b) in Hupp and gives the same condition on a, b as above, that is, b=f(a) for a{q¯,,}, with the same value γ2a of ΦG(g)a,f(a). Calculating all of xq¯gx1,xgx1,xgx1 and their y-versions one obtains matrices with main diagonal 1, 1, implying the second item of (b).

Regarding the function f, mutual equivalence of the three equations ξa+ξb+1=0,ξb+ξab+1=0, and ξa+b+ξa+1=0 implies that f(a) = b, f(b)=ab, and f(ab)=a for a{q¯,,}. Since raising in qth power is an automorphism of F=GF(q2) one also has f(qa)=qf(a). It follows that if q is not a power of 3 then f as a permutation of the undashed indices not in {q¯,,} has two fixed points, namely, the two primitive 3rd roots of 1 in F, and for the remaining values, f consists of 3-cycles of the form (a, b, c) where b=f(a) and a+b+c=0; if q is a power of 3 then f has 0 as its unique fixed point and the remaining orbits of f are 3-cycles as above. An analogous statement applies to dashed indices.

The second item of part (c). For g=g3(j)=[dia(ξj,1),1] finding the value of yagxb1 by the calculation rules in G reduces to evaluating the product X=m(a)·dia(ξjq,1)·m(b)1, which results in X=(ξ0ξa1)(ξjq001)(10ξb1)=(ξjq+10ξjq+aξb1). Here one has XHupp if and only if b=a+jq mod (q21). To evaluate Φ at such an X one needs to represent it in the form [h(u,d),0], which reduces to looking for an element zF such that zX has a pair of mutually inverse entries ξu and ξu in the main diagonal. The obvious choice here is z=ξ(jq+1)/2, giving by (Equation3) the value ΦG(g)a,a+jq=γjq+1.

The second item of part (d). Here, in xagyb1=[m(a)·off(ξj,1)·n(b),0] the product X of the three matrices is X=(10ξa1)(0ξj10)(ξ10ξb11)=(ξb+j1ξjξ1ξa+b+j1ξa+j). Now, X belongs to Hupp if and only if b=aj mod (q21), and to determine the corresponding value of Φ we again need to represent this element as [h(u,d),0] by finding some zF such that the main diagonal of zX has the form (ξu,ξu). Using b+j=a and (1)=ξ(q21)/2, multiplication by z=ξ(2j2+q21)/4 produces the required diagonal entries in zX for u=a(j+1)/2(q21)/4. By (Equation3), the value of Φ at [h(u,d),0]Hupp for h(u,d)=zX is then equal to exp(4πiu/(q21)). Finally, with the help of γ=exp(2πi/(q21)) and the substitution for u the value of Φ simplifies to (1)γj2a1, which is as claimed if a = t and b=(tj).

It will be of advantage to restate the above results in form of block matrices. For L let P=P be the matrix of dimension q2+1 indexed by the ordered set Ind with P,=γ1 and P,=Pa,a+1=γ for every aInd{,} mod (q21), and with all the remaining entries equal to zero. We will leave out the subscript if no confusion is likely. Further, let Aupp,Alow, B, Cupp=Cupp(j) and Clow=Clow(j) for odd positive j be square matrices of dimension q2+1, indexed by the ordered set Ind, and defined as follows: (6) Aupp=P2,Alow=P2q=AuppqBa,f(a)=γ2aifaInd{q¯,,},Ba,b=1if(a,b){(q¯,),(,),(,q¯)}Cupp=Cupp(j)=γ1Pj,Clow=Clow(j)=γPjq(6) where entries of B not listed are assumed to be zero; the parameter j will be omitted if no loss of clarity is likely. From now on we will also extend the symbols dia and off also to 2 × 2 blocks of dimension q2+1. One may check that, in this new notation, the matrices A=ΦG(g1),B=ΦG(g2) and C=C(j)=ΦG(g3(j)) from Proposition 3 can be displayed in the form (7) A=dia(Aupp,Alow),B=dia(B,B),C=off(Cupp,Clow)(7) where we assume a natural extension of indexation from Ind to Ind′ in the bottom q2+1 coordinates of the “larger” matrices of dimension 2(q2+1); this also justifies the usage of the subscripts “upp” and “low” for the upper and lower non-zero blocks of the matrices of (Equation6) which appear in (Equation7).

