Difference between revisions of "Conjugacy class size formula in general linear group over a finite field"

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| Size of conjugacy class || complicated polynomial, obtained by dividing <math>|GL(n,q)|</math> by the order of the centralizer as given above. || <math>n^2 - \sum_{i,j} ij^2s_{i,j} </math> || <math>\binom{n}{2} - \sum_{i,j} i\binom{j}{2}s_{i,j}</math> || <math>\sum_{i,j} js_{i,j}</math> || 1
 
| Size of conjugacy class || complicated polynomial, obtained by dividing <math>|GL(n,q)|</math> by the order of the centralizer as given above. || <math>n^2 - \sum_{i,j} ij^2s_{i,j} </math> || <math>\binom{n}{2} - \sum_{i,j} i\binom{j}{2}s_{i,j}</math> || <math>\sum_{i,j} js_{i,j}</math> || 1
 
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| Number of conjugacy classes || <math>\prod_{i=1}^n \binom{M(q,i)}{s_{i,1}, s_{i,2},\dots,s_{i,n}} = \prod_{i=1}^n \left( \binom{M(q,i)}{\sum_j s_{i,j}} \binom{\sum_j s_{i,j}}{s_{i,1}, \dots, s_{i,n}} \right)</math> where <math>M(q,i)</math> is the necklace polynomial || <math>\sum_{i,j} is_{i,j}</math> || ? || ? || ?
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| Number of conjugacy classes || <math>\prod_{i=1}^n \binom{M(q,i)}{s_{i,1}, s_{i,2},\dots,s_{i,n}} = \prod_{i=1}^n \left( \binom{M(q,i)}{\sum_j s_{i,j}} \binom{\sum_j s_{i,j}}{s_{i,1}, \dots, s_{i,n}} \right)</math> where <math>M(q,i)</math> is the necklace polynomial || <math>n - \sum_{i,j} is_{i,j}</math> || ? || ? || ?
 
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Revision as of 14:30, 9 July 2019

This article gives formula(s) for the conjugacy class sizes in a general linear group of finite degree n over a finite field with q elements, which we denote by GL(n,q).

See also element structure of general linear group over a finite field.

Case of semisimple elements

Elements diagonalizable over \mathbb{F}_q

Setup

Suppose g \in GL(n,q) is diagonalizable over \mathbb{F}_q, with eigenvalues \lambda_1, \dots, \lambda_k having multiplicities r_1,r_2,\dots,r_k respectively (the \lambda_is are all distinct). Note that \sum_{i=1}^k r_i = n.

Further, let's say that among the r_is, there are s_1 1s, s_2 2s, and so on till s_n ns.

Then, the centralizer of the diagonal representative of this conjugacy class is isomorphic to:

GL(r_1,q) \times GL(r_2,q) \times GL(r_3,q) \times \dots \times GL(r_k,q)

In fact, if the diagonal entries are arranged so that all the \lambda_1s occur first, then the \lambda_2s, and so on, then the centralizer is the set of invertible block diagonal matrices with blocks of sizes r_1, r_2, \dots, r_k.

In particular, we have:

\sum_{i=1}^k r_i = n

\sum_{j=1}^n js_j = n

Summary information

Item Value Degree as polynomial in q Largest power of q in polynomial for value Largest power of q - 1 in polynomial for value Leading coefficient
Order of centralizer q^{\sum_{i=1}^k \binom{r_i}{2}} \prod_{i=1}^k \prod_{j=0}^{r_i} (q^{r_i - j} - 1) \sum_{i=1}^k r_i^2 \sum_{i=1}^k \binom{r_i}{2} = \frac{1}{2} \left(\sum_{i=1}^k r_i^2 \right) - \frac{n}{2} n 1
Size of conjugacy class (obtained as order of GL(n,q) divided by order of centralizer) complicated polynomial, but it is the q-analogue of the binomial coefficient and is written as \binom{n}{r_1,r_2,\dots,r_k}_q n^2 - \sum_{i=1}^k r_i^2 \frac{1}{2}\left(n^2 - \sum_{i=1}^k r_i^2 \right) 0 1
Number of such conjugacy classes \binom{q - 1}{k} \binom{k}{s_1, s_2, \dots, s_n} k 0 1 \frac{1}{\prod_{j=1}^n s_j!}

