Number of conjugacy classes in symplectic group of fixed degree over a finite field is PORC function of field size

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Statement

Suppose n is an even natural number, i.e., n = 2m for some natural number m. Then, there exists a PORC function f of degree m = n/2 such that, for any prime power q, the number of conjugacy classes in the symplectic group Sp(n,q) (i.e., the symplectic group of degree n over the finite field of size q) is f(q).

A PORC function is a polynomial on residue classes -- it looks like different polynomial functions on different congruence classes modulo a particular number. In this case, we only need to consider congruence classes modulo n to define the PORC function. In fact, for a field size of q, the polynomial depends only on the value \operatorname{gcd}(\operatorname{lcm}(2,m),q - 1).

Particular cases

n (degree of symplectic group) m = n/2 (degree of PORC function, also number used in Chevalley notation) \operatorname{lcm}(2,m) Possibilities for \operatorname{gcd}(\operatorname{lcm}(2,m),q - 1) Corresponding congruence classes mod \operatorname{lcm}(2,m) for q Corresponding polynomials in PORC function of q giving number of conjugacy classes in Sp(n,q) More information
2 1 2 1
2
0
1
q + 1
q + 4
See element structure of special linear group of degree two over a finite field (note that Sp(2,q) = SL(2,q)).
4 2 2 1
2
0
1
q^2 + 2q + 3
q^2 + 5q + 10
See element structure of symplectic group of degree four over a finite field

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