Since the matrices A and B of dimension 2(q2+1) assigned to the generators g1 and g2 of HPSL(2,q2) in the representation ΦG are block-diagonal, it follows that all matrices of ΦG assigned to elements of H are block-diagonal (with diagonal blocks of dimension q2+1). Thus, the restriction ΦH of the unitary representation ΦG to H is a sum of two irreducible (q2+1)-dimensional unitary representations, say, Φ(1) and Φ(2), of the group H; formally, ΦH=Φ(1)Φ(2).

This is a good point to recall that our aim is to determine the 2(q2) characters denoted χ and χλ at the beginning of the previous section (which are the characters of part (Equation6) of Proposition 2) for L with |L|=q2. As the restrictions (χ)H and (χλ)H coincide and are equal to the character ρ of HPSL(2,q2), it follows that in the sum ΦH=Φ(1)Φ(2) the constituents Φ(1) and Φ(2), generated, respectively, by the pairs Aupp,B and Alow,B, must be equivalent unitary irreducible representations of H. This implies the existence of an “intertwining” unitary matrix M=M of dimension q2+1 with the property that Φ(1)(h)M=MΦ(2)(h) for every hH. In particular, using the common notation YM=M1YM for conjugation, for the constituents of A and B in (Equation7) one has (8) Alow=Auppq=AuppMandB=BM.(8) Further, using Aupp=P2,Alow=P2q=Auppq from (Equation6) and realizing that Pq2=P one obtains AuppM2=AlowM=(Auppq)M=(AuppM)q=Alowq=Auppq2=Aupp. Since from (Equation8) we obviously have BM2=B, it follows that M2 commutes with the representation Φ(1). By Schur’s theorem, M2 is a constant multiple of the identity matrix. However, for the purpose of intertwining the constant multiple may be an arbitrary non-zero complex number, which we will henceforth choose to be equal to 1. Thus, without loss of generality we may assume that M2=I, which means that M is a unitary Hermitian matrix; now we also have det(M)=±1.

6. Determination of the remaining characters

We begin this section by determining all the non-trivial G-invariant (and hence (q2+1)-dimensional) subspaces of our 2(q2+1)-dimensional representation ΦG. The method relies on the following result which extends [Citation15, Lemma 2.6, p. 56] and is likely to be folklore in representation theory; we therefore include only a short proof.

Lemma 5.

Let Ψ1Ψ2 be a direct sum of a pair of equivalent non-real irreducible representations of a group K with associated disjoint vector spaces V1 and V2 and with an intertwining matrix N. If a non-trivial subspace V of V1V2 is K-invariant, then either the projection of V onto exactly one of V1, V2 is zero, or there is a non-zero cC such that V={(u,u(cN));uV1}.

Proof.

By default, all of Ψi and Vi for i = 1, 2 as well as V and N must have the same dimension. Letting πi be the projection of V1V2 onto Vi for i = 1, 2, by irreducibility the intersection Vker(πi) is either Vi or trivial. Leaving the possibility when (exactly) one of πi(V) is equal to Vi we are left with the case when both intersections Vker(πi) are trivial. But then, letting πi,V denote the restriction of πi to V, it follows that Vi=πi,V(V)V for i = 1, 2; in particular, the mappings πi,V are K-invariant isomorphisms. The composition Ψ=π1,V1π2,V is a K-invariant isomorphism V1V2, which, by Schur’s theorem, must be given by uΨ=u(cN) for an intertwining matrix N, unique up to a non-zero multiplicative constant. Now, if vV is an arbitrary vector, letting u=vπ1|V we obtain v=(vπ1|V,vπ2|V)=(u,uΨ)=(u,u(cN)).