Examples of conjugacy classes diagonalizable over \mathbb{F}_q

Some particular cases for the partition of n as a sum of r_is, and the corresponding sizes, are given below.

n Partition of n k |GL(n,q)| Order of centralizer of diagonal element Degree as polynomial of q (equals \sum_{i=1}^k r_i^2) Size of conjugacy class Degree as polynomial of q (equals n^2 - \sum_{i=1}^k r_i^2) Number of conjugacy classes (degree k polynomial in q)
1 1 1 q - 1 q - 1 1 1 0 q - 1
2 2 1 q(q^2 - 1)(q - 1) q(q^2 - 1)(q - 1) 4 1 0 q - 1
2 1 + 1 2 q(q^2 - 1)(q - 1) (q - 1)^2 2 q(q + 1) 2 \binom{q - 1}{2} = \frac{(q - 1)(q - 2)}{2}
3 3 1 q^3(q^3 - 1)(q^2 - 1)(q - 1) q^3(q^3 - 1)(q^2 - 1)(q - 1) 9 1 0 q - 1
3 2 + 1 2 q^3(q^3 - 1)(q^2 - 1)(q - 1) q(q^2 - 1)(q - 1)^2 5 q^2(q^2 + q + 1) 4 2\binom{q-1}{2} = (q - 1)(q - 2)
3 1 + 1 + 1 3 q^3(q^3 - 1)(q^2 - 1)(q - 1) (q - 1)^3 3 q^3(q^2 + q + 1)(q + 1) 6 \binom{q-1}{3} = \frac{(q - 1)(q - 2)(q - 3)}{6}
n n 1 q^{\binom{n}{2}} \prod_{i=0}^{n-1} (q^{n - i} - 1) q^{\binom{n}{2}} \prod_{i=0}^{n-1} (q^{n-i} - 1) n^2 1 0 q - 1
n 1 + 1 + \dots + 1 n q^{\binom{n}{2}} \prod_{i=0}^{n-1} (q^{n - i} - 1) (q - 1)^n n q^{\binom{n}{2}}\prod_{i=1}^{n-1} \left[\left(\sum_{j=0}^i q^j \right)\right] n(n - 1) \binom{q - 1}{n}

Regular semisimple elements not necessarily diagonalizable over the original field

Setup

Some elements may be semisimple but not diagonalizable over \mathbb{F}_q, i.e., they can be diagonalized over a suitable field extension of \mathbb{F}_q. We begin by considering the regular semisimple case -- elements that can be diagonalized over some field extension of \mathbb{F}_q such that all their diagonal entries are pairwise distinct. We can show that these elements are precisely the ones that can be converted over \mathbb{F}_q to a block diagonal form for some partition r_1 + r_2 + \dots + r_k = n, where the entry in block r_i is diagonalizable with distinct diagonal entries over the field \mathbb{F}_{q^{r_i}} and no smaller field.

Further, let's say that among the r_is, there are s_1 1s, s_2 2s, and so on till s_n ns.

In particular, we have:

\sum_{i=1}^k r_i = n

\sum_{j=1}^n js_j = n

Summary information

In this case, the centralizer of the element in this block diagonal form is:

\mathbb{F}_{q^{r_1}}^\ast \times \mathbb{F}_{q^{r_2}}^\ast \times \dots \times \mathbb{F}_{q^{r_k}}^\ast

Item Value Degree as polynomial in q Largest power of q in polynomial for value Largest power of q - 1 in polynomial for value Leading coefficient
Order of centralizer \prod_{i=1}^k (q^{r_i} - 1) n 0 k 1
Size of conjugacy class (obtained as order of GL(n,q) divided by order of centralizer) (complicated polynomial) n(n - 1) \binom{n}{2} n - k 1
Number of such conjugacy classes \prod_{j=1}^n \binom{M(q,j)- E(j)}{s_j} (where M(q,j) is a necklace polynomial and E is the characteristic function at 1; see explanation below) n \sum_{j=2}^n s_j \frac{j}{\operatorname{square-free part of } j} number of j with nonzero s_j, i.e., the number of distinct parts in the partition of n. It is less than or equal to k \prod_{j=1}^n \frac{1}{j^{s_j}(s_j)!} (this is the fraction of elements in the symmetric group S_n whose cycle decomposition matches the partition)

