90LECTURE 8.THE MODULAR CURVEProof.We know this already when Γ = Γ(1).So we may assume thatμΓ>1.ByTheorem 8.5 we can identify the spaceM6(Γ) withL(D), wheredegD= 6 degKX+ 6r∞+ 3r2+ 4r3= 12g-12 + 6r∞+ 3r2+ 4r3.I claim that degD >2g+ 1.Ifg≥0 this is obvious. Ifg= 0 we use the formula forthe genus from Theorem 8.4. It easily gives that-12 + 6r∞+ 3r2+ 4r3=μ >2g+ 1 = 1.It follows from the proof of Theorem 8.8 thatνx(fi) =νx(fi/j6) +D(x).(8.28)Now we use the standard argument from the theory of algebraic curves. First of all themap is well-defined. In fact, if all functionsfivanish at the same pointx, we obtainνx(fi)>0 for alli= 0, . . . , N, and henceνx(fi/j6) +D(x)-1≥0 fori= 0, . . . , N.This implies thatL(D) =L(D-x).However, this contradicts the Riemann-Rochtheorem: sincedeg(KX-D)<deg(KX-(D-x)) = 2g-2-degD+ 1<0,it gives dimL(D) = degD+ 1-g >dimL(D-x) = degD-1 + 1-g.Supposef(x) =f(x) =p∈PN(C) for somex=x. Without loss of generality we may assumethatp= (1,0, . . . ,0) (to achieve this we make a linear transformation of coordinates).
This is the end of the preview.
access the rest of the document.