Applying Lemma 5 to our restricted representation ΦH of the subgroup H < G one sees that an invariant subspace W for ΦH has either zero projection onto the first (undashed) q2+1 coordinates in the set Ind, or a zero projection onto the second (dashed) coordinates in Ind, or else has the form W={(u,cuM);uCq2+1}, where M=M is the intertwining involutory matrix of (Equation8). One of these then must form an invariant subspace for the entire representation ΦG of the group G. Realizing that the representation ΦG is obtained from ΦH by adjoining any of the matrices C from (Equation7) with zero diagonal blocks, the first two possibilities are immediately ruled out, and we obtain:

Corollary 1.

Every non-trivial invariant subspace of ΦG has the form W={(u,cuM); uCq2+1}, where M is the intertwining involutory matrix of the representations Φ(1) and Φ(2) of H generated by Aupp,B and Alow,B, and c is a non-zero complex constant.

By the corollary, for every vector wW, i.e., of the form w=(u,cuM) for uCq2+1, the vector wC=(u,cuM)off(Cupp,Clow)=(c(uM)Clow,uCupp) must also belong to W. This means that uCupp=u(cM)Clow(cM), and as u was arbitrary and M=M1, it follows that (9) Clow=c2CuppM.(9)

With the help of Corollary 1 and the knowledge of Cupp=γ1Pj,Clow=γPjq and Aupp=P2 by (Equation6) we now determine the possible values of the constant c. Namely, squaring the expressions for Cupp and Clow one obtains Cupp2=γ2P2j=γ2AuppjandClow2=γ2P2jq=γ2Auppjq, and substituting these into the square of (Equation9) gives γ2Auppjq=c4γ2(Auppj)M. The last equation reduces by (Equation8), i.e., by AuppM=Auppq, to c4=γ4, so that c=±iδ()γ1 for some δ(){0,1}, where i is the complex imaginary unit. Hence, (Equation9) can equivalently be written in the form (10) Clow=(1)δ()γ2CuppM.(10)

Observe that if c is one of the four values determined above for which the subspace W={(u,cuM);uCq2+1} is G-invariant, then W={(v,cvM);vCq2+1} is another such subspace and the pair (W,W) forms an orthogonal decomposition of C2(q2+1). Indeed, letting * denote the complex conjugate transpose and using the fact that M is unitary (MM*=I) together with cc*=1, evaluating the standard inner product of (complex) vectors from W and W gives (u,c(uM))·(v,c(vM))=uv*cuM((cvM)*=uv*cc*uMM*v*=0.

Consider such a pair W,W of G-invariant subspaces. From the way our induced representation ΦG was introduced before the statement of Proposition 3 and from the calculations in its proof it follows that for every gGH the unitary matrix ΦG(g) has the form E=E(g)=off(Eupp,Elow) with off-diagonal (and necessarily unitary) blocks of dimension q2+1 each. Consider a complex eigenvalue x of E associated with a non-zero eigenvector w, splitting uniquely as w=w1+w2 for w1W and w2W. Now, (w1+w2)E=wE=xw=x(w1+w2) and by C2(q2+1)=WW and the G-invariance of the two subspaces we obtain w1E=xw1 and w2E=xw2. It follows that the eigenspace Eig(E,x) of E associated with the eigenvalue x is a direct sum (Eig(E,x)W)(Eig(E,x)W) of the corresponding spaces of eigenvectors of E that belong to W and W. By diagonalizability of unitary matrices this also means that the spectrum of E, of size 2(q2+1), is a concatenation of the spectra Spec(E,W) and Spec(E,W) of size q2+1 consisting, respectively, of the eigenvalues corresponding to eigenspaces contained in W and W.

Let now w=(u,cuM)W be an eigenvector of E for an eigenvalue x. Since W is G-invariant and hence preserved by E, one has (xu,xcuM)=xw=wE=(cuMElow,uEupp)W. Membership of the last vector in W means that uEupp=cuMElow(cM), and since this holds for every eigenvector u corresponding to any eigenvalue x, we obtain Eupp=c2ElowM, or, equivalently, c*EuppM=cMElow. The chain of displayed equations further gives xu=ucMElow and xcuM=(cuM)cElowM, which means that u and cuM are eigenvectors of the matrices cMElow=c*EuppM and cElowM, respectively, for the same eigenvalue x. Conversely, if u is an eigenvector of c*EuppM=cMElow for an eigenvalue x, then right multiplication by cM shows that cuM is an eigenvector of cMElowcM=Eupp for x and hence x(u,cuM)=(cuMElow,uEupp), that is, w=(u,cuM)W is an eigenvector of E for the same eigenvalue x.

