This means that J(τ) = g3 2(τ) ∆(τ) = 1 + 720x+I 123x(1−24x+I) = 1 + 720x+I 123x(1−24x+I)· 1 + 24x+I 1 + 24x+I = 1 123x (1+720x+I)(1+24x+I), which implies that 123J(τ) = 1 x + 744 + ∞ X n=1 c(n)xn. In this formula, the c(n) are integers [1]. QED Thus, we have our expansion of the J-function. The coefficients have been calculated for n≤100. The first few are c(0) = 744, c(1) = 196 884, c(2) = 21 493 760, and c(3) = 864 299 970. In some versions of this formula, the 744 is subtracted from both sides, and the result is called J(τ). Letting q =e2πiτ, that gives us 123J(τ)−744 =q−1 + 196 884q+ 21 493 760q2 + 864 299 970q3 +· · · . From now on we will refer to this version of the formula as J(τ). Going back to modular functions in general for a moment, for a group G, we can expand its unique modular function JG(τ) as JG(τ) =q−1 + ∞ X n=1 anq n. “This function JG is called the (normalised) Hauptmodul for the genus 0 group G” [7]. If we let G=SL(2, Z) we get J(τ) =q−1 + 196 884q + 21 493 760q2 + 864 299 970q3 +· · · [7]. At this point, we return to the Monster. Recall that the first few dimensions of irreducible representations of the Monster are (rn)n=1,··· ,194 = (1, 196 883, 21 296 876, 842 609 326, 18 538 750 076, · · · ). We now notice that 196 884 = 1 + 196 883 21 493 760 = 1 + 196 883 + 21 296 876 864 299 970 = 2· 1 + 2· 196 883 + 21 296 876 + 842 609 326, where the numbers on the left hand side of the equation are coefficients from J(τ) and the numbers on the right are from the dimensions of the irreducible representations of the Monster. So we see that there is a connection between modular functions and a sporadic finite simple group [7]. Channels • 2022 • Volume 6 • Number 2 Page 45
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