Most elements are regular semisimple

As noted in the summary table above:

  • For each partition of n, the number of conjugacy classes is a polynomial of degree n and the size of each conjugacy class is a polynomial of degree n(n - 1). In total, therefore, the number of elements corresponding to that partition is a polynomial of degree n^2 in q.
  • The leading coefficient of that polynomial is the product of the leading coefficient for the size of conjugacy class polynomial (which is 1) and the number of conjugacy classes (which is the fraction of elements in the symmetric group S_n whose cycle decomposition corresponds to that partition).
  • Therefore, the total number of regular semisimple elements is a polynomial of degree n^2 whose leading coefficient is 1 (this is the sum of "the fraction of elements in the symmetric group S_n whose cycle decomposition corresponds to that partition" over all partitions).
  • The polynomial giving the size of GL(n,q) is also a polynomial of degree n^2 with leading coefficient 1.
  • Thus, the number of elements that are not regular semisimple is the difference of the two polynomials, which is a polynomial of degree strictly less than n^2.

Therefore, as q \to \infty, holding n constant, the fraction of elements that are regular semisimple goes to 1. Or, in simple terms, most elements are regular semisimple.

This statement also has an algebraic interpretation when we consider the analogous statement over an infinite algebraic group. Being regular semisimple corresponds to the discriminant of the characteristic polynomial being nonzero. The discriminant of the characteristic polynomial can be written as a polynomial function of the coefficients of the characteristic polynomial (which are the elementary symmetric functions) and hence of the matrix entries. Therefore, the set of regular semisimple elements is an open and hence dense subset, and therefore comprises most elements.

For large enough q and n > 2, the following is true: among the regular semisimple elements, the partition n = n, which corresponds to the case of elements with irreducible characteristic polynomial, gives rise to the largest number of conjugacy classes. Its leading coefficient is 1/n. In contrast, the partition n = 1 + 1 + \dots + 1, corresponding to the case of elements diagonalizable over \mathbb{F}_q, gives the fewest number of conjugacy classes; the leading coefficient is 1/(n!).

Finally, note that not only are most elements regular semisimple, most conjugacy classes are also regular semisimple, for very similar reasons.

Derivation of number of conjugacy classes in the regular semisimple case

Step no. Statement Justification
1 The necklace polynomial M(q,j) is the number of irreducible monic polynomials of degree j over \mathbb{F}_q. The formula for M(q,j) is as follows: M(q, j) = \frac{1}{j} \sum_{d \mid j} \mu\left( \frac{j}{d} \right) q^d Define L(q,j) = jM(q,j). We have, by element-counting, that q^j = \sum_{d \mid j} L(q,d). Apply Mobius inversion to get L(q,j) = \sum_{d \mid j} \mu\left( \frac{j}{d} \right) q^d and then rewrite L(q,j) in terms of M(q,j)
2 The number of irreducible polynomials of degree j over \mathbb{F}_q with invertible roots is M(q,j) - E(j), where M(q,j) is the necklace polynomial as defined in step (1), and E(j) is the function that is 1 if j = 1 and zero otherwise. When j = 1, the irreducible polynomials include the polynomial with 0 as a root. This is not invertible, so in this case we need to subtract 1 from M(q,1) to get the number of polynomials with invertible roots. For j > 1, the roots are outside \mathbb{F}_q, hence all nonzero, and hence invertible, so nothing needs to be subtracted.
3 The total number of conjugacy classes is \prod_{j=1}^n \binom{M(q,j)- E(j)}{s_j} For each j, there are s_j distinct irreducible polynomials with invertible roots (distinctness follows from the regularity assumption). In other words, we need to select s_j elements out of a pool of M(q,j) - E(j) possibilities. The number of ways to select these elements is \binom{M(q,j)- E(j)}{s_j}. Multiplying this expression over j gives the number of possible conjugacy classes.