This correspondence between the eigenspaces of E that are subspaces of W and eigenspaces of the (unitary and hence diagonalizable) matrix cMElow=c*EuppM together with the earlier established facts about spectra lead to the conclusion that the multi-set Spec(E,W) is equal to the spectrum of cMElow=c*EuppM. This way we have arrived at the following conclusion for our missing character χ for L, which we state as a summary of the above considerations together with (Equation8) and (Equation10).

Proposition 4.

Let M be a unitary Hermitian matrix of dimension q2+1 with determinant ±1 such that (11) Alow=AuppM,B=BMandClow=c2CuppM=(1)δ()γ2CuppM(11) for some c{±iδ()γ1}. For an arbitrary gGH let ΦG(g)=off(Eupp(g),Elow(g)). Then, the pair of characters χ and χλ for each L is determined by letting χ(g)=tr(cMElow(g))=tr(c*Eupp(g)M) for every gGH.

Proposition 4 allows for a quick determination of the missing pair of characters χ and χλ for L+={r(q+1)/2;1r(q3)/2}. The key observation now is that for =r(q+1)/2L+ (1r(q3)/2) one has γq1=(1)r by Proposition 3. Let Q be the permutation matrix corresponding to the permutation of the set Ind fixing * and and sending aaq for every aInd{,} mod (q21). Observe that, due to γq1=(1)r, for the matrix P introduced immediately after the end of the proof of Proposition 3 one has Pq=(1)rPQ. With this in hand and using oddness of j, an inspection of (Equation6) shows that (12) Alow=AuppQ,B=BQandClow=(1)rγ(Pj)Q=(1)rγ2CuppQ.(12) Comparing (Equation12) with (Equation11) implies that for L+ one can simply take M = Q for the intertwining matrix, with (1)δ()=(1)r and c1=c*=irγ.

Applying Proposition 4 further, for L+ the value of the “missing” character χ at the element g=g3(j) of part (c) of Proposition 3 may be taken to be the trace of the product c*Cupp(j)Q. By (Equation6) one has Cupp(j)=γ1Pj and one may check that the only non-zero diagonal elements of the product Cupp(j)Q are those in positions * and , which are γj1 and γj1. Therefore χ(g3(j))=c*(γj1+γj1)=ir(γj+γj) for =r(q+1)/2L+,1r(q3)/2. By the same token, letting ΦG(g4(j))=off(Dupp,Dlow) for the element g4(j) of part (d) of Proposition 3, for L+ one has χ(g4(j))=tr(c*Dupp(j)Q) and as this matrix has a zero diagonal by inspection it follows that χ(g4(j))=0.

Using the notation τ+(j)=ir(αrj+αrj) for =r(q+1)/2, with αr=exp(iπr/(q1))=γ, the missing pair of characters χ,χλ may be given as follows.

Corollary 2.

For =r(q+1)/2L+,1r(q3)/2, the pair of characters χ and χλ are determined by χ(g3(j))=τ+(j) for odd positive jq2, and χ(g4(j))=0 for odd positive jq.

We continue by calculating the values of χ for L on conjugacy classes of twisted elements of G. Out of the previous results it is easy to extract, for each of the (q+1)/2 values of L, the (q1)/2-dimensional vector of values of (χ(g3(j);oddjq2). Observe that, for any fixed L, the restriction (χ)H is orthogonal to each of the (q3)/2 restrictions (χ)H for L+ and to the trivial character ι. It follows that, for our fixed L, the (q1)/2-dimensional vector w=(χ(g3(j);oddjq2) is orthogonal to the system of (q1)/2 mutually orthogonal vectors consisting of the all-one vector and the (q3)/2 vectors (γj+γj;oddjq2) for the (q3)/2 values of L+, all of dimension (q1)/2. But such a w must then be the zero vector, which implies that χ is zero at all elements g3(j) for every L.

It remains to determine the values of χ for L at the elements g4(j) for odd positive jq, which will take considerably more space. To this end it is sufficient to determine the matrix M for this situation, and although general methods for calculating intertwining matrices are available (see e.g. [Citation11]) we adopt here a more direct approach. Summing up some of the facts established so far, the unitary Hermitian matrix M with det(M)=±1 is, up to the sign, determined by the equations from Proposition 4, the second and third of which are equivalent to BM = MB and (1)δ()γ2CuppM=MClow. Written in terms of coordinates Ma,b for a,bInd, the first equation gives (13) Mf(a),f(b)=γ2(|a||b|)Ma,bwith|a|={aifa,0otherwise.(13) To deal with the second equation we will assume that =s(q1)/2L,1s(q1)/2, and we will choose δ() such that (1)δ()=(1)s; as we shall see, this choice will be consistent with the forthcoming calculations. Taking this into the account and realizing that now γq+1=(1)s, it can be checked that the equation (1)δ()γ2CuppM=MClow, taken first for j = 1 and then extended by induction for every t mod q2+1, translates into the following linear system, where a,b{,}: (14) Ma+t,b+tq=γ2tMa,bMa+t,=γ2tMa,,M,b+tq=γ2tM,b,M,=0Ma+t,=Ma,,M,b+tq=M,b,M,=0(14) It can also be verified that the system (Equation14) for t = 2 implies the first equation Alow=AuppM from Proposition 4.

In what follows we will use the fact that q¯=(q21)/2 is the only non-trivial involution mod (q21), with qq¯q¯ mod (q21); we will also frequently use the cycle (q¯,,) of the permutation f. Let y=M0,. The leftmost equation in the last row of (Equation14) shows that Ma,=y for every aInd0=Ind{,}. Applying first (Equation13) with γ2q=γ2, followed by using the middle equation of (Equation14) for t = q and finally by (Equation13) again one obtains M,f(q¯+1)=γ2M,q¯+1=γ2(γ2M,q¯)=Mq¯,=y. But from M=M* it also follows that M,f(q¯+1)=(Mf(q¯+1),)*=y*, which in combination with the above implies y=y*, that is, y is a real number. Using (Equation13) it also follows that y=Mq¯,=M,=M,*=M,, so that the entries of the column of M marked are constantly equal to y except for M,=0. Due to M=M* the same conclusion is valid for the row of M marked . By MM*=M2=I, the dot product of the row and the column of M marked must be equal to 1, that is, q2y2=1, from which y=±q1.

The entries of M in the row and column marked * are determined by the equations of (Equation14) containing asterisks. To determine the remaining entries, observe first that, by (Equation14) and (Equation13), one has M,=Mq¯,q¯=M0,0=0, and hence by the first equation of (Equation14) one also has Mt,tq=Mtq,t=0 for every t mod (q21). The remaining entries of M turn out to be non-zero and can be determined as follows. For every aInd{q¯,,} by (Equation13) one obtains Mf(a),q¯=γ2aMa,=yγ2a. Combining this with application of the first equation of (Equation14) yields (15) Mf(a)+t,q¯+tq=γ2tMf(a),q¯=yγ2a2t.(15) Letting now a=f(a)+t and b=q¯+tq and evaluating t and a in terms of a and b gives t=bq+q¯ and a=f1(abq+q¯) and hence (Equation15) with γ2q=γ2 reduces to (16) Ma,b=yγ2f1(abq+q¯)+2bforabq;(16) the condition abq is a consequence of aq¯. Now, (Equation16) together with the information about zero entries of M and about rows and columns marked * and determine the matrix M=M completely. Conversely, one may check that the entries of M=M that have just been determined satisfy EquationEquations (Equation13) and Equation(Equation14) with (1)δ()=(1)s.