Example necklace polynomial values

Value of j M(q,j) M(q,j) - E(j)
1 q q - 1
a prime number \frac{q^i - q}{i} \frac{q^i - q}{i}
a power of a prime t \frac{q^i - q^{i/t}}{i} \frac{q^i - q^{i/t}}{i}

Examples of regular semisimple conjugacy classes

Note that in the particular cases of the partition 1 +1 + \dots + 1, this simplifies to  \binom{q - 1}{n}; also this special case is the only case that overlaps with the previous section.

n Partition of n |GL(n,q)| Order of centralizer of representative (degree n polynomial in q) Size of conjugacy class (degree n(n - 1) polynomial in q) Number of conjugacy classes (degree n polynomial in q)
1 1 q - 1 q - 1 1 q - 1
2 2 q(q^2 - 1)(q - 1) q^2 - 1 q(q - 1) M(q, 2) = \frac{q^2 - q}{2} = \frac{q(q - 1)}{2}
2 1 + 1 q(q^2 - 1)(q - 1) (q - 1)^2 q(q + 1) \binom{M(q,1) - 1}{2} = \binom{q - 1}{2} = \frac{(q - 1)(q - 2)}{2}
3 3 q^3(q^3 - 1)(q^2 - 1)(q - 1) q^3 - 1 q^3(q - 1)^2(q + 1) M(q, 3) = \frac{q^3 - q}{3} = \frac{q(q^2 - 1)}{3}
3 2 + 1 q^3(q^3 - 1)(q^2 - 1)(q - 1) (q^2 - 1)(q - 1) q^3(q^3 - 1) M(q, 2)(M(q,1) - 1) = \frac{q^2 - q}{2}(q - 1) = \frac{q(q - 1)^2}{2}
3 1 + 1 + 1 q^3(q^3 - 1)(q^2 - 1)(q - 1) (q - 1)^3 q^3(q^2 + q + 1)(q + 1) \binom{M(q,1) - 1}{3} = \binom{q - 1}{3} = \frac{(q - 1)(q - 2)(q - 3)}{6}
4 4 q^6(q^4 - 1)(q^3 - 1)(q^2 - 1)(q - 1) q^4 - 1 q^6(q^3 - 1)(q^2 - 1)(q - 1) M(q,4) = \frac{q^4 - q^2}{4} = q^2(q^2 - 1)/4
4 3 + 1 q^6(q^4 - 1)(q^3 - 1)(q^2 - 1)(q - 1) (q^3 - 1)(q - 1) q^6(q^4 - 1)(q^2 - 1) M(q,3)(M(q,1) - 1) = \frac{q^3 - q}{3}(q - 1) = \frac{q(q^2 - 1)(q - 1)}{3}
4 2 + 2 q^6(q^4 - 1)(q^3 - 1)(q^2 - 1)(q - 1) (q^2 - 1)^2 q^6(q^3 - 1)(q^2 + 1)(q - 1) \binom{M(q,2)}{2} = \frac{q(q - 1)(q + 1)(q - 2)}{8}
4 2 + 1 + 1 q^6(q^4 - 1)(q^3 - 1)(q^2 - 1)(q - 1) (q^2 - 1)(q - 1)^2 q^6(q^4 - 1)(q^2 + q + 1) M(q,2)\binom{M(q,1) - 1}{2} = \frac{q^2 - q}{2} \binom{q - 1}{2} = q(q - 1)^2(q - 2)/2
4 1 + 1 + 1 + 1 q^6(q^4 - 1)(q^3 - 1)(q^2 - 1)(q - 1) (q - 1)^4 q^6(q^3 + q^2 + q + 1)(q^2 + q + 1)(q + 1) \binom{M(q,1) - 1}{4} = \binom{q - 1}{4} = \frac{(q - 1)(q - 2)(q - 3)(q - 4)}{24}

General semisimple case

The general semisimple case combines ideas from both the preceding cases (the diagonalizable over \mathbb{F}_q case, and the regular semisimple case), and generalizes both of them.

For a given conjugacy class in GL(n,q), and for i,j \in \{ 1, 2, \dots, n \} define s_{i,j} to be the number of distinct cases where the conjugacy class contains exactly j conjugate elements of \mathbb{F}_{q^i}^\ast (i.e., a multiplicity of j for an extension of degree i over \mathbb{F}_q).

We have:

\sum_{i, j} ijs_{i,j} = n

The centralizer is then described as follows:

\prod_{i,j} \left(GL(j,q^i) \right)^{s_{i,j}}

How the previous two cases are special cases of this general case

  • The "diagonalizable over \mathbb{F}_q" case is the special case where s_{i,j} = 0 for i > 1. Moreover, s_{1,j} in this general notation matches up with s_j in the notation for that case.
  • The "regular semisimple" case is the special case where s_{i,j} = 0 for j > 1. Moreover, s_{i,1} in this general notation matches s_i in the notation for that case.