With this in hand we are in position to calculate the remaining values of χ for L at the elements g4(j), represented by the matrix ΦG(g4(j))=off(Dupp(j),Dlow(j)) given in part (d) of Proposition 3; in particular, entries of Dupp(j) are zero except for those indexed (,),(,), and (a,aj) for aInd0, equal, respectively, to (1)γj1,(1)γj1, and (1)γj2a1. By Proposition 4 we have χ(g4(j))=tr(c*Dupp(j)M), and a straightforward evaluation of the trace gives χ(g4(j))=is(γjaInd0γ2aMaj,a+(γj+γj)M,). With the help of (Equation16) and y=±q1, and also by changing the summation variable from a to t=a+(q1)/2, the trace equation transforms to (17) χ(g4(j))=±isq1(γjtInd0γ2f1(t(q+1)j)+(γj+γj)).(17) Again the key is to use the fact that now γ2(q+1)=1, so that to understand the sum in (Equation17) one just needs to study the residues of f1(t(q+1)j) mod (q+1), and it is sufficient to do this only for t{0,1,,q2} as the values of f1 depend only on the residue of t but this time mod (q1). The following auxiliary result will help sort the situation.

Lemma 6.

For t{0,1,,q2} the values of f1(t(q+1)j) are in distinct congruence classes mod (q+1); moreover, f1(t(q+1)j)0 and j mod (q+1).

Proof.

Let e(t)=f1(t(q+1)j). Suppose that for t1,t2{0,1,,q2} one has e(t1)=u1(q+1)+v and e(t2)=u2(q+1)+v for the same residue v mod (q+1). Observe that the power z=ξq+1 of the primitive element ξFGF(q2) that has been used throughout is an element of the subfield F0FG(q) of F. By the defining equation for f in part (b) of Proposition 3 one then has 1=ξe(t1)+ξt1(q+1)j=zu1ξv+zt1ξj and, similarly, 1=zu2ξv+zt2ξj. Eliminating ξv from the two equations gives ξj(zt1u1zt2u2)=zu2zu1. But as ξjF0 for 1jq while all the powers of z are in F0, it follows that u1 = u2 and hence also t1 = t2, proving the first part of the result. The second part is proved similarly, letting e(t) = j and then e(t) = 0 in the first part of the calculation above. □

The set Ind0={0,1,,q22} contains, for each u{0,1,,q2}, exactly q + 1 distinct elements with residue mod (q1) equal to u, and each of these q + 1 elements of the form t=u(q1)+u for u{0,1,,q} determine the same value of the argument of f1 in (Equation17). Therefore the sum in (Equation17) evaluates to (18) tInd0γ2f1(t(q+1)j)=(q+1)0uq2γ2f1(u(q+1)j).(18) By Lemma 6, the values of f1 appearing in (Equation18) cover exactly q − 1 out of the possible q + 1 residue classes mod (q+1), except for the classes 0 and j. But the sum of powers of γ2 with exponents ranging over all the q + 1 distinct residue classes mod (q+1) is equal to zero, because it is precisely the sum of all the q + 1 distinct complex (q+1)th roots of unity. It follows that the sum of (Equation18) is equal to the negative of the “missing” two terms corresponding to residue classes j and 0, that is, (19) 0uq2γ2f1(u(q+1)j)=γ2jγ0.(19) Substituting now (Equation19) into (Equation18) and then the resulting sum into (Equation17) finally gives the values of χ(g4(j)) equal to ±is(γj+γj). We state the corresponding result as a corollary, using the notation τ(j)=is(βsj+βsj) for =s(q1)/2,1s(q1)/2, where βs=exp(iπs/(q+1))=γ; we will also take into account the earlier observation about zero values of χ(g3(j)) for L:

Corollary 3.

For =s(q1)/2L,1s(q1)/2, the pair of characters χ and χλ are determined by χ(g3(j))=0 for odd positive jq2, and χ(g4(j))=τ(j) for odd positive jq.

7. The character table of M(q2)

It remains to translate the facts derived in Propositions 2 and Corollaries 2 and 3 into a tabular form. In the interest of saving space and displaying all the irreducible characters of M(q2) in form of a single table one needs to use a number of abbreviations. The following is a legend for reading .

Table 5. Table of irreducible characters of G=M(q2).

Characters. We use the symbols ι, ρ, and ρ in the first column of in their meaning as in of characters of PSL(2,q2), and Σ for the Steinberg character. In the explanations below we will also use the characters ρ and πm of PSL(2,q2). Continuing in the description of character designations in the first column of , we let ρ;q=ρ+ρq for U and πm;q=πm+πmq for mV; the sets U with |U|=18(q24q+3) and V with |V|=18(q21) have been introduced in Section 2 before . The last two entries in the first column of are based on the characters χ from Corollaries Citation2 and Citation3, and we let χ+=χ for L+={r(q+1)/2;1r(q3)/2} and χ=χ for L={s(q1)/2;1s(q1)/2}. In rows marked ψ,ψλ for ψ{ι, Σ, χ+,χ} where double signs appear on some values, those with the bottom signs are assumed to be values of the product of ψ with the alternating character λ. The number of characters totals to 2+2+1+18(q24q+3)+18(q2+1)+2×12(q3)+2×12(q1)=(q+1)(q+5)/4, which is the number of conjugacy classes as stated at the beginning of Section 3.

Conjugacy classes. Again, in the interest of saving space, in displaying representatives of conjugacy classes of M(q2) in the first row of we will omit square brackets, and the second coordinate, 0 or 1, will appear as a subscript; thus, for example, instead of [a(ζj),0] and [off(ξj),1] we simply write a(ζj)0 and off(ξj)1. The second and the third row of display the number of conjugacy classes (# cl) and orders of the corresponding centralizers (|Cent|); the size of a conjugacy class can be obtained by dividing the order q2(q41) of M(q2) by the order of the centralizer. The column a(ξj)0 (U) corresponds to those representatives a(ξj)0 for which jU. The column headed by a(ξj)0 actually comprises four columns, namely, those corresponding to the total of q − 2 values of j =j(q±1) and j =j(q±1)/2 from , including the bounds and restrictions on j as given in this table. (Observe that although entries in these columns have the same shape, they take different values of j and j  as input.) The column headed a(ζk)0 applies for kV, and the in the last two columns corresponding to dia(ξj)1 and off(ξj)1 the variable j is odd and bounded by 1jq2 and 1jq, respectively.

Entries. We have used the following in : ωρ(j)=αj+αj+αjq+αjq, where α=exp(4πi/(q21)); ωπ(km)=βkm+βkm+βkmq+βkmq, where β=exp(4πi/(q2+1)); τ+(j)=ir(αrj+αrj) for =r(q+1)/2,1r(q3)/2, where αr=exp(iπr/(q1)); and finally τ(j)=is(βsj+βsj) for =s(q1)/2,1s(q1)/2, where βs=exp(iπs/(q+1)). (Note that the complex numbers α and β have also been used in , the character table of PSL(2,q2).)

In the terms introduced above we are finally in position to present the character table of the twisted linear group M(q2).

Theorem 1.

The character table of the group M(q2) is given by .

In a forthcoming paper [Citation10], we apply the character table from Theorem 1 to obtain an enumeration of regular maps of any given type (i.e., of given vertex valency and face length) supported by the groups M(q2). This refines the earlier enumeration of [Citation3] where types were not considered, and extends the available collection of classes of regular maps of given type [Citation8] enumerated with the help of Frobenius’ character formula [Citation4].

Acknowledgments

The authors thank the anonymous referee for pointing out an inaccuracy in the original handling of conjugacy classes and for very helpful and illuminating comments and suggestions that led to improvements in the presentation of our results.

We also thank G. A. Jones for helpful discussions and G. Erskine for computational results that enabled to verify validity of the character table of M(q2) for q13.

Additional information

Funding

The authors gratefully acknowledge support from the APVV Research Grants 17-0428 and 19-0308, as well as from the VEGA Research Grants 1/0206/20 and 1/0567/22.

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