Summary table

Item Value Degree as polynomial in q Largest power of q in polynomial for value Largest power of q - 1 in polynomial for value Leading coefficient
Order of centralizer \prod_{i=1}^n \left( q^{i \binom{j}{2}} \prod_{k=0}^{j-1} \left(q^{i(j - k)} - 1 \right) \right)^{s_{i,j}} \sum_{i,j} ij^2s_{i,j} \sum_{i,j} i\binom{j}{2} s_{i,j} \sum_{i,j} js_{i,j} 1
Size of conjugacy class complicated polynomial, obtained by dividing |GL(n,q)| by the order of the centralizer as given above. n^2 - \sum_{i,j} ij^2s_{i,j} \binom{n}{2} - \sum_{i,j} i\binom{j}{2}s_{i,j} \sum_{i,j} js_{i,j} 1
Number of conjugacy classes \prod_{i=1}^n \binom{M(q,i)}{s_{i,1}, s_{i,2},\dots,s_{i,n}} = \prod_{i=1}^n \left( \binom{M(q,i)}{\sum_j s_{i,j}} \binom{\sum_j s_{i,j}}{s_{i,1}, \dots, s_{i,n}} \right) where M(q,i) is the necklace polynomial n - \sum_{i,j} is_{i,j}  ?  ?  ?

Case of non-semisimple elements

Regular elements with all eigenvalues over \mathbb{F}_q

We begin by considering a very easy class of non-semisimple elements: those where all the eigenvalues are over \mathbb{F}_q, and where all distinct Jordan blocks correspond to distinct eigenvalues, i.e., they are regular elements. This means that the minimal polynomial coincides with the characteristic polynomial.

Suppose the Jordan block sizes are r_1, r_2, \dots, r_k.

The centralizer for each Jordan block of size r_i is the invertible matrices in the subalgebra generated by the Jordan block. This corresponds to the units in \mathbb{F}_q[t]/(t^{r_i}), and the size of this group is q^{r_i - 1}(q - 1).

The order of the centralizer is therefore:

\prod_{i=1}^k (q^{r_i - 1}(q - 1)) = q^{n - k}(q - 1)^k

The size of the conjugacy class is obtained by dividing the order of the group by this expression.

For each partition:

n = r_1 + r_2 + \dots + r_k

The number of conjugacy classes associated with that partition can be calculated as follows. Suppose that, among the r_js, there are s_1 1s, s_2 2s, and so on. Then, the total number of conjugacy classes is:

\binom{q - 1}{k}\binom{k}{s_1, s_2, \dots, s_n}

The total number of conjugacy classes corresponding to a choice of k is therefore:

\binom{q - 1}{k} \sum \binom{k}{s_1, s_2, \dots, s_n}

where the sum is across all the possible partitions.

n Partition of n k (number of parts) |GL(n,q)| Order of centralizer of representative (equals q^{n-k}(q - 1)^k, a degree n polynomial over q) Size of conjugacy class (degree n(n - 1) polynomial over q) Number of conjugacy classes (degree k polynomial over q)
1 1 1 q - 1 q - 1 1 q - 1
2 2 1 q(q^2 - 1)(q - 1) q(q - 1) q^2 - 1 q - 1
2 1 + 1 2 q(q^2 - 1)(q - 1) (q - 1)^2 q(q + 1) \binom{q - 1}{2} = \frac{(q - 1)(q - 2)}{2}
3 3 1 q^3(q^3 - 1)(q^2 - 1)(q - 1) q^2(q - 1) q(q^3 - 1)(q^2 - 1) q - 1
3 2 + 1 2 q^3(q^3 - 1)(q^2 - 1)(q - 1) q(q - 1)^2 q^2(q^3 - 1)(q + 1) 2 \binom{q - 1}{2} = (q - 1)(q - 2)
3 1 + 1 + 1 3 q^3(q^3 - 1)(q^2 - 1)(q - 1) (q - 1)^3 q^3(q^2 + q + 1)(q + 1) \binom{q - 1}{3} = \frac{(q - 1)(q - 2)(q - 3)}{6}

General case with all eigenvalues over \mathbb{F}_q

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Most